Absolutely appreciated for this video! Professor Wolfgang Rindler was one of my favorite professor when I was working on my Ph.D in Physics. He was the one recruited me to chase my dream when I was young ... This video helped the rest of the people to understand "How to bridge the General relativity with Special relativity with gauge invariance. I learned when I was young and glad to see this video to resurface the beauty of Rindler's Horizon.
@eigenchris your courses on Tensors(algebra+calculus) along with these Relativity series deserve to be put out in form of a Relativity Introduction book. Neither of books out there on relativity I have ever read, introduce in ways your courses do. Great Job!!!
Great videos, I watched your Tensor Calculus series to get myself caught up because I know Einstein Field Equations have Tensors. Extremely great videos💯
Minor correction: at 27:00 where you show Rindler's acceleration stopping, his curve should not be a straight line vertically, but rather a straight line (or ray) tangent to the green curve at that point. It will still be true that all the light beams will cross this line eventually. Thanks.
I'll have to watch that when you're done. You can feel free to keep the intro if you want. It doesn't hurt to have multiple explanations of things online. Unless you feel it's a waste of time.
@@caseyglick5957 yeah, in terms of the dopamine circuits those are really similar! but, it seems to me that evolution cares about sex much better than physics! )
23:57 an easy way to see this without hyberbolic identities is to note that d(S.S) =dS.S+S.dS. Now, in the frame of accelerating Rindler, position S is purely spacelike (since S is parallel to acceleration A in Rindler coordinates), and dS is purely timelike (since in Rindler rest frame the velocity does not have space components). Therefore the dot products are zero and S.S is constant.
28:15 one can also think the horizon effect in terms of shifting simultaneity: as the velocity increases the "plane of simultaneity" tilts more an more, so that einstein's past becomes present for Rindler. The faster the farther away from Rindler Einstein "looks" . And there is a point, the origin, where "Rindler's now" totally ceases to move in Einstein's time. Left of this point the Rindler's "present" seems to move backwards in Einstein's time. I think this also goes to show how we should be cousious to give "simultaneity" too much actual meaning...even in flat spacetime. Lastly: Excellent video! Thanks.
Trying to finish a course on General Relativity through Coursera from HSE University in Russia before I lose access to it, glad to have videos like these to further explain the material.
Correction: @26:58 Rindler stopping accelerating would have a tangent world line to the hyperbola, so you must mean stopping velocity to have a world line parallel to the home frame world line. This would be a very high deceleration of course. But what I think you mean is stopping acceleration the new tangent world line would then be able to see the light since the tangent world line would enter all remaining light cones.
Great series about relativity, even I didn't learn differential geometry or vector calculus. One question : is there a reason why all movements are described only in x direction and y, z are usually not considered? Is it because to keep it simple or would you need to transform coordinates to make y and z direction to zero?
Thank you for making these videos available on RUclips for free. Love your generous work. I had a doubt: In the video 'relativity 104f', to derive 4-velocity, we had used the fact that derivative of basis vectors (with respect to proper time) is zero as they are constant throughout. Hence we could show that Minkowski squared length of U is c*c. But while dealing with non inertial frames, and accelerated motion, shouldn't the derivative of basis vectors be non zero? How can we still say that Minkowski squared length of U is c*c?
The length of a vector should not depend on the coordinate system. It should be the same in all coordinate systems. If you like you can review video 103d for showing how the length of a vector doesn't change when you change basis.
your videos are really understandable. Could you tell any book or reference to study more about Rindler coordinates ?? That seems very interesting topic !!
The concept of the Rindler horizon is set to become one of the most mind-blowing concepts in all of physics. imagine you are chasing your friend at a constant speed, and although your friend is trying to accelerate to match your speed, they are still running slower than you. It seems inevitable that you will eventually catch them, since no matter how hard and how long they try, their speed will never be the same as yours. However, astonishingly, no matter how long you pursue them, you will never actually be able to catch them. Just think about it.
Makes sense. If they accelerate indefinitely they will eventually outspeed you and you surely won't catch them from that point on. The real weird thing is how even if they are racing against your laser pointer (the literal universal speed limit), they can win the race if given the appropriate headstart, no matter how long you prolong the race.
2:54 Both a cylinder and a cone can be "rolled up" and, in reverse, "unwrapped" into a straight sheet. That's two dimensional surfaces, which we can see, as they can be embedded in a three dimensional space. Then, there are surfaces with some curvature and of these only an S^2 sphere can be embedded in three dimensional space, but, for instance, hyperbolic surface can only be projected. In other words, we can make a three dimensional model of a cylindrical space using just a sheet of paper. If we ourselves were capable of seeing higher dimensions (we only have a pseudo - spatial flat vision based on two projections, which are both identical, bar shifting and close to a stereographic one. Had we spatial vision, we would see, say, the other side of a monitor, on screen of which you see this message, without moving our head) In the same way Minkowski space has zero intrinsic curvature - it can be rolled up from a 4d euclidean space and there will be no "crease" left after "unwrapping".
The only warning I'll give is that the 4D Eulcidean metric is fundamentally different than the 4D Minkowski metric, so that does make them fundamentally different, even if they are both flat.
Dear Chris, are you planning to discuss General relativity ? I am waiting for Relativistic tensors like EM field and so on. You told that can be uploaded after X mas.
As I said in the video, the 4-velocity vectors U (which are tangent to the worldline) are always minkowski-orthogonal to the 4-acceleration vectors, and the 4-acceleration vectors A are always parallel to the e_x basis vector for any given instantaneous frame. You can see at 23:23 that 4-acceleration A is parallel to 4-position S, so S is also parallel to the e_x basis vectors. S vectors always start at the origin by definition, since they measure the spacetime displacement from the origin. So this explains why all the S vectors start at the origin, and are also parallel to the e_x basis vectors everywhere along the curve.
I'm loving this series (the foregoing episodes are the best explanation of SR that I've seen). But I'm having difficulty understanding part of the world line of the accelerated frame. What is the physical meaning when t < 0? Does it look to an observer at t=0 that Rinder is moving (accelerating) towards him and then Rindler starts to move away again?
Yes. Loosely speaking, you could imagine an Rindler starts out at near-infinity, travelling left at near-the speed of light, with constant acceleration to the right. Eventually he will slow to a stop (this is the t=0 event on his worldline) and then speed up back to infinity, approaching the speed of light in the opposite direction.
@@eigenchris Ah. That explains it. I didn't realize that the acceleration vector was pointing to the right. But now I see that you mention that at 9:30.
Also, Thanks for your quick response. I must say, I really enjoy all of your videos; you are one of the best physics communicators I know of. I really appreciate your very methodical approach. What a wonderful service you provide to the world! BTW: Have you ever thought of making a series on string theory?
In the case of non constant acceleration, would the 4 velocity change at different points in the accelerating frame world line. Would that mean that between instantaneous points on the curved world line that time is relative?
Everyone always thinks their own 4-velocity is c*e_t (this is true whether you are inertial or non-inertial). I'm not sure what your question mean about "time is relative" means. The proper time for a given worldline is objective, but outside observers in different frames will measure different values for the time between the start event and end event.
Einstein (from his How I Created Relativity lecture): "I was dissatisfied with the special theory of relativity, since the theory was restricted to frames of reference moving with constant velocity relative to each other and could not be applied to the general motion of a reference frame." I looked elsewhere for his opinion on this subject, since this quote makes its seem almost certain he did not believe special relativity could be extended to accelerated frames. In other quotes though he is careful not to say that the postulates or principles don't extend to accelerated frames, rather only the equivalency of frames, so this particular quote may only have been a thoughtless imprecision on his part. However, if you read his intro to his 1916 GR paper, or his explanation of the Twin Paradox where he invokes GR, both of these heavily imply that he was not happy with the idea of treating acceleration in SR. Likely this is due to his Machian influence and his belief, which I quoted in our comments below, that absolute acceleration made no kinematical sense.
I guess I can't know for sure what he meant when he said that. I'll point out that in the next paragraph of that lecture he says: "all the natural laws except the law of gravity could be discussed within the framework of the special theory of relativity". That sentence is one I would more firmly agree with. The laws of electricity and magnetism result in accelerations of charged particles and I believe these accelerations are perfectly well-handled by SR. Is this the 1916 paper you're talking about?en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity This is a quote from the 2nd section: "Can any observer, at rest relative to K' then conclude that he is in an actually accelerated reference-system? This is to be answered in the negative; the above-named behaviour of the freely moving masses relative to K' can be explained in as good a manner in the following way. The reference-system K' has no acceleration. In the space-time region considered there is a gravitation-field which generates the accelerated motion relative to K'." I agree with this. It's basically just the statement that an accelerometer measures zero when it's dropped from a plane and freely falling to earth, because both the mass and container move at the same rate and there is therefore no spring compression. The worldline of a freely falling object is a "geodesic" in GR, and my spring-and-mass accelerometer would measure how much an observer is deviating from a geodesic curve. I'm going to talk about this idea in the context of SR in 105f, because the basic idea still applies in SR... accelerometers in SR just measure deviation from inertial worldlines. My explanation for the twin paradox is coming in 105d and it's nothing really deep... it's just that one twin has an inertial frame (straight line) and the other has a non-intertial frame (there-and-back zig-zag). In theory you could graph this so that the inertial frame is a zig-zag and the non-inertial frame is a straight line, but you can't do this transformation using a Lorentz transformation, so there is no reason to believe both observers are on equivalent footing. The non-inertial frame (which looks like a zig-zig from the inertial frame, and experiences ficticious forces) is fundamentally different and counts different proper time on its clock. I don't believe any GR needs to be used to explain this. Sean Carol says this even more simply... He says that just as distance depends on your path through space, proper time depends on path in spacetime: ruclips.net/video/Cg2tOUTE2F4/видео.html
@@eigenchris The quote from the 2nd section, where Einstein states a version of the EQ principle I certainly don't have any disagreement with. But I think my issue here again is, what defines the accelerometer? Having an accelerometer with us in a uniform gravitational field may allow us to measure that we are in an inertial frame, but once spring compression occurs we still have to determine whether it is the mass or container which is accelerating, which brings us straight back to the discussion we had below. You can make the claim, for instance, that an object's deviation from a geodesic path on the spacetime manifold, if taken in reference to an object not deviating, would properly define an accelerometer. But then we have to determine which object is on the geodesic and which one is deviating. To do this we would have to know the curvature, or the value of Riemann tensor, of the spacetime locale we are in. This requires knowing the mass/energy distribution of the surrounding environment. (And if you want the curvature to be absolutely accurate, you have to account for the mass/energy of the whole universe, I believe.) Thus, you come to the conundrum that is bothering me: the accelerometer is supposed to provide an objective "local" measurement of proper acceleration. But it hinges on having access to lots of information about your surrounding system, environment, etc. Thus I can't think of any way to come up with a "local" definition of one that is consistent. Maybe this was why Einstein was so uncomfortable with it, and why he writes in that same 1916 paper: "We have to take it that the general law of motion... is partly conditioned, in quite essential respects, by distant masses which we have not included in the system under consideration."
@@WSFeuer Sorry, this second comment slipped by me. When you say the following: "but once spring compression occurs we still have to determine whether it is the mass or container which is accelerating"... Sorry if we're going in circles, but let's go one step at a time. With my definition of accelerometer, given that the mass is electrically/magnetically neutral, can you list all possible physical scenarios you can come up with that would cause the spring to compress? (Or at least, list a few?)
@@WSFeuer Einstein's initial motivation for developing a more general theory was indeed to include accelerated frames. But while doing so, he came up with the equivalence principle, linking accelerated frames to gravitational fields. And of course, he then figured out that curved spacetime was needed to explain tidal forces in gravitational fields around massive objects. I think it was later discovered that you can indeed describe accelerated frames in flat spacetime using special relativity, so long as you add some extra assumptions (like the clock hypothesis). So now the more modern view is that special relativity can describe any kind of motion in flat spacetime, whereas general relativity is for curved spacetime. But this wasn't the view during Einstein's time (when I think it was believed that special relativity only can be used for inertial frames in flat spacetime).
is there an intuitive reason for why the position basis vector for the spaceship is always parallel to the separation vector from the origin in the "stationary" frame / why do lines of simultaneity radiate from the origin? Or rather, why does the spaceship measure the proper length to the origin when using it's own basis vectors?
They represent initial conditions, by they are not exactly the initial coordinates. Since sinh(0)=0, kt is the initial time value. Since cosh(0) = 1, the initial position is c^2/alpha + kx. Setting kt = kt = 0, the initial coordinates at tau=0 are ct=0 and x=c^2/alpha.
Hi Chris, at 27:00 if acceleration stops will the x position go vertically up and not go in the tangent direction of hyperbola at that point (continue with same velocity)? I did not get this part
Can entangled quantum particles transmit information across the Rindler horizon? Will the particles remai n in an entangled relation even after the observers of each is separated by more thjan the Rindler critical distance? If so, is that a transmission of info > c violateing special relativity?
I don't know how quantum mechanics interacts with horizons in GR. I know that the "Hawking Radiation" related to black holes has an equivalent called "Unruh Radiation" for accelerating observers. I'm not sure if that's related.
I love these videos and in really looking forward to GR! But I think I might have spotted a tiny mistake. With the three rockets whose worldlines are tangents to Rindler's worldlines, the first and third pass through the origin; but if Rindler's worldline is a hyperbola I don't think that's possible. The only tangent that would pass through the origin would be the asymptote, surely? Perhaps it doesn't matter here because we haven't yet proven that the path is a hyperbola, but I just want check I'm understanding this correctly.
For each of those 3 points, I was taking the point itself to be the origin of its own coordinate system. So I showed 3 separate coordinate systems there with 3 separate origins. But the choice of origin in spacetime doesn't matter since all points should be on equal footing.
@@eigenchris You're right! Thanks for replying. And I have to say this is the first set of lectures on Relativity where I haven't had to keep doing to try to understand every single point. Thank you for taking the time to explain everything clearly and thoroughly, with only a handful of places where you say "it can be shown that ...". Your approach is fresh and exciting! I look forward to every single video, and I can't wait to get on to GR!
26:55 f: _As soon as Rindler stops accelerating..._ Your diagram shows Rindler _stop moving_ which actually is a kind of infinite acceleration in the −x direction. Rindler stopping accelerating would look like the hyperbola at this point goes over into a straight line with the hyperbola's current slopes, without any sharp bend. This would still work since the slope is steeper than 45°, and the RINDLER horizon would vanish.
How close can the Rindler Horizon be located, if alpha is very, very small? I've never noticed the universe behind me freeze when I walk across the room.
The Rindler horizon distance is c^2/alpha. So if alpha is 1m/s^2, the Rindler horizon would be 9*10^16m behind you. Rindler horizons are usually very very far away.
Why the accelerated motion in spacetime is drawn in spacelike region? Isn’t the motion of particle in spacelike region is greater than light? Or, I'm missing something there? Please explain.
It's the slope (4-velocity vector) of the worldline that indicates the speed. The slope of the hyperbola is always time-like (it's never more horizontal than a light beam at 45 degrees). The "timelike" and "spacelike" regions of the spacetime diagram only make sense from the perspective of the origin event point. You could try re-drawing the spacetime diagram with the origin at various points along the hyperbola curve to confirm this.
In my electrodynamics book, it considers the constant force and F=dP/dt=m dU/dt=const. to get the result. From your video, I think constant proper acceleration should be A=dU/dτ=γ dU/dt=const. and they are different because of the gamma factor, which is varying because of acceleration. But both of them get the same result for hyperbolic motion. What is the difference between those two assumption?
Proper acceleration is just the length of the 4-acceleration vector A. If you keep this length constant, and maintain the acceleration in a single direction, this is equivalent to your criteria, where the 4-velocity vector U doesn't change.
@17.48 you say that because we are observing from an inertial frame, basis vectors are constant in spacetime so their derivatives go to 0 - please clarify why does an inertial frame imply that the basis vectors are constant in spacetime?
An "intertial frame" is just a constant orthonormal coordinate system. You can take a personal ruler (x basis vector) and a personal clock (t basis vector) and lay them out consistently over spacetime by repeating the basis vectors in a grid pattern. Since all the basis vectors are constant over spacetime, their derivative is zero.
True gravity is the result of curbed spacetime. Rindler coordinates are for flat spacetime, so any gravity-like effects are just the result of acceleration.
Dear Chris. I see that the initial position for any constant accelerated system cannot be the origin. I cannot find a reason for that. Another observation that is related to the former question is that the hyperbola branch cannot be over the 45 degrees line, it seems that for initial positions nearby the origin the systems cannot possess a low acceleration, because in that case, the hyperbola could have some part of the hyperbola branch cruising the 45 degrees line. Why the system seems to have only high acceleration nearby the origin and prevent low acceleration? And finally, the U vector is always normal to A vector, not only for constant acceleration? Again, I am very grateful for your time.
It's perfectly fine to have acceleration at the origin. You just shift the hyperbola worldline to the left so it is on the origin. The Rindler horizon would shift left by the same amount. I will explain why the different hyperbolas require different accelerations in the next video, 105b. And yes, U dot A = 0 is a result of U having a constant length. It is true 100% of the time, constant acceleration or non-constant acceleration.
Hey chris, I am struggling to understand why the velocity and acceleration vector should be orthogonal to each other. Isnt the entire point of acceleration to increase the velocity and thus change its length, especially when we're dealing with one space coordinate only in the minkowski diagram?
There are 2 types of acceleration: linear acceleration and angular acceleration. Linear acceleration is when the acceleration vector is parallel with the velocity vector. This would indeed make the velocity vector longer (or shorter, if they point in opposite directions). Angular acceleration is when the acceleration vector is orthogonal to the velocity vector. Angular acceleration doesn't change the length of the velocity vector, it just changes its direction. In 4D spacetime, 4-acceleration is always 100% angular. Does this make sense? I'm not sure what you mean by "we're doing with one space coordinate only".
The magnitude (and dot product) of 4 velocity is always constant (c) . Thus it can only change in orthogonal direction (in Minkowski sense) to itself. Think about an analogy in euclidean space. If a vector has constant length, it can only change perpendicular to itself.
I am having a bit of trouble understanding acceleration without time? What is purely spatial acceleration? How would an observer in the MCRF think they are stationary if they are measuring their own acceleration with their accelerometer?
The variable that we take derivatives with respect to for 4-vectors is always proper time tau. By "purely spatial acceleration", I mean the acceleration 4-vector A has x,y,z components only (these would be d^2 x/dtau^2, d^2 y/dtau^2, d^2 z/dtau^2 ) and that the time component of the 4-acceleration vector is zero (d^2 ct / dtau^2 = 0). The observer in the MCRF thinks they are stationary because they are the origin of their own coordinate system. Any time they measure the position of an object (car, ship, planet, etc.), they always measure position relative to themselves. By definition, their position with respect to themselves is zero, so they always have a 4-velocity that only has a time component (d ct/dtau) and no space components (dx/dtau = dy/tau = dz/dtau = 0). However, if they are in a non-inertial frame, an accelerometer that they're holding would measure a non-zero value. From their point of view, they'd feel a strange "pull" or "weight" in a direction, similar to how you'd feel in a car of someone stepped hard on the gas pedal.
In the example, Rindler is accelerating in the +x direction, except the observer in the accelerating frame perceives himself to be stationary in space. And the observer in the frame has only spatial acceleration. Which seems to contradict the other perception of being stationary in space. Rindler measures a proper acceleration, a force?. But in his reference frame, he is stationary. Does Rindler then perceive that other frames are accelerating to the left? In the equation for 4 acceleration (at time 13:10), assuming only x dimension, the component, a, is constant? And the spatial bases constantly change?
I think you missed an opportunity at 7:45. The reason U dot A is zero has an important interpretation. Acceleration in special relativity is always a rotation. All objects are always moving at the same speed through spacetime. We can only ever “rotate” the direction. If U dot A we’re nonzero, it would imply an object moving “faster” or “slower” than c through spacetime, which can’t happen. A perfect analogy in Newtonian physics would be rotation as well. It is as if the rules say we can only ever travel at the same speed, and we are only allowed to change direction. In that interpretation, U dot A is necessarily zero as well.
Hello eigenchris I have another question! Recently I've been watching Dialect (another RUclipsr that talks about Special Relativity) and then some other videos of other creators, and it got me confused by the notion of absolute vs relative acceleration in Special Relativity (lets not include General Relativity)... I know that in Special Relativity proper acceleration is an absolute. Any observer no matter what he does he will always know that someone is undergoing proper acceleration. For example a spaceship firing its rocket engines - that's something that everyone can see and deduce that this spaceship is infact undergoing proper acceleration. But my question is about coordinate acceleration. Is that always relative? Do you have any specific video that talks more about the relative nature of coordinate acceleration?
Coordinate acceleration is relative. If you have an observer inside a rocket, you can always put the origin of your coordinate system.inside the rocket. This makes the coordinate position, velocity, and acceleration all equal to zero. I don't have too much else to say, as coordinate acceleration is rarely a useful concept in relativity.
The 2 postulates of special relativity are that (1) the laws of physics are the same in all inertial frames, and (2) the speed of light is constant in all inertial frames. Neither of these apply to non-inertial frames. Non-inertial frames don't need to follow either of these rules, so it's fine for the speed of light to be not c.
@@eigenchris But if every specific point in a non-inertial reference frame can be considered inertial, why isn't the speed of light constant at every point?
@@pedrokrause7553 The speed of light is c at every point, when measured in an inertial frame at that point. I go into more detail regarding this in 105b.
27:25 I take it that in Rindler's frames, S has no et~ component because A is parallel to S, (A=(c^2)S) and A has no et~ component. Since S.S is constant in all inertial frames, and no t' component ever appears in calculation of S.S, then |S|, (the distance to Einstein's origin) is constant. The distance to Einsten's origin is lorentz contracted by increasing amounts that are exactly in tandem with Rindler's increasing speeds.
@@eamon_concannon Rindler coordinates are a bit tricky because the basis vectors change from point to point. Doing integration of vectors along a curve is also a bit tricky because we need to use the Rindler metric (video 105d, which isn't out yet) to ensure the changing lengths and directions of the basis vectors are taken into account. (et~).(et~) does not equal +1 for Rindler, it equals (alpha x~/c^2)^2. I'll be explaining this in future videos.
It's difficult for me to understand, though I haven't watched the whole duration, but some issues are bugging me. In the beginning, like in 5:33, you drew the worldline in the spacelike section of the light cone. In my current understanding, that means the events are unobservable from the current wordline. Shouldn't the trajectory be drawn in the timelike section, but the trajectory changes as you move forward in your proper time? (because the velocity vector changes).
The timelike and spacelike portions of the diagram only make sense for the origin point. Einstein is able to send light beams to Rindler earlier on in his journey just fine. And any light beams that Rindler sends out will be received by Einstein.
Can you help with a confusion? At 13:00 you show the equations for U and A. U has a time component but zeros for spatial components, while A has non-zero spatial components. But if A is the derivative of U with respect to tau (proper time) how can the derivative of zero-valued components with respect to proper time yield non-zero components? I understand the accelerometer argument and I understand the dot-product proof of orthogonality, but this part isn’t carrying through with me.
You'll have to keep watching until Relativity 105e, when I discuss the covariant derivative. The short answer is that the vector components are staying the same, but the basis vectors are changing. The required terms pop out of the basis vector derivatives.
Dear Chris. My 2nd question In minute 12:21 you show graphically, that U must be orthogonal to A, but in minute 7:29 you try to show the same idea, but in this last graph, the U vectors are as world lines of different observers at different velocities, but representing the same time. I feel confused My question is this. Did you make the diagram in 7:29 just to show that U is constant because they are in a hyperbola of constant radio? Also, the axes do not represent components of vector U, and because of that, I cannot see all the U vectors concentrated in one origin.
The diagram at 7:29 is meant to show a collection of different U vectors, all with the same minkowski length (they all have their tips on the same hyperbola). I realize I labelled the axes as ct and x. I probably should have labelled them as U^t and U^x.
Rindler Horizon has been a long time struggle for me because I have never understood intuitively the reason light from behind the origin, Lo =C^2/α never catches up to the accelerating observer. I am not sure if the following is close to correct, but I think it satisfies my struggles. It is what I have recently worked out. Just not sure how close to correct this is. Lo is S = c^2/α (which has meters as a dimension). That distance is constant from the accelerating frame, regardless of the acceleration and any change to γ or the frame velocity. L' would be the distance from the inertial frame, with orthonormal bases. Lo is proper distance and is the longest measurement of that distance. L' is shorter, by length contraction, and gets shorter with each new MCRF because of increasing γ. As L' gets shorter, light itself slows to a stop. Is light immediately, at T=0 seen to be stopped. Or does it slow to a stop as y increases. For light origin closer than S, does it also continue to slow as y increases? Does it eventually come to a stop?
Looking at the graph at 25:43. Does the speed of light increase as the accelerating observer decelerates before T=0. Then after T=0 does the speed of light decrease until it slows to a stop, as the accelerating observer accelerates away from the origin?
The reason is because at each point on the hyperbolic world-line, the four-position vector is tangent to the hyperbola, and the acceleration vector is Minkowski Orthogonal to the four-position vector. Minkowski orthogonal refers to the time and space axes being at equal angles with a beam of light. When the ct and x axes grow closer together, with the angle between the beam of light becoming less than 90 degrees, this indicates greater time dilation, and lenght contraction. In the next video, in the first half, the dot product of the four-position vectors for the Rindler coordinates are shown using the hyperbolic trignometric idenity, and dot products are all invariant. You need to remember that it is the momentary comoving inertial frame of reference that we are concerned with, in which the time and space axes are becoming closer and closer together.
Sorry, I missed this comment. Basically, everyone in relativity considers themselves to be stationary, with the world moving around them. However, if I'm in a car that's accelerating with respect to the ground, I'll measure non-zero values on my accelerometer. I'll believe myself to be stationary, with the road accelerating beneath me, but I will also feel a "force" pulling me back into my seat (this is essentially me feeling my own proper acceleration).
Hi eigenchris, may I ask whether there is anything similar to “Rotating reference frame” in special relativity? Just like the “Rindler coordinates” in special relativity corresponding to the “uniform acceleration frame” in Newtonian physics ? If such thing does not exist, what is the difficulty to define it? Thanks
You can't really invent a sensible coordinate system for a rotating frame in special relativity. You can look up "ehrenfest paradox" or "rotating frame special relativity" to try to learn more. It ends up being impossible to to define "simultaneity" for a group of observers in a rotating frame. In the Rindler coordinates, I can define a set of rockets and draw lines of simultaneity in the spacetime diagram to indicates times when all rockets agree is "now" (basically, you can draw a line that is spacetime-orthogonal to all the rocket tangent vectors). You can't define these lines of simultaneity for a rotating frame in special relativity, so it becomes very difficult to use common sense intuition when describing the physics, or to make a coordinate system for it. Your spacetime diagram would have to be 3D: a 2D plane of space, and an extra dimension of time. Your set of observers would form a ring in the 2D plane and they would all start spiraling up along the time axis in a helix as they rotate together. Their tangent vectors along their worldlines all point in directions that don't share a common orthogonal plane of simultaneity. So I don't see how you could make a sensible coordinate system without isolating one of the rotating observers as "special".
@@eigenchris thanks soooo much !!! Btw I really love your videos, it is really clear. I would love to see if there is a quantum field theory videos like this if you have time to make it - just an idea no pressure.
@@vincentxu2637 I won't be making QFT videos sadly. The main reason is I don't understand QFT, so you won't be getting any "really clear" explanations from me. The other reason is that QFT is around 3 semesters worth of material, so it would have to be much longer than this GR playlist. I don't think I have the energy for that. I will be covering a little bit of the Dirac equation in a couple months in my spinors playlist, which I'm working on now.
Chris, I calculated the rindler horizon for earth for plugging in **g** as acceleration and I got 0.97ly. This means the distance between earth and its rindler horizon is 0.97ly so nothing beyond that can affect earth in the reference frame of the earth. With this logic, we should never see light past 0.97ly and yet we do. What’s going on here?
I'm still learning the exact way General Relativity connects to Special Relativity, so I can't answer this with 100% scientific accuracy, but you need to keep in mind that the Rindler Horizon is *behind* the accelerating observer, in the same direction as the ficticious force they feel. So the equivalent horizon for gravity would be inside the Earth, in the same direction as the gravitational pull. Earth does have a very very small "Schartzschild Radius" of about 1cm. This would be the radius of Earth's gravitational event horizon if Earth were a black hole with the same mass, but since this radius is inside the Earth's interior, I don't think it's very physically meaningful.
The Rindler Horizon seems to me to be more like the Cosmic Horizon... the beams of light emitted from it and forward can no more reach our earth... Thank you.
Dear Chris. This video made me ask myself several questions, that I hope to be able to formulate after some time thinking about them. I also hope not to be a nuisance. This is my 1st question In minute 3:09. What do represent all those hyperbolas on the right? Are they systems with different constant accelerations? I also wonder, if the system starts its movement far from the origin, how will be its trajectory according to different values of constant acceleration? And what mean the straight lines over and below the (x) axis? I guess I need to see different space-time diagrams for different accelerated systems viewed from an inertial frame
The coordinate system on the right at 3:09 is Rindler Coordinates, which I will talk about in the next video. Each hyperbola is an accelerating observer with constant distance (1,2,3,...) from the origin. Each hyperbola to the right accelerates less hard than hyperbolas on the left. The diagonal line on top of the x-axis is the rindler horizon. The diagonal line below is not a horizon but it is the edge of the coordinate system. I will cover all this in future videos.
I think that's called the "cosmological horizon", but it's the same idea. Horizons are given different names depending on the context that they appear in. en.wikipedia.org/wiki/Cosmological_horizon
Isn't it weird that in flat spacetime constant acceleration has some kind of relation to the position of the hyperbola? Isn't that almost like some acceleration field?
@@eigenchris I mean an acceleration field. Some sort of function that assigns an acceleration to points in a space. In this case, a field which assigns a constant acceleration to any point on the x axis.
Why is U dot U equal to U(t)^2 - U(x)^2? (at time 15:30). Is that related to the equation for an hyperbola, or some how to the dot products of the bases?
It's due to the Minkowski metric I introduced in 104e, where et.et = +1 and ex.ex = -1. This is what gives us the definition of "Minkowski Orthogonal". A set of vectors with the same Minkowski length will have tips that sit on the same hyperbola.
The t-tilde coordinate is the time measured by Rindler as he travels along the green hyperbolic wolrdline. This is also Rindler's proper time (tau). However, if we picked another hyperbola (one that was closer to or farther away from the origin), and measured time along it, it would not be equal to t-tilde. It would have its own proper time tau which is different than Rindler's.
Wait, I'm confused... If I'm moving towards a light beam in space, say that I travel at 50% C towards an incoming lightbeam. What is going to be the speed of the lightbeam from my perspective? Is it going to be C? Or is it going to be C + my speed, so 1.5C? I thought that no matter what you did (accelerate or not) you will always measure the same speed for light, no? Also, what if I perform the same experiment, but I'm now accelerating towards the lightbeam? Will that change anything? Wil I measure a different speed for light other than C in any given moment?
I explain this a bit more in the next video, but for inertial frames, the speed of light anywhere in spacetime, as measured by you, is globally equal to C. When we look at non-inertial frames, the LOCAL speed of light is always C. This means that a light beam right next to you will always have a speed of C in your reference frame. However, if you are in a non-inertial frame and look at a beam of light that is far away from you (non-local), its apparent speed might not be C. In this Rindler horizon example, light "behind" the accelerating observer will have an apparent speed that's less than C, and light "ahead" of the accelerating observer will have an apparent speed that's more than C. Light that's right next to the accelerating observer will have a speed of C.
@@eigenchris Fascinating! So how far is "far away from you"? Is it like 1 light year? So... Inside a lightyear bubble from my perspective, I will always measure the speed of light to be C regardless of my motion or acceleration, but beyond that 1 lightyear bubble I will start measuring light to have progressively slower speeds depending on my motion or acceleration?
@@-_Nuke_- Technically speaking, when I saw "non-local", I mean any point that isn't the exact point you're currently at. So even 1 cm away from you the speed of light will be slightly different, if you're in a non-inertial frame. You've watched 105b by now so you've seen the conversion factor for getting the non-local speed of light is "x * c/D". The speed of light changes linearly with distance from the observer. Although this is due to measuring the speed of light with a non-local clock. With your local clock and rulers, the speed of light will always be "c".
There’s an issue with the accelerometer definition: imagine we simply transform into the frame of the accelerometer. Surely the accelerometer doesn’t have its own accelerometer (if so, apply process iteratively). Therefore, we have a class of frames whose state of motion is technically undefinable from the outset - meaning SR cannot treat all types of motion.
@@eigenchris I guess this depends on what you define as an "accelerometer". The term is somewhat ambiguous to begin with, so let's take a simple set-up: a mass attached to a string (like you would have in your car) that swings back and forth to indicate acceleration. Now transform into the frame where the attached mass is at rest. Obviously, it has no accelerometer by which to judge the reference of its own acceleration. So, by the definition you've provided for accelerated frames (i.e. having and observing an accelerometer) the mass would not be able to determine whether or not it was in an accelerated frame.
Try this, imagine we have a ball inside but detached inside a larger hollow ball. Have the outer ball be attached to the accelerating frame then if we change frames of reference we'll know we accelerated (there was a force exerted on our frame of reference) because the inner ball is assumed to be detached as well as inertial. But at the start of our motion a fictitious force would arise to accelerate the inner ball cause it moves so the conclusion is that we accelerated because it will later impact the outer ball. Use this accelerometer if you desire to determine whether we are or are not accelerating due to a force. .
@@substantivalism6787 Dialect, in this case, wouldn't the mass on the string need some sort of explanation as to why it is "floating" in the air instead of being suspended on the string completely vertically (due to gravity)? Or conversely, need an explanation why the car is moving around it? If no forces have been applied, I would assume the change in the mass's motion can only be attributed to the existence of non-inertial frame. I think Justin has the right idea: an accelerometer is basically like a mass on a spring sitting inside a rigid container. If the container accelerates, the mass will want to "stay where it is", and compress or extend the spring parallel to the acceleration. This spring compression is observable whether you are in the frame of the mass or in the frame of the container. Justin' description is basically this situation, except the spring is removed, and so acceleration is determined by in mass colliding with the container walls instead.
Hi Chris, can you please explain why U.U = c*c is still true even when you are accelerating? In Relativity 104f video, you showed U.U = c*c when you are NOT accelerating by using dt/dtau=gamma. Gamma, in this case, is a constant because beta is a constant (i.e. inertial). It is not clear to me on why U.U = c*c is still true when you are accelerating. Thanks.
Hi Wen. I think the formula I gave for U still works just fine, even if dt/dtau = gamma is a changing function of the 3-velocity u instead of being a constant. Even when I calculated the 4-acceleration in Relativity 104f, I allowed for the possibility for a non-constant gamma; you'll see there's a quick slide where I take the derivative of gamma with respect to a time coordinate and get a non-zero result. As long as you use a standard Minkowski-orthonormal spacetime basis that's constant everywhere, the formulas in 104f should work no matter how the target object is moving through spacetime. When we have a non-inertial basis that changes from point to point, taking derivatives can get more complicated, but I will discuss this in future videos.
@@eigenchris Thanks Chris, I did not realize gamma can be a function of the changing 3-velocity u. This is little subtle to me. The 4-acceleration in Relativity 104f video always confuse me bit.
@@wenchang8241 I wouldn't worry about the 4-acceleration in 104f too much. I showed it mainly for completeness. The acceleration in 105 will be more important.
Just to help clarify for me, I think that the motion is Rindler initially approaching Einstein, but slowing down. Then Rindler turns around and moves away from Einstein with increasing speed, all along the x axis. All the time, the acceleration is constant.
Is there an intuitive way to understand what the hyperbolic angle physically represents in terms of the area of the sector in front of the hyperbola? Sector angle as (α/2c) T? T=0, sector area is 0 T=1, sector area is α/2c
Sorry, I didn't see your 2nd question. I discuss cosh and sinh and the areas in the final 10 minutes of this video: ruclips.net/video/km7WTO_6K5s/видео.html
@@eigenchris It is like the argument (α/c*Τ) is a dimensionless number reflecting the ratio of the proper acceleration to c, then multiplied by T. Not unlike β. Or perhaps, better (αΤ)/c. How great the proper acceleration is wrt to c, times how long, T, the acceleration occurred.
I have a question. I see how rindler horizons work in the spacetime diagram, but I don’t see how they work in reality. You’re saying that if I were to release a light beam right now that travels in the x direction, and I was at a position far away but accelerated towards it, I still wouldn’t see it even after billions of years? It still wouldn’t make sense to me in I picture it in real life. I would surely have intersected with the photon no? Also, I’ve learned that the rindler horizon actually outputs out particles which can heat up the accelerated object (called the unruh effect). Will you be explaining this too in a future video?
One thing to keep in mind is that distances to Rindler Horizons are usually extremely large (note the c^2 in the numerator). Another thing to note is that constant acceleration is an extremely difficult thing to maintain for a long time. When you're driving a car, you normally accelerate up to a given speed and then maintain that speed for your car trip (no more acceleration). To accelerate forever, you'd need an infinite amount of fuel/energy. All you really need to know, mathematically, is that a hyperbola (ct)^2 - x^2 = -1 will never intersect the beam of light x = ct (you'll see that you get 0 = -1, because there are no solutions). Also, I've seen the PBS spacetime video on the Unruh effect but I know very little about it. I won't be covering it because I don't know quantum very well. I'm going to focus on relativity without quantum.
1. You are accelerating *away* from it (and the origin) - all in the upper right quadrant. The wave fronts of light follow you at a constant distance (when you look back over your shoulder), with E- and B-fields frozen into one static spacial sine pattern, I imagine. 2. The effect of seeing additional real particles on top of the virtual vacuum fluctuations due to acceleration is a result of quantum field theory.
The rule for measuring distances in relativity is to measure the length along a line of simultaneity. On the Rindler hyperbola, all the lines of simultaneity point back to the origin, so that's the point we measure to.
@@eigenchris After all this time this finally makes complete sense. Basically the hyperbola that is the worldline is always mapped to itself by Lorentz boosts, so if we do an active LT from an event B on the hyperbola to an event A (on the horizontal x axis) on the hyperbola, we are now in the rest frame of the observer at B, and B is now where A was before, so the proper distance to the Rindler horizon remains the same
I'm pretty sure Einstein said that you couldn't handle non-inertial frames in SR. An accelerating object doesn't magically "know" it's accelerating but rather sees a uniform gravitational field in accordance with the equivalence principle. I agree with the earlier commenter, an accelerometer seems like a pretty flimsy and ill-defined pretext for distinguishing acceleration.
Do you have a source for when Einstein said that? And can you explain why you think an accelerometer is flimsy for this case? I think the concept of hyperbolic motion was discussed only a few years after Einstein's original 1905 paper on special relativity. These 1909 and 1910 papers by Born and Sommerfield discuss hyperbolic motion: en.wikisource.org/wiki/Translation:The_Theory_of_the_Rigid_Electron_in_the_Kinematics_of_the_Principle_of_Relativity en.wikisource.org/wiki/Translation:The_Theory_of_the_Rigid_Electron_in_the_Kinematics_of_the_Principle_of_Relativity The textbook "Gravitation" by Misner, Thorne and Wheeler also discusses it.
@@eigenchris In his 1916 GR paper he goes on a big spiel about how special relativity, like Newtonian mechanics, arbitrarily privileges acceleration and so is inadequate for accelerated frames of reference. Re-reading over it now though, it seems more like he just found SR's treatment of acceleration epistemically unsatisfying. So I might just be confusing how to interpret his stance. As to the accelerometer question, I feel its flimsy because I've never come across a rigorous formulation of what exactly constitutes an accelerometer (texts always just give comparative examples)
@@WSFeuer I have another 5 videos planned where I will continue talking about acceleration in SR (things like how to keep extended bodies rigid during acceleration and how to get proper time along curves). You might feel better about the concept after watching those. I'm repeating myself from my other comment, but I think an accelerometer basically just measures that feeling you get in a car when the driver slams on the gas pedal. It uses a test mass on a spring. If acceleration is parallel to the spring, the spring either extend or compress. The amount the test mass is displaced gives you a way to measure the acceleration in that direction. A full 3D accelerometer would be 3 of these test masses in mutually perpendicular directions (or the same test mass with 3 perpendicular springs attached to it). Do you feel this is inadequate?
@@eigenchris Hey I love your videos, I watch them religiously :-) Sorry to make you repeat yourself, but yeah, I did read your earlier response from the other commenter and here's my issue with it: to establish a rigorous definition of an accelerometer, say, using a mass on a spring as you suggested, you have to measure a displacement. But measuring a displacement means first fixing your coordinate system. But the whole point of the accelerometer is to determine whether or not your system is fixed in the first place! Do you see the issue?
@@WSFeuer I don't see the problem with choosing a coordinate system to do the measurement. The spring compression could be measured in either the frame of the car/container or the frame of the mass, and it would come out to be the same number in either case. I think you'd have to conclude that either (1) there's a force acting on the mass or (2) the container is accelerating. If we exclude gravity since we're in SR, and assume the mass has no charge or magnetism, I'm don't see how a real force could be acting on the mass given known laws a physics, so I'd have to conclude the container is accelerating.
Chris, have you or anyone considered non constant accelerating frames of reference? In real life this can happen. People have observed acceleration on acceleration. Also what about deacceleration? Please respond. Jack Rubin
I don't plan on talking about non-constant acceleration, but the "U . A = 0" and "U . U = c^2" formulas would hold for those as well, so if you invent your own worldlines, they must obey these equations. Deceleration is just negative acceleration (the bottom half of the hyperbola is just decelerating before the turning-around point on the x-axis).
If the accelerated observer is charged and is accelerated in an electric field, how would the observer feel it's acceleration and measure it, because it looks as everything around the observer is accelerating. So how can they measure their own acceleration from their own clock?
If the observer was carrying an accelerometer, they should be able to detect that acceleration is happening. I think on a basic level an accelerometer is like a mass on a spring, and if the observer is accelerating, the mass's neutral position on the spring will shift, indicating acceleration.
@@eigenchris But if the spring is also charged and is accelerating with the mass wouldn't it detect no acceleration since the position between the mass and the start of the spring will always stay the same. A bit like if you drop your accelerometer in a gravitational field
@@joshuapasa4229 I don't think a charged accelerometer would function properly as an accelerometer, because it would respond to changes in electric fields even if it is in an inertial frame. I think the issue is that there's a fundamental difference between gravity and electric fields: Newton's 2nd law is F = mass*a, not F = charge*a. This means that a body being "pulled" by gravity (in free-fall) is travelling along an inertial frame (there's no actual force, just the curvature of spacetime), whereas a body being pulled by an electric field is undergoing a force.
@@eigenchris Why can't charge be represented as a curvature in space-time, but the strength of the curvature depends on the charge of the object in the curved space time?
@@joshuapasa4229 This is starting to get outside my knowledge. I think Electric and Magnetic fields can contribute to the curvature of spacetime because they contain energy. You can google the Reissner-Nordström stress energy tensor to see how the electric field comes into play in the Einstein field equations. (The Reissner-Nordström solution gives the spacetime solution for the curvature around a charged spherical body.)
When accelerations and velocities on on the scale of our everyday experience, we're only looking at a very small portion of the hyperbola near the origin. In this very small region, a hyperbola and parabola are basically impossible to tell apart.
@@eigenchris Thanks a lot for your reply... Can we deduce those hyperbolic trajectories by selecting suitable metric and solving geodesic equation using einstein equation..?
@@keshavshrestha1688 These hyperbolas are not geodesics, because they are the result of being on a path with constant proper acceleration. Geodesics are paths where the proper acceleration is always zero, so the hyperbolas are not geodesics. I talk about geodesics in video 105f.
Hi Eigenchris, 23:58 Sorry, but did I misunderstand something? Shouldn't the spacetime interval lead to a constant proper time as well? Since I learned that proper time is defined as the spacetime interval divided by c.
In this case, tau is measuring the proper time along the hyperbola, which increases as you travel along it. The S-vector is measuring the interval between the origin and any given point on the hyperbola. This S-vectors turns out to have a constant interval value for any point on the hyperbola. Does that clear things up? Nice profile pic, by the way.
Something bothers me about about your 4-acceleration vector because at the upper part when τ>0 rindler has A pointing up. This means At and Ax are both possitive and this feels wrong. The temporal component of 4Acceleration should be pointing down to slow down the temporal component of 4velocity and give room for the spatial component of the 4Velocity to increace. also if the temporal component of 4Acceleration was facing down this would make the A and V geometrically orthogonal and the dot product A*V=0 would feel totally normal. maybe I am thinking weird because of this geometry is not familiar to me. I ve learned so many things through your channel and I can't THANK YOU enough for the amazing work you do!
I think you're still thinking in "Euclidean space" too much. When you take 4-velocity vector for a stationary particle and boost it, both its time component and space component increase. This is what's required to keep the formula (ct)^2 - x^2 a constant: both (ct)^2 and x^2 need to get bigger. The idea that one needs to decrease in order for the other to increase comes from the circle formula in euclidean space y^2 + x^2 = constant. Here if y^2 gets bigger, then x^2 must get smaller. Orthogonality in Minkowski space is also about 2 vectors making equal angles with a beam of light. This is the relationship that U and A have. It's not about "right angles" like in Euclidean space.
@@eigenchris Thank you. Yes you are right. What confused me is that I tried to picture those vectors in 3d-space .I used 1 spatial dimension for the 4V and v and 1 more spatial dimension for α so it can vary reltive to v. This gave rise to A spanning all three dimensions but I was thinking "too euclidean" in this "model". It seems to work btw because A is always at euclidean right angle to V but my mistake is hidden in my eucledean dot products so the geometry follows...I ll fix it because it misleads my intuition. I liked it because i saw that when α vector traces a circle in its spatial dimentions ,the A-spatial traces an ellipse but i guess in the corrected version it will do something similar. And btw this "ugly" equation of A is actually very elegant geometrically. It is just A=αγ^2 minus the projection of αγ^2 on V
@@eigenchris most classes just go by saying we need something about curvature which is a rank 2 tensor and has covariant derivate zero. Since Ricci doesnt work we go with a correction to Ricci and call it einstein tensor
Which btw is a shitty way of explaining it. Its almost as if, since LHS =0 we can equate it to RHS =0. Physics is kinda lost in this way of explaining it
@@eigenchris I did't expect a reply from the author. Thank you. You made excellent videos. I have not watched all the previous videos, but watched those 4-vectors related multiple times. I don't think my courses will be that detailed currently. I am here just for my curiosity.
@@146fallon9 Watching the previous videos might help, but if there's anything specific you're confused about, let me know and I can try to explain it better.
If A moves relative to B, B moves relative to A. There is no such thing as "stationary." Therefore both special and general relativity are WRONG! Idiots will disagree with me.
True. But IF you consider yourself to be stationary (just saying IF you do this) then SR and GR will tell you what you will measure... And that's the topic of relativity... IF now, you DON'T do that, then you don't have to worry about relativity since you somehow have a "bird's eye view" of spacetime... If you know how to do that, then you might as well be the next Einstein... Cuz so far we don't know a way to "escape" spacetime.
@@-_Nuke_- Imagine a "photon" moving away from you and say you are chasing it at 99999/100000ths the speed of light. If you could somehow "measure" that photon, you're saying it would be going c because your time is slowed down? Help me understand Einstein.
@@CandidDate Everything is moving inside *spacetime* at the speed of C But not always in space. Light is moving at C in space (because it doesn't have mass) but we Humans (massive objects) move at C mostly in time and a little bit in space... (movement in time is called "aging" and in space is called velocity) So if you combine my space speed and my time speed using the Lorentz factor γ=1/sqr(1-(V^2)^2/(c^2)) you should always get C. So yeah, everytime you measure the speed of anything, IF you know how to measure speeds in spacetime CORRECTLY you should always find that everything is moving at the speed of C in spacetime, not just light... Light isn't anything special, it just has all it's speed in space, most of the things in the Universe have a big portion of their C speed mostly in time and a lesser portion in space... An object moving at 1% C in space (about 10800 km/h) you are still moving at 99.9949% C in time... As you can see from here www.emc2-explained.info/Dilation-Calc/ if you put 1 in the first box, that's 1% C in space and 99.9949 in time... Whether or not space contracts or time dilates is irrelevant... Everything is moving at C in spacetime.
Absolutely appreciated for this video! Professor Wolfgang Rindler was one of my favorite professor when I was working on my Ph.D in Physics. He was the one recruited me to chase my dream when I was young ... This video helped the rest of the people to understand "How to bridge the General relativity with Special relativity with gauge invariance. I learned when I was young and glad to see this video to resurface the beauty of Rindler's Horizon.
The best set of lectures I've ever seen. I'm waiting anxiously for the ones to follow on GR. Great job.
@eigenchris your courses on Tensors(algebra+calculus) along with these Relativity series deserve to be put out in form of a Relativity Introduction book. Neither of books out there on relativity I have ever read, introduce in ways your courses do. Great Job!!!
Extremely great videos. The full serie have helped me to understand the Special Relativity in great details
Great videos, I watched your Tensor Calculus series to get myself caught up because I know Einstein Field Equations have Tensors. Extremely great videos💯
Minor correction: at 27:00 where you show Rindler's acceleration stopping, his curve should not be a straight line vertically, but rather a straight line (or ray) tangent to the green curve at that point. It will still be true that all the light beams will cross this line eventually. Thanks.
yeah i just noticed it
and was going to comment but then saw this
I was waiting for this Chris,thank u.
This is so much educative that every student on phyics must learn this lecture.
Excellent video! I was just working on a video about the Unruh effect; now I can skip its intro and just link to this video.
I'll have to watch that when you're done. You can feel free to keep the intro if you want. It doesn't hurt to have multiple explanations of things online. Unless you feel it's a waste of time.
I was going to say "this is like porn" but I take that back, this is better
It's not actually a good analogy for porn. Porn has kinks and spacetime is smooth everywhere.
@@caseyglick5957 yeah, in terms of the dopamine circuits those are really similar! but, it seems to me that evolution cares about sex much better than physics! )
@@karkunow sorry that was intended as a joke, not actual commentary.
@@caseyglick5957 doesn't matter :)
@@karkunow And the universe has more regards for general physics then evolution, it seems (I know. I know... evolution is physics as well :))
honestly these videos are dope, thank you
yo i live for your vids man
23:57 an easy way to see this without hyberbolic identities is to note that d(S.S) =dS.S+S.dS.
Now, in the frame of accelerating Rindler, position S is purely spacelike (since S is parallel to acceleration A in Rindler coordinates), and dS is purely timelike (since in Rindler rest frame the velocity does not have space components). Therefore the dot products are zero and S.S is constant.
28:15 one can also think the horizon effect in terms of shifting simultaneity: as the velocity increases the "plane of simultaneity" tilts more an more, so that einstein's past becomes present for Rindler. The faster the farther away from Rindler Einstein "looks" .
And there is a point, the origin, where "Rindler's now" totally ceases to move in Einstein's time.
Left of this point the Rindler's "present" seems to move backwards in Einstein's time.
I think this also goes to show how we should be cousious to give "simultaneity" too much actual meaning...even in flat spacetime.
Lastly: Excellent video! Thanks.
Trying to finish a course on General Relativity through Coursera from HSE University in Russia before I lose access to it, glad to have videos like these to further explain the material.
The Coursera HSE Russia course on GR is fine if you know the material already. Otherwise, you get lost really fast.
Fabulous, Thank You.
Correction: @26:58 Rindler stopping accelerating would have a tangent world line to the hyperbola, so you must mean stopping velocity to have a world line parallel to the home frame world line. This would be a very high deceleration of course. But what I think you mean is stopping acceleration the new tangent world line would then be able to see the light since the tangent world line would enter all remaining light cones.
One mention, at 7:17 the dot product of a vector is not c^2 but the magnitude of the vector squared.
Great series about relativity, even I didn't learn differential geometry or vector calculus.
One question : is there a reason why all movements are described only in x direction and y, z are usually not considered?
Is it because to keep it simple or would you need to transform coordinates to make y and z direction to zero?
It's just to keep it simple. For Lorentz and Rindler transformations, the y and z coordinates don't change.
Thank you for making these videos available on RUclips for free. Love your generous work.
I had a doubt:
In the video 'relativity 104f', to derive 4-velocity, we had used the fact that derivative of basis vectors (with respect to proper time) is zero as they are constant throughout. Hence we could show that Minkowski squared length of U is c*c.
But while dealing with non inertial frames, and accelerated motion, shouldn't the derivative of basis vectors be non zero? How can we still say that Minkowski squared length of U is c*c?
The length of a vector should not depend on the coordinate system. It should be the same in all coordinate systems. If you like you can review video 103d for showing how the length of a vector doesn't change when you change basis.
your videos are really understandable. Could you tell any book or reference to study more about Rindler coordinates ?? That seems very interesting topic !!
I remember there was one short section on them in "Gravitation" by Misner, Thorne an Wheeler. It wasn't very long, but it gave the basic idea.
The concept of the Rindler horizon is set to become one of the most mind-blowing concepts in all of physics.
imagine you are chasing your friend at a constant speed, and although your friend is trying to accelerate to match your speed, they are still running slower than you. It seems inevitable that you will eventually catch them, since no matter how hard and how long they try, their speed will never be the same as yours. However, astonishingly, no matter how long you pursue them, you will never actually be able to catch them. Just think about it.
Makes sense. If they accelerate indefinitely they will eventually outspeed you and you surely won't catch them from that point on.
The real weird thing is how even if they are racing against your laser pointer (the literal universal speed limit), they can win the race if given the appropriate headstart, no matter how long you prolong the race.
Thanks a lot , Dear Chris.
N Das
Great explanation
2:54 Both a cylinder and a cone can be "rolled up" and, in reverse, "unwrapped" into a straight sheet. That's two dimensional surfaces, which we can see, as they can be embedded in a three dimensional space. Then, there are surfaces with some curvature and of these only an S^2 sphere can be embedded in three dimensional space, but, for instance, hyperbolic surface can only be projected. In other words, we can make a three dimensional model of a cylindrical space using just a sheet of paper. If we ourselves were capable of seeing higher dimensions (we only have a pseudo - spatial flat vision based on two projections, which are both identical, bar shifting and close to a stereographic one. Had we spatial vision, we would see, say, the other side of a monitor, on screen of which you see this message, without moving our head)
In the same way Minkowski space has zero intrinsic curvature - it can be rolled up from a 4d euclidean space and there will be no "crease" left after "unwrapping".
The only warning I'll give is that the 4D Eulcidean metric is fundamentally different than the 4D Minkowski metric, so that does make them fundamentally different, even if they are both flat.
Dear Chris, are you planning to discuss General relativity ?
I am waiting for Relativistic tensors like EM field and so on. You told that can be uploaded after X mas.
I expect the 105 videos to be done by December, and GR videos will come after.
@@eigenchris Thanks , Dear Chris.
27:31 could you please tell me why position vectors meet at the origin? Anything I missed from the previous videos?
As I said in the video, the 4-velocity vectors U (which are tangent to the worldline) are always minkowski-orthogonal to the 4-acceleration vectors, and the 4-acceleration vectors A are always parallel to the e_x basis vector for any given instantaneous frame. You can see at 23:23 that 4-acceleration A is parallel to 4-position S, so S is also parallel to the e_x basis vectors. S vectors always start at the origin by definition, since they measure the spacetime displacement from the origin. So this explains why all the S vectors start at the origin, and are also parallel to the e_x basis vectors everywhere along the curve.
I'm loving this series (the foregoing episodes are the best explanation of SR that I've seen). But I'm having difficulty understanding part of the world line of the accelerated frame. What is the physical meaning when t < 0? Does it look to an observer at t=0 that Rinder is moving (accelerating) towards him and then Rindler starts to move away again?
Yes. Loosely speaking, you could imagine an Rindler starts out at near-infinity, travelling left at near-the speed of light, with constant acceleration to the right. Eventually he will slow to a stop (this is the t=0 event on his worldline) and then speed up back to infinity, approaching the speed of light in the opposite direction.
@@eigenchris Ah. That explains it. I didn't realize that the acceleration vector was pointing to the right. But now I see that you mention that at 9:30.
Also, Thanks for your quick response. I must say, I really enjoy all of your videos; you are one of the best physics communicators I know of. I really appreciate your very methodical approach. What a wonderful service you provide to the world!
BTW: Have you ever thought of making a series on string theory?
In the case of non constant acceleration, would the 4 velocity change at different points in the accelerating frame world line. Would that mean that between instantaneous points on the curved world line that time is relative?
Everyone always thinks their own 4-velocity is c*e_t (this is true whether you are inertial or non-inertial). I'm not sure what your question mean about "time is relative" means. The proper time for a given worldline is objective, but outside observers in different frames will measure different values for the time between the start event and end event.
Einstein (from his How I Created Relativity lecture): "I was dissatisfied with the special theory of relativity, since the theory was restricted to frames of reference moving with constant velocity relative to each other and could not be applied to the general motion of a reference frame."
I looked elsewhere for his opinion on this subject, since this quote makes its seem almost certain he did not believe special relativity could be extended to accelerated frames. In other quotes though he is careful not to say that the postulates or principles don't extend to accelerated frames, rather only the equivalency of frames, so this particular quote may only have been a thoughtless imprecision on his part. However, if you read his intro to his 1916 GR paper, or his explanation of the Twin Paradox where he invokes GR, both of these heavily imply that he was not happy with the idea of treating acceleration in SR. Likely this is due to his Machian influence and his belief, which I quoted in our comments below, that absolute acceleration made no kinematical sense.
I guess I can't know for sure what he meant when he said that. I'll point out that in the next paragraph of that lecture he says: "all the natural laws except the law of gravity could be discussed within the framework of the special theory of relativity". That sentence is one I would more firmly agree with. The laws of electricity and magnetism result in accelerations of charged particles and I believe these accelerations are perfectly well-handled by SR.
Is this the 1916 paper you're talking about?en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity
This is a quote from the 2nd section: "Can any observer, at rest relative to K' then conclude that he is in an actually accelerated reference-system? This is to be answered in the negative; the above-named behaviour of the freely moving masses relative to K' can be explained in as good a manner in the following way. The reference-system K' has no acceleration. In the space-time region considered there is a gravitation-field which generates the accelerated motion relative to K'."
I agree with this. It's basically just the statement that an accelerometer measures zero when it's dropped from a plane and freely falling to earth, because both the mass and container move at the same rate and there is therefore no spring compression. The worldline of a freely falling object is a "geodesic" in GR, and my spring-and-mass accelerometer would measure how much an observer is deviating from a geodesic curve. I'm going to talk about this idea in the context of SR in 105f, because the basic idea still applies in SR... accelerometers in SR just measure deviation from inertial worldlines.
My explanation for the twin paradox is coming in 105d and it's nothing really deep... it's just that one twin has an inertial frame (straight line) and the other has a non-intertial frame (there-and-back zig-zag). In theory you could graph this so that the inertial frame is a zig-zag and the non-inertial frame is a straight line, but you can't do this transformation using a Lorentz transformation, so there is no reason to believe both observers are on equivalent footing. The non-inertial frame (which looks like a zig-zig from the inertial frame, and experiences ficticious forces) is fundamentally different and counts different proper time on its clock. I don't believe any GR needs to be used to explain this. Sean Carol says this even more simply... He says that just as distance depends on your path through space, proper time depends on path in spacetime: ruclips.net/video/Cg2tOUTE2F4/видео.html
@@eigenchris The quote from the 2nd section, where Einstein states a version of the EQ principle I certainly don't have any disagreement with. But I think my issue here again is, what defines the accelerometer? Having an accelerometer with us in a uniform gravitational field may allow us to measure that we are in an inertial frame, but once spring compression occurs we still have to determine whether it is the mass or container which is accelerating, which brings us straight back to the discussion we had below.
You can make the claim, for instance, that an object's deviation from a geodesic path on the spacetime manifold, if taken in reference to an object not deviating, would properly define an accelerometer. But then we have to determine which object is on the geodesic and which one is deviating. To do this we would have to know the curvature, or the value of Riemann tensor, of the spacetime locale we are in. This requires knowing the mass/energy distribution of the surrounding environment. (And if you want the curvature to be absolutely accurate, you have to account for the mass/energy of the whole universe, I believe.)
Thus, you come to the conundrum that is bothering me: the accelerometer is supposed to provide an objective "local" measurement of proper acceleration. But it hinges on having access to lots of information about your surrounding system, environment, etc. Thus I can't think of any way to come up with a "local" definition of one that is consistent.
Maybe this was why Einstein was so uncomfortable with it, and why he writes in that same 1916 paper: "We have to take it that the general law of motion... is partly conditioned, in quite essential respects, by distant masses which we have not included in the system under consideration."
@@WSFeuer Sorry, this second comment slipped by me. When you say the following: "but once spring compression occurs we still have to determine whether it is the mass or container which is accelerating"...
Sorry if we're going in circles, but let's go one step at a time. With my definition of accelerometer, given that the mass is electrically/magnetically neutral, can you list all possible physical scenarios you can come up with that would cause the spring to compress? (Or at least, list a few?)
@@WSFeuer Einstein's initial motivation for developing a more general theory was indeed to include accelerated frames. But while doing so, he came up with the equivalence principle, linking accelerated frames to gravitational fields. And of course, he then figured out that curved spacetime was needed to explain tidal forces in gravitational fields around massive objects. I think it was later discovered that you can indeed describe accelerated frames in flat spacetime using special relativity, so long as you add some extra assumptions (like the clock hypothesis). So now the more modern view is that special relativity can describe any kind of motion in flat spacetime, whereas general relativity is for curved spacetime. But this wasn't the view during Einstein's time (when I think it was believed that special relativity only can be used for inertial frames in flat spacetime).
is there an intuitive reason for why the position basis vector for the spaceship is always parallel to the separation vector from the origin in the "stationary" frame / why do lines of simultaneity radiate from the origin? Or rather, why does the spaceship measure the proper length to the origin when using it's own basis vectors?
23:53 does kt and kx represent the initial coordinates of the inertial frame? I am confused(what does it mean when we put them to zero?)
They represent initial conditions, by they are not exactly the initial coordinates. Since sinh(0)=0, kt is the initial time value. Since cosh(0) = 1, the initial position is c^2/alpha + kx.
Setting kt = kt = 0, the initial coordinates at tau=0 are ct=0 and x=c^2/alpha.
Hi Chris, at 27:00 if acceleration stops will the x position go vertically up and not go in the tangent direction of hyperbola at that point (continue with same velocity)? I did not get this part
Yeah, it would go in the tangent direction, not go vertical.
@@eigenchris Ok thank you for clarification and thanks a lot for all video series I find them very helpful as a newbie ..
Can entangled quantum particles transmit information across the Rindler horizon? Will the particles remai n in an entangled relation even after the observers of each is separated by more thjan the Rindler critical distance? If so, is that a transmission of info > c violateing special relativity?
I don't know how quantum mechanics interacts with horizons in GR. I know that the "Hawking Radiation" related to black holes has an equivalent called "Unruh Radiation" for accelerating observers. I'm not sure if that's related.
I love these videos and in really looking forward to GR!
But I think I might have spotted a tiny mistake. With the three rockets whose worldlines are tangents to Rindler's worldlines, the first and third pass through the origin; but if Rindler's worldline is a hyperbola I don't think that's possible. The only tangent that would pass through the origin would be the asymptote, surely?
Perhaps it doesn't matter here because we haven't yet proven that the path is a hyperbola, but I just want check I'm understanding this correctly.
For each of those 3 points, I was taking the point itself to be the origin of its own coordinate system. So I showed 3 separate coordinate systems there with 3 separate origins. But the choice of origin in spacetime doesn't matter since all points should be on equal footing.
@@eigenchris You're right! Thanks for replying. And I have to say this is the first set of lectures on Relativity where I haven't had to keep doing to try to understand every single point. Thank you for taking the time to explain everything clearly and thoroughly, with only a handful of places where you say "it can be shown that ...". Your approach is fresh and exciting! I look forward to every single video, and I can't wait to get on to GR!
26:55 f: _As soon as Rindler stops accelerating..._
Your diagram shows Rindler _stop moving_ which actually is a kind of infinite acceleration in the −x direction. Rindler stopping accelerating would look like the hyperbola at this point goes over into a straight line with the hyperbola's current slopes, without any sharp bend. This would still work since the slope is steeper than 45°, and the RINDLER horizon would vanish.
How is the 3 aceleration vector in this example? Does it still parallel to the four aceleration?
Rindler's 3-acceleration vector is parallel the spatial part of his 4-acceleration, because he has no y or z velocity, only x-velocity.
a lot of thanks !!
How close can the Rindler Horizon be located, if alpha is very, very small? I've never noticed the universe behind me freeze when I walk across the room.
The Rindler horizon distance is c^2/alpha. So if alpha is 1m/s^2, the Rindler horizon would be 9*10^16m behind you. Rindler horizons are usually very very far away.
Why the accelerated motion in spacetime is drawn in spacelike region? Isn’t the motion of particle in spacelike region is greater than light?
Or, I'm missing something there? Please explain.
It's the slope (4-velocity vector) of the worldline that indicates the speed. The slope of the hyperbola is always time-like (it's never more horizontal than a light beam at 45 degrees). The "timelike" and "spacelike" regions of the spacetime diagram only make sense from the perspective of the origin event point. You could try re-drawing the spacetime diagram with the origin at various points along the hyperbola curve to confirm this.
In my electrodynamics book, it considers the constant force and F=dP/dt=m dU/dt=const. to get the result. From your video, I think constant proper acceleration should be A=dU/dτ=γ dU/dt=const. and they are different because of the gamma factor, which is varying because of acceleration. But both of them get the same result for hyperbolic motion. What is the difference between those two assumption?
Proper acceleration is just the length of the 4-acceleration vector A. If you keep this length constant, and maintain the acceleration in a single direction, this is equivalent to your criteria, where the 4-velocity vector U doesn't change.
@17.48 you say that because we are observing from an inertial frame, basis vectors are constant in spacetime so their derivatives go to 0 - please clarify why does an inertial frame imply that the basis vectors are constant in spacetime?
An "intertial frame" is just a constant orthonormal coordinate system. You can take a personal ruler (x basis vector) and a personal clock (t basis vector) and lay them out consistently over spacetime by repeating the basis vectors in a grid pattern. Since all the basis vectors are constant over spacetime, their derivative is zero.
question- can rindler coordinates describe gravity or it's just any special case of general theory of relativity ?
True gravity is the result of curbed spacetime. Rindler coordinates are for flat spacetime, so any gravity-like effects are just the result of acceleration.
Dear Chris.
I see that the initial position for any constant accelerated system cannot be the origin. I cannot find a reason for that.
Another observation that is related to the former question is that the hyperbola branch cannot be over the 45 degrees line, it seems that for initial positions nearby the origin the systems cannot possess a low acceleration, because in that case, the hyperbola could have some part of the hyperbola branch cruising the 45 degrees line. Why the system seems to have only high acceleration nearby the origin and prevent low acceleration?
And finally, the U vector is always normal to A vector, not only for constant acceleration?
Again, I am very grateful for your time.
It's perfectly fine to have acceleration at the origin. You just shift the hyperbola worldline to the left so it is on the origin. The Rindler horizon would shift left by the same amount.
I will explain why the different hyperbolas require different accelerations in the next video, 105b.
And yes, U dot A = 0 is a result of U having a constant length. It is true 100% of the time, constant acceleration or non-constant acceleration.
Hey chris, I am struggling to understand why the velocity and acceleration vector should be orthogonal to each other. Isnt the entire point of acceleration to increase the velocity and thus change its length, especially when we're dealing with one space coordinate only in the minkowski diagram?
There are 2 types of acceleration: linear acceleration and angular acceleration. Linear acceleration is when the acceleration vector is parallel with the velocity vector. This would indeed make the velocity vector longer (or shorter, if they point in opposite directions). Angular acceleration is when the acceleration vector is orthogonal to the velocity vector. Angular acceleration doesn't change the length of the velocity vector, it just changes its direction. In 4D spacetime, 4-acceleration is always 100% angular. Does this make sense? I'm not sure what you mean by "we're doing with one space coordinate only".
The magnitude (and dot product) of 4 velocity is always constant (c) . Thus it can only change in orthogonal direction (in Minkowski sense) to itself.
Think about an analogy in euclidean space. If a vector has constant length, it can only change perpendicular to itself.
I am having a bit of trouble understanding acceleration without time? What is purely spatial acceleration?
How would an observer in the MCRF think they are stationary if they are measuring their own acceleration with their accelerometer?
The variable that we take derivatives with respect to for 4-vectors is always proper time tau. By "purely spatial acceleration", I mean the acceleration 4-vector A has x,y,z components only (these would be d^2 x/dtau^2, d^2 y/dtau^2, d^2 z/dtau^2 ) and that the time component of the 4-acceleration vector is zero (d^2 ct / dtau^2 = 0).
The observer in the MCRF thinks they are stationary because they are the origin of their own coordinate system. Any time they measure the position of an object (car, ship, planet, etc.), they always measure position relative to themselves. By definition, their position with respect to themselves is zero, so they always have a 4-velocity that only has a time component (d ct/dtau) and no space components (dx/dtau = dy/tau = dz/dtau = 0). However, if they are in a non-inertial frame, an accelerometer that they're holding would measure a non-zero value. From their point of view, they'd feel a strange "pull" or "weight" in a direction, similar to how you'd feel in a car of someone stepped hard on the gas pedal.
In the example, Rindler is accelerating in the +x direction, except the observer in the accelerating frame perceives himself to be stationary in space.
And the observer in the frame has only spatial acceleration. Which seems to contradict the other perception of being stationary in space.
Rindler measures a proper acceleration, a force?. But in his reference frame, he is stationary. Does Rindler then perceive that other frames are accelerating to the left?
In the equation for 4 acceleration (at time 13:10), assuming only x dimension, the component, a, is constant? And the spatial bases constantly change?
What about the light rays infront of Rindler? Will it appear faster or slower or same speed?
They will appear to go faster than light, according to Rindler's coordinates. I talk about this more in the next video (105b)
I think you missed an opportunity at 7:45. The reason U dot A is zero has an important interpretation. Acceleration in special relativity is always a rotation. All objects are always moving at the same speed through spacetime. We can only ever “rotate” the direction. If U dot A we’re nonzero, it would imply an object moving “faster” or “slower” than c through spacetime, which can’t happen.
A perfect analogy in Newtonian physics would be rotation as well. It is as if the rules say we can only ever travel at the same speed, and we are only allowed to change direction. In that interpretation, U dot A is necessarily zero as well.
Hello eigenchris I have another question!
Recently I've been watching Dialect (another RUclipsr that talks about Special Relativity) and then some other videos of other creators, and it got me confused by the notion of absolute vs relative acceleration in Special Relativity (lets not include General Relativity)...
I know that in Special Relativity proper acceleration is an absolute. Any observer no matter what he does he will always know that someone is undergoing proper acceleration. For example a spaceship firing its rocket engines - that's something that everyone can see and deduce that this spaceship is infact undergoing proper acceleration.
But my question is about coordinate acceleration. Is that always relative? Do you have any specific video that talks more about the relative nature of coordinate acceleration?
Coordinate acceleration is relative. If you have an observer inside a rocket, you can always put the origin of your coordinate system.inside the rocket. This makes the coordinate position, velocity, and acceleration all equal to zero. I don't have too much else to say, as coordinate acceleration is rarely a useful concept in relativity.
How can special relativity deal with non-inertial reference frames since speed of light doesn't remain constant in those frames?
The 2 postulates of special relativity are that (1) the laws of physics are the same in all inertial frames, and (2) the speed of light is constant in all inertial frames. Neither of these apply to non-inertial frames. Non-inertial frames don't need to follow either of these rules, so it's fine for the speed of light to be not c.
@@eigenchris But if every specific point in a non-inertial reference frame can be considered inertial, why isn't the speed of light constant at every point?
@@pedrokrause7553 The speed of light is c at every point, when measured in an inertial frame at that point. I go into more detail regarding this in 105b.
27:25 I take it that in Rindler's frames, S has no et~ component because A is parallel to S, (A=(c^2)S) and A has no et~ component. Since S.S is constant in all inertial frames, and no t' component ever appears in calculation of S.S, then |S|, (the distance to Einstein's origin) is constant. The distance to Einsten's origin is lorentz contracted by increasing amounts that are exactly in tandem with Rindler's increasing speeds.
Yes, I'll get to this eventually, but for Rindler, S = (x~) (ex~); there's no time component. The spacetime interval s^2 = S.S = - (x~)^2.
@@eigenchris so ex~ basis vector is always constant for Rindler and hence we can integrate acceleration twice wrt tau to get S= (x ~) (ex~) .
@@eamon_concannon Rindler coordinates are a bit tricky because the basis vectors change from point to point. Doing integration of vectors along a curve is also a bit tricky because we need to use the Rindler metric (video 105d, which isn't out yet) to ensure the changing lengths and directions of the basis vectors are taken into account. (et~).(et~) does not equal +1 for Rindler, it equals (alpha x~/c^2)^2. I'll be explaining this in future videos.
@@eigenchris OK. I'll come back to this point about why S is parallel to ex~ in Rindler's frame after I've seen more videos. Thanks.
It's difficult for me to understand, though I haven't watched the whole duration, but some issues are bugging me.
In the beginning, like in 5:33, you drew the worldline in the spacelike section of the light cone.
In my current understanding, that means the events are unobservable from the current wordline.
Shouldn't the trajectory be drawn in the timelike section, but the trajectory changes as you move forward in your proper time? (because the velocity vector changes).
The timelike and spacelike portions of the diagram only make sense for the origin point. Einstein is able to send light beams to Rindler earlier on in his journey just fine. And any light beams that Rindler sends out will be received by Einstein.
What shape is the Rindler horizon?
I have to be honest and say I've never calculated it in 2D or 3D, so I don't know. Sorry! If I find out I'll respond to this comment.
Can you help with a confusion? At 13:00 you show the equations for U and A. U has a time component but zeros for spatial components, while A has non-zero spatial components. But if A is the derivative of U with respect to tau (proper time) how can the derivative of zero-valued components with respect to proper time yield non-zero components? I understand the accelerometer argument and I understand the dot-product proof of orthogonality, but this part isn’t carrying through with me.
You'll have to keep watching until Relativity 105e, when I discuss the covariant derivative. The short answer is that the vector components are staying the same, but the basis vectors are changing. The required terms pop out of the basis vector derivatives.
Dear Chris. My 2nd question
In minute 12:21 you show graphically, that U must be orthogonal to A, but in minute 7:29 you try to show the same idea, but in this last graph, the U vectors are as world lines of different observers at different velocities, but representing the same time. I feel confused
My question is this.
Did you make the diagram in 7:29 just to show that U is constant because they are in a hyperbola of constant radio? Also, the axes do not represent components of vector U, and because of that, I cannot see all the U vectors concentrated in one origin.
The diagram at 7:29 is meant to show a collection of different U vectors, all with the same minkowski length (they all have their tips on the same hyperbola). I realize I labelled the axes as ct and x. I probably should have labelled them as U^t and U^x.
Rindler Horizon has been a long time struggle for me because I have never understood intuitively the reason light from behind the origin, Lo =C^2/α never catches up to the accelerating observer.
I am not sure if the following is close to correct, but I think it satisfies my struggles. It is what I have recently worked out. Just not sure how close to correct this is.
Lo is S = c^2/α (which has meters as a dimension). That distance is constant from the accelerating frame, regardless of the acceleration and any change to γ or the frame velocity.
L' would be the distance from the inertial frame, with orthonormal bases.
Lo is proper distance and is the longest measurement of that distance.
L' is shorter, by length contraction, and gets shorter with each new MCRF because of increasing γ.
As L' gets shorter, light itself slows to a stop.
Is light immediately, at T=0 seen to be stopped. Or does it slow to a stop as y increases.
For light origin closer than S, does it also continue to slow as y increases? Does it eventually come to a stop?
Looking at the graph at 25:43. Does the speed of light increase as the accelerating observer decelerates before T=0. Then after T=0 does the speed of light decrease until it slows to a stop, as the accelerating observer accelerates away from the origin?
The reason is because at each point on the hyperbolic world-line, the four-position vector is tangent to the hyperbola, and the acceleration vector is Minkowski Orthogonal to the four-position vector. Minkowski orthogonal refers to the time and space axes being at equal angles with a beam of light. When the ct and x axes grow closer together, with the angle between the beam of light becoming less than 90 degrees, this indicates greater time dilation, and lenght contraction. In the next video, in the first half, the dot product of the four-position vectors for the Rindler coordinates are shown using the hyperbolic trignometric idenity, and dot products are all invariant. You need to remember that it is the momentary comoving inertial frame of reference that we are concerned with, in which the time and space axes are becoming closer and closer together.
At 12'47'' how can the observer think he is stationary while his accelerometer is showing a nonzero acceleration?
Sorry, I missed this comment. Basically, everyone in relativity considers themselves to be stationary, with the world moving around them. However, if I'm in a car that's accelerating with respect to the ground, I'll measure non-zero values on my accelerometer. I'll believe myself to be stationary, with the road accelerating beneath me, but I will also feel a "force" pulling me back into my seat (this is essentially me feeling my own proper acceleration).
Looks like if you're accelerating at a constant 1g, the Rindler horizon is almost exactly 1 light year:
(3e8 m/s)^2 / (9.8 m/s^2) = 9e15 m = 0.97 LY
Interesting! That's a fun fact I'll have to keep in mind.
Hi eigenchris, may I ask whether there is anything similar to “Rotating reference frame” in special relativity? Just like the “Rindler coordinates” in special relativity corresponding to the “uniform acceleration frame” in Newtonian physics ? If such thing does not exist, what is the difficulty to define it? Thanks
You can't really invent a sensible coordinate system for a rotating frame in special relativity. You can look up "ehrenfest paradox" or "rotating frame special relativity" to try to learn more. It ends up being impossible to to define "simultaneity" for a group of observers in a rotating frame.
In the Rindler coordinates, I can define a set of rockets and draw lines of simultaneity in the spacetime diagram to indicates times when all rockets agree is "now" (basically, you can draw a line that is spacetime-orthogonal to all the rocket tangent vectors). You can't define these lines of simultaneity for a rotating frame in special relativity, so it becomes very difficult to use common sense intuition when describing the physics, or to make a coordinate system for it.
Your spacetime diagram would have to be 3D: a 2D plane of space, and an extra dimension of time. Your set of observers would form a ring in the 2D plane and they would all start spiraling up along the time axis in a helix as they rotate together. Their tangent vectors along their worldlines all point in directions that don't share a common orthogonal plane of simultaneity. So I don't see how you could make a sensible coordinate system without isolating one of the rotating observers as "special".
@@eigenchris thanks soooo much !!! Btw I really love your videos, it is really clear. I would love to see if there is a quantum field theory videos like this if you have time to make it - just an idea no pressure.
@@vincentxu2637 I won't be making QFT videos sadly. The main reason is I don't understand QFT, so you won't be getting any "really clear" explanations from me. The other reason is that QFT is around 3 semesters worth of material, so it would have to be much longer than this GR playlist. I don't think I have the energy for that.
I will be covering a little bit of the Dirac equation in a couple months in my spinors playlist, which I'm working on now.
@@eigenchris no problems. Thanks for all your hard work!
Chris, I calculated the rindler horizon for earth for plugging in **g** as acceleration and I got 0.97ly. This means the distance between earth and its rindler horizon is 0.97ly so nothing beyond that can affect earth in the reference frame of the earth. With this logic, we should never see light past 0.97ly and yet we do. What’s going on here?
I'm still learning the exact way General Relativity connects to Special Relativity, so I can't answer this with 100% scientific accuracy, but you need to keep in mind that the Rindler Horizon is *behind* the accelerating observer, in the same direction as the ficticious force they feel. So the equivalent horizon for gravity would be inside the Earth, in the same direction as the gravitational pull. Earth does have a very very small "Schartzschild Radius" of about 1cm. This would be the radius of Earth's gravitational event horizon if Earth were a black hole with the same mass, but since this radius is inside the Earth's interior, I don't think it's very physically meaningful.
The Rindler Horizon seems to me to be more like the Cosmic Horizon... the beams of light emitted from it and forward can no more reach our earth... Thank you.
awsome vid
Thanks alot💙💙💙
Dear Chris. This video made me ask myself several questions, that I hope to be able to formulate after some time thinking about them. I also hope not to be a nuisance.
This is my 1st question
In minute 3:09.
What do represent all those hyperbolas on the right?
Are they systems with different constant accelerations?
I also wonder, if the system starts its movement far from the origin, how will be its trajectory according to different values of constant acceleration?
And what mean the straight lines over and below the (x) axis?
I guess I need to see different space-time diagrams for different accelerated systems viewed from an inertial frame
The coordinate system on the right at 3:09 is Rindler Coordinates, which I will talk about in the next video. Each hyperbola is an accelerating observer with constant distance (1,2,3,...) from the origin. Each hyperbola to the right accelerates less hard than hyperbolas on the left. The diagonal line on top of the x-axis is the rindler horizon. The diagonal line below is not a horizon but it is the edge of the coordinate system. I will cover all this in future videos.
@@eigenchris Thank you very much, Chris. I will wait for the new videos, and still thinking about this one
@@eigenchris How much time it will take to the next video?
Is the horizon of the observable universe also a Rindler Horizon? Beyond which no light can reach us.
I think that's called the "cosmological horizon", but it's the same idea. Horizons are given different names depending on the context that they appear in. en.wikipedia.org/wiki/Cosmological_horizon
Isn't it weird that in flat spacetime constant acceleration has some kind of relation to the position of the hyperbola? Isn't that almost like some acceleration field?
What do you mean by "acceleration field"?
@@eigenchris I mean an acceleration field. Some sort of function that assigns an acceleration to points in a space. In this case, a field which assigns a constant acceleration to any point on the x axis.
Why is U dot U equal to U(t)^2 - U(x)^2? (at time 15:30).
Is that related to the equation for an hyperbola, or some how to the dot products of the bases?
It's due to the Minkowski metric I introduced in 104e, where et.et = +1 and ex.ex = -1. This is what gives us the definition of "Minkowski Orthogonal". A set of vectors with the same Minkowski length will have tips that sit on the same hyperbola.
Eigenchris replied (below) to review 104e; I think a better reference to U dot U = C^2 is 104f @ 14:03.
Why is the worldline "spacelike" in Rindler coordinates? Doesn't that imply it is travelling faster than the speed of light?
Sorry, I don't understand the question. The worldline's tangent vectors are always timelike, so they are never travelling faster than light.
@@eigenchris I understand now. Rindler's worldline lies inside the Rindler Horizon region. 26:21
Question sir. What is the difference between tau (proper time) and t_tilda (time in rindler's coordinate) ?
The t-tilde coordinate is the time measured by Rindler as he travels along the green hyperbolic wolrdline. This is also Rindler's proper time (tau). However, if we picked another hyperbola (one that was closer to or farther away from the origin), and measured time along it, it would not be equal to t-tilde. It would have its own proper time tau which is different than Rindler's.
Ok sir thank you so much. One more thing. Do you have a vid about expansion of universe gen relativity?
@@nellvincervantes6233 I am working towards it. It will probably be out in the summer.
Thanks in advanced sir! You make this subject easier. 😊
Wait, I'm confused...
If I'm moving towards a light beam in space, say that I travel at 50% C towards an incoming lightbeam. What is going to be the speed of the lightbeam from my perspective?
Is it going to be C? Or is it going to be C + my speed, so 1.5C?
I thought that no matter what you did (accelerate or not) you will always measure the same speed for light, no?
Also, what if I perform the same experiment, but I'm now accelerating towards the lightbeam? Will that change anything? Wil I measure a different speed for light other than C in any given moment?
I explain this a bit more in the next video, but for inertial frames, the speed of light anywhere in spacetime, as measured by you, is globally equal to C. When we look at non-inertial frames, the LOCAL speed of light is always C. This means that a light beam right next to you will always have a speed of C in your reference frame. However, if you are in a non-inertial frame and look at a beam of light that is far away from you (non-local), its apparent speed might not be C. In this Rindler horizon example, light "behind" the accelerating observer will have an apparent speed that's less than C, and light "ahead" of the accelerating observer will have an apparent speed that's more than C. Light that's right next to the accelerating observer will have a speed of C.
@@eigenchris Fascinating!
So how far is "far away from you"?
Is it like 1 light year?
So... Inside a lightyear bubble from my perspective, I will always measure the speed of light to be C regardless of my motion or acceleration, but beyond that 1 lightyear bubble I will start measuring light to have progressively slower speeds depending on my motion or acceleration?
@@-_Nuke_- Technically speaking, when I saw "non-local", I mean any point that isn't the exact point you're currently at. So even 1 cm away from you the speed of light will be slightly different, if you're in a non-inertial frame. You've watched 105b by now so you've seen the conversion factor for getting the non-local speed of light is "x * c/D". The speed of light changes linearly with distance from the observer. Although this is due to measuring the speed of light with a non-local clock. With your local clock and rulers, the speed of light will always be "c".
There’s an issue with the accelerometer definition: imagine we simply transform into the frame of the accelerometer. Surely the accelerometer doesn’t have its own accelerometer (if so, apply process iteratively). Therefore, we have a class of frames whose state of motion is technically undefinable from the outset - meaning SR cannot treat all types of motion.
What do you mean by "change to the frame of the accelerometer"?
If you go into the frame of the accelerometer then you'll see that it's accelerating because of the value given on the device.
@@eigenchris I guess this depends on what you define as an "accelerometer". The term is somewhat ambiguous to begin with, so let's take a simple set-up: a mass attached to a string (like you would have in your car) that swings back and forth to indicate acceleration. Now transform into the frame where the attached mass is at rest. Obviously, it has no accelerometer by which to judge the reference of its own acceleration. So, by the definition you've provided for accelerated frames (i.e. having and observing an accelerometer) the mass would not be able to determine whether or not it was in an accelerated frame.
Try this, imagine we have a ball inside but detached inside a larger hollow ball. Have the outer ball be attached to the accelerating frame then if we change frames of reference we'll know we accelerated (there was a force exerted on our frame of reference) because the inner ball is assumed to be detached as well as inertial. But at the start of our motion a fictitious force would arise to accelerate the inner ball cause it moves so the conclusion is that we accelerated because it will later impact the outer ball. Use this accelerometer if you desire to determine whether we are or are not accelerating due to a force. .
@@substantivalism6787 Dialect, in this case, wouldn't the mass on the string need some sort of explanation as to why it is "floating" in the air instead of being suspended on the string completely vertically (due to gravity)? Or conversely, need an explanation why the car is moving around it? If no forces have been applied, I would assume the change in the mass's motion can only be attributed to the existence of non-inertial frame.
I think Justin has the right idea: an accelerometer is basically like a mass on a spring sitting inside a rigid container. If the container accelerates, the mass will want to "stay where it is", and compress or extend the spring parallel to the acceleration. This spring compression is observable whether you are in the frame of the mass or in the frame of the container. Justin' description is basically this situation, except the spring is removed, and so acceleration is determined by in mass colliding with the container walls instead.
Hi Chris, can you please explain why U.U = c*c is still true even when you are accelerating? In Relativity 104f video, you showed U.U = c*c when you are NOT accelerating by using dt/dtau=gamma. Gamma, in this case, is a constant because beta is a constant (i.e. inertial). It is not clear to me on why U.U = c*c is still true when you are accelerating. Thanks.
Hi Wen. I think the formula I gave for U still works just fine, even if dt/dtau = gamma is a changing function of the 3-velocity u instead of being a constant. Even when I calculated the 4-acceleration in Relativity 104f, I allowed for the possibility for a non-constant gamma; you'll see there's a quick slide where I take the derivative of gamma with respect to a time coordinate and get a non-zero result. As long as you use a standard Minkowski-orthonormal spacetime basis that's constant everywhere, the formulas in 104f should work no matter how the target object is moving through spacetime. When we have a non-inertial basis that changes from point to point, taking derivatives can get more complicated, but I will discuss this in future videos.
@@eigenchris Thanks Chris, I did not realize gamma can be a function of the changing 3-velocity u. This is little subtle to me. The 4-acceleration in Relativity 104f video always confuse me bit.
@@wenchang8241 I wouldn't worry about the 4-acceleration in 104f too much. I showed it mainly for completeness. The acceleration in 105 will be more important.
Just to help clarify for me, I think that the motion is Rindler initially approaching Einstein, but slowing down. Then Rindler turns around and moves away from Einstein with increasing speed, all along the x axis.
All the time, the acceleration is constant.
Is there an intuitive way to understand what the hyperbolic angle physically represents in terms of the area of the sector in front of the hyperbola?
Sector angle as (α/2c) T?
T=0, sector area is 0
T=1, sector area is α/2c
Yes, exactly.
Sorry, I didn't see your 2nd question. I discuss cosh and sinh and the areas in the final 10 minutes of this video: ruclips.net/video/km7WTO_6K5s/видео.html
@@eigenchris Thank you.
@@eigenchris It is like the argument (α/c*Τ) is a dimensionless number reflecting the ratio of the proper acceleration to c, then multiplied by T.
Not unlike β.
Or perhaps, better (αΤ)/c.
How great the proper acceleration is wrt to c, times how long, T, the acceleration occurred.
wonderful !!!!
I understand that the earth is acceleraiting but why is he doing it why shouldnt small things like an Apple accelerate?
Vector field is not rasnal function. But complex field are rational function.
I have a question. I see how rindler horizons work in the spacetime diagram, but I don’t see how they work in reality. You’re saying that if I were to release a light beam right now that travels in the x direction, and I was at a position far away but accelerated towards it, I still wouldn’t see it even after billions of years? It still wouldn’t make sense to me in I picture it in real life. I would surely have intersected with the photon no?
Also, I’ve learned that the rindler horizon actually outputs out particles which can heat up the accelerated object (called the unruh effect). Will you be explaining this too in a future video?
One thing to keep in mind is that distances to Rindler Horizons are usually extremely large (note the c^2 in the numerator). Another thing to note is that constant acceleration is an extremely difficult thing to maintain for a long time. When you're driving a car, you normally accelerate up to a given speed and then maintain that speed for your car trip (no more acceleration). To accelerate forever, you'd need an infinite amount of fuel/energy. All you really need to know, mathematically, is that a hyperbola (ct)^2 - x^2 = -1 will never intersect the beam of light x = ct (you'll see that you get 0 = -1, because there are no solutions).
Also, I've seen the PBS spacetime video on the Unruh effect but I know very little about it. I won't be covering it because I don't know quantum very well. I'm going to focus on relativity without quantum.
1. You are accelerating *away* from it (and the origin) - all in the upper right quadrant. The wave fronts of light follow you at a constant distance (when you look back over your shoulder), with E- and B-fields frozen into one static spacial sine pattern, I imagine.
2. The effect of seeing additional real particles on top of the virtual vacuum fluctuations due to acceleration is a result of quantum field theory.
Why use the origin to measure the distance to the horizon?
The rule for measuring distances in relativity is to measure the length along a line of simultaneity. On the Rindler hyperbola, all the lines of simultaneity point back to the origin, so that's the point we measure to.
@@eigenchris I see. Thanks for the answer
@@eigenchris After all this time this finally makes complete sense. Basically the hyperbola that is the worldline is always mapped to itself by Lorentz boosts, so if we do an active LT from an event B on the hyperbola to an event A (on the horizontal x axis) on the hyperbola, we are now in the rest frame of the observer at B, and B is now where A was before, so the proper distance to the Rindler horizon remains the same
@@charliewu4110 Yes, that's correct.
I'm pretty sure Einstein said that you couldn't handle non-inertial frames in SR. An accelerating object doesn't magically "know" it's accelerating but rather sees a uniform gravitational field in accordance with the equivalence principle. I agree with the earlier commenter, an accelerometer seems like a pretty flimsy and ill-defined pretext for distinguishing acceleration.
Do you have a source for when Einstein said that? And can you explain why you think an accelerometer is flimsy for this case? I think the concept of hyperbolic motion was discussed only a few years after Einstein's original 1905 paper on special relativity. These 1909 and 1910 papers by Born and Sommerfield discuss hyperbolic motion:
en.wikisource.org/wiki/Translation:The_Theory_of_the_Rigid_Electron_in_the_Kinematics_of_the_Principle_of_Relativity
en.wikisource.org/wiki/Translation:The_Theory_of_the_Rigid_Electron_in_the_Kinematics_of_the_Principle_of_Relativity
The textbook "Gravitation" by Misner, Thorne and Wheeler also discusses it.
@@eigenchris In his 1916 GR paper he goes on a big spiel about how special relativity, like Newtonian mechanics, arbitrarily privileges acceleration and so is inadequate for accelerated frames of reference. Re-reading over it now though, it seems more like he just found SR's treatment of acceleration epistemically unsatisfying. So I might just be confusing how to interpret his stance. As to the accelerometer question, I feel its flimsy because I've never come across a rigorous formulation of what exactly constitutes an accelerometer (texts always just give comparative examples)
@@WSFeuer I have another 5 videos planned where I will continue talking about acceleration in SR (things like how to keep extended bodies rigid during acceleration and how to get proper time along curves). You might feel better about the concept after watching those. I'm repeating myself from my other comment, but I think an accelerometer basically just measures that feeling you get in a car when the driver slams on the gas pedal. It uses a test mass on a spring. If acceleration is parallel to the spring, the spring either extend or compress. The amount the test mass is displaced gives you a way to measure the acceleration in that direction. A full 3D accelerometer would be 3 of these test masses in mutually perpendicular directions (or the same test mass with 3 perpendicular springs attached to it). Do you feel this is inadequate?
@@eigenchris Hey I love your videos, I watch them religiously :-) Sorry to make you repeat yourself, but yeah, I did read your earlier response from the other commenter and here's my issue with it: to establish a rigorous definition of an accelerometer, say, using a mass on a spring as you suggested, you have to measure a displacement. But measuring a displacement means first fixing your coordinate system. But the whole point of the accelerometer is to determine whether or not your system is fixed in the first place! Do you see the issue?
@@WSFeuer I don't see the problem with choosing a coordinate system to do the measurement. The spring compression could be measured in either the frame of the car/container or the frame of the mass, and it would come out to be the same number in either case. I think you'd have to conclude that either (1) there's a force acting on the mass or (2) the container is accelerating. If we exclude gravity since we're in SR, and assume the mass has no charge or magnetism, I'm don't see how a real force could be acting on the mass given known laws a physics, so I'd have to conclude the container is accelerating.
Chris, have you or anyone considered non constant accelerating frames of reference? In real life this can happen. People have observed acceleration on acceleration. Also what about deacceleration? Please respond. Jack Rubin
I don't plan on talking about non-constant acceleration, but the "U . A = 0" and "U . U = c^2" formulas would hold for those as well, so if you invent your own worldlines, they must obey these equations. Deceleration is just negative acceleration (the bottom half of the hyperbola is just decelerating before the turning-around point on the x-axis).
Thank you from all of us.
If the accelerated observer is charged and is accelerated in an electric field, how would the observer feel it's acceleration and measure it, because it looks as everything around the observer is accelerating. So how can they measure their own acceleration from their own clock?
If the observer was carrying an accelerometer, they should be able to detect that acceleration is happening. I think on a basic level an accelerometer is like a mass on a spring, and if the observer is accelerating, the mass's neutral position on the spring will shift, indicating acceleration.
@@eigenchris But if the spring is also charged and is accelerating with the mass wouldn't it detect no acceleration since the position between the mass and the start of the spring will always stay the same. A bit like if you drop your accelerometer in a gravitational field
@@joshuapasa4229 I don't think a charged accelerometer would function properly as an accelerometer, because it would respond to changes in electric fields even if it is in an inertial frame. I think the issue is that there's a fundamental difference between gravity and electric fields: Newton's 2nd law is F = mass*a, not F = charge*a. This means that a body being "pulled" by gravity (in free-fall) is travelling along an inertial frame (there's no actual force, just the curvature of spacetime), whereas a body being pulled by an electric field is undergoing a force.
@@eigenchris Why can't charge be represented as a curvature in space-time, but the strength of the curvature depends on the charge of the object in the curved space time?
@@joshuapasa4229 This is starting to get outside my knowledge. I think Electric and Magnetic fields can contribute to the curvature of spacetime because they contain energy. You can google the Reissner-Nordström stress energy tensor to see how the electric field comes into play in the Einstein field equations. (The Reissner-Nordström solution gives the spacetime solution for the curvature around a charged spherical body.)
Then Why do we see all objects following parabolic trajectories under constant gravitational acceleration in our daily life..?
When accelerations and velocities on on the scale of our everyday experience, we're only looking at a very small portion of the hyperbola near the origin. In this very small region, a hyperbola and parabola are basically impossible to tell apart.
@@eigenchris Thanks a lot for your reply... Can we deduce those hyperbolic trajectories by selecting suitable metric and solving geodesic equation using einstein equation..?
@@keshavshrestha1688 These hyperbolas are not geodesics, because they are the result of being on a path with constant proper acceleration. Geodesics are paths where the proper acceleration is always zero, so the hyperbolas are not geodesics. I talk about geodesics in video 105f.
@@eigenchris Really appreciate your answer.. I hope and request that you will reply my comment in your relativity lecture.
Hi Eigenchris, 23:58 Sorry, but did I misunderstand something? Shouldn't the spacetime interval lead to a constant proper time as well? Since I learned that proper time is defined as the spacetime interval divided by c.
In this case, tau is measuring the proper time along the hyperbola, which increases as you travel along it. The S-vector is measuring the interval between the origin and any given point on the hyperbola. This S-vectors turns out to have a constant interval value for any point on the hyperbola. Does that clear things up?
Nice profile pic, by the way.
@@eigenchris Thank you! Now I understand.
Yes, I also love the lambda and greek letters, they looks so cool!
Thanks!
plz upload next video ASAP
Wow 😲
Something bothers me about about your 4-acceleration vector because at the upper part when τ>0 rindler has A pointing up. This means At and Ax are both possitive and this feels wrong. The temporal component of 4Acceleration should be pointing down to slow down the temporal component of 4velocity and give room for the spatial component of the 4Velocity to increace. also if the temporal component of 4Acceleration was facing down this would make the A and V geometrically orthogonal and the dot product A*V=0 would feel totally normal. maybe I am thinking weird because of this geometry is not familiar to me. I ve learned so many things through your channel and I can't THANK YOU enough for the amazing work you do!
I think you're still thinking in "Euclidean space" too much. When you take 4-velocity vector for a stationary particle and boost it, both its time component and space component increase. This is what's required to keep the formula (ct)^2 - x^2 a constant: both (ct)^2 and x^2 need to get bigger. The idea that one needs to decrease in order for the other to increase comes from the circle formula in euclidean space y^2 + x^2 = constant. Here if y^2 gets bigger, then x^2 must get smaller.
Orthogonality in Minkowski space is also about 2 vectors making equal angles with a beam of light. This is the relationship that U and A have. It's not about "right angles" like in Euclidean space.
@@eigenchris Thank you. Yes you are right. What confused me is that I tried to picture those vectors in 3d-space .I used 1 spatial dimension for the 4V and v and 1 more spatial dimension for α so it can vary reltive to v. This gave rise to A spanning all three dimensions but I was thinking "too euclidean" in this "model". It seems to work btw because A is always at euclidean right angle to V but my mistake is hidden in my eucledean dot products so the geometry follows...I ll fix it because it misleads my intuition. I liked it because i saw that when α vector traces a circle in its spatial dimentions ,the A-spatial traces an ellipse but i guess in the corrected version it will do something similar. And btw this "ugly" equation of A is actually very elegant geometrically. It is just A=αγ^2 minus the projection of αγ^2 on V
Bro plzz upload next video ASAP
Duuuuuuude, I would like to chance to THANK THE FUCK OUT OF YOU. I always liked your videos, mate.
Having a rather slow brain I found the comparison between Rindler's view and Einstein's view difficult to follow.
Sorry to hear. Is there a certain point in the video I can explain better?
267 days huh, its gonna be interesting
It would be better to explain some paradoxes of str
I'll be explaining Bell's Spaceship paradox in 105b and the twin paradox in 105d.
@@eigenchris
And also make a separate video on e=mc^2
@@minhaskhan9164 I talk about E=mc^2 in video 104f
whatever happened to GR videos?
GR will start in Relativity 106. I felt I needed to lay the groundwork of SR first.
@@eigenchris i would very much like to understand the physical meaning of bianchi identities
I don't really understand them myself, unfortunately. I've read you can summarize the bianchi identities using the exterior derivative somehow.
@@eigenchris most classes just go by saying we need something about curvature which is a rank 2 tensor and has covariant derivate zero. Since Ricci doesnt work we go with a correction to Ricci and call it einstein tensor
Which btw is a shitty way of explaining it. Its almost as if, since LHS =0 we can equate it to RHS =0. Physics is kinda lost in this way of explaining it
6:12 a v would more accurate than an f
I
( ia) *(ia)=-a^२
5:11 v_0 + 9.81t> 300 000 000 ehhh.... ok....
Sorry, I don't follow your comment?
I can feel that this video is very well-made but my brain cannot understand it.😇
Sorry to hear. Have you watched all the previous videos in this series?
@@eigenchris I did't expect a reply from the author. Thank you. You made excellent videos. I have not watched all the previous videos, but watched those 4-vectors related multiple times. I don't think my courses will be that detailed currently. I am here just for my curiosity.
@@146fallon9 Watching the previous videos might help, but if there's anything specific you're confused about, let me know and I can try to explain it better.
Black background please, it’s easier to learn
If A moves relative to B, B moves relative to A. There is no such thing as "stationary." Therefore both special and general relativity are WRONG! Idiots will disagree with me.
True. But IF you consider yourself to be stationary (just saying IF you do this) then SR and GR will tell you what you will measure... And that's the topic of relativity...
IF now, you DON'T do that, then you don't have to worry about relativity since you somehow have a "bird's eye view" of spacetime...
If you know how to do that, then you might as well be the next Einstein... Cuz so far we don't know a way to "escape" spacetime.
@@-_Nuke_- Imagine a "photon" moving away from you and say you are chasing it at 99999/100000ths the speed of light. If you could somehow "measure" that photon, you're saying it would be going c because your time is slowed down? Help me understand Einstein.
@@CandidDate Everything is moving inside *spacetime* at the speed of C
But not always in space. Light is moving at C in space (because it doesn't have mass) but we Humans (massive objects) move at C mostly in time and a little bit in space...
(movement in time is called "aging" and in space is called velocity)
So if you combine my space speed and my time speed using the Lorentz factor γ=1/sqr(1-(V^2)^2/(c^2)) you should always get C.
So yeah, everytime you measure the speed of anything, IF you know how to measure speeds in spacetime CORRECTLY you should always find that everything is moving at the speed of C in spacetime, not just light...
Light isn't anything special, it just has all it's speed in space, most of the things in the Universe have a big portion of their C speed mostly in time and a lesser portion in space...
An object moving at 1% C in space (about 10800 km/h) you are still moving at 99.9949% C in time...
As you can see from here www.emc2-explained.info/Dilation-Calc/ if you put 1 in the first box, that's 1% C in space and 99.9949 in time...
Whether or not space contracts or time dilates is irrelevant... Everything is moving at C in spacetime.
Are you human or God?
17:30