According to my Euclidean Relativistic Slide Rule the light signals are redshifted when they are moving away from each other by gamma plus proper velocity (1.25 + .75 = 2) and blueshifted when moving towards each other (1.25 - .75 = .5) as you have shown. At the beginning of the video you point out that in Steve’s frame, only the moving spaceship is length contracted by .8 (which would make no difference to the distance between earth and star) whereas in Trevor’s frame the moving distance between earth and star is length contracted by .8 (.8 x 6 = 4.8) which accounts for the time difference on both outward and inward journeys. (2 x 4.8/.6 = 16) verses (2 x 6/.6 = 20). No one ever experiences time dilation or length contraction in there OWN rest frame, it’s only observed in the other (moving) frame as an explanation for the asymmetry.
For the Twin Paradox, what if the Rocket traveled in a Large Circular Path then came back to the same start point, then there is no accelerate/decelerate
The Twin paradox compares the worldline arc lengths of two travelers connecting a pair of spacetime events so in a theory of general frames either twin can end up being older depending on how the paths are drawn up. It should be easy enough to google the twin paradox in general relativity.
Time dilation, the projection of the traveler worldline onto an arbitrary set of basis vectors, always has the traveler clock running slow. The observation of a clock couples time dilation with the Doppler effect and clocks will be observed to be running either faster or slower.
@@kylelochlann5053 Ok thanks. The relationship of simultaneity changes in a discontinuous manner (i.e., abruptly) when the traveler changes direction, and therefore inertial reference frames, at the star. And this causes the traveling twin to age less.
9:30 - Both clocks slow... that is the clock/twin paradox. This is what makes no logical sense about the Galilean Inertial Frame symmetry proposed by Einstein's relativity theory. There does not seem to be any resolution to this issue, and this "clock" issue is caused by removing the Aether reference frame for EM radiation/light. If we are always required to set a preferred reference frame to resolve the paradox, then we are back to Lorentz-Poincare Ether Theory based relativity.
Special Relativity is completely logically self-consistent, and if you work in Minkowski Space (M4): there are no contradictions, hence no apparent contradictions, and thus: no paradoxes. When you pick a frame and slice M4 into Euclidean space (E3) plus time, you get apparent contradictions (aka: paradoxes). The Twin Paradox exists on several levels: 0) Time dilation. It's a paradox from any Newtonian point of view, but that is considered 'resolved' before starting the Twin Paradox. 1) the naive one is caused by thinking the twins are symmetric. They are manifestly not, as this video points out. Most treatments stop here. 2) the deeper one is cause by the fact that when space twin is at the turn around point, there is no unique "now" back on Earth. If he changes his speed while 10 light years away, the Earth clock can change, at will, by +/-10 years. This accounts for the "missing time" caused by the fact that both twins see the other clock ticking slowly. This is the source of the "Andromeda Paradox"....and that is, and always will be: a paradox.
@@DrDeuteron *>* This is the problem ^^^. You ALWAYS have to invalidate Einstein's Principle of Relativity via a non-inertial frame to resolve who's clock is moving slower, and "who's clock is moving slower" is really just an abstraction of "who is really moving faster." Take the clocks/twins out of it. *Who is actually moving faster?* Because the Principle of Relativity has them BOTH "at rest" and BOTH moving at the same speed simultaneously. Which Lorentz math answer do you throw away, without invoking a non-inertial frame, i.e. break the symmetry declared in SR? This is the illogic bug in SR and it's source lives in the Relative Simultaneity argument in Section 2 of his 1905 paper. The equations in Section 2 are just _Distance = Rate * Time,_ so anyone can understand them.
@@Hexnilium ** Yes, if you understand where the math came from (Voigt Transformation), you can see that it was derived from the Doppler Effect and a moving light source relative to a stationary observer. The Doppler Effect is changing the TIMING and SPACING of the wave oscillations of light. If you think of that as CLOCKs and RULERs, like Einstein, you get into his weird, fantasy physics.
@@DrDeuteron ** But you just jumped way ahead and into the fantasy physics that does not represent how nature/reality works. If you take Einstein's 1st postulate of the Principle Of Relativity and combine it with the 2nd postulate of Voigt/Lorentz Math (otherwise known as the speed of light postulate), they are incompatible for inertial frames. Inertial frames is what Special Relativity is about. With 2 inertial frames, you always end up with the clock paradox of BOTH frames having a t'. This is not only illogical, it has NEVER been proven in any experiment. SR is defined by the Principle Of Relativity (not the Voigt/Lorentz math) and this Principle of Relativity is NEVER applied in ANY Special Relativity experiments. Basically, all the "time dilation" experiments just prove Lorentz Ether Theory correct and invalidate Einstein's SR, because they all require a preferred reference frame. Otherwise, they would all fail.
I disagree, THRE IS NO SYMMETRY BETWEEN THE TWINS. The clock of the astronaut twin slows down compared to that of the other twin on Earth, even if the spaceship continues to travel. In my opinion (and according to Mathematics), both clocks cannot slow down. (as predicted by SR) I consider the two Lorentz transformations: a) x_1 = gamma * (x - v * t) b) x = gamma * (x_1 + v * t_1) IF WE REPRESENT THE LORENTZ TRANSFORMATIONS IN THIS FORM, IT IS EASIER TO UNDERSTAND. With gamma I obviously indicated the Lorentz factor and I do not consider the other two Lorentz transformations because they depend on a) and b). The two Lorentz transformations indicate that: ............................................................................................................................................................................. (x_1 = 0) c) the frame F_1 moves with speed v with respect to the frame F, and the frame F_1 travels a certain distance d in the frame F. .............................................................................................................................................................................. (x = 0) d) the frame F moves with speed -v with respect to the frame F_1, and the frame F travels a certain distance d_1 in the frame F_1. .............................................................................................................................................................................. c) and d) are MUTUALLY EXCLUSIVE! IF IT IS TRUE c), IT IS NOT TRUE d) IF IT IS TRUE d), IT IS NOT TRUE c) What does c) mean? If the frame F_1 travels a distance d while moving with respect to the frame F (x = v * t = d, t = d / v, x_1 = 0), we obtain: d = gamma * v * t_1, t_1 = d / (v * gamma), t_1 < t. What does d) mean? If the frame F travels a distance d_1 while moving with respect to the frame F_1 (x_1 = - v * t_1 = - d_1, t_1 = d_1 / v, x = 0), we obtain: - d_1 = - gamma * v * t, t = d_1 / (v * gamma), t < t_1. If the frame F_1 travels a distance d in the frame F, it is a mistake to think that frame F travels any distance in the frame F_1, the Lorentz transformations represent a system of two equations in four unknowns x, t, x_1, t_1. When we solve a system of two equations and we use some initial conditions, those initial conditions we cannot change! If a frame F_1 is moving with speed v with respect to a second frame F, it is the contracted frame of F (where the distances between any two points are contracted) that moves with speed -v with respect to the frame F_1. If we have chosen x_1 = 0 in the system resolution, then it makes no sense at all to go back and consider x = 0. If we choose x_1 = 0, then we don't consider x = 0. If we choose x = 0, then we don't consider x_1 = 0. If we make a choice (and we don't go back to change the choice made), THERE ARE NO CONTRADICTIONS! If x_1 = 0, then t_1 < t and THE FRAME F IS AT REST! If x = 0, then t < t_1 and THE FRAME F_1 IS AT REST! If (x = 0) and (x_1 = 0), then x = x_1 = t = t_1 = 0! In an empty space the twin paradox cannot be solved, but if we know that one twin is in a spaceship ("the spaceship reaches a star") and the other twin is on Earth, the astronaut twin is younger than Earth. (The frame of the Earth is "AT REST") The frame of the Earth is the frame F, the frame of the spaceship is the frame F_1. THE FRAME OF THE EARTH IS NOT THE PRIVILEGED FRAME, but in the frame of the Earth a spaceship moves with uniform rectilinear motion (v * t = d, t = d / v and the elapsed time cannot be different! That's all) If the astronaut twin launches a rocket, then the frame of the spaceship is ALSO "AT REST" (obviously not with respect to Earth, but with respect to the rocket) The frame of the Earth and the frame of the spaceship are both inertial frames. (The two postulates of Relativity apply to both the frame of the Earth and the frame of the spaceship, even if the frame of the spaceship is younger than the frame of the Earth) THE CLOCK OF THE SPACESHIP SLOWS DOWN, THE FRAME OF THE SPACESHIP IS YOUNGER THAN EARTH (EVEN IF THE SPACESHIP CONTINUES TO TRAVEL), AND THE FRAME OF THE EARTH IS OLDER THAN THE FRAME OF THE SPACESHIP! IF THE TWINS KNOW THE LORENTZ TRANSFORMATIONS, ONLY ONE OF THE TWO CLOCKS CAN SLOW DOWN.(COMPARED TO THE OTHER CLOCK) The twins may be the same age (imagine two twins traveling in two spaceships, one to the left and one to the right with respect to the Earth), each twin "sees" the other twin in motion but it is wrong to consider the time dilation!
That is incorrect. Time dilation is just the projection of a traveler's worldline onto the observer's worldline, and it's arbitrary which worldline is considered which. It is effectively the same symmetry you find with a dot product.
@@kylelochlann5053 Ok, but this is the resolution strategy for me. I try to summarize one of my articles Consider two frames F and F_1 in relative motion to each other with speed v, the frames F and F_1 are coincident at the initial time. (t = t_1 = 0 s) The point P (-d; 0) of the frame F_1 reaches the origin O of the frame F at time t = d / (gamma *v) The point Q (d; 0) of the frame F also reaches the origin O_1 of the frame F_1 at time t_1 = d / (gamma * v) F: frame of the Earth F_1 : frame of the spaceship Imagine the spaceship has a tail, the length of the tail is d in the frame of the spaceship. At time t = d / (gamma * v), the end of the tail reaches the Earth in the frame of the Earth. Meantime, the point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1 at time t_1 = d / (gamma * v) t = t_1 = d / (gamma * v), for any value of d. At the end of the tail of the spaceship is John, at time t = d / (gamma * v) the twin on Earth sees John and, in the frame of the spaceship, at time t_1 = d / (gamma * v), John sees the Earth. At time t = d / (gamma * v), the spaceship did not reach point Q (a star, for example) If we know that the spaceship moves with uniform rectilinear motion in the frame of the Earth at speed v (x = v * t), then: a) the point Q (d; 0) reaches the spaceship at time t_1 = d / (gamma * v), and b) the Earth also reaches the point P (-d; 0) at time t_1 = d / (gamma * v). The whole frame F_1 moves with uniform rectilinear motion in the frame F, "there is another spaceship in the point P(-d; 0)". Do not consider a single spaceship only in the origin O_1 (0; 0) If we analyze the motion of the Earth in the frame of the spaceship, the mathematical equation x_1 = - v * t_1 is wrong. We must consider x_1 = - gamma * v * t_1. Thus, at time t_1 = d / (gamma * v), x_1 = -d. And that's right, because the Earth reaches point P at time t_1 = d / (gamma * v). If we know that a frame moves with uniform rectilinear motion in a second frame with speed v, the second frame cannot move with speed -v with respect to the first frame. The spaceship moves in the frame of the Earth, we know that x = v * t! If instead x_1 = - v * t_1, then: c) the point P(-d; 0) reaches the origin O of the frame F at time t = d / (gamma * v), and d) the origin O_1 of the frame F_1 also reaches the point Q (d; 0) at time t = d / (gamma * v).
@@massimilianodellaguzzo8571 That is completely unclear. There is a need to specify what the quantities refer to in the calculations. I gather we are considering a spacetime, S=[M,g], that is a smooth semi-Riemannian manifold equipped with metric g, where g is the Minkowski metric, or g_{\mu u}=\eta_{\mu u}. So far so good, but what metric signature are you using? What makes a reading of your scenario impossible is that you're not making the distinction between physical and coordinate values. Let's take the first thing you wrote as an example: "The point P (-d; 0) of the frame F_1 reaches the origin O of the frame F at time t = d / (gamma *v)"
@@kylelochlann5053 I try to repeat myself, adding more information. Consider two frames F and F_1 in relative motion to each other with speed v, the frames F and F_1 are coincident at the initial time. (t = t_1 = 0 s) Frame F_1 moves to the right (with respect to frame F) and frame F moves to the left (with respect to frame F_1) The point P (- d; 0) of the frame F_1 reaches the origin O of the frame F at time t = d / (gamma * v) The point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1 at time t_1 = d / (gamma * v) t = t_1 = d / (gamma * v), for any value of d. (THIS IS THE SYMMETRY OF THE LORENTZ TRANSFORMATIONS !) This happens both if the frame F_1 moves at speed v with uniform rectilinear motion in the frame F, and if the frame F moves at speed - v with uniform rectilinear motion in the frame F_1. The contracted distances of the frame F_1 move at speed v (with respect to frame F), and the contracted distances of the frame F move at speed - v. (with repect to frame F_1) F: frame of the Earth F_1 : frame of the spaceship Imagine the spaceship has a tail, the length of the tail is d in the frame of the spaceship. At time t = d / (gamma * v), the end of the tail reaches the Earth in the frame of the Earth. Meantime, the point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1 at time t_1 = d / (gamma * v) in the frame of the spaceship. t = t_1 = d / (gamma * v), for any value of d. (THE SYMMETRY OF THE LORENTZ TRANSFORMATIONS, THE TRUE SYMMETRY) At the end of the tail of the spaceship is John, at time t = d / (gamma * v) the twin on Earth "sees" John and, in the frame of the spaceship, at time t_1 = d / (gamma * v), John "sees" the Earth and point Q reaches the spaceship. At time t = d / (gamma * v), in the frame of the Earth the spaceship did not reach point Q (a star, for example) Furthermore at time t = d / (v * gamma) in the Earth's frame, the time is t_1 = d / (v * gamma * gamma) in the frame of the spaceship (If x = v * t) Imagine the point R (d / gamma; 0) of the frame of the Earth and consider the two events A and B A: Earth reaches point P. B: point R reaches the spaceship. The two events A and B are simultaneous in the frame of the Earth, the point P reaches the Earth at time t = d / (gamma * v) and the spaceship reaches the point R at time t = d / (gamma * v) In the frame of the Earth, the length of the tail is contracted. If the spaceship travels a distance equal to d / gamma, the twin on Earth "sees" John. The two events A and B are not simultaneous in the frame of the spaceship, point R reaches the spaceship at time t_1 = d / (gamma * gamma * v) and the Earth reaches point P at time t_1 = d / (gamma * v) Now consider event C C: point Q reaches the spaceship Event C represents: the spaceship reaches point Q (in the frame of the Earth) Event C represents: point Q reaches the spaceship (in the frame of the spaceship) In the frame of the Earth the spaceship reaches the point Q at time t = d / v and, in the frame of the spaceship, the point Q reaches the spaceship at time t_1 = d / (gamma * v), exactly when the Earth reaches the point P. If we know that the spaceship moves with uniform rectilinear motion in the frame of the Earth at speed v (x = v * t), then: a) the point Q (d; 0) reaches the spaceship at time t_1 = d / (gamma * v), and b) the Earth also reaches the point P (- d; 0) at time t_1 = d / (gamma * v). The whole frame F_1 moves with uniform rectilinear motion in the frame F, "there is another spaceship in the point P(-d; 0)". Do not consider a single spaceship only in the origin O_1 (0; 0). Everyone considers a single spaceship in the origin O_1 (0; 0), the number of spaceships is infinite! If we analyze the motion of the Earth in the frame of the spaceship, the mathematical equation x_1 = - v * t_1 IS WRONG. We must consider x_1 = - gamma * v * t_1. Thus, at time t_1 = d / (gamma * v), x_1 = - d. And that's right, because the Earth reaches point P at time t_1 = d / (gamma * v). If instead x_1 = - v * t_1, then: c) the point P(-d; 0) reaches the origin O of the frame F at time t = d / (gamma * v), and d) the origin O_1 of the frame F_1 also reaches the point Q (d; 0) at time t = d / (gamma * v). Let me know if anything is unclear. (I don't know the English language perfectly)
@@massimilianodellaguzzo8571 Awesome, thank you. I have some family coming in for the weekend and will work through what you have there when I can, but you can expect a reply in a day or two. I do see in the wording that you have a ship of some length and I'm not seeing calculations for the clock desynchronization (the leading clock runs deeper in time) so there is already some red flags, but it might be just my superficial glance of what I see. It might be nothing, but I'll let you know.
How this will look for an outside observer watching the whole scene from their frame of reference ?
According to my Euclidean Relativistic Slide Rule the light signals are redshifted when they are moving away from each other by gamma plus proper velocity (1.25 + .75 = 2) and blueshifted when moving towards each other (1.25 - .75 = .5) as you have shown. At the beginning of the video you point out that in Steve’s frame, only the moving spaceship is length contracted by .8 (which would make no difference to the distance between earth and star) whereas in Trevor’s frame the moving distance between earth and star is length contracted by .8 (.8 x 6 = 4.8) which accounts for the time difference on both outward and inward journeys. (2 x 4.8/.6 = 16) verses (2 x 6/.6 = 20). No one ever experiences time dilation or length contraction in there OWN rest frame, it’s only observed in the other (moving) frame as an explanation for the asymmetry.
For the Twin Paradox, what if the Rocket traveled in a Large Circular Path then came back to the same start point, then there is no accelerate/decelerate
The Twin paradox compares the worldline arc lengths of two travelers connecting a pair of spacetime events so in a theory of general frames either twin can end up being older depending on how the paths are drawn up. It should be easy enough to google the twin paradox in general relativity.
On the return journey, do the moving clocks really run faster? I thought they always ran slower.
Time dilation, the projection of the traveler worldline onto an arbitrary set of basis vectors, always has the traveler clock running slow. The observation of a clock couples time dilation with the Doppler effect and clocks will be observed to be running either faster or slower.
@@kylelochlann5053 Ok thanks. The relationship of simultaneity changes in a discontinuous manner (i.e., abruptly) when the traveler changes direction, and therefore inertial reference frames, at the star. And this causes the traveling twin to age less.
9:30 - Both clocks slow... that is the clock/twin paradox. This is what makes no logical sense about the Galilean Inertial Frame symmetry proposed by Einstein's relativity theory. There does not seem to be any resolution to this issue, and this "clock" issue is caused by removing the Aether reference frame for EM radiation/light. If we are always required to set a preferred reference frame to resolve the paradox, then we are back to Lorentz-Poincare Ether Theory based relativity.
Special Relativity is completely logically self-consistent, and if you work in Minkowski Space (M4): there are no contradictions, hence no apparent contradictions, and thus: no paradoxes. When you pick a frame and slice M4 into Euclidean space (E3) plus time, you get apparent contradictions (aka: paradoxes).
The Twin Paradox exists on several levels:
0) Time dilation. It's a paradox from any Newtonian point of view, but that is considered 'resolved' before starting the Twin Paradox.
1) the naive one is caused by thinking the twins are symmetric. They are manifestly not, as this video points out. Most treatments stop here.
2) the deeper one is cause by the fact that when space twin is at the turn around point, there is no unique "now" back on Earth. If he changes his speed while 10 light years away, the Earth clock can change, at will, by +/-10 years. This accounts for the "missing time" caused by the fact that both twins see the other clock ticking slowly. This is the source of the "Andromeda Paradox"....and that is, and always will be: a paradox.
@@DrDeuteron *>*
This is the problem ^^^. You ALWAYS have to invalidate Einstein's Principle of Relativity via a non-inertial frame to resolve who's clock is moving slower, and "who's clock is moving slower" is really just an abstraction of "who is really moving faster." Take the clocks/twins out of it. *Who is actually moving faster?* Because the Principle of Relativity has them BOTH "at rest" and BOTH moving at the same speed simultaneously. Which Lorentz math answer do you throw away, without invoking a non-inertial frame, i.e. break the symmetry declared in SR? This is the illogic bug in SR and it's source lives in the Relative Simultaneity argument in Section 2 of his 1905 paper. The equations in Section 2 are just _Distance = Rate * Time,_ so anyone can understand them.
I personally think there is no SR/GR and that the time dilation effects after sourced from something else.
@@Hexnilium **
Yes, if you understand where the math came from (Voigt Transformation), you can see that it was derived from the Doppler Effect and a moving light source relative to a stationary observer. The Doppler Effect is changing the TIMING and SPACING of the wave oscillations of light. If you think of that as CLOCKs and RULERs, like Einstein, you get into his weird, fantasy physics.
@@DrDeuteron **
But you just jumped way ahead and into the fantasy physics that does not represent how nature/reality works. If you take Einstein's 1st postulate of the Principle Of Relativity and combine it with the 2nd postulate of Voigt/Lorentz Math (otherwise known as the speed of light postulate), they are incompatible for inertial frames. Inertial frames is what Special Relativity is about. With 2 inertial frames, you always end up with the clock paradox of BOTH frames having a t'. This is not only illogical, it has NEVER been proven in any experiment. SR is defined by the Principle Of Relativity (not the Voigt/Lorentz math) and this Principle of Relativity is NEVER applied in ANY Special Relativity experiments. Basically, all the "time dilation" experiments just prove Lorentz Ether Theory correct and invalidate Einstein's SR, because they all require a preferred reference frame. Otherwise, they would all fail.
I disagree, THRE IS NO SYMMETRY BETWEEN THE TWINS.
The clock of the astronaut twin slows down compared to that of the other twin on Earth, even if the spaceship continues to travel.
In my opinion (and according to Mathematics), both clocks cannot slow down. (as predicted by SR)
I consider the two Lorentz transformations:
a) x_1 = gamma * (x - v * t)
b) x = gamma * (x_1 + v * t_1)
IF WE REPRESENT THE LORENTZ TRANSFORMATIONS IN THIS FORM, IT IS EASIER TO UNDERSTAND.
With gamma I obviously indicated the Lorentz factor and I do not consider the other two Lorentz transformations because they depend on a) and b).
The two Lorentz transformations indicate that:
.............................................................................................................................................................................
(x_1 = 0)
c) the frame F_1 moves with speed v with respect to the frame F, and the frame F_1 travels a certain distance d in the frame F.
..............................................................................................................................................................................
(x = 0)
d) the frame F moves with speed -v with respect to the frame F_1, and the frame F travels a certain distance d_1 in the frame F_1.
..............................................................................................................................................................................
c) and d) are MUTUALLY EXCLUSIVE!
IF IT IS TRUE c), IT IS NOT TRUE d)
IF IT IS TRUE d), IT IS NOT TRUE c)
What does c) mean?
If the frame F_1 travels a distance d while moving with respect to the frame F (x = v * t = d, t = d / v,
x_1 = 0), we obtain:
d = gamma * v * t_1, t_1 = d / (v * gamma), t_1 < t.
What does d) mean?
If the frame F travels a distance d_1 while moving with respect to the frame F_1 (x_1 = - v * t_1 = - d_1,
t_1 = d_1 / v, x = 0), we obtain:
- d_1 = - gamma * v * t, t = d_1 / (v * gamma), t < t_1.
If the frame F_1 travels a distance d in the frame F, it is a mistake to think that frame F travels any distance in the frame F_1, the Lorentz transformations represent a system of two equations in four unknowns x, t, x_1, t_1.
When we solve a system of two equations and we use some initial conditions, those initial conditions we cannot change!
If a frame F_1 is moving with speed v with respect to a second frame F, it is the contracted frame of F (where the distances between any two points are contracted) that moves with speed -v with respect to the frame F_1.
If we have chosen x_1 = 0 in the system resolution, then it makes no sense at all to go back and consider x = 0.
If we choose x_1 = 0, then we don't consider x = 0.
If we choose x = 0, then we don't consider x_1 = 0.
If we make a choice (and we don't go back to change the choice made), THERE ARE NO CONTRADICTIONS!
If x_1 = 0, then t_1 < t and THE FRAME F IS AT REST!
If x = 0, then t < t_1 and THE FRAME F_1 IS AT REST!
If (x = 0) and (x_1 = 0), then x = x_1 = t = t_1 = 0!
In an empty space the twin paradox cannot be solved, but if we know that one twin is in a spaceship ("the spaceship reaches a star") and the other twin is on Earth, the astronaut twin is younger than Earth. (The frame of the Earth is "AT REST")
The frame of the Earth is the frame F,
the frame of the spaceship is the frame F_1.
THE FRAME OF THE EARTH IS NOT THE PRIVILEGED FRAME, but in the frame of the Earth a spaceship moves with uniform rectilinear motion (v * t = d, t = d / v and the elapsed time cannot be different! That's all)
If the astronaut twin launches a rocket, then the frame of the spaceship is ALSO "AT REST" (obviously not with respect to Earth, but with respect to the rocket)
The frame of the Earth and the frame of the spaceship are both inertial frames. (The two postulates of Relativity apply to both the frame of the Earth and the frame of the spaceship, even if the frame of the spaceship is younger than the frame of the Earth)
THE CLOCK OF THE SPACESHIP SLOWS DOWN, THE FRAME OF THE SPACESHIP IS YOUNGER THAN EARTH (EVEN IF THE SPACESHIP CONTINUES TO TRAVEL), AND THE FRAME OF THE EARTH IS OLDER THAN THE FRAME OF THE SPACESHIP!
IF THE TWINS KNOW THE LORENTZ TRANSFORMATIONS, ONLY ONE OF THE TWO CLOCKS CAN SLOW DOWN.(COMPARED TO THE OTHER CLOCK)
The twins may be the same age (imagine two twins traveling in two spaceships, one to the left and one to the right with respect to the Earth), each twin "sees" the other twin in motion but it is wrong to consider the time dilation!
That is incorrect. Time dilation is just the projection of a traveler's worldline onto the observer's worldline, and it's arbitrary which worldline is considered which. It is effectively the same symmetry you find with a dot product.
@@kylelochlann5053 Ok, but this is the resolution strategy for me. I try to summarize one of my articles
Consider two frames F and F_1 in relative motion to each other with speed v, the frames F and F_1 are coincident at the initial time. (t = t_1 = 0 s)
The point P (-d; 0) of the frame F_1 reaches the origin O of the frame F at time t = d / (gamma *v)
The point Q (d; 0) of the frame F also reaches the origin O_1 of the frame F_1
at time t_1 = d / (gamma * v)
F: frame of the Earth
F_1 : frame of the spaceship
Imagine the spaceship has a tail, the length of the tail is d in the frame of the spaceship.
At time t = d / (gamma * v), the end of the tail reaches the Earth in the frame of the Earth.
Meantime, the point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1
at time t_1 = d / (gamma * v)
t = t_1 = d / (gamma * v), for any value of d.
At the end of the tail of the spaceship is John, at time t = d / (gamma * v) the twin on Earth sees John and, in the frame of the spaceship, at time t_1 = d / (gamma * v), John sees the Earth.
At time t = d / (gamma * v), the spaceship did not reach point Q (a star, for example)
If we know that the spaceship moves with uniform rectilinear motion in the frame of the Earth at speed v
(x = v * t), then:
a) the point Q (d; 0) reaches the spaceship at time t_1 = d / (gamma * v), and
b) the Earth also reaches the point P (-d; 0) at time t_1 = d / (gamma * v).
The whole frame F_1 moves with uniform rectilinear motion in the frame F, "there is another spaceship in the point P(-d; 0)". Do not consider a single spaceship only in the origin O_1 (0; 0)
If we analyze the motion of the Earth in the frame of the spaceship, the mathematical equation
x_1 = - v * t_1 is wrong.
We must consider x_1 = - gamma * v * t_1.
Thus, at time t_1 = d / (gamma * v), x_1 = -d.
And that's right, because the Earth reaches point P at time t_1 = d / (gamma * v).
If we know that a frame moves with uniform rectilinear motion in a second frame with speed v, the second frame cannot move with speed -v with respect to the first frame.
The spaceship moves in the frame of the Earth, we know that x = v * t!
If instead x_1 = - v * t_1, then:
c) the point P(-d; 0) reaches the origin O of the frame F at time t = d / (gamma * v), and
d) the origin O_1 of the frame F_1 also reaches the point Q (d; 0) at time t = d / (gamma * v).
@@massimilianodellaguzzo8571 That is completely unclear. There is a need to specify what the quantities refer to in the calculations.
I gather we are considering a spacetime, S=[M,g], that is a smooth semi-Riemannian manifold equipped with metric g, where g is the Minkowski metric, or g_{\mu
u}=\eta_{\mu
u}. So far so good, but what metric signature are you using?
What makes a reading of your scenario impossible is that you're not making the distinction between physical and coordinate values. Let's take the first thing you wrote as an example:
"The point P (-d; 0) of the frame F_1 reaches the origin O of the frame F at time t = d / (gamma *v)"
@@kylelochlann5053 I try to repeat myself, adding more information.
Consider two frames F and F_1 in relative motion to each other with speed v, the frames F and F_1 are coincident at the initial time. (t = t_1 = 0 s) Frame F_1 moves to the right (with respect to frame F) and frame F moves to the left (with respect to frame F_1)
The point P (- d; 0) of the frame F_1 reaches the origin O of the frame F at time t = d / (gamma * v)
The point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1
at time t_1 = d / (gamma * v)
t = t_1 = d / (gamma * v), for any value of d.
(THIS IS THE SYMMETRY OF THE LORENTZ TRANSFORMATIONS !)
This happens both if the frame F_1 moves at speed v with uniform rectilinear motion in the frame F, and if the frame F moves at speed - v with uniform rectilinear motion in the frame F_1.
The contracted distances of the frame F_1 move at speed v (with respect to frame F), and the contracted distances of the frame F move at speed - v. (with repect to frame F_1)
F: frame of the Earth
F_1 : frame of the spaceship
Imagine the spaceship has a tail, the length of the tail is d in the frame of the spaceship.
At time t = d / (gamma * v), the end of the tail reaches the Earth in the frame of the Earth.
Meantime, the point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1
at time t_1 = d / (gamma * v) in the frame of the spaceship.
t = t_1 = d / (gamma * v), for any value of d.
(THE SYMMETRY OF THE LORENTZ TRANSFORMATIONS, THE TRUE SYMMETRY)
At the end of the tail of the spaceship is John, at time t = d / (gamma * v) the twin on Earth "sees" John and, in the frame of the spaceship, at time t_1 = d / (gamma * v), John "sees" the Earth and point Q reaches the spaceship.
At time t = d / (gamma * v), in the frame of the Earth the spaceship did not reach point Q
(a star, for example)
Furthermore at time t = d / (v * gamma) in the Earth's frame, the time is t_1 = d / (v * gamma * gamma) in the frame of the spaceship
(If x = v * t)
Imagine the point R (d / gamma; 0) of the frame of the Earth and consider the two events A and B
A: Earth reaches point P.
B: point R reaches the spaceship.
The two events A and B are simultaneous in the frame of the Earth, the point P reaches the Earth
at time t = d / (gamma * v) and the spaceship reaches the point R at time t = d / (gamma * v)
In the frame of the Earth, the length of the tail is contracted.
If the spaceship travels a distance equal to d / gamma, the twin on Earth "sees" John.
The two events A and B are not simultaneous in the frame of the spaceship, point R reaches the spaceship
at time t_1 = d / (gamma * gamma * v) and the Earth reaches point P
at time t_1 = d / (gamma * v)
Now consider event C
C: point Q reaches the spaceship
Event C represents: the spaceship reaches point Q (in the frame of the Earth)
Event C represents: point Q reaches the spaceship (in the frame of the spaceship)
In the frame of the Earth the spaceship reaches the point Q at time t = d / v and, in the frame of the spaceship, the point Q reaches the spaceship at time t_1 = d / (gamma * v), exactly when the Earth reaches the point P.
If we know that the spaceship moves with uniform rectilinear motion in the frame of the Earth at speed v
(x = v * t), then:
a) the point Q (d; 0) reaches the spaceship at time t_1 = d / (gamma * v), and
b) the Earth also reaches the point P (- d; 0) at time t_1 = d / (gamma * v).
The whole frame F_1 moves with uniform rectilinear motion in the frame F, "there is another spaceship
in the point P(-d; 0)". Do not consider a single spaceship only in the origin O_1 (0; 0). Everyone considers a single spaceship in the origin O_1 (0; 0), the number of spaceships is infinite!
If we analyze the motion of the Earth in the frame of the spaceship, the mathematical equation
x_1 = - v * t_1 IS WRONG. We must consider x_1 = - gamma * v * t_1. Thus, at time
t_1 = d / (gamma * v), x_1 = - d.
And that's right, because the Earth reaches point P at time t_1 = d / (gamma * v).
If instead x_1 = - v * t_1, then:
c) the point P(-d; 0) reaches the origin O of the frame F at time t = d / (gamma * v), and
d) the origin O_1 of the frame F_1 also reaches the point Q (d; 0) at time t = d / (gamma * v).
Let me know if anything is unclear. (I don't know the English language perfectly)
@@massimilianodellaguzzo8571 Awesome, thank you. I have some family coming in for the weekend and will work through what you have there when I can, but you can expect a reply in a day or two.
I do see in the wording that you have a ship of some length and I'm not seeing calculations for the clock desynchronization (the leading clock runs deeper in time) so there is already some red flags, but it might be just my superficial glance of what I see. It might be nothing, but I'll let you know.