# I Simultaneity: Train and Lightning Thought Experiment

1. Jan 31, 2018

### PeroK

It's also a question of by how much the seesaw is out of sync. In the frame of the train, clocks at either end of the seesaw will be out of sync by $\frac{Lv}{c^2}$, where $L$ is the rest length of the seesaw and $v$ is the speed of the train. The train must be travelling at less than $c$, so an upper limit on this is $\frac{L}{c}$.

Now, for a seesaw of even $100m$, say, this is a very small time difference, less than a micro-second. The woman on the train will still measure the seesaw as being essentially in sync and observe nothing unusual.

Note that the seesaw will be curved in the platform frame as well, due to the forces along its length.

2. Jan 31, 2018

### Ziang

We don't have to use a seesaw moving up and down. Let us use a horizontal seesaw that its two ends move closer and farther from the railway. I mean the horizontal seesaw rotates freely an small angle during the experiment.

3. Jan 31, 2018

### Ziang

You can imagine a horizontal seesaw. And the seesaw rotates an angle freely during the experiment.

4. Jan 31, 2018

### pervect

Staff Emeritus
I would tend to believe the conclusion that the see-saw is curved, as "rigid objects" simply aren't compatible with special relativity. One can define Born-rigid motions in special relativity, but objects satisfying the necessary criterion to be Born-rigid can't change their state of rotation. Your see-saw is changing it's state of rotation, so it can't be Born-rigid.

You may not be familiar with Born rigidity. I don't see how you can learn about it before you learn about the relativity of simultaneity, though. So the process of learning special relativity involves first realizing that simultaneity is relative, then exploring all the logical consequences of this fact (along with the other aspects of SR such as length contraction and time dilation, though the relativity of simultaneity seems to be the hardest thing for people to learn). The lack of rigid objects is one of the logical consequences of special relativity, I'm unsure if it can be formally deduced solely from the relativity of simultaneity however.

I don't think your exposition isn't quite complete, a drawing of the seesaw from the perspective of the ground and from the train using the Lorentz transform would be interesting. I believe your conclusion is probably right, but the argument isn't quite rock solid yet.

5. Jan 31, 2018

### Janus

Staff Emeritus
This doesn't change anything about any of the arguments already made. For the seesaw to swing back and forth on any axis, some force must be applied to it at some point, and that force cannot propagate through the seesaw faster than c.

6. Feb 1, 2018

### Ziang

Let me say the seesaw is rotating at a constant angular speed (like the earth is spinning).
At the time point that the train is passing it, it is parallel with the railway (and is still rotating)

7. Feb 1, 2018

### Ibix

The seesaw will not appear straight. Also its length will vary as it rotates. You may wish to Google for "relativistic wheel" and look at the shapes of the spokes.

8. Feb 1, 2018

### Janus

Staff Emeritus
To build on Ibix's point, Let's consider the following two illustrations:

On the left we have our rotating pole (light blue bar), The tracks (black and brown lines), the train( green line) and our observers on the embankment and on the train (red circles), according to the embankment frame. This is the moment the observers pass each other. The pole is parallel to the tracks, and its ends line up with both the end of the train and particular points on the track (the white lines). The pole is rotating counter-clockwise as shown by the blue arrows.

One thing needs to be noted in this image, the train, since it is moving relative to this frame is length contracted. In other words, the length of the train as measured in this frame is shorter than the length of the train as measured in its own frame.

This becomes apparent when we look at the right image, which is drawn from the frame of the train. We are still dealing with the moment the two observers pass each other. In this frame the train measures its length as its proper length, while it measures the tracks and embankment, which is moving relative to it as being length contracted. As a result, the train no longer fits neatly between the white lines, but extends quite a bit beyond is both directions. This also means that one end ( the right end in this case) has already passed its white line before the other end hasn't reached its white line yet.

Also note that the bar appears as being curved, as per Ibex's post, and the right end of the pole has already gone past the point where it is adjacent to the tracks, while the left end has not yet reached that point. If this were an animation that we could run backwards and forwards, you would see the ends of the train lining up with a white line at the same moment that an end of the rod was adjacent to the same spot. This just happens at different times for each white line. Another thing to note here is that as the rod rotates in this frame, the curvature of the rod does not remain constant but changes from being curved as shown here to being straight when aligned vertically.

9. Feb 1, 2018

### Ziang

That is what I said: "the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?"

10. Feb 1, 2018

### Janus

Staff Emeritus
Just because the See-saw is curved in her frame does not mean that it is under stress in her frame.
In the following image we see the same diamond shape, in both its rest frame and according to a frame in which it is moving at 0.8c.
In the top image all the corner angles are 90 degrees, in the bottom image, one pair of corners is more than 90 degrees and the other pair is less than 90 degrees. But this does not mean that the shape is under some type of stress at the corner. Even if the shape was rotating, and thus constantly changing shape in the bottom frame, it would not be undergoing any stress.

11. Feb 1, 2018

### Mister T

Yes, but this is not about how the seesaw appears, it's about its shape. Observations differ in different frames. The shape of the seesaw is one such observation.

12. Feb 1, 2018

### Mister T

Because in its rest frame it's not bent (enough for it to break).

13. Feb 2, 2018

### Ibix

That's what a freely rotating beam looks like when viewed from a moving frame. It can't be straight - as you've already observed the relativity of simultaneity shows that immediately. Any other maths you do will support this. For example you can transform the centripetal force to show that it's not centripetal in the moving frame, and its direction depends on the radius, which is consistent with a curved beam. Or you can draw a 2+1 dimensional Minkowski diagram and consider the intersection of a sloped simultaneity plane with the helical worldtube of the spinning beam.

The question is, why do you think it ought to be straight? The short answer is that you are trying to use common sense in an extremely uncommon situation. That doesn't work well.

14. Feb 2, 2018

### Ziang

The laws of physics are invariant in all inertial frames.
If the seesaw gets stress when it gets curved in one inertial frames then it would get stress when it gets curved in all inertial frames.

15. Feb 2, 2018

### pervect

Staff Emeritus
The laws of physics are invariant in all inertial frames. However, "being curved", which I interpret as a statement about the purely spatial geometry of the object, is a frame dependent statement, due to the relativity of simultaneity. If parts of an object have a non-zero proper acceleration - such as a rotating stick, the see-saw, or an accelerating elevator - the spatial projection of the object may be flat in some frames of reference and "curved" in other frames of reference.

Note that if an object is not accelrating, if all parts of the object have a zero proper acceleration, then the property of "being flat" does turn out to be independent of the chocie of frame, as the Lorentz transformation is linear, and linear functions map straight lines to straight lines.

However, the worldline of an accelerating point is not a straight line, so this argument does not apply.

16. Feb 2, 2018

### Mister T

Okay. And if the stress is not enough to break it in one frame, then it doesn't break in all frames. It will be bent by different amounts in different frames, though.

17. Feb 3, 2018

### pervect

Staff Emeritus
The way I would put the stress situation is that the stress-energy tensor is a rank 2 tensor, and transforms as such. I do have some questions on the relationship between the physicists usage of the stress energy tensor, which I'm familiar with, and engineering usage, which I'm not familiar with and which may be slightly different.

I'm tempted to say that if the stress-energy tensor is zero in one frame, it's zero in all frames. While this is a correct statement, it's misleading, because the energy part of the stress-energy tensor isn't zero if one has matter present. So in the cases under consideration, one would actually need to carry out the appropriate transformations to figure out how it transforms.

The ability to handle rank 2 tensors and their transformations comes well after one learns introductory special relativity, however. If one doesn't learn the basics of SR, one will never get this far.

18. Feb 3, 2018

### pervect

Staff Emeritus
I did work out the shape of a rotating bar in a moving reference frame via the Lorentz transform. This is simpler than the see-saw case to analyze, but one can gain some insight of the see-saw case from the rotating bar case.

It's convenient to write the coordinate of any point on the bar as a function of r, which we set to zero at the origin of the bar (the origin being the point on the bar with zero proper acceleration). We will call the inertial frame of reference in which the origin of the bar is at rest "the rest frame of the bar", even though only the origin of the bar is at rest in this frame. r is a parameter that picks out a specific point on the bar. We also need a time coordinate, which we will take as the proper time $\tau$ of the point we just specified by specifying r. We will let the angular frequency of the rotation be $\omega$ in the bar's rest frame. Then in the bar's rest frame, we write:

$$x = r \cos \frac{\omega \tau}{\sqrt{1-r^2\omega^2}} \quad y = r \sin \frac{\omega \tau}{\sqrt{1-r^2\omega^2}} \quad t = \frac{\tau}{\sqrt{1-r^2\omega^2}}$$

Applying the Lorentz transform, we can transform these coordinates to a moving frame. Let the velocity be determined by the dimensionless parameter $\beta$, so that v = \beta c. And let $\gamma = 1/\sqrt{1-\beta^2}$. Then we can find the position of a point on the bar in a moving frame with coordinates x1, y1, t1 via the Lorentz transform.

$$x1 = \gamma \left(x + \beta\,c\,t \right) \quad y1=y \quad t1 = \gamma \left( t +\beta x/c \right)$$

We can substitute the expressions for $x(r,\tau)$, $y(r,\tau)$, and $t(r,\tau)$ from the first set of equations into the second to find $x1(r, \tau), y1(r,\tau), t1(r,\tau)$. It's also convenient to set c=1 at this point, unless one really wants to keep tract of it (I did not).

We now wish to find and plot x1 and y1 as a function of t1, rather than as a function of $\tau$. To do this, we need to solve the equation

$$t1 = \frac{1}{\sqrt{1-\beta^2}} \left( {\frac {\tau}{\sqrt {1-{r}^{2}{\omega}^{2}}}}+\frac{\beta}{c}\,r\cos \left( {\frac {\omega\,\tau}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) \right)$$

for $\tau$ as a function of r and t1. This is not something that has a closed form solution. We will call this $\tau = f^{-1}(r, t1)$

Then we can write the coordinates x1 and y1 of the bar at time t1 as a function of r, where I have now omitted the factors of c:

$$x1 = \left( r\cos \left( {\frac {\omega\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) +{\frac {\beta\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) { \frac {1}{\sqrt {1-{\beta}^{2}}}}$$

$$y1=r\sin \left( {\frac {\omega\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right)$$

This would be quite messy to carry out by hand - I used a computer algebra package to do the algebraic manipulations and to do some graphs for $\beta = .9$ and $\omega = .5$ with r varying from 0 to 1. While I don't have a convenient way of posting the results, I can say that the bar is generally not straight (there's an exception when it's vertical), that one can see the effects of time dilation (the bar rotates more slowly), and that the bar is length contracted as well as bent when it's not vertical.

As I mentioned in a previous post, this is more or less to be expected. While the Lorentz transformation is a linear transformation and must map straight lines into straight lines, the worldlines of points on the bar (a fixed value of r as $\tau$ varies) are not straight lines for r>0. So we don't expect the bar to be straight, and we don't expect the bar to have a constant length, and the calculations demonstrate this.

I like this approach because a) I'm used to it, and b) the coordinates r and $\tau$ have physical significance, being the proper distance of a point on the bar from the origin in the former case, and being the proper time of a point on the bar in the later case. Both the proper distance and the proper times are independent of any coordinate choices, so it's a convenient representation of the spinning bar.

Last edited: Feb 3, 2018
19. Feb 3, 2018

### Mister T

Nice analysis @pervect

Suppose the bar is made of some delicate material such as glass so that a very small bend will cause it to break. What the OP is asking about, I believe, is why it doesn't break when it bends in the moving frame. This is just one of those things that's part of special relativity, like time dilation and length contraction. It's like asking how, when a moving rod is Lorentz-contracted, it can withstand the compression without breaking. Relativity contradicts common sense in many ways, and this is just one of those ways. Understanding them expands common sense.

20. Feb 3, 2018

### Ibix

This might be another way to look at it.

In the rest frame of its centre of mass it's straight, and rotated at each subsequent time. So a rotating rod can be viewed as a helical worldsheet. This can be illustrated with a (2+1)d Minkowski diagram - I've drawn one below. One spatial plane of the rest frame is marked as a blue plane (time is perpendicular to this), and naturally the intersection of the green helical worldsheet and the plane is a straight line. So the rod is straight in this frame.

But what happens in a moving frame? In a moving frame, the spatial plane is tilted compared to the one drawn above. That looks like this:

Now you can clearly see that the intersection of the helix and the plane is a curve. That's why the rod is curved in this frame (although as @pervect notes, there's a special case when the rod is "across" the slope where it'll be straight). You can also see immediately that I've changed nothing about the rod by changing the frame. So it can't be broken; it's just the intuitive notion that "if it isn't straight it ought to be broken" that doesn't work properly.

21. Feb 3, 2018

### Staff: Mentor

Is this the rest frame of the bar's center of mass (but non-rotating), or a (rotating) frame in which the bar as a whole is at rest? It looks like the former, since you have coordinates varying with time. But I think clarity would be helpful.

22. Feb 3, 2018

### pervect

Staff Emeritus
WIth my setup, it's the frame in which the bar's origin is at rest. The origin is defined as the point on the bar with zero proper acceleration. Points not at the origin will have a non-zero proper acceleration. I'd rather avoid the center of mass, I seem to recall there are potential issues with it's frame dependence in SR, though I'm hazy on the details.

I've slightly ammended my original post to clarify this point.

Last edited: Feb 3, 2018
23. Feb 4, 2018

### Ebeb

Great visuals, Ibix!
The 2D straight and 2D curved rod are only different 2D measurements/observations of one and the same 3D helix.

24. Feb 8, 2018

### Peter Martin

The answer can be explained without any reference to Special Relativity, Einstein, or inertial reference frames. Ask: How could the two lightening strikes occur so that the woman on the train sees them as simultaneous? We can assume a conventional view of reality: that the landscape is "at rest" and the train is "in motion".

Since the flash from the rear strike must travel a longer distance to reach the woman, due to her "forward motion", it must occur before the front strike. Well, that is exactly what the man sees when the woman sees the strikes as simultaneous. So this explanation makes perfect sense even with the conventional view that the landscape is "at rest" and the train is "moving".

No need for Einstein!

25. Feb 8, 2018

### Ibix

A "view of reality" in the sense you are using it here is a reference frame - in this case, the rest frame of the landscape.

That explanation works fine in the landscape frame. But what is the explanation in the train frame? If you formulate a coherent answer to this, you will find that you have developed special relativity.