# Doubt in using the train example used to show simultaneity

The classic example used to demonstrate the breakdown of simultaneity is a train car in which a source $$C$$ is at the centre of the train, with a receptor $$A$$ at the back of the car, and receptor $$B$$ at the front, located equidistant from $$C$$ in the train frame. The source emits a couple of photons towards the receptors.

From the point of view of a platform observer, the following spacetime diagram is presented:

And it's apparent from the diagram that there's some time separation between the events of $$A$$ and $$B$$ receiving the signal. But I'm not sure how this definitively proves that the events aren't simultaneous for the platform observer. I could've come up with this diagram, for example:

and I could've claimed that lengths in the train are affected in such a way from the POV of a platform observer that time separation is still preserved between frames (obviously I know this is wrong, but I could hypothetically make this ridiculous assumption nonetheless).

I could've said that the infinitesimal lengths contract more and more (kind of like a gradient of stretchability) as we move further in the direction of the train reference frame's motion (all this from the POV of the platform reference frame).

So my question is: what allows us to definitively say that the top picture is correct and the bottom picture is wrong? Of course, experimental evidence is one thing, but I'm guessing it's possible to prove it theoretically as well?

what allows us to definitively say that the top picture is correct and the bottom picture is wrong?

Recall that the Lorentz transform is specifically a transformation between two inertial frames. The term “inertial” refers to Newton’s first law which describes inertia, or the tendency for a body to move in a straight line at constant speed unless forced to do otherwise.

In spacetime, inertial objects have worldlines that are straight lines. So a transform between inertial frames must map straight lines in one frame to straight lines in the other frame. Such transformations are called “affine” and are quite well studied. In particular, if an event is the midpoint of a segment in one frame it is the midpoint of the segment under any affine transform.

The transform you described would transform a straight line to a bent one as the line goes through that gradient of stretchability you mentioned. Such a transformation can not represent a transformation between inertial frames because an inertial object would be accelerating in that frame.

• Very clear answer! A couple of doubts: the first one is probably silly - why do we demand that Newton's 1st law be followed in all inertial frames? What would go wrong if it weren't followed in all inertial frames? Second doubt: how do we prove that the transform I described would transform a straight line to a bent one? My current thought is that if we were to do markings on the train at regular intervals, then in the platform's frame, the spacial distances between those markings would get incrementally shorter towards the front in the platform frame...(cont'd) – user9343456 May 9 '20 at 9:22
• (cont'd) So I could argue that the particle has to cover incrementally shorter distances (as per the platform frame) as it moves towards the front of the train. But then I could argue that even the time intervals start getting longer ("gradient of time dilation" loosely speaking) the more the particle moves towards the front. Incrementally shorter distances covered in incrementally longer times could allow for constant velocity even in the platform frame. So Newton's 1st Law still seems to be followed - that's where I'm stuck. – user9343456 May 9 '20 at 9:26
• @u23 for your first comment, that is just the definition. An inertial frame is defined as one where Newton’s first law is valid. It is useful to know if N1 is valid in a frame so we give it a name. For your second comment, you could indeed do what you describe for some lines, but not for all lines. If the transform is not affine then there will be some straight lines that get bent by the transform, even if there are some that happen to remain straight. It would help for you to actually sit down and explicitly write down such a transform as what you have in mind. Its harder than you might think – Dale May 9 '20 at 11:08
• So even if a potentially nonlinear transformation may transform some straight worldlines into straight worldlines, it still cannot do so with all straight worldlines. It's clear to me now, thanks a lot! – user9343456 May 9 '20 at 13:42
• @u23 yes that is correct. However, there is a little caveat. A free falling frame is considered inertial in general relativity (gravity is not a force) but a free falling frame is considered non-inertial in typical Newtonian physics (gravity is considered a force). I prefer the general relativity classification, but Newtonian physics is valid for most “everyday” uses. – Dale May 12 '20 at 17:55

For the platform frame, any position in the train is:

$$x' = \gamma(x - vt)$$

If $$C$$ is middle point of $$A$$ and $$B$$ in the train's frame:

$$x'_A + x'_B = 2x'_C$$

So, for the same time $$t$$ measured in platform's frame: $$\gamma(x_A - vt) + \gamma(x_B - vt) = 2\gamma(x_C - vt)$$
$$(x_A + x_B = 2x_C)$$

What is middle point for the train is middle point for the platform. The second drawing shows for $$t = 0$$ a $$C$$ that is not at the middle.

• So if I understood it correctly, we need the additional bit of info that $x'=\gamma(x-vt)$ - without that, we wouldn't have been able to conclude which of the drawings was correct. Thanks a lot! – user9343456 May 8 '20 at 22:50
• Note that the $\gamma$ plays no role in that result. That is, the result is not particular to special relativity. Under affine transformations, parallelism of pairs of lines and midpoints of segments are preserved. – robphy May 9 '20 at 0:51