Special Relativity- Length 'expansion' possible? [closed]

In Special Relativity, I know that a Lorentz length contraction and Lorenz time dilation are possible.

If only length contraction is possible, then, why is the distance greater than 100m (the trains' rest length)?

• Why do you think it could be possible? (The short answer is no, it isn't.) Often this misconception occurs because people don't realise exactly what a length measurement is in different frames, my answer to one such question might be helpful. Aug 22, 2020 at 12:57
• @Philip I have updated my answer with all context I have, note the pink box. If only length contraction is possible, then, why is the distance greater than 100m (the trains' rest length)? Aug 22, 2020 at 14:10
• This is correct, except that no one will consider this length to truly be the length of the train. Consider this from the point of view of someone on the platform: the two events (Rick and Lucy dropping the cats) will not be considered as simultaneous. They will see one of them dropping the cat first, and then the other. As a result, the train will appear to move in between these "drops" and so the measured distance from the platform frame will be larger. However the people on the platform will certainly not count the distance between these dropped cats as being the train's length. Aug 22, 2020 at 14:19
• Remember, in order for the difference in the two endpoints' coordinates to be the "length" of a moving object, these endpoints have to be measured simultaneously (this makes sense if you think about it). However, the events are simultaneous in the train's reference frame, and we know that this means they are not simultaneous in the platform's reference frame. This is in all ways identical to the question I linked, and I believe my answer there explains it. Aug 22, 2020 at 14:21
• @Philip "This is correct" What do you mean? The thought experiment is correct and the distance really is length 'expanded'? Please clarify. This thought experiment was given to us before non-simultaneity was taught to us. Aug 22, 2020 at 14:22

"Length contraction" refers to the result of measuring a length by a certain standard procedure. In the second half of this problem, they aren't following that procedure, which is why they can get a different answer.

In the first half, the clocks used to coordinate the cat-dropping are at rest relative to the tracks, and the distance beween the cats is measured using the tracks (or metersticks laid at rest on the tracks).

In the second half, the clocks are at rest relative to L and R, but the the distance between the cats is again measured using the tracks. The clocks and tracks are in relative motion so their distance and time coordinates don't form an inertial reference frame. If they'd dropped the cats on the train and measured the distance with respect to the train then they would have gotten 100m.

The notes you have attached to your question are, in my opinion, very badly formulated. I see the point that they are trying to make, but for beginners they will serve no purpose other than to confuse them. I will try to make it clearer, and to do so:

1. I'm going to use all "primed" quantities to represent quantities measured in the $$S^\prime$$ frame, and all unprimed quantities to represent the same quantities in the $$S$$ frame.

2. We will use the Lorentz Transformations are the basis for our analysis:

\begin{aligned} &\text{(A)}\quad\Delta x^\prime = \gamma \left(\Delta x - v \Delta t\right)\\ &\text{(B)}\quad \Delta t^\prime = \gamma \left( \Delta t - \frac{v}{c^2}\Delta x\right)\\ \\ &\text{(C)}\quad\Delta x = \gamma \left(\Delta x^\prime + v \Delta t^\prime \right)\\ &\text{(D)}\quad \Delta t = \gamma \left( \Delta t^\prime + \frac{v}{c^2}\Delta x^\prime \right)\\ \end{aligned} \label{LT}

The reason for the confusion is because the notion of length is inextricably linked to the notion of simultaneous events, and in special relativity (unlike in Galilean relativity) people will not agree on whether two spatially separated events are simultaneous or not. I have explained the intricacies of length measurements in this answer here:

[Consider an object that is at rest in $$S'$$.] For an observer sitting in $$S^\prime$$, since the object is at rest with respect to him, its length $$L^\prime$$ is simply the difference in the coordinates, irrespective of when $$x_B^\prime$$ and $$x_A^\prime$$ are measured. He could measure $$x_B^\prime$$, have a coffee, and then measure $$x_A^\prime$$ and the difference would give him the length. However, for an observer sitting in $$S$$, since the object is moving with respect to her, both the endpoints $$x_B$$ and $$x_A$$ need to be measured simultaneously in her frame of reference ($$S$$) in order for the difference to be the length $$L$$. (In other words, if she has a coffee between measuring $$x_B$$ and $$x_A$$, the object would have moved between measurements!) So, we have

\begin{aligned} L^\prime &= x_B^\prime - x_A^\prime |_\text{ for any \Delta t^\prime}\\ L\,\, &= x_B - x_A |_\text{ only when \Delta t=0}\end{aligned}

Given this, let's examine your two scenarios. In both scenarios, Lucy and Rick are on the the train, and their friends (let's call them Alice and Bob) are on the platform.

Scenario 1:

Alice is assigned Lucy and Bob is assigned Rick. They simultaneously mark the positions of the person they are assigned to as the train passes by them. According to Alice and Bob, the distance between these marks in the length of the train, since the train is moving with respect to them and they have measured the coordinates of its endpoints simultaneously. What would this length be? Well, what do we know?

1. Alice and Bob marked the points simultaneously (i.e. $$\Delta t = 0$$).
2. They marked the positions of Lucy and Rick who (in $$S'$$) are separated by 100m (i.e. $$\Delta x' = 100$$m).

We need to find the distance between the points marked by Alice and Bob (which is $$\Delta x$$) given that $$\Delta t = 0$$ and $$\Delta x' = 100$$m. This can easily be done using Equation $$(A)$$ given above, and we see that

$$\Delta x' = \gamma \Delta x \quad \quad \implies \quad \quad \Delta x = \frac{100\text{m}}{\gamma} < 100\text{m}$$

Good? Let's move on the problematic scenario then.

Scenario 2:

In this scenario, Alice and Bob are on the platform, but Lucy and Rick are the ones marking the positions. What do we know?

1. Lucy and Rick mark the positions simultaneously in their frame of reference (and so $$\Delta t' = 0$$).
2. They continue to be at either end of the train (and so $$\Delta x'=100$$).

We need to find $$\Delta x$$ (the coordinates between the marks that Lucy and Rick made in the platform's frame of reference), and some inspection will tell you that Equation $$(C)$$ is the useful one, and so

$$\Delta x = \gamma \Delta x' = \gamma \times 100\text{m} > 100\text{m}$$

All of this so far is correct. So what's "wrong"? The mistake comes in the interpretation that length has "expanded" in the second scenario. Let's examine this closely:

The interpretation:

Lucy and Rick believe that they have marked 100m on the platform, since they marked the two endpoints simultaneously. When they come back later, they find that (according to Alice and Bob standing on the platform) the distance they marked on the platform is $$\gamma 100$$m.

Rick immediately says that the length of the train must have been larger than 100m according to someone on the platform. However, Alice, having studied a little more relativity says:

"No. We measured a distance on the platform (which was moving with respect to us) that we thought was $$100$$m. According to someone on the platform, this distance was actually larger! Therefore, what we actually saw was that the (platform) distance of $$\gamma 100$$m was contracted to $$100$$m. In other words, we saw lengths on the platform contract, as should be expected, since the platform was moving relative to us."

They argue a little, and then decide to ask Alice and Bob what they saw. Alice and Bob agree with Lucy. According to them, the actions of Rick and Lucy "marking" points on the platform was not simultaneous. They first saw Rick mark a point, and then a short amount of time later saw Lucy mark a point. In the meantime, the train had moved some distance. But this is a natural consequence of relativity: observers in different inertial frames do not agree upon simultaneous events.

"So what did you measure as the length of the train?", Rick asks Alice and Bob. "Well," says Alice, "since the train was moving with respect to us, we measured its length by measuring its endpoints simultaneously (just like in Scenario 1)."

So in conclusion, distances on the train seemed to contract according to someone on the platform, and distances on the platform seemed to contract according to someone on the train. However, due to the way that simultaneous events transform between reference frames, no contradictions arise.

It is impossible to answer your question because you haven't specified how the train starts moving.

1. In a frame where the front of the train starts moving forward before the back starts moving, the train expands. In a frame where the back starts moving before the front, the train contracts. In a frame where the front and back start moving at the same time, the length of the train does not change.

2. Once the train has expanded or contracted, internal forces might or might not cause it to revert to something closer to its original length (and/or to break apart) but this would depend very much on the details of how the train is constructed and I think it is appropriate to ignore it for a question at this level.

3. Following the expansion/contraction, the train will always be longer in its own ("moving") frame than in the station frame. This might be because the train expanded in its own frame but did not change length in the station frame. Or it might be because the train contracted in the station frame but did not change length in its own frame. Or many other things. If you tell me how the train started moving, I can choose among these. If not, your question is ill-posed.

4. Point 3) above has to do with relativity but your actual question is not a relativity question, because (once you specify how the train starts moving) you can do all your analysis in the station frame, and never have to worry about figuring out what happens in the moving train's frame. In other words, point 3) is irrelevant to your question, which is really very simple, and by tagging it with "relativity" you have implicitly way overcomplicated it.

• " the train will always be longer in its own (moving) frame than in the station frame. " You mean 'shorter' right? Aug 22, 2020 at 14:26
• No, I mean longer. Aug 22, 2020 at 14:28
• Longer is correct. An object's length in the frame where it is at rest will always be the largest length one can measure. That being said, I have grave issues with this answer, as I don't think it matters at all how the train started or stopped, this could be a problem even for a train moving at a constant speed eternally. This problem certainly seems to me to be one of special relativity. Aug 22, 2020 at 14:31
• @WillO I'm sorry but I can't specify how the train started moving since it was not specified to me, those 3 images I posted is all the information I have been given. Aug 22, 2020 at 14:31
• @Philip: "I dont think it matters at all how the train started or stopped" --- but of course it does. Surely in any frame where the front and back follow the same acceleration path (as a function of time) the length can't change. Surely this can't be true in more than one frame. Surely we haven't been told in which frame (if any) it is true. Which of those statements do you doubt? Aug 22, 2020 at 14:33