The Two Light Beam Simultaneity Conundrum

[Otto Nomicus]

Let a 10 m square train car (as seen when stationary) be traveling left to right at 0.866 c. Let a light beam be emitted horizontally from the back and travel to the front, and a second beam be emitted from the upper left back corner and travel to the lower right front corner. A combination of time dilation and length contraction would make time on the train car pass at half the rate of a stationary observer's clock and contract the length of the car, and horizontal beam, from 10 m to 5 m long and the length of the diagonal beam from 14.1421 m to 11.1803 m long. For simplicity, let's just say that it takes 1 second for light to travel 10 m in the stationary frame in this thought experiment.

We would see it take 1 second for the horizontal beam to travel 5 m, in the train car frame, due to the time dilation and length contraction, and we need to get it down to 0.5 second, so I suppose we would need to see the front clock being half a second behind the back clock. We would see the diagonal beam take 2.23606 seconds to travel its 11.1421 m path and we need to get it down to 1.11421 seconds, so I suppose we would need the front clock to be 1.11421 seconds behind the back clock. Now we appear to have a conundrum, because 0.5 second and 1.11421 seconds are significantly different and clocks can't be desynchronized by two different amounts at the same time.

[Lorentz Jr]

WARNING: still editing, please don't comment yet.....

[math]c = 10m/s [/math]

[math]\gamma = 2 [/math]

The time on the clock in back, as seen from the ground frame, is [math]t_B = \gamma (t' + v * 0) = \gamma t' [/math]

The time on the clock in front is [math]t_F = \gamma (t' + v L) = t_B + 2 ( .866c * 10m/c^2) =  = t_B + 1.732 [/math]

So the desynchronization is 1.732 seconds.

Assuming the length and height of the car are both L, and having the diagonal beam go from y = 0 up to y = L instead of downward,

[math]x_{i}' = 0[/math]

[math]x_{f}' = L[/math]

[math]x_{1}'(t') = c t'[/math]

[math]s_{2}'(t') = c t'[/math]

[math]y_{2}'(t') = x_{2}'(t') = s_{2}'(t')/\sqrt{2} = c t'/\sqrt{2}[/math]

[math]T_{1}' = L/c[/math]

[math]T_{2}' = \sqrt{2} t_{1}' = \sqrt{2} L/c[/math]

In the ground frame:

[math]x_{i} = vt = 0[/math]

[math]x_{1f} = \gamma ( x_{1f}' - v T_{1}') = \gamma (L + v L/c)[/math]

[math]x_{2f} = \gamma ( x_{2f}' - v T_{2}') = \gamma (L + v \sqrt{2} L/c)[/math]

Edited February 6, 2023 by Lorentz Jr

[studiot]

1 hour ago, Otto Nomicus said:

Let a 10 m square train car (as seen when stationary) be traveling left to right at 0.866 c. Let a light beam be emitted horizontally from the back and travel to the front, and a second beam be emitted from the upper left back corner and travel to the lower right front corner. A combination of time dilation and length contraction would make time on the train car pass at half the rate of a stationary observer's clock and contract the length of the car, and horizontal beam, from 10 m to 5 m long and the length of the diagonal beam from 14.1421 m to 11.1803 m long. For simplicity, let's just say that it takes 1 second for light to travel 10 m in the stationary frame in this thought experiment.

We would see it take 1 second for the horizontal beam to travel 5 m, in the train car frame, due to the time dilation and length contraction, and we need to get it down to 0.5 second, so I suppose we would need to see the front clock being half a second behind the back clock. We would see the diagonal beam take 2.23606 seconds to travel its 11.1421 m path and we need to get it down to 1.11421 seconds, so I suppose we would need the front clock to be 1.11421 seconds behind the back clock. Now we appear to have a conundrum, because 0.5 second and 1.11421 seconds are significantly different and clocks can't be desynchronized by two different amounts at the same time.

I repeat swansont's comment from your previous thread.

You seem to be mixing frames up.

Who are the 'we' at the bginning of paragraph 2 ?

You have specified a square train. What shape do you think it is in the 'stationary' frame ?

Please clarify your description before you confuse everybody.

 

Edited February 6, 2023 by studiot

[swansont]

You also need to look at round trip times, since the time it takes to traverse the train car in the station’s frame depends on the direction of motion. Light travels at c, but the end of the car is moving toward or away from the photon. The station observer sees it travel more or less than 5m, depending on direction.

“We would see it take 1 second for the horizontal beam to travel 5 m, in the train car frame”

No. In the train car frame, it travels 10m. In the train car frame, the car is stationary. I concur with studiot: you are mixing frames, and that destroys the validity of any analysis.

 

(There’s also the conundrum of why people who are trying to understand a difficult topic choose to make the examples inordinately complicated. It’s math, and the math is internally consistent. Simplicity is a friend. Making the math harder just makes the math harder)

[Lorentz Jr]

Oops! Sorry about that.

With [math]c = 10[/math]m/s, and as seen from the ground frame, the clock in the back of the car is ahead of the one in front by

[math]\displaystyle{\frac{vL}{c^2} = .866}[/math] seconds.

Edited February 7, 2023 by Lorentz Jr

[Genady]

3 minutes ago, Lorentz Jr said:

Oops! Sorry about that.

As seen from the ground frame, the clock in the back of the car is ahead of the one in front by

vLc2=.866 seconds.

"The leading clock lags" is imprinted in my brain since my very first class in SR  

[Lorentz Jr]

3 hours ago, Otto Nomicus said:

I suppose we would need to see the front clock being half a second behind the back clock.

Don't suppose anything except the given parameters of the problem. Work everything out with the Lorentz transformations.

 

1 hour ago, Genady said:

"The leading clock lags" is imprinted in my brain since my very first class in SR  

I haven't gotten to that point yet. I have to keep going back to Wikipedia to remind myself that the sign is negative when you're going from the ground to the primed frame and positive going the other way. 🙄

The leader lags. That's a good way to remember. 🙂

 

To calculate the difference between the two clocks on the train, we'll read them simultaneously at a time t in the ground frame.

The time on the clock in back is [math]\displaystyle{t_{B}' = \gamma \left(t - \frac{v x_B}{c^2}\right)}[/math].

The time on the clock in front is [math]\displaystyle{t_{F}' = \gamma \left(t - \frac{v x_F}{c^2}\right)}[/math].

So [math]c^2(t_{B}'-t_{F}') = \gamma v(x_F - x_B)[/math].

Now we need to calculate [math]x_F[/math] and [math]x_B[/math], taking [math]x' = 0[/math] at the back of the car.

[math]x_B = \gamma(x_{B}' + v t_{B}') = \gamma v t_{B}'[/math].

[math]x_F = \gamma(x_{F}' + v t_{F}') = \gamma(L + v t_{F}')[/math].

So [math]x_F - x_B = \gamma(L + v(t_{F}' - t_{B}'))[/math].

Now we combine the first and last equations to get the desynchronization [math]\Delta = t_{B}'-t_{F}'[/math]:

[math]c^2(t_{B}'-t_{F}') = \gamma v(\gamma(L + v(t_{F}' - t_{B}')))[/math]

[math]c^2\Delta = \gamma^2 v(L - v\Delta)[/math]

[math](1 - \beta^2)\Delta = \beta(L/c - \beta\Delta)[/math], where [math]\beta \equiv v/c[/math]

[math]\Delta = \beta L/c = vL/c^2[/math]

Edited February 7, 2023 by Lorentz Jr

[Lorentz Jr]

Actually, that last step was unnecessarily complicated. After dividing both sides of the previous equation by [math]\gamma^2[/math], it should have been

[math](c^2 - v^2)\Delta = v(L - v\Delta)[/math],

and then you just add [math]v^2\Delta[/math] to both sides.

Edited February 7, 2023 by Lorentz Jr

[Markus Hanke]

8 hours ago, Otto Nomicus said:

Now we appear to have a conundrum

So let me ask you this straight - what is the purpose of these threads? Are you seeking to genuinely understand how to correctly analyse situations such as these, or are you just trying to show that relativity must be wrong by presenting “conundrums”? I’m asking this because thus far you haven’t shown yourself receptive to anything that was explained to you, and I therefore see no reason to assume that this thread will go any differently than your last one.

Based on what I have seen so far, I personally suspect it is the latter of the two options - but please do correct me if I misjudged. Thus, instead of getting endlessly entangled in highly specific scenarios that do nothing to illuminate the actual issue, I suggest it would be far better if you were to simply state clearly and directly what your contention with regards to Special Relativity actually is, and we can all save ourselves the needless beating around the bush and cut straight to the chase instead.

[Otto Nomicus]

41 minutes ago, Markus Hanke said:

So let me ask you this straight - what is the purpose of these threads? Are you seeking to genuinely understand how to correctly analyse situations such as these, or are you just trying to show that relativity must be wrong by presenting “conundrums”? I’m asking this because thus far you haven’t shown yourself receptive to anything that was explained to you, and I therefore see no reason to assume that this thread will go any differently than your last one.

Based on what I have seen so far, I personally suspect it is the latter of the two options - but please do correct me if I misjudged. Thus, instead of getting endlessly entangled in highly specific scenarios that do nothing to illuminate the actual issue, I suggest it would be far better if you were to simply state clearly and directly what your contention with regards to Special Relativity actually is, and we can all save ourselves the needless beating around the bush and cut straight to the chase instead.

I'm genuinely trying to find the resolution to what I perceive as a conundrum, I thought the wise people here might be able to help. Why are you so cynical? I'll try to clarify the scenario more with a diagram in a little while. So far, one person said 1.732 seconds and another said half that. I'll try to see if either works in a diagram. Thanks to all who replied so far, even if cynical.

[Lorentz Jr]

1 hour ago, Otto Nomicus said:

one person said 1.732 seconds

Sorry, that first one is wrong. It corresponds to measurements that are simultaneous in the train, so, in the ground frame, the distance between the two measurements is [math]\gamma L = 20[/math]m, i.e. twice the length of the car. The second result (.866) is the right one.

11 hours ago, Otto Nomicus said:

We would see it take 1 second for the horizontal beam to travel 5 m, in the train car frame

Light travels at the speed of light in all frames. In the train, it travels 10m in one second. In the ground frame, it travels a longer distance in a longer period of time to catch up to the front of the car, because the front is moving away from the back at almost the speed of light, but the beam of light still travels at the speed of light.

So the source of the conundrum is your incorrect suppositions. The only way to solve SR problems is to apply the Lorentz transformations to the specific times and locations of the events in each individual problem.

Edited February 7, 2023 by Lorentz Jr

[Markus Hanke]

1 hour ago, Otto Nomicus said:

I'm genuinely trying to find the resolution to what I perceive as a conundrum

Ok, fair enough, that’s a valid motivation.
So do you feel that the conundrum, as you call it, is due to not having analysed the specific scenario correctly, or do you feel that SR as a model is not internally self-consistent? The latter is easy to address in a very general manner; whereas disentangling the former may be tedious and time-consuming, and (IMHO) not very illuminating so far as the actual underlying physics are concerned.

1 hour ago, Otto Nomicus said:

Why are you so cynical?

Not cynical so much as sceptical. For two reasons, mainly:

1. I’ve been participating on various science forums for a long time, and sadly it’s a very common tactic for people to try and covertly smuggle “anti-relativity” type of sentiments onto the main physics sections (where such things aren’t supposed to go) by dressing them up as specific scenarios that purport to show some sort of paradox or contradiction. Once other posters clarify the mistake that was made in analysing the scenario - which lead to the apparent contradiction in the first place -, the OP then refuses to take on board anything that is explained to them, and will stick hand tooth and nail to their contention that relativity is self-contradictory and thus wrong. Usually such threads end up locked or abandoned. Sorry to say, but this is an exceedingly common modus operandi.

2. The general impression you gave on your other thread was not one of someone having genuinely come here to learn; it also had some of the hallmarks mentioned under (1). Perhaps I’ve gotten the wrong impression somehow, but as an uninvolved reader of said thread I’ve got to tell you that the vibes weren’t good.

But maybe I got you all wrong, and maybe you are right in that I have become cynical by having spent too many years on science forums…if so, I’ll take responsibility for that. So let’s see how this thread goes  

[Otto Nomicus]

Okay here's a snapshot taken of the train car as it passes directly in front of a ground observer with an instantaneous camera. The red and green lines are continuous laser beams. The velocity is known to the ground observer to be 0.866 c because it was previously set up in the lab to move at that velocity in the lab frame. Length contraction has reduced the apparent length from 10 to 5 m, as the observer knows is predicted by SR, and the observer knows that time is slowed down to half the rate of his because the Lorentz factor for that velocity is 2 Through trigonometry, he knows that the diagonal line has gone from being 14.1421 m when the car was stationary to 11.1803399 m in the snapshot.

Knowing that the speed of light in his world is 1 m/s, the ground observer knows that from his point of view a light beam traveling that distance would take 0.5 second but he knows that observers in the car must see it take 1 second of their time, because they see it travel 10 m instead of 5, the car being uncontracted in their frame. How does he make 0.5 second of his time look like 1 second in the car? He imagines a timer at the back wall of the car showing 0 and a timer at the front wall showing 0.75 second when the light beam is emitted from the back wall, so that by the time the beam reaches the front wall the timer will have progressed by 0.25 second in that frame, while 0.5 second has passed in his, and will then show 1 second. That would seem to reconcile things but he recalls that Einstein stated that clocks at the back of moving frames will be seen by a stationary observer to be ahead of ones at the front so he finds this perplexing, because he had to imagine the front clock being ahead of the back one instead. 

The ground observer then sees what would happen with the diagonal beam with that same timer desynchronization. It would take 1.11803399 seconds in his frame for the green laser to travel the path so, with the timers in the car running at half speed, it would take 2.23606798 seconds on their timers so, with the front timer being 0.25 second ahead of the back timer, the front timer would show 2.48606798 had passed but in the uncontracted car frame the distance would be 14.1421 m and should have taken 1.41412 second on their timers. Now the ground observer is wondering how Einstein was so wrong, apparently, because nothing at all turned out the way he stated that it should and, thus, he perceived that a conundrum was afoot and he needed some people on a physics forum to try to explain how that happened.

 

1 minute ago, Markus Hanke said:

Ok, fair enough, that’s a valid motivation.
So do you feel that the conundrum, as you call it, is due to not having analysed the specific scenario correctly, or do you feel that SR as a model is not internally self-consistent? The latter is easy to address in a very general manner; whereas disentangling the former may be tedious and time-consuming, and (IMHO) not very illuminating so far as the actual underlying physics are concerned.

Not cynical so much as sceptical. For two reasons, mainly:

1. I’ve been participating on various science forums for a long time, and sadly it’s a very common tactic for people to try and covertly smuggle “anti-relativity” type of sentiments onto the main physics sections (where such things aren’t supposed to go) by dressing them up as specific scenarios that purport to show some sort of paradox or contradiction. Once other posters clarify the mistake that was made in analysing the scenario - which lead to the apparent contradiction in the first place -, the OP then refuses to take on board anything that is explained to them, and will stick hand tooth and nail to their contention that relativity is self-contradictory and thus wrong. Usually such threads end up locked or abandoned. Sorry to say, but this is an exceedingly common modus operandi.

2. The general impression you gave on your other thread was not one of someone having genuinely come here to learn; it also had some of the hallmarks mentioned under (1). Perhaps I’ve gotten the wrong impression somehow, but as an uninvolved reader of said thread I’ve got to tell you that the vibes weren’t good.

But maybe I got you all wrong, and maybe you are right in that I have become cynical by having spent too many years on science forums…if so, I’ll take responsibility for that. So let’s see how this thread goes  

I would never do such a thing as to smuggle anti-relativity sentiments into a physics forum, I'm offended that you would even think that. I am merely trying to resolve apparent conundrums, that's all.

[studiot]

30 minutes ago, Otto Nomicus said:

he perceived that a conundrum was afoot and he needed some people on a physics forum to try to explain how that happened.

 

Well this somebody is even more confused by what you are trying to say than before.

What you say now seems to confirm what I thought you said in your opening paragraph in your opening post, and I agree with that.

What follows in both posts is what is confusing since you now seem to have introduced new terminology which to me is even more muddled than before.

Please explain again what you think is happening the problem is, using only one name or better symbol)  for each variable.

 

Where did this 'conundrum' come from ?

Can you give a reference or did you think it up yourself ?

[Otto Nomicus]

I made a mistake in the diagram post, but can't edit it. The ground observer would see the 11.1803399 m beam take 1.11803399 second in his frame but since time passes at half the rate in the car frame he would see it take 0.559016995 second in that frame, I mistakenly doubled it instead of halving it. So with the front timer being ahead of the back one by 0.75 second, it would show 1.309016995 for the travel time, but it was supposed to show 1.41421 second in the car frame so it was short by 0.105193005 second. For that beam to work out right, the front clock would need to be 0.855193005 second ahead of the back clock, instead of 0.75 That's the conundrum, in addition to the timers being desynchronized in reverse order from what Einstein stated.

Edited February 7, 2023 by Otto Nomicus

[Markus Hanke]

1 hour ago, Otto Nomicus said:

I am merely trying to resolve apparent conundrums, that's all.

Wouldn’t it be easier then to just write down a general proof that there can be no physical paradoxes based on any pair of inertial frames related via Lorentz transformations? Because that’s easy enough to do, and it would apply to all possible such scenarios.

[Otto Nomicus]

2 minutes ago, Markus Hanke said:

Wouldn’t it be easier then to just write down a general proof that there can be no physical paradoxes based on any pair of inertial frames related via Lorentz transformations? Because that’s easy enough to do, and it would apply to all possible such scenarios.

But I don't have such a proof, because the reality of the diagram appears to indicate that in fact there can be physical paradoxes.

[studiot]

13 minutes ago, Otto Nomicus said:

I made a mistake in the diagram post, but can't edit it. The ground observer would see the 11.1803399 m beam take 1.11803399 second in his frame but since time passes at half the rate in the car frame he would see it take 0.559016995 second in that frame, I mistakenly doubled it instead of halving it. So with the front timer being ahead of the back one by 0.75 second, it would show 1.309016995 for the travel time, but it was supposed to show 1.41421 second in the car frame so it was short by 0.105193005 second. For that beam to work out right, the front clock would need to be 0.855193005 second ahead of the back clock, instead of 0.75 That's the conundrum, in addition to the timers being desynchronized in reverse order from what Einstein stated.

So you managed to confuse even yourself.

Diagrams themselves however are always good to have.

What about addressing swansont's excellent point about direction ?

 

 

[Otto Nomicus]

14 minutes ago, studiot said:

So you managed to confuse even yourself.

Diagrams themselves however are always good to have.

What about addressing swansont's excellent point about direction ?

 

 

I already fixed my original first post description up in the diagram post. The text in that post had an error in the last paragraph but I corrected it in a later post. To avoid confusion please disregard that post and use the revised version which I will post right after this. Sadly, I can neither edit nor delete the first version.

REVISED VERSION

Here's a snapshot taken of the train car as it passes directly in front of a ground observer with an instantaneous camera. The red and green lines are continuous laser beams. The velocity is known to the ground observer to be 0.866 c because it was previously set up in the lab to move at that velocity in the lab frame. Length contraction has reduced the apparent length from 10 to 5 m, as the observer knows is predicted by SR, and the observer knows that time is slowed down to half the rate of his because the Lorentz factor for that velocity is 2 Through trigonometry, he knows that the diagonal line has gone from being 14.1421 m when the car was stationary to 11.1803399 m in the snapshot.

Knowing that the speed of light in his world is 1 m/s, the ground observer knows that from his point of view a light beam traveling that distance would take 0.5 second but he knows that observers in the car must see it take 1 second of their time, because they see it travel 10 m instead of 5, the car being uncontracted in their frame. How does he make 0.5 second of his time look like 1 second in the car? He imagines a timer at the back wall of the car showing 0 and a timer at the front wall showing 0.75 second when the light beam is emitted from the back wall, so that by the time the beam reaches the front wall the timer will have progressed by 0.25 second in that frame, while 0.5 second has passed in his, and will then show 1 second. That would seem to reconcile things but he recalls that Einstein stated that clocks at the back of moving frames will be seen by a stationary observer to be ahead of ones at the front so he finds this perplexing, because he had to imagine the front clock being ahead of the back one instead. 

The ground observer then sees what would happen with the diagonal beam with that same timer desynchronization. He would see the 11.1803399 m beam take 1.11803399 second in his frame but since time passes at half the rate in the car frame he would see it take 0.559016995 second in that frame. So with the front timer being ahead of the back one by 0.75 second, it would show 1.309016995 for the travel time, but it was supposed to show 1.41421 second in the car frame so it was short by 0.105193005 second. That's the conundrum, in addition to the timers being desynchronized in reverse order from what Einstein stated.

Edited February 7, 2023 by Otto Nomicus

[studiot]

16 minutes ago, Otto Nomicus said:

Here's a snapshot taken of the train car as it passes directly in front of a ground observer with an instantaneous camera.

That is a confusing way to look at it because your 'snapshot' is far from instantaneous. The snapshot is 5m wide in the frame of the ground observer. And tkes times to pass.

[joigus]

2 hours ago, Otto Nomicus said:

I would never do such a thing as to smuggle anti-relativity sentiments into a physics forum,

This is not a chauvinistic society. It's a science-discussion network.

You're thinking Euclidean. Your diagram shows only too clearly that you are. Because you're thinking Euclidean, there's a rigidity of sorts in your mind that forces you to conflate different reference frames --as you've been told here and elsewhere-- in order to shoehorn your picture into the Pythagorean theorem. Here's a series of pictures that tell you why:

(from Ray D'Inverno, Introducing Einstein's Relativity, Oxford University Press.)

So, again,

13 hours ago, studiot said:

You seem to be mixing frames up.

 

13 hours ago, studiot said:

Who are the 'we' at the beginning of paragraph 2 ?

IOW, who are the S and S' on diagram 3.6 and 3.7 from D'Inverno? If you dodge the question again and throw brand new Euclidean diagram, there isn't much else to talk about, is there?

One final word about 'sentiments.' This is a recurring theme in science discussions. People often mistake dead-seriousness about the subject with grumpiness, or even anger. Nothing of the kind. I personally sympathise very much with everyone who's interested in science. People who know about the subject would be doing you no favour by indulging in a logic they know to be flawed. They would be doing themselves no favour either.

Sorry, I forgot:

My sentiments are in the right place. No sentiments about space-time, really. Only perhaps longing for a space-time that allows to go backwards in time. I have some unfinished business in the past.

[Lorentz Jr]

3 hours ago, Otto Nomicus said:

the ground observer knows that from his point of view a light beam traveling that distance would take 0.5 second but he knows that observers in the car must see it take 1 second of their time, because they see it travel 10 m instead of 5

You're ignoring the motion of the car. In the ground frame, the front of the car is moving away from the light beam, so the beam has to catch up to it.

Light beam: [math]x = c t[/math]

Front of the car: [math]x = L/\gamma + v t[/math]

When they meet:

[math]c t = L/\gamma + v t[/math]

[math](c-v) t = L/\gamma[/math]

[math]\displaystyle{ t = \frac{ L}{\gamma(c-v)} }[/math]

[math]\displaystyle{ x = \frac{ L}{\gamma(1-v/c)} = \frac{10m}{2(1 - .866)} }[/math] = 37.3m.

t = 3.73 sec.

These numbers are much larger than 5m and 0.5sec because the front of your train is running away from the light beam almost as fast as the beam is chasing it.

 

3 hours ago, studiot said:

That is a confusing way to look at it because your 'snapshot' is far from instantaneous.

The snapshot is instantaneous (i.e. the diagram is correct) from the ground's perspective.

3 hours ago, studiot said:

The snapshot ... tkes times to pass.

Snapshots don't move. The train passes the ground observer while the light beam approaches the front of the car from behind.

Edited February 7, 2023 by Lorentz Jr

[studiot]

1 hour ago, Lorentz Jr said:

4 hours ago, studiot said:

That is a confusing way to look at it because your 'snapshot' is far from instantaneous.

The snapshot is instantaneous (i.e. the diagram is correct) from the ground's perspective.

4 hours ago, studiot said:

The snapshot ... tkes times to pass.

Snapshots don't move. The train passes the ground observer while the light beam approaches the front of the car from behind.

Earlier in you reply to Otto you offer an abstract deduction in reply to his abstract deduction.

A snapshot in my view is different since it implies observing all the light reaching one point (the ground observer) at one time, as with any photograph.

This of course must take into account the difference in transit tiem of the light from each end and the middle of the train.

You see the same effect when drawing diagrams of spaceships outward and return journeys sending time markers every hour, day, whatever in their time, but arriving non linearly at the base observer.

 

[Lorentz Jr]

1 hour ago, studiot said:

A snapshot in my view is different since it implies observing all the light reaching one point (the ground observer) at one time, as with any photograph.

Right, that would be a literal photograph taken by a single camera. But interpreting the word "snapshot" literally isn't very useful in relativity. The intent in taking a snapshot is to record the state of some system at one instant in time, and I think most people understand that. So many diagrams in relativity problems are reconstructions that show the positions of the various parts of a system at the same time in the specified frame.

1 hour ago, studiot said:

This of course must take into account the difference in transit tiem of the light from each end and the middle of the train.

And that's what Otto's diagram does. I don't know how other people think about this (although I thought it was the standard procedure), but I imagine that there's a whole string of cameras or detectors along the train track, pre-synchronized and pre-set in the ground frame to each simultaneously record what's in its immediate vicinity. Or they can all just keep recording constantly and mark each image with a timestamp once they've been synchronized.

Anyway, a simpler and more useful "snapshot" in relativity (because it eliminates the transit times) is to reconstruct the moving object from all of those local detections. Likewise, a "snapshot" from the train's perspective would be reconstructed from cameras lined up along the train's length (and maybe even beyond it) and synchronized in the train's frame. Or you can reconstruct it in either frame from video taken at one location, sort of "reverse engineering" the video by calculating all the transit times and taking them into account after the images have arrived. But that's kinda complicated! 😨

And of course there are other uses of the word "snapshot" in other contexts, like analyzing a business or an economy or other social, biological, or ecological system. The common factor of all these uses is that they approximate a picture of some system at some point in time, even if the tools and information used are surveys, data from medical monitoring devices, or any other source, rather than literal photographs from a camera.

Edited February 7, 2023 by Lorentz Jr

[Otto Nomicus]

The problem with showing a moving version is that t would take different amounts of time for the straight and the slanted beams to reach the end of the car, it would be rather complicated. The snapshot version is equally valid without the complications. The beams are considered to be continuously emitted, not pulses. All you need to do is calculate how long they would take to complete their paths in both frames, I already contracted the length for you, you know the time dilation factor is 2, easy peasy. If somebody wants to make a moving version themselves and make things a lot more complicated, go for it, I'm just not going to do it myself because there's no good reason to.

If anyone has a problem with my instantaneous camera, it's a thought experiment, instant cameras are available in thought experiments. Regarding time taken for the light to reach the camera, Einstein never included the time taken for the light to reach the stationary observer's eyes. Again, it just complicates things. Just pretend I'm Einstein, nobody griped about him doing it that way. I suspect that most of his thought experiments would have quite different results if he had included that. My advice is to simply accept the obvious fact that his theory doesn't work when there is horizontal beam and a slanted beam at the same time, nobody's perfect, including Einstein.