# In c-v what is v, the train's speed relative to?

1. Aug 21, 2015

### DAC

Hello PF.
Re.Einstein's thought experiment in which he asks, what is the speed of light as seen from a moving train. He concludes it can't be c-v as the speed of light is always c. Hence time and length must vary. My question is, If the light's speed is relative to the train, why isn't the train's speed relative to the train? In which case v relative to itself is zero and c-v is c.

2. Aug 21, 2015

### Staff: Mentor

From the point of view of the train: the speed of the train is zero and the speed of light is $c$.

From the point of view of the platform: the speed of the train is $v$ and the speed of light is $c$.

$c-v$ is the difference between the speed of the flash of light and the speed of the train, from the point of view of the platform. From that point of view, the distance between the front of the train and the flash of light will be always be $T_p(c-v)$ where $T_p$ is the amount of time that has passed since the flash was emitted.

$c-0=c$ is the difference between the speed of the flash of light and the speed of the train, from the point of view of the train. From that point of view, the distance between the front of the train and the flash of light will always be $T_t(c-0)=T_tc$ where $T_t$ is the amount of time that has passed since the flash was emitted.

3. Aug 21, 2015

### loislane

I've seen many people with questions about this point, enough to think there must be a common source of bafflement. Maybe what nonpluses students is that the usual explanation lends itself to mixing-up the Lorentz transformation in the train direction x, between the frame of the platform and the frame of the train, that uses the relative velocity $v$ between the train and the platform, with the calculation of the distance between the point the flash of light is sent out and the front of the train performed on each frame separately, this last calculation is done by postulate of the invariance of the speed of light, and it can not be derived from the Lorentz transformations (while the transformations can of course be derived from the postulates).

So if we are restricted to one frame K, say the platform's frame the OP asked about, the distance is calculated between 2 events being one labeled at $t_1$ in point $x_1$ when the flash of light is sent out, and the other at $t_2$ when it reaches the front of the train($x_2$), being $t_2-t_1$ what is called above $T_p$. Now the distance calculation follows the simple Newtonian equation $x=VT_p$ and $V$ is calculated following Poincare translations that allow the substraction of velocity vectors including c as $c-v$ with $v$ the speed of the train relative to the platform.

The postulates of SR require $V$ to be an absolute 4-velocity while the equation used to calculate the distance remains to be the newtonian one for 3-velocities with absolute time parameter , wich might allow for some ambiguity but the invariance of $c$ so requires it(for the 4-velocity formula with proper time $\tau$ is not defined for objects travelling at light speed) .

Last edited: Aug 21, 2015
4. Aug 22, 2015

### loislane

Could you clarify exactly if you are calculating distance travelled by the light flash as $x=ct$, implying $\frac{dx}{dt}=c$ in the train's frame and as $x'=t'(c-v)$ -the distance $T_p(c-v)$ in your post-, implying $\frac{dx'}{dt'}=c-v$ in the platform's frame? Because that would seem to contradict the second sentence in your post since the latter formula is galilean relativistic. The correct formula is $x'=ct'$ with $t'=(t-\frac{vt}{c})\gamma$.

Last edited: Aug 22, 2015
5. Aug 23, 2015

### Staff: Mentor

Yes, although it would be more precise to say that I'm calculating that distance to be $\Delta{x}=c\Delta{t}$. $x=ct$ looks more like an expression for position as a function of time.
Yes, if you intend $x$ to be position of the light flash as a function of $t$.
No, I am not saying/implying that. The quantity $T_p(c-v)$ is a distance measured by the platform observer. Connecting it to any observation made by the train observer takes more work; the thought experiment OP is asking about is setting the stage for doing that work.
Indeed it is. But it's your formula, not mine.

Last edited: Aug 23, 2015
6. Aug 23, 2015

### loislane

Ok, however...
Regardless of whether you are trying to make that connection or not, it is hard not to conclude that the distance $T_p(c-v)$ covered in time $T_p$ all in the platform's frame doesn't correspond with a galilei's speed $c-v$.

7. Aug 24, 2015

### DAC

A supplementary question. Assume the speed of light is 100, and the speed of the train is 70.
The speed of light as seen from the train would be 30. ( c-v ), which is unacceptable.
But the question could also be answered, the speed of light as seen from the train is 30 faster than the train, i.e. 100.

8. Aug 25, 2015

### Staff: Mentor

Yes, it is unacceptable. The speed of light is the same for all observers, so the speed as seen from the train must be the same as the speed of light as seen from the platform.
100 is the right answer, but that's the wrong way of coming up with it. From the point of view of the train, the train is at rest while the platform is moving backwards at speed 70. Why should the train guy care about the platform when he's measuring the speed of light? He'd get the same result (100) even if he never looked at platform and didn't know it was there. The right way of thinking about it is to say that the speed of light as seen by the train is 100 and the speed of the platform relative to the train (70) has nothing to do with it.

(As an aside, once we accept that the speed of light is the same for all observers, it is easier to describe this situation by saying that the speed of the train relative to the platform is .7c instead of 70 with the speed of light being 100).

Last edited: Aug 25, 2015
9. Aug 25, 2015

### harrylin

A velocity difference is not "galilean"; instead it is fundamental in mathematical physics. Coincidentally it was used for deriving the Lorentz transformations (surely you will agree that those are not "galilean"!). A common mix-up which may be at play here, is to confound a system transformation with a velocity difference (also called "relative velocity" or "closing speed"). A velocity difference is invariant in classical physics but not in relativistic physics.
For example, in classical physics one may confound the following question without consequence:
- "a car goes at 100 km/h to the west and an airplane at 1000 km/h to the west. What is the velocity difference between the airplane and the car in the ground coordinate system?",
with the question:
- "what is the velocity of the airplane in the coordinate system that is co-moving with the car?".
In SR the answers two those different questions are only approximately the same (although for these speeds the approximation is extremely good).

10. Aug 25, 2015

### loislane

In the sentence you quoted there is no mention of any relative velocity(unless you consider light speed c, wich is alluded to, a relative velocity wich would be odd in this forum). There is just a distance and a time in a certain frame and a very simple algebraic conclusion.

11. Aug 26, 2015

### harrylin

I think that Nugatory explained it already very well, so I won't elaborate much further, but it does sound as if the confusion that I mentioned is at play here. Velocities in physics are always defined relative to something else and with respect to certain reference standards of length and time. There is therefore nothing "galilean" about Einstein's derivation of the Lorentz transformations, in which he also used the velocity difference c-v, as follows (when referring to a "moving" system k):
"the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v".
- http://fourmilab.ch/etexts/einstein/specrel/www/

If people have difficulty with such mathematical physics descriptions, then this may be due to a conceptual misunderstanding.

PS: in fact the OP was correct, only more clarification was needed.

Last edited: Aug 26, 2015
12. Aug 26, 2015

### loislane

Right, that is why I devoted post #3 to clarify the confusion.

What I'm saying , that is not caught by your replies AFAICS, is that if we agree that light travels at c from both the stationary frame and the moving frame(do you agree so far? If not just say so and I can explain it further) a clarification might be needed for the OP to understand(and his question in #7 confirms he is still confused about it) why "the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v" if in both frames the ray moves at c.

The fact he asks about it might indeed reflect a conceptual misunderstanding between reference system transformations and relative velocities in SR, in wich $\vec{v}_\mathrm{B|A}=-\vec{v}_\mathrm{A|B}$ doesn't hold in general(but note it does hold for the magnitude-speed- of the relative velocity), unlike in classical mechanics where it holds, but that confusion is not addressed by Nugatory AFAICS and that is what I was pointing out in #4 and #6.

The clarification is offered by Einstein himself in his foundational paper in a previous section to the one quoted above, when he defines simultaneity, where he is forced to use a 2-way speed for light, and subsequently the discussion is restricted to motion in just one axis wich leaves an inherent ambiguity about one-way speed(source of many discussions to this day),this is what originates the ambiguity I referred to in #3.

13. Aug 26, 2015

### vanhees71

14. Aug 26, 2015

### loislane

The issue of clock synchronization and definition of simultaneity is not even mentioned in that FAQ.

15. Aug 27, 2015

### vanhees71

Good point. It's perhaps a bit implicit in simply stating the two postulates a la Einstein in 1905 and then analyse it in terms of Minkowski space (and also Minkowski diagrams, which some people find helpful in understanding particularly the issue of relativity of simultaneity). What would you think, should I add, to make the issue of clock synchronization more explicit?

16. Aug 27, 2015

### loislane

As I commented in #3 I've run into enough students confused about Einstein's train thought experiment(in sections 7-11 of Einstein's popularization book "Relativity:the special and the general" ) to conclude there is some common misunderstanding, that I think lies in the definition of simultaneity.

The objection usually goes ike this The lightning strikes both the engine and caboose of the train in any frame(the damages to the train can be ascertained by all after the fact). The passenger is right in the middle of the points hit by lightning and she sees the flash from the front before than the rear while the person in the embankment observes them simultaneously.

This is of course explained by the relativity of simultaneity. But some have trouble to reconcile this with the postulate of light speed constancy in all frames. They reason that if there is the same distance from the points in the train hit by lightning to the passenger but light takes different times for one distance than for the other the speed of light cannot be constant in both directions.

The straight-forward answer is that Einstein was careful(or sly) enough to define his synchronization in purely conventional terms so that it is not necessary that the 2-way speed of light and the one-way coincide, the latter speed cannot be experimentally measured independently of a convention as to how to synchronize the clocks at the source and the detector.

There is an ambiguity associated to this especially when going to GR( see i.e. : "The problem of clock synchronization - A relativistic approach" by Klioner in Celestial Mechanics and Dynamical Astronomy , vol. 53, no. 1, 1992, p. 81-109 ) but perhaps that is better discussed in a different thread.

17. Aug 27, 2015

### vanhees71

Ok, let's take the train thought experiment as in Wikipedia. Alice (inertial frame $(ct,x)$) is standing in the middle of the compartment in the train, which moves in positive $x'$ direction from the point of view of Bob (reference frame $(ct',x')$) on the platform. At $t=t'=0$ Alice sends out a spherical light signal. If $L$ is the length of the compartment as measured by Alice, the light front hits the front and the rear end simultaneously, i.e., these events from the point of view of Alice are
$$(x_{\pm}^{\mu})=(L/2,\pm L/2).$$
For Bob the two events are given by the Lorentz boost
$$\begin{pmatrix} c \tilde{t} \\ \tilde{x} \end{pmatrix} = \gamma \begin{pmatrix} 1 & \beta \\ \beta & 1 \end{pmatrix} \begin{pmatrix} c t \\ x \end{pmatrix}.$$
Plugging in the above space-time vectors for the events of the light front reaching the front and rear end of the compartment, one gets
$$(\tilde{x}_{\pm}^{\mu})=\gamma[(1 \pm \beta) L,\pm (1 \pm \beta)L/2].$$
So the from the point of view of Bob, the light front reaches to front of the compartment later than the rear of the compartment, but the speed of light is of course still the same, i.e., the events are also light-like relative to the origin. This is, of course, so by construction of the Lorentz transform from the principle of relativity and the invariance of the vacuum-speed of light for all inertial observers. Keeping this in mind it's qualitatively quite clear that the light signal needs less time to reach the rear end than the front end, because of the motion of the train relative to Bob: the front moves away and the rear towards the light front and thus it needs more (less) time to reach the respective ends of the compartment since the speed of light is still the same for Bob.

In the attachment you find the Minkowski diagram depicting the situation, where this becomes also quite clear:

Last edited: Aug 28, 2015
18. Aug 27, 2015

### loislane

Yes, of course it is quite clear.
The point I was raising is clearer in the original Einstein's train thought experiment(wich I believe it is the one the OP was concerned with) that is slightly different than the wikipedia one. For beginners in relativity to fully understand it when they object as above, the conventionality of Einstein synchronization is a useful pedagogical tool.

By the way I see in your FAQ how to some extent the simultaneity issue is implicit when you write: "Tacitly, he also made another assumption: Any inertial observer will find that in the physical space the laws of Euclidean geometry is valid, which is true in Newtonian mechanics too. In a way this is also implicit in the assumption of the
invariance of Maxwell’s equations, which is formulated in terms of vector calculus in three-dimensional Euclidean space."

It is kind of ironic that with that tacit assumption Einstein introduced absolute Euclidean space in his theory(at least for EM waves), wich he was trying to get rid of at all costs.

19. Aug 28, 2015

### vanhees71

The original example is not much different. There two light signals are sent from the ends of the compartment towards the observer (Alice) in the middle of the train simultaneously from the point of view of the observer on the platform (Bob). For A the two emission events are not simultaneous and that's why for her the wave fronts are not arriving simultaneously (see attached Minkowski diagram).

There is no absolute Euclidean space, because what each inertial observer considers his "space" is not always the same. For an accelerated observer his space does not appear Euclidean. Of course, then you have to define distance measurements with the two-way ligth-signal travel time to define distances in "space" according to this observer. In general this is possible only on a part of spacetime around the accelerated observer. Here, we are already pretty close to the full general relativistic treatment of "time slicing". The only difference is that in GR spacetime itself is not flat, while it is in SR's Minkowski space.

20. Aug 28, 2015

### loislane

Here's an excerpt of the original text:"When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment. But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A —> B on the travelling train. Just when the flashes of lightning occur as judged from the embankment , this point M' naturally coincides with the point M..."

This means that regardless of whether the lightnings are observed as simultaneous or not, they strike the same points in the embankment and the train("events A and B also correspond to positions A and B on the train"), and this can actually be confirmed a posteriori by looking at the marks of lightning at the rails and train. Do you agree that the passenger in the train at M' is at the midpoint from the points where the lightnings caused the damages on the train(emission events)?

An accelerated observer requires the use of non-inertial coordinates. You were writing about inertial observers in the quote:"Any inertial observer will find that in the physical space the laws of Euclidean geometry is valid".

21. Aug 28, 2015

### vanhees71

Concerning the Einstein train example you can read off everything from the Minkowski diagram attached to #19. Alice is in the restframe of the train (black Minkowski coordinates), the world lines of the endpoints of the compartment are the blue vertical lines. Bob's Minkowski coordinates, who is at rest on the embarkment is depicted by the red Minkowski axes. He is moving with $\beta<0$ wrt. Alice, i.e., the train is moving in positive $x'$-direction seen from Bob's point of view.

Now the lightnings hit the ends of the compartment exactly simultaneously at Bob's time $t'=0$, where Alice and Bob are right at the same spot at $x'=0$. Alice is at rest in the middle of the compartment. The light signals from the lightnings are depicted by the orange light-like world lines, and of course they reach Bob simultaneously, because his distance from the two sources is the same and the light signals have started simultaneously for Bob from their sources.

For Alice the lightnings have hit the ends of the compartment not simultaneously but as the horizontal lines of constant $t$ (Alice's time) show the lightning hit the front end earlier than the one at the rear end, and consequently the light signals arrive at Alice's place also not simultaneously. That's only, because Alice moves with constant relative to Bob, and both are in inertial frames and thus for both space (i.e., in this 1+1-dimensional example the black and red spacial axis) is Euclidean, but it's clear that necessarily the spatial axis of Alice and Bob cannot be the same, because the speed of light must be the same for both Alice and Bob. So it also depends on the observer, which spatial submanifold of Minkowski space is taken as "space", i.e., the hypersurfaces of simultaneous events.

22. Aug 28, 2015

### loislane

We obviously agree about the above. What I'm trying to get across is what I think is the best way to explain it to relativity newbies.
As you say at the end of your post, there is something not completely captured here by the Minkowski diagram due to its being restricted to one dimensional space. You are forced to invoke a non-euclidean spatial hypersurface. That's one way to explain it but I tend to believe that it might be too much when first encountering SR, thus I find more useful to take advantage of the conventionality of synchronization that allows to choose a new synchronization instead of Einstein's, in wich the 2-way speed of light is different from the one-way. This also opens the way to a quicker understanding of non-inertial frames in GR in wich the speed of light may be different than c.

23. Aug 29, 2015

### loislane

In fact my point is reflected in the diagram, the obliqueness of the axes in Bob's frame(a visual license, the axis should be also orthogonal but it is a graphical way to make up for the effect of noneuclidean space of velocities) allows to represent the relativity of simultaneity for objects with timelike paths.

On the other hand, restricting to just one frame, for a ray of light propagating between events P1 and P2 according to the equation $r=cΔt$(usually used to derive the null interval) the only synchronization valid for that equation to hold is Einstein's synchronization, according to the postulate of lighrt speed invariance.

24. Aug 29, 2015

### vanhees71

Sure, I don't know, what point you want to make. It's the point of the entire space-time structure in special relativity to be a Minkowski space. Also note that the coordinates I use are Minkowski-orthonormal coordinates. You must not think about the diagram in terms of Euclidean geometry, which is mentally difficult, because we are used to use the plane of a piece of paper to depict a Euclidean plane, but Minkowski diagrams are, who'd have guessed it, depicting Minkowski geometry (for more details, see my SRT writeup).

From the point of group theory the difference between the Euclidean plane and the Minkowski plane is that the symmetry groups are $\mathrm{ISO}(2)$, which is a semidirect product of translations and rotations and $\mathrm{ISO}(1,1)$, which is a semidirect product of the proper orthochronous Lorentz transformations, the group $\mathrm{SO}(1,1)^{\uparrow}$, and translations.

The analogous groups of course exist for the Eudlidean $\mathbb{R}^4$ and the (1+3)-dimensional Minkowski space.

25. Aug 29, 2015

### loislane

Yes, that's why I said the use of oblique axes is a visual license that allows the depiction of the relativity of simultaneity. Orthonormality in the plane diagram would imply absolute simultaneity.

Of course the geometry of a hyperbola in a plane is as Euclidean as that of a circle, but I catch your meaning.