1:05 Covariant derivatives of vectors with covariant components 8:05 Covariant derivatives of vectors with contravariant components 14:44 Parallel Transport 24:37 The curve with all of its tangent vector parallel to each other is geodesic curve 29:44 Notion of the tangent vectors 31:35 Parallel transport of tangent vectors of geodesic curve is constant. This gives equation of motion of geodesics. 36:43 Trajectories in Space-Time (Minkowski geometry and Riemannian geometry together) 40:58 In general relativity metric becomes a function of space and time. 41:15 The metric of space time always has one negative eigen value and three positive eigen values. 46:23 Problems with special theory of Relativity 48:55 Polar coordinates as the analog of uniformly accelerated coordinate system. 1:05:44 Our aim is to introduce an arbitrary set of coordinates and write the equation of motion of geodesic in it and see if it looks like the particle falling in a uniform gravitational field. 1:06:47 Metric in polar coordinates, its analogy to uniformly accelerated coordinate system. 1:09:12 Expressing Metric of uniformly accelerated coordinate systems interms of ‘g’ 1:22:10 Finding how the particle moves in this obtained metric (Equation of motion in geodesics) which is Newton’s equation. 1:33:49 Equation of motion in real gravitational field 1:35:25 The metric for the horizon of black hole
Strictly speaking, Susskind is more of a String theorist and a QFT specialist than a General Relativist. But I’m absolutely amazed at his incredible abilities and insightful approach in teaching GR. He is a gifted lecturer who has a special knack for communicating the heart of the matter behind the seemingly hairy-intimidating equations. By the time he is done teaching a subject you feel like you’ve always had an intuitive feel for it; and that’s the trait of a supremely gifted professor.
I studied physics in college for a few years 55 years ago and then moved on to other things. The last few years I’ve been returning to physics through these lectures. It’s been a real gift. I hope these lectures stay around for me to return to as I move on to books and other sources of study. Just want it known they don’t grow old.
@23:00 Vector fields are not parallel transported, elements of the tangent space are. Generally, if you have a path from the point p to the point q, then parallel transport gives you a map from the tangent space at p to the tangent space at q.
The (really good!) exposition about "fictitious" gravity in an "accelerated elevator" complements a discussion of a "uniformly accelerated rocket" in Hartle's giant introductory GR tome. It was very useful to have thoroughly understood how a constant proper acceleration appears in a "rest" frame in SR--how each one of the coordinate hyperbolas happens.
If you're a newcomer to GR wanting to really get into it, and watching this series: after going through this route myself, I can definitely tell you Susskind's presentation is very accessible on the one hand, but *extremely* dated on the other hand. There is a much more geometrical approach to GR today that sure, requires developing some familiarity with differential geometry, but boy after you know this stuff all those indices don't look so arbitrary and unmotivated. The simplest example is the "A tensor is a thing that transforms like a tensor" line is one that really bothered me for a long time. Just my own opinion. Nevertheless, coming to his lectures after some additional familiarity with the background math is very refreshing and anyone can learn something new from it, because after all... it's Susskind :)
Are you talking about approaching tensors as multi-linear maps between vector spaces? That is the way I got it from Frederic Schuller on RUclips as well as XylyXylyX What is a tensor as an intro to his 75+ What is GR All this "It transforms as ...." is pre 1970, maybe 1960.
While watching these lectures I found it useful to pause and look up things like the definition of a tensor field in terms of tensor products of copies of the tangent bundle and cotangent bundle. This helped me to get a better geometric understanding of what is going on, and relate tensors to concepts that I am more familiar with.
My dream is one day I can master the general relativity. Thank you so much Mr. Leonard Susskind for these lectures, you help me to make my dream come true.
You will. It takes patience and practice. But trust me, at some point it will all click together. This lecture in particular is crucial, as it demonstrates how the geodesic equation and the metric expressed in an accelerated frame suddenly produce gravity. This is exactly what the equivalence principle is all about. Rewatch it as often as you need to understand it completely.
Funny that the professor did not understand the question at 1:39:45. You don't need a big R for a small A in general. You can certainly translate the world line with a big R left along the x-axis and make it go through the origin. The professor chose a big R only because he insisted, for mathematical simplicity, that the equations have the form x = R coshω and x = R sinhω.
Apparently the official terminology is "geodesic normal coordinates". But honestly, to get hung up on a just a NAME, when Susskind clearly and unambiguously defined what he meant in the previous lecture, is just showing off for no reason...
Almost no one in the scientific field uses the imperial system in the U.S., we hate it too :p Unless you mean the tensor... that's something else entirely
@@orientaldagger6920 It was a joke. He's intentionally confusing the metric tensor field with the metric system, which shows a lack of comprehension, and is making a valid point about the preference of the metric system. Now, physicists often use natural units, as you stated, so I understand what you're mentioning. :)
Anyone else severely under-qualified and still following along to his GR lectures? I'm watching these, reading through Einstein's works, and starting a journal of the whole derivation of GR. To be honest, i kinda want to write it out building up from high school algebra to GR like you would normally, but then expand GR and represent it in a way that only uses notations from a high school algebra level. See how many pages that takes up... lol
Quite a bit of hand waving in this lecture. For example, he replaced Rω by t at 1:14:06, without ever explaining why. It would be beneficial to derive the equation of constant acceleration first. Then all will make sense. It can be derived by following these steps. 1) In the rest frame of the particle (rest at a single moment), acceleration A will change its four-velocity from (1,0,0,0) to (γ,vγ,0,0) after an infinitesimal amount of time dτ. Here v = A dτ, and γ = 1/sqrt(v). 2) Its four-acceleration, again in the instantaneous rest frame, is therefore [(γ,vγ,0,0) - (1,0,0,0)] / dτ = (0,A,0,0). 3) Since the four-acceleration is a four-vector, it obeys Lorenz transformation. So in our "true rest frame", the four-acceleration is (γvA, γA, 0, 0). Here γ = 1/sqrt(v), and v is the speed of the particle in our frame. 4) In the four-velocity (U0, U1, 0, 0) of the particle, U0 = γ, U1 = γv. Therefore, the four-acceleration can be written as (A U1, A U0, 0,0) 5) So we have the following differential equations. dU0 / dτ = A U1, and dU1 / dτ = AU2. 6) The general solution is U0 = α exp(A τ) + β exp(-A τ), and U1 = α exp(A τ) - β exp(-A τ). 7) We can set U0 = 1, and U1 = 0 at τ = 0, i.e., the particle has zero velocity at time 0. Then α = β = 1/2. And the solution becomes U0 = cosh(A τ), U1 = sinh(A τ). 8) Note U0 = dt / dτ, and U1 = dx / dτ. Therefore t = t0 + 1/A (sinh(A τ), and x = x0 + 1/A cosh(A τ). This result is exactly the equation given in the lecture with t0 = x0 = 0, R = 1/A, and ω = A τ. Now it should be clear why Rω can be considered t, which is the same as τ in the particle's own frame.
The fact that the covariant derivative of the metric tensor vanishes everywhere (not only at a single point) cannot be deduced without the assumption that the parallel transport preserves the scalar product of vectors. This assumption leads to the celebrated Levi-Civita theorem of Riemannian geometry, which states that the affine connection components must be the Christoffel symbols. See, for instance, "Riemannian geometry" by M. P. do Carmo (Chapter 2).
The formula for Christoffer (?) symbol he writes out on the board in the beginning is different from what he derived (wrote down, actually) in Lecture 3. Which of the formulae is correct? Thanks.
In lecture 3 he simplified "partial derivative with respect to x(index)" to just "partial derivative(index)." The indices of g are switched from one video to the other; g contains implicitly a Kronecker delta (1 if indices are equal, 0 if unequal) and is symmetric, so the order of g's indices do not matter (www.ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf section 5.2). Compare to the Christoffel equation from ruclips.net/video/foRPKAKZWx8/видео.htmlh11m1s , which does contain an error - can you spot it?
I know 4 years is a while to wait for a reply, but Emil Akhmedov has a course over at Coursera, www.coursera.org/learn/general-relativity, which has peer reviewed exercises. It's a bit more mathematical than LS's lectures, so it would be a good one to do next.
1. "Flat" applies also to 8-ary space ("octonion"), in which case it's not limited, but for convention, to just-1-or-3 negative-1's, but to 1-or-7 negative-1's... Or any other n-ary. 2. To suppose that the string would not,-break because it's merely a Minkowski space transformation, would be comparable to saying the returning-relativistic-Twin was not-younger... Some of the SR results are going to yield "true" changes, others virtual... (I know, Susskind says we don't-say "SR" but do-say "GR".)
The Gaussian Normal Coordinates he keeps mentioning refer to coordinates obtained from flattening the space time around a point. In a Smooth Riemannian Manifold we can select a neighborhood around each point P small enough so that the metric can be written as the Cartesian metric (i.e. no curvature of any kind) at point P and these coordinates are what he is calling Gaussian Normal coordinates. In terms of Math it is difficult to precisely define what infinesimal means but in Physics if you take an infinitesimal neighborhood around a point P in a curved space time you can choose coordinates such as the metric in that infinitely small space around the point is precisely the Kronecker delta (i.e. Minskowki metric).
the big R means small acceleration in this case he can assume that the dervative of time with respect to prober time is equal to one ,since the velocity is very low compared to the speed of light
So, R works out to be about 1 light year... where acceleration starts resembling Earth gravity. So what the heck would be the interpretation of distances near zero? What sort of mass would it take to create 1g 1ly away?
sharpfang There is no mass-the geometry of spacetime being considered was flat, which means no mass present to distort it. It was the COORDINATES which were chosen to be curved, representing an observer in a uniformly-accelerated frame.
nice observation :) and what that means is: if you accellerate with g in free space (forever and ever... constant accelleration), an Event Horizon forms 1 lightyear behind you. You would outrun light from the closest stars that is emitted there at the time you start the accelleration.
I'd agree but only in a minor way because it's basically Riemannian geometry considerations... I'd fix the notation and move the counterclockwise "upstairs" contravariant superscript to the base symbol leftside evenly with the rightside.
yea heres my doubt ! - well apart from R being equal to 1 over A, isnt R physically the distance of "something" from Origin ? if so what is that "something" and what does a really large R mean ??
Consider the hyperbola at r=R, but forget about the origin just for a moment. The hyperbola is essentially the worldline of a particle accelerating with constant proper acceleration a=1/R (just some number; R has no interpretation as a distance - yet). Notice that the particle doesn't gain infinite speed. It approaches a maximum asymptotically (the speed of light) without ever exceeding it. Now consider the asymptote to the hyperbola. It is also a wordline, but for a particle with constant velocity, c, the speed of light, i.e. it is the wordline of a photon. This photon, traveling on this asymptote NEVER catches up with the hyperbola (the hyperbola and the asymptote never intersect). Notice that this asymptote actually goes through the origin at r=0. In other words, at time T=0, the photon was a distance R away from the accelerating particle. So here you have the clearest physical interpretation of R I can think of: consider yourself to be a uniformly accelerating observer. The distance R then, is the MINIMUM distance away that a photon needs to start, in order to NEVER catch up with you. Notice that, as you accelerate, the proper distance you measure to that photon, stays R, always. The photon doesn't ever appear to move toward you. The distance R, again, is the minimum distance such a photon needs to be from you, for it it appear to never move toward you. Indeed, as you accelerate, your "axis of simultaneity", that is the line along which you measure proper distances, "rotates" (in the Lorentz sense of rotation). But it always goes through the origin. So you can think of this origin as "the unique event in spacetime, which is common to ALL my axis of simultaneity". In order words, "the origin is an event, which from my perspective is ALWAYS "now", no matter how much proper time I experience". You cannot see beyond it, either in space (r < 0), or in time (T > 0, X < T). For you, as long as you are accelerating, it constitutes an event horizon. So that's the second physical interpretation for R: it is the distance to the event horizon, which forms because of your uniform acceleration. In a truly flat spacetime, the event horizon would be 10^16m, which is about a lightyear, away from a uniformly accelerating observer with acceleration 1g, as Susskind calculates.
OK I was hoping at the end when he asked that question, I would get an answer. I know that A = 1/R, to get a modest A = g (w/ c = 1) we should use very big R. But what is the inuition of a big R? If modest A correlates to our everyday feel of g, what does R correspond to in such experience? An alien very far away seeing the effects of our gravity? Because the radius of the earth isn't that big!
See my answer to Satosh Kumar's question above, for a physical interpretation of R in FLAT spacetime. Any alien at a distance R or greater from you, can never catch up with you, or influence you in any way, even if he travels at the speed of light toward you. However, remember that true spacetime near the Earth is NOT flat. The flat spacetime with uniform acceleration is just some APPROXIMATION, valid in a small region close to the Earth's surface. Now imagine the Earth were a single point (i.e. a black hole) with the same mass, rather than an extended body. The gravitational field out at the usual Earth radius would be exactly the same. But as you move further in, the gravitational field, that is the acceleration felt by an observer at fixed radius, gets bigger. So at that smaller radius, the flat spacetime approximation is a DIFFERENT one. Indeed closer in to the black-hole Earth, the effective a is greater, and therefore the effective R=c2/a gets smaller. The effective flat spacetime approximation of real spacetime, keeps changing as you move radially in. That's what it means for real spacetime to be curved. After some distance, you come to a point where the effective a of the flat spacetime approximation, is infinite, and hence R=0. You have now reached the black hole event horizon, at which you appear to be at the ORIGIN of a flat spacetime with uniformly accelerated coordinates. Indeed, in lectures 5 and 6, Susskind analyzes the spacetime close to the horizon, again by studying the flat spacetime approximation appropriate to that region. And in that uniformly accelerated spacetime, the "real" gravity-induced event horizon corresponds to the event horizon at the origin of that flat approximate spacetime. The distance from your original position, where a=g to the real event horizon, is much sorter than 10^16m, because the real spacetime around the hypothetical earth-mass black hole, is NOT flat, and hence over longer distances the flat approximation doesn't hold. The effective R decreases by MUCH more than the distance you travel inward (due to the increasing gravitational field strength), until you eventually reach the black hole event horizon, where the effective R=0.
No. If there your coordinate system is curvilinear whilst describing a flat space then the christoffel symbol can have nonzero values that still imply there is a flat space. He went over this.
I think there is a mistake; in calculating the parallel transport we multiply the covariant derivative by dx_m and put it equal to zero and because there is a minus sign in the definition of covariant derivative that gives the condition for parallel transport = dV^m = gamma*dx_m so see there is no minus sign in this equation, but while using the same equation in calculating the parallel transport of tangent vector the minus sign pops up and thus the equation of motion has a minus sign which should not be there if we are using the definition of the covariant derivative with a minus sign
There is a minus sign for the expression of the covariant derivative for COVARIANT vectors. For CONTRAVARIANT vectors, there's a plus sign in the covariant derivate. Therefore when setting it to 0, and deriving the geodesic equation for dx^m (which is contravariant), or equivalently the tangent vector, you get the extra minus sign.
So let me get this straight.. No matter which coordinates you use the covariant derivative is written as D_r V_m=d_r V_m - (Gamma)_rm^t V_t. If we choose a point and use Guassian Normal Coordiantes defined at that point, then covariant derivative of V_m at that point just reduces to d_r V_m as all gammas are zero. So the covariant derivative transforms like a vector but It always takes the form D_r V_m=d_r V_m - (Gamma)_rm^t V_t no matter which frame we are in. Am I right? Anyone who knows?
Hi there One comment. You drew the hyperbolas to the left of the space-time plane. But should be directing upward as they should be inside the light cone which should going upward to be be in the region of velocities less than speed of lights. The drawn hyperbolas in the lecture lie in the region of speeds higher than the speed of light. Alternatively, you can interchange the “time” abs “space axes and it would be correct
All the hyperbolas lie entirely outside the lightcone of the origin, that is correct. In other words, no signal can ever reach any of the accelerating observers from the origin: the origin constitutes an event horizon for all the accelerating observers. Notice, that the hyperbolas, which lie "close" to the horizon, i.e. have a small R, must correspondingly have huge acceleration, as Susskind points out. Put another way: you need to have an increasingly large acceleration to outrun a photon which starts increasingly close to you.
Its just a choice of coordinates. In 2D euclidean, that motion could be represented using polar coordinates..Likewise for Minskowski (x,y,z,t) system, he has used hyperbolic coordinates to describe the motion.
whenever somebody brings up in class "I read something", the professor should say "see me after class and bring your receipts". in my experience people who say this can't bring receipts. those who do, it turns out to be mass media or something fluffed up for general consumption.
around 31:00 he says that dx^m is equal to the distance between the 2 points, but he goes on to saying that you devide that by ds, which is the distance between the 2 points. what?
dx is the difference between the coordinates of two points, ds is the "real" distance between the two points because it is calculated using the metric of the spacetime g_mn
I asked one of my professors this once. His response was that essentially you could, but it isn't the best way to do it. My impression was that for reasonable systems one could extract correct answers to a sufficient precision this way. At extreme scales, however, (notably those where QED is necessary) the approximation would fail. So you have two cases: where classical EM is a sufficiently good approximation, and thus there is no reason to complicate the problem with the difficult machinery of GR, and where it isn't, in which case GR isn't good enough either because EM is a quantum phenomenon. I wish I had asked via email rather than during a lecture so that I could properly quote or even remember his response. I may be misrepresenting what he told me. But it was clearly something he had put some thought into previously.
Very simple: because of the principle of equivalence. Because intertial mass is the same as gravitational mass, gravity doesn't have to be a force, it can be simply seen as an acceleration field. Whereas electromagnetic forces are not proportional to the inertial mass, so they do not affect all bodies the same way, but depending on their electric charge. So how can the EM field be an intrinsic property of space, if it is not the same for all bodies in space?
Lenny, A0620-00, the nearest black hole, is 3,000 light-years away. The Fermipause is the end of the influence of Fermi Bubbles and sweeps out a sphere that comes within 3,000 light-years of our solar system. Fermi Bubbles protruding from the Galactic Center span 25,000 light-years. Our Solar System is 28,000 light-years from the Galactic Center. Black holes are an integral part of a unified theory of cosmic plasma physics. Very Respectfully, Eric D'Aleo
What is at the origin of the light cone? The observer or the thing accelerating? And if the points on those curves all have the same distance from the origin why does it look like X will increase as you evolve up the time axis?
The origin of the light cone is the exterior observer; that is, the person stationary outside of the "moving elevator" or whatever you want to call it. X looks like it increases up the time axis because it does; when time increases, the spatial component has to as well to make the distance (remember, pseudo-euclidean) a constant (since distance^2 = t^2 - x^2)
Theoreticians use something called Natural units to simplify their calculations, setting c or mass = 1 for example. It just makes it more convenient as long as you don't forget about them!
Yea I forgot about those. Those constants should really be included in derivations like these though. Otherwise you never know what you can and cannot do mathematically. Thanks
To complete that statement about the string breaking or-not, Note that time in the relative frame, by SR-calculation, remains zero-offset: so, at a given time, the string is, being stretched the greater distance... Then, the only question is whether or not the string might 'stretch' due to 'spooky action at a distance...' But the answer to that is, QM requires the stretched atom to want to fall back and so break the string; Also constant acceleration changes wavelengths to lock-in possibilities.
He is not a mathematician. He just says it. Something I think was missed in his notes. If you go back about a minute he is differentiating a product with the metric... WHY??. The metric allows a lower index vector (tensor) to be converted to an upper index (contravariant and covariant are for Lit majors). I think he meant to show that after taking the derivative at 11:53, he then raises the index. The plus minus on the Gamma, I think is a whole other trick.
No, dx^m refers to the change in the coordinates between the two points, while ds refers to the distance (or rather, displacement) travelled by the vector.
@@bhuvansv3701 i;m afraid you’re incorrect, sweetie. “dx*m is CLEARLY a veiled reference to dextromethorphan (aka the active ingredient in most cough syrups). it is a highly recommended substance that many aid you in undestanding more difficult equations. its a ‘dissociative’, which means it will allow you to shed your grammar school biases. hope this helps!
And for a little more clarity I'll bi-clarify that last clarification and point out that before the stretched string breaks it emits heat F×D=W by Hook's Law which has to be an interesting result from SR that accelerated objects heat-up.... 2. Only if accelerating the string in the co-moving frame converts, its rest mass, to kinetic energy, and thereby lengthens its momentum wavelength, should it stretch without breaking; But such 'affectation' is not expected, of the greater mass pulling it....
Reposted from lecture 3: I'm still confused about the Gaussian normal coordinates. Let's say I'm looking at the simplest example of curvilinear coordinates: polar coordinates (r,phi) in the plane with the metric tensor g=diag(1,r^2). Now let's say I want to write down a transformation x^i = y^i + c^i_j,k y^j y^k that takes the metric tensor to the Kronecker delta, where the y^i are (r - r_0) and (phi - phi_0) respectively. The Jacobian of this tranformation is always the unit matrix, leaving the metric tensor invariant. What I can do is Taylor expand u=r*cos(phi) and v=r*sin(phi) around a point, and then use the expanded u and v as coordinates. This makes the first derivatives of g vanish, but is in general not of the desired form above. - I'm guessing that the least I need to do is include the scale factors in the definition of the y^i, i.e. consider polynomial functions of (r - r0) and r_0*(phi - phi_0).
I think this is great mathematics, personally it makes me feel like I'm wasting my time though; I don't think its the best way to model gravity. I am working on a different model which uses QFT, m-theory and some interesting observations from the LHC.
The subject to our surrounding says different than my teacher .. what do we do when circumstances is he is cut out or did not want to participate ... We feel alone in space don't we that's exactly where we are .... If you find yourself not the only one reading or participating because circumstances of our involvement of our surrounding is included . . The potential value for you is your surroundings space .... It Is So into You I don't know what could be a stranged out there But I like things holding certain amounts of my feelings It sets what is a danger zone out there ... I would like to know the strength of my fight For who you are Eventually I bet I get my bundles .... I bet I do Now the teacher who did not want to include himself Knows nothing about it because he is a smart computer program Smart meaning sticking around and f****** with your head .. there is nothing potential in something that can't actually be there If it can't be there it should not be talking that's a general fact ... If it was there it should not be pretending it was Because it's totally not ... 😂 Now that same little smart teacher Might get very mad at you because he's not participating but you are That's the strangest thing to encounter about your activity or yourself😮
We can't help you at what is your life I can only help you at what is mine A little further away from the pyramids but nobody told you to build them ... From today and to the day tomorrow
Bricks were meant to stay exactly where they were meant to stay in the Earth .... Your labor is a voice But my time is a break. 😂🎉. . What you have you already take .... Keep it well in the Kingdom of Heaven
😮 nothing means nothing if the Earth already looks the way you made it then it looks the way you made it I don't pay extra because somebody made it You move on top of the pyramids I don't pay for landslides If it wasn't a landslide you would still be there .... I start at zero that way I know exactly what I'm getting And not figuring out what you were worth .... It's nothing to me and I don't need your ancestry and I don't need your things ..... I don't care if people think they have the right to remove everything and make it normal ... Do what you have to do because I ain't paying for these things ..... 😂 It comes at ground zero Unless you are circumstantially staking higher values in your area and I don't have to make them except for participating Then yeah sure we'll see what it's worth we'll see what you give out
😢 you can do thousands of things and it doesn't mean nothing ... If you were on the right strand I would know it 😊 Trying to pay them back for it is a different story .....😂 Sometimes it's like leaving the pyramids Why would they do that Screw that Somebody left the Washington monument here Let's go see what's inside the White House😂
1:05 Covariant derivatives of vectors with covariant components
8:05 Covariant derivatives of vectors with contravariant components
14:44 Parallel Transport
24:37 The curve with all of its tangent vector parallel to each other is geodesic curve
29:44 Notion of the tangent vectors
31:35 Parallel transport of tangent vectors of geodesic curve is constant. This gives equation of motion of geodesics.
36:43 Trajectories in Space-Time (Minkowski geometry and Riemannian geometry together)
40:58 In general relativity metric becomes a function of space and time.
41:15 The metric of space time always has one negative eigen value and three positive eigen values.
46:23 Problems with special theory of Relativity
48:55 Polar coordinates as the analog of uniformly accelerated coordinate system.
1:05:44 Our aim is to introduce an arbitrary set of coordinates and write the equation of motion of geodesic in it and see if it looks like the particle falling in a uniform gravitational field.
1:06:47 Metric in polar coordinates, its analogy to uniformly accelerated coordinate system.
1:09:12 Expressing Metric of uniformly accelerated coordinate systems interms of ‘g’
1:22:10 Finding how the particle moves in this obtained metric (Equation of motion in geodesics) which is Newton’s equation.
1:33:49 Equation of motion in real gravitational field
1:35:25 The metric for the horizon of black hole
Strictly speaking, Susskind is more of a String theorist and a QFT specialist than a General Relativist. But I’m absolutely amazed at his incredible abilities and insightful approach in teaching GR. He is a gifted lecturer who has a special knack for communicating the heart of the matter behind the seemingly hairy-intimidating equations. By the time he is done teaching a subject you feel like you’ve always had an intuitive feel for it; and that’s the trait of a supremely gifted professor.
couldn't agree more!
Chip 3
Congrats everyone! We made it to lecture 4!
❤thank you very much
I studied physics in college for a few years 55 years ago and then moved on to other things. The last few years I’ve been returning to physics through these lectures. It’s been a real gift. I hope these lectures stay around for me to return to as I move on to books and other sources of study. Just want it known they don’t grow old.
this is inspiring!
@23:00 Vector fields are not parallel transported, elements of the tangent space are. Generally, if you have a path from the point p to the point q, then parallel transport gives you a map from the tangent space at p to the tangent space at q.
It's so f... satisfying when after 4 lectures he shows such a nice result. Congrats to all that came this far 🎉
The (really good!) exposition about "fictitious" gravity in an "accelerated elevator" complements a discussion of a "uniformly accelerated rocket" in Hartle's giant introductory GR tome. It was very useful to have thoroughly understood how a constant proper acceleration appears in a "rest" frame in SR--how each one of the coordinate hyperbolas happens.
If you're a newcomer to GR wanting to really get into it, and watching this series: after going through this route myself, I can definitely tell you Susskind's presentation is very accessible on the one hand, but *extremely* dated on the other hand. There is a much more geometrical approach to GR today that sure, requires developing some familiarity with differential geometry, but boy after you know this stuff all those indices don't look so arbitrary and unmotivated. The simplest example is the "A tensor is a thing that transforms like a tensor" line is one that really bothered me for a long time. Just my own opinion. Nevertheless, coming to his lectures after some additional familiarity with the background math is very refreshing and anyone can learn something new from it, because after all... it's Susskind :)
Are you talking about approaching tensors as multi-linear maps between vector spaces? That is the way I got it from Frederic Schuller on RUclips as well as XylyXylyX What is a tensor as an intro to his 75+ What is GR
All this "It transforms as ...." is pre 1970, maybe 1960.
While watching these lectures I found it useful to pause and look up things like the definition of a tensor field in terms of tensor products of copies of the tangent bundle and cotangent bundle. This helped me to get a better geometric understanding of what is going on, and relate tensors to concepts that I am more familiar with.
@@howieg Exactly
@1:10:00"3 times 3 is ten, so c squared is 10 to the 17th."
HA! Physicists like estimates...
That is hardly an estimate as c is very very close to 3x10^8 m/s
@@orientaldagger6920 3x3=9 not 10. 3.16^2 = 10.0 so this is the estimate
My dream is one day I can master the general relativity. Thank you so much Mr. Leonard Susskind for these lectures, you help me to make my dream come true.
You will. It takes patience and practice. But trust me, at some point it will all click together. This lecture in particular is crucial, as it demonstrates how the geodesic equation and the metric expressed in an accelerated frame suddenly produce gravity. This is exactly what the equivalence principle is all about. Rewatch it as often as you need to understand it completely.
Funny that the professor did not understand the question at 1:39:45.
You don't need a big R for a small A in general. You can certainly translate the world line with a big R left along the x-axis and make it go through the origin.
The professor chose a big R only because he insisted, for mathematical simplicity, that the equations have the form x = R coshω and x = R sinhω.
Thank you so much for sharing these lectures
2:14 dude jumps at a quick opportunity to ask a question
Apparently the official terminology is "geodesic normal coordinates". But honestly, to get hung up on a just a NAME, when Susskind clearly and unambiguously defined what he meant in the previous lecture, is just showing off for no reason...
It's such a privilege being able to learn from Leonard Susskind. I could not have been able to follow along this subject otherwise
In a beer in a cigarillo with some 8:16 Marijuana each 8 mili: later 4:57 Timecode: 6:27 timecode Human Resources: 6:37 6:39 6:40 6:41 6:42 6:42 6:42 6:42 6:42 6:43 6:43 6:43 6:43 6:43 6:44 6:44 6:44 stop count 6:48 6:48 6:49 6:49 6:49 6:49 6:49 6:49 6:49 6:51 6:51 6:51 6:51 6:51 Tim code logline: 7:08 but the only of one cigarillo with some Marijuana 8:05
Because you ain’t gone leave the car on 5:58
7:57 gives me fits of laughter. "G blah blah with respect to X flah flah" poetic.
Great lectures though, currently on number 5.
cybermoose17
Me: looks at lecture number
My brain: “liar!”
i am happy this professor is working in metric, i cant stand US customary units
Almost no one in the scientific field uses the imperial system in the U.S., we hate it too :p
Unless you mean the tensor... that's something else entirely
Lol
What??? Physicists work in units where all the constants are set to 1 preferably LOL !
@@orientaldagger6920 It was a joke. He's intentionally confusing the metric tensor field with the metric system, which shows a lack of comprehension, and is making a valid point about the preference of the metric system. Now, physicists often use natural units, as you stated, so I understand what you're mentioning. :)
@@logansimon6653 There is a running joke about how many constants you can set equal to one and not have things completely fall apart.
Lecture 1, 324k views, Lecture 2, 108k, Lecture 3, 88k, Lecture 4, 64k, Lecture 5, 52K, Lecture 6, 46K, Lecture 7 34K, ... :)
Christian Farina hi now lecture 1 got over 2 million views☺
Exponential decay of views 😂
Anyone else severely under-qualified and still following along to his GR lectures? I'm watching these, reading through Einstein's works, and starting a journal of the whole derivation of GR. To be honest, i kinda want to write it out building up from high school algebra to GR like you would normally, but then expand GR and represent it in a way that only uses notations from a high school algebra level. See how many pages that takes up... lol
Many pages. But why would you do that? You can probably formulate it purely in terms of addition and multiplications and so on but why would you?
Absolutely well done and definitely keep it up!!! 👍👍👍👍👍👍
Quite a bit of hand waving in this lecture. For example, he replaced Rω by t at 1:14:06, without ever explaining why.
It would be beneficial to derive the equation of constant acceleration first. Then all will make sense.
It can be derived by following these steps.
1) In the rest frame of the particle (rest at a single moment), acceleration A will change its four-velocity from (1,0,0,0) to (γ,vγ,0,0) after an infinitesimal amount of time dτ. Here v = A dτ, and γ = 1/sqrt(v).
2) Its four-acceleration, again in the instantaneous rest frame, is therefore [(γ,vγ,0,0) - (1,0,0,0)] / dτ = (0,A,0,0).
3) Since the four-acceleration is a four-vector, it obeys Lorenz transformation. So in our "true rest frame", the four-acceleration is (γvA, γA, 0, 0). Here γ = 1/sqrt(v), and v is the speed of the particle in our frame.
4) In the four-velocity (U0, U1, 0, 0) of the particle, U0 = γ, U1 = γv. Therefore, the four-acceleration can be written as (A U1, A U0, 0,0)
5) So we have the following differential equations. dU0 / dτ = A U1, and dU1 / dτ = AU2.
6) The general solution is U0 = α exp(A τ) + β exp(-A τ), and U1 = α exp(A τ) - β exp(-A τ).
7) We can set U0 = 1, and U1 = 0 at τ = 0, i.e., the particle has zero velocity at time 0. Then α = β = 1/2. And the solution becomes U0 = cosh(A τ), U1 = sinh(A τ).
8) Note U0 = dt / dτ, and U1 = dx / dτ. Therefore t = t0 + 1/A (sinh(A τ), and x = x0 + 1/A cosh(A τ).
This result is exactly the equation given in the lecture with t0 = x0 = 0, R = 1/A, and ω = A τ.
Now it should be clear why Rω can be considered t, which is the same as τ in the particle's own frame.
The fact that the covariant derivative of the metric tensor vanishes everywhere (not only at a single point) cannot be deduced without the assumption that the parallel transport preserves the scalar product of vectors. This assumption leads to the celebrated Levi-Civita theorem of Riemannian geometry, which states that the affine connection components must be the Christoffel symbols. See, for instance, "Riemannian geometry" by M. P. do Carmo (Chapter 2).
Its just disgusting how easy he makes GR look lol. Good Work!
It shows how clearly he thinks and how accurately he can visualize abstract things. It's fascinating, really.
The formula for Christoffer (?) symbol he writes out on the board in the beginning is different from what he derived (wrote down, actually) in Lecture 3. Which of the formulae is correct?
Thanks.
In lecture 3 he simplified "partial derivative with respect to x(index)" to just "partial derivative(index)." The indices of g are switched from one video to the other; g contains implicitly a Kronecker delta (1 if indices are equal, 0 if unequal) and is symmetric, so the order of g's indices do not matter (www.ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf section 5.2). Compare to the Christoffel equation from ruclips.net/video/foRPKAKZWx8/видео.htmlh11m1s , which does contain an error - can you spot it?
Is it possible to find tutorials and assignments for these lectures as well somewhere?
I know 4 years is a while to wait for a reply, but Emil Akhmedov has a course over at Coursera, www.coursera.org/learn/general-relativity, which has peer reviewed exercises. It's a bit more mathematical than LS's lectures, so it would be a good one to do next.
Simon Crase Lmao rhat was a fast reply
Yes but they cost $30,000 a semester.
@@anyuru Not for 1 course.
1. "Flat" applies also to 8-ary space ("octonion"), in which case it's not limited, but for convention, to just-1-or-3 negative-1's, but to 1-or-7 negative-1's... Or any other n-ary.
2. To suppose that the string would not,-break because it's merely a Minkowski space transformation, would be comparable to saying the returning-relativistic-Twin was not-younger... Some of the SR results are going to yield "true" changes, others virtual... (I know, Susskind says we don't-say "SR" but do-say "GR".)
The Gaussian Normal Coordinates he keeps mentioning refer to coordinates obtained from flattening the space time around a point. In a Smooth Riemannian Manifold we can select a neighborhood around each point P small enough so that the metric can be written as the Cartesian metric (i.e. no curvature of any kind) at point P and these coordinates are what he is calling Gaussian Normal coordinates. In terms of Math it is difficult to precisely define what infinesimal means but in Physics if you take an infinitesimal neighborhood around a point P in a curved space time you can choose coordinates such as the metric in that infinitely small space around the point is precisely the Kronecker delta (i.e. Minskowki metric).
the big R means small acceleration
in this case he can assume that the dervative of time with respect to prober time is equal to one ,since the velocity is very low compared to the speed of light
So, R works out to be about 1 light year... where acceleration starts resembling Earth gravity. So what the heck would be the interpretation of distances near zero? What sort of mass would it take to create 1g 1ly away?
sharpfang There is no mass-the geometry of spacetime being considered was flat, which means no mass present to distort it. It was the COORDINATES which were chosen to be curved, representing an observer in a uniformly-accelerated frame.
nice observation :)
and what that means is: if you accellerate with g in free space (forever and ever... constant accelleration), an Event Horizon forms 1 lightyear behind you. You would outrun light from the closest stars that is emitted there at the time you start the accelleration.
No, it just means your coordinates aren't Cartesian. Try it - work out the Christoffel symbols in 2 dimensions for ordinary polar coordinates.
I'd agree but only in a minor way because it's basically Riemannian geometry considerations... I'd fix the notation and move the counterclockwise "upstairs" contravariant superscript to the base symbol leftside evenly with the rightside.
yea heres my doubt ! - well apart from R being equal to 1 over A, isnt R physically the distance of "something" from Origin ? if so what is that "something" and what does a really large R mean ??
Consider the hyperbola at r=R, but forget about the origin just for a moment. The hyperbola is essentially the worldline of a particle accelerating with constant proper acceleration a=1/R (just some number; R has no interpretation as a distance - yet). Notice that the particle doesn't gain infinite speed. It approaches a maximum asymptotically (the speed of light) without ever exceeding it. Now consider the asymptote to the hyperbola. It is also a wordline, but for a particle with constant velocity, c, the speed of light, i.e. it is the wordline of a photon. This photon, traveling on this asymptote NEVER catches up with the hyperbola (the hyperbola and the asymptote never intersect). Notice that this asymptote actually goes through the origin at r=0. In other words, at time T=0, the photon was a distance R away from the accelerating particle.
So here you have the clearest physical interpretation of R I can think of: consider yourself to be a uniformly accelerating observer. The distance R then, is the MINIMUM distance away that a photon needs to start, in order to NEVER catch up with you. Notice that, as you accelerate, the proper distance you measure to that photon, stays R, always. The photon doesn't ever appear to move toward you. The distance R, again, is the minimum distance such a photon needs to be from you, for it it appear to never move toward you.
Indeed, as you accelerate, your "axis of simultaneity", that is the line along which you measure proper distances, "rotates" (in the Lorentz sense of rotation). But it always goes through the origin. So you can think of this origin as "the unique event in spacetime, which is common to ALL my axis of simultaneity". In order words, "the origin is an event, which from my perspective is ALWAYS "now", no matter how much proper time I experience". You cannot see beyond it, either in space (r < 0), or in time (T > 0, X < T). For you, as long as you are accelerating, it constitutes an event horizon.
So that's the second physical interpretation for R: it is the distance to the event horizon, which forms because of your uniform acceleration.
In a truly flat spacetime, the event horizon would be 10^16m, which is about a lightyear, away from a uniformly accelerating observer with acceleration 1g, as Susskind calculates.
@@jimmyb998 this is an excellent explanation, thanks :)
@@tamarravon6638 thanks :)
OK I was hoping at the end when he asked that question, I would get an answer. I know that A = 1/R, to get a modest A = g (w/ c = 1) we should use very big R. But what is the inuition of a big R? If modest A correlates to our everyday feel of g, what does R correspond to in such experience? An alien very far away seeing the effects of our gravity? Because the radius of the earth isn't that big!
See my answer to Satosh Kumar's question above, for a physical interpretation of R in FLAT spacetime. Any alien at a distance R or greater from you, can never catch up with you, or influence you in any way, even if he travels at the speed of light toward you.
However, remember that true spacetime near the Earth is NOT flat. The flat spacetime with uniform acceleration is just some APPROXIMATION, valid in a small region close to the Earth's surface.
Now imagine the Earth were a single point (i.e. a black hole) with the same mass, rather than an extended body. The gravitational field out at the usual Earth radius would be exactly the same. But as you move further in, the gravitational field, that is the acceleration felt by an observer at fixed radius, gets bigger. So at that smaller radius, the flat spacetime approximation is a DIFFERENT one. Indeed closer in to the black-hole Earth, the effective a is greater, and therefore the effective R=c2/a gets smaller. The effective flat spacetime approximation of real spacetime, keeps changing as you move radially in. That's what it means for real spacetime to be curved.
After some distance, you come to a point where the effective a of the flat spacetime approximation, is infinite, and hence R=0. You have now reached the black hole event horizon, at which you appear to be at the ORIGIN of a flat spacetime with uniformly accelerated coordinates. Indeed, in lectures 5 and 6, Susskind analyzes the spacetime close to the horizon, again by studying the flat spacetime approximation appropriate to that region. And in that uniformly accelerated spacetime, the "real" gravity-induced event horizon corresponds to the event horizon at the origin of that flat approximate spacetime.
The distance from your original position, where a=g to the real event horizon, is much sorter than 10^16m, because the real spacetime around the hypothetical earth-mass black hole, is NOT flat, and hence over longer distances the flat approximation doesn't hold. The effective R decreases by MUCH more than the distance you travel inward (due to the increasing gravitational field strength), until you eventually reach the black hole event horizon, where the effective R=0.
Gaussian Normal Coordinates are also known as Riemannian Normal Cordinates (according to S.Carroll)
No. If there your coordinate system is curvilinear whilst describing a flat space then the christoffel symbol can have nonzero values that still imply there is a flat space. He went over this.
Anyone considered to make a transcript of this?
Or anyone would be interested?
I think there is a mistake; in calculating the parallel transport we multiply the covariant derivative by dx_m and put it equal to zero and because there is a minus sign in the definition of covariant derivative that gives the condition for parallel transport = dV^m = gamma*dx_m so see there is no minus sign in this equation, but while using the same equation in calculating the parallel transport of tangent vector the minus sign pops up and thus the equation of motion has a minus sign which should not be there if we are using the definition of the covariant derivative with a minus sign
There is a minus sign for the expression of the covariant derivative for COVARIANT vectors. For CONTRAVARIANT vectors, there's a plus sign in the covariant derivate. Therefore when setting it to 0, and deriving the geodesic equation for dx^m (which is contravariant), or equivalently the tangent vector, you get the extra minus sign.
So let me get this straight.. No matter which coordinates you use the covariant derivative is written as D_r V_m=d_r V_m - (Gamma)_rm^t V_t. If we choose a point and use Guassian Normal Coordiantes defined at that point, then covariant derivative of V_m at that point just reduces to d_r V_m as all gammas are zero. So the covariant derivative transforms like a vector but It always takes the form D_r V_m=d_r V_m - (Gamma)_rm^t V_t no matter which frame we are in. Am I right? Anyone who knows?
correct.
Hi there
One comment. You drew the hyperbolas to the left of the space-time plane. But should be directing upward as they should be inside the light cone which should going upward to be be in the region of velocities less than speed of lights. The drawn hyperbolas in the lecture lie in the region of speeds higher than the speed of light. Alternatively, you can interchange the “time” abs “space axes and it would be correct
All the hyperbolas lie entirely outside the lightcone of the origin, that is correct. In other words, no signal can ever reach any of the accelerating observers from the origin: the origin constitutes an event horizon for all the accelerating observers. Notice, that the hyperbolas, which lie "close" to the horizon, i.e. have a small R, must correspondingly have huge acceleration, as Susskind points out. Put another way: you need to have an increasingly large acceleration to outrun a photon which starts increasingly close to you.
This gentleman is so engaging.
how does he know that the metric of uniform acceleration can be derived from hyperbola curves?
Its just a choice of coordinates. In 2D euclidean, that motion could be represented using polar coordinates..Likewise for Minskowski (x,y,z,t) system, he has used hyperbolic coordinates to describe the motion.
Where is the guy who was listing up the content till now?
in this lesson we are missing subtitles. You can insert them? can you insert subtitles?Thank you
I thoroughly enjoyed this lecture thank you
whenever somebody brings up in class "I read something", the professor should say "see me after class and bring your receipts". in my experience people who say this can't bring receipts. those who do, it turns out to be mass media or something fluffed up for general consumption.
Can someone explain how the g_yr and g_ys terms are zeroes at 1:29:04
He's working out Gamma^y_{tt}, so r=t and s=t. The expression of the metric contains no cross-terms in dydt, and therefore g_yt=0
around 31:00 he says that dx^m is equal to the distance between the 2 points, but he goes on to saying that you devide that by ds, which is the distance between the 2 points. what?
dx is the difference between the coordinates of two points, ds is the "real" distance between the two points because it is calculated using the metric of the spacetime g_mn
dx^m is a vector which goes STRAIGHT through the second point from the first point. dS is the distance between those two points ALONG the curve.
+David O'Neill No, delta S is the distance along the curve between two distinct points, dS is the magnitude of the vector dxi.
Consider unitarity of the vectors along that curve.
Why is it not possible to express electric field as space curvature at the same way we can do with gravitational field?
because it gives wrong values.
I asked one of my professors this once. His response was that essentially you could, but it isn't the best way to do it.
My impression was that for reasonable systems one could extract correct answers to a sufficient precision this way. At extreme scales, however, (notably those where QED is necessary) the approximation would fail. So you have two cases: where classical EM is a sufficiently good approximation, and thus there is no reason to complicate the problem with the difficult machinery of GR, and where it isn't, in which case GR isn't good enough either because EM is a quantum phenomenon.
I wish I had asked via email rather than during a lecture so that I could properly quote or even remember his response. I may be misrepresenting what he told me. But it was clearly something he had put some thought into previously.
Very simple: because of the principle of equivalence. Because intertial mass is the same as gravitational mass, gravity doesn't have to be a force, it can be simply seen as an acceleration field. Whereas electromagnetic forces are not proportional to the inertial mass, so they do not affect all bodies the same way, but depending on their electric charge. So how can the EM field be an intrinsic property of space, if it is not the same for all bodies in space?
tl;dr: because EM forces are not proportional to mass.
Lenny,
A0620-00, the nearest black hole, is 3,000 light-years away. The Fermipause is the end of the influence of Fermi Bubbles and sweeps out a sphere that comes within 3,000 light-years of our solar system. Fermi Bubbles protruding from the Galactic Center span 25,000 light-years. Our Solar System is 28,000 light-years from the Galactic Center. Black holes are an integral part of a unified theory of cosmic plasma physics.
Very Respectfully,
Eric D'Aleo
What is at the origin of the light cone? The observer or the thing accelerating? And if the points on those curves all have the same distance from the origin why does it look like X will increase as you evolve up the time axis?
The origin of the light cone is the exterior observer; that is, the person stationary outside of the "moving elevator" or whatever you want to call it. X looks like it increases up the time axis because it does; when time increases, the spatial component has to as well to make the distance (remember, pseudo-euclidean) a constant (since distance^2 = t^2 - x^2)
now i get it, what Einstein said:
‘If you can’t explain it simply, you don’t understand it well enough’.
(great job done sir!)
So you don't understand it?😂
Tensor and matrix?
1:17:45 How can you do that when you change the units? 1 is unitless but 2yg has units. How does this work? O_o
Theoreticians use something called Natural units to simplify their calculations, setting c or mass = 1 for example. It just makes it more convenient as long as you don't forget about them!
Yea I forgot about those. Those constants should really be included in derivations like these though. Otherwise you never know what you can and cannot do mathematically. Thanks
if c=1, the units of acceleration are length^(-1), so 2yg is indeed dimensionless. In conventional units 2yg/c^2 would be dimensionless.
To complete that statement about the string breaking or-not, Note that time in the relative frame, by SR-calculation, remains zero-offset: so, at a given time, the string is, being stretched the greater distance... Then, the only question is whether or not the string might 'stretch' due to 'spooky action at a distance...' But the answer to that is, QM requires the stretched atom to want to fall back and so break the string; Also constant acceleration changes wavelengths to lock-in possibilities.
12:27 How did he get this result? I'm not good at tensor index manipulation, so I couldn't figure out. Can someone please explain ?
He got that from the earlier lecture. It's the same equation as he derived in the earlier lecture.
He is not a mathematician. He just says it. Something I think was missed in his notes. If you go back about a minute he is differentiating a product with the metric... WHY??.
The metric allows a lower index vector (tensor) to be converted to an upper index (contravariant and covariant are for Lit majors). I think he meant to show that after taking the derivative at 11:53, he then raises the index. The plus minus on the Gamma, I think is a whole other trick.
at 30:58, is ds same as dx^m?
No, dx^m refers to the change in the coordinates between the two points, while ds refers to the distance (or rather, displacement) travelled by the vector.
@@bhuvansv3701 i;m afraid you’re incorrect, sweetie. “dx*m is CLEARLY a veiled reference to dextromethorphan (aka the active ingredient in most cough syrups). it is a highly recommended substance that many aid you in undestanding more difficult equations. its a ‘dissociative’, which means it will allow you to shed your grammar school biases. hope this helps!
A d as an eyeball 5:16 ????? Too much potus now ???? 5:25 5:26 5:26 5:26 5:26 5:26 5:27 5:27 5:27 5:27 5:27 5:27 5:28 5:28 5:28 5:28 5:28 5:28 5:29 5:29 5:29 5:29 5:29 5:30 5:30 5:30 5:30 5:30
McCAT spinnin' up the Best Coordinates!
looking for the timestamp guy
And for a little more clarity I'll bi-clarify that last clarification and point out that before the stretched string breaks it emits heat F×D=W by Hook's Law which has to be an interesting result from SR that accelerated objects heat-up.... 2. Only if accelerating the string in the co-moving frame converts, its rest mass, to kinetic energy, and thereby lengthens its momentum wavelength, should it stretch without breaking; But such 'affectation' is not expected, of the greater mass pulling it....
why that 2yg term is -gtt?
1+2yg=-gtt, because it's the factor in front of dt^2 in the expression of the metric in the (t,y) coordinates. That's what gtt means.
Einstein at first couldn’t grasp what Minkowski had done but when he did used it ( line elements)to start his quest for GR .
nice lecture, very inspirational, although I haven't fully understood.
are there any lectures on "dark matter" and "dark energy".?
Uniform gravitational Force
Tensors
Metric 0
Coverent
I fell asleep and woke up to this
when he wrote cosine square plus sine square equals 1 I lost it
First level big 👹 finally comes up! 😁🙀
@00:53 he missed a spot. am I the only one annoyed by that.
It annoys me too. And he always does that!
Reposted from lecture 3: I'm still confused about the Gaussian normal coordinates. Let's say I'm looking at the simplest example of curvilinear coordinates: polar coordinates (r,phi) in the plane with the metric tensor g=diag(1,r^2). Now let's say I want to write down a transformation x^i = y^i + c^i_j,k y^j y^k that takes the metric tensor to the Kronecker delta, where the y^i are (r - r_0) and (phi - phi_0) respectively. The Jacobian of this tranformation is always the unit matrix, leaving the metric tensor invariant. What I can do is Taylor expand u=r*cos(phi) and v=r*sin(phi) around a point, and then use the expanded u and v as coordinates. This makes the first derivatives of g vanish, but is in general not of the desired form above. - I'm guessing that the least I need to do is include the scale factors in the definition of the y^i, i.e. consider polynomial functions of (r - r0) and r_0*(phi - phi_0).
starting to make sense :) thanks for these
Calculus, Advanced calculus, Physics, Advanced Physics
❤❤
I think this is great mathematics, personally it makes me feel like I'm wasting my time though; I don't think its the best way to model gravity. I am working on a different model which uses QFT, m-theory and some interesting observations from the LHC.
Uh? Are you talking about Maldacena?
This won't slove world hunger
R.U.N.C.L.E. L.E.O.
I think I lost him 4 lectures ago
i am play testing version 1.14d, i don't want to publish the builds. i want it uniquely for these aristocrats to use. that's the plan at least
Gold!
1:35:40 Savage
Sleep, Party and get your GED. And you shall be endowed with knowledge.
Thomas Steven Thomas Christopher Harris Christopher
susskind may get it first!
somewheres.
BCAAP
how noble
During my stay at Harvard in 2006, I learned that Yale is beneath us. Generally and relatively speaking.
your "NOBLE PRIZE" is on its way ;p
Lol who literally would dislike this???
Leonard Susskind=I+M+N/nirvana
Whats now? Die nerven 🙄 und ihr langsam auch
I think relativity should be revised in order to survive the next 20 years.
Ro
1st!
The subject to our surrounding says different than my teacher
.. what do we do when circumstances is he is cut out or did not want to participate
... We feel alone in space don't we that's exactly where we are
....
If you find yourself not the only one reading or participating because circumstances of our involvement of our surrounding is included
. . The potential value for you is your surroundings space
.... It Is So into You
I don't know what could be a stranged out there
But I like things holding certain amounts of my feelings
It sets what is a danger zone out there
... I would like to know the strength of my fight
For who you are
Eventually I bet I get my bundles
.... I bet I do
Now the teacher who did not want to include himself
Knows nothing about it because he is a smart computer program
Smart meaning sticking around and f****** with your head
.. there is nothing potential in something that can't actually be there
If it can't be there it should not be talking that's a general fact
... If it was there it should not be pretending it was
Because it's totally not
... 😂 Now that same little smart teacher
Might get very mad at you because he's not participating but you are
That's the strangest thing to encounter about your activity or yourself😮
We can't help you at what is your life I can only help you at what is mine
A little further away from the pyramids but nobody told you to build them
... From today and to the day tomorrow
Bricks were meant to stay exactly where they were meant to stay in the Earth
.... Your labor is a voice
But my time is a break.
😂🎉. . What you have you already take
.... Keep it well in the Kingdom of Heaven
I don't pay more for landslides
😮 nothing means nothing if the Earth already looks the way you made it then it looks the way you made it I don't pay extra because somebody made it
You move on top of the pyramids
I don't pay for landslides
If it wasn't a landslide you would still be there
.... I start at zero that way I know exactly what I'm getting
And not figuring out what you were worth
.... It's nothing to me and I don't need your ancestry and I don't need your things
..... I don't care if people think they have the right to remove everything and make it normal
... Do what you have to do because I ain't paying for these things
..... 😂 It comes at ground zero
Unless you are circumstantially staking higher values in your area and I don't have to make them except for participating
Then yeah sure we'll see what it's worth we'll see what you give out
😢 you can do thousands of things and it doesn't mean nothing
... If you were on the right strand I would know it
😊 Trying to pay them back for it is a different story
.....😂 Sometimes it's like leaving the pyramids
Why would they do that
Screw that
Somebody left the Washington monument here
Let's go see what's inside the White House😂