Simultaneity Check for Light Flash in Train and Platform Frames

[morrobay]

If the below algebra is correct then a light flash originating from the center of a train with .6c will be observed inside both train and platform reference frames as striking the back in less time.
Given : train length= 20ls
v = .6c
lorentz factor .8

As observed from the platform the front train clock=t0 and back train clock = vx/c sq
=t12
As observed from platform clock the light strikes the front of the train at 1/1-.6 (.8)(10)
=20s and strikes the back of the train at 1/1+.6(.8)(10) =5s
this is in accord with (t)front- (t)back= vL/lorentz = .6(20)/.8 =15s
As observed from inside the trains reference frame the light strikes the front at t-vx/LZ
20-.6(12)/.8 =16s
And strikes the back at 5-.6(3)/.8 = 4s
Also both train clocks as observed from platform read t16
So is the above correct. that is from inside both train and platform reference frames the
light strike event at back of train is first ?
morrobay B.S. biology, chemistry minor San Diego State

 

[JesseM]

If the below algebra is correct then a light flash originating from the center of a train with .6c will be observed inside both train and platform reference frames as striking the back in less time.
Given : train length= 20ls
v = .6c
lorentz factor .8

As observed from the platform the front train clock=t0 and back train clock = vx/c sq
=t12

Yes, in the platform frame, at the moment the clock at the front reads 0 seconds the back clock reads 12 seconds.

As observed from platform clock the light strikes the front of the train at 1/1-.6 (.8)(10)
=20s

If the flash is emitted at x=0 and t=0 in the platform frame, and the front of the train is at x=8 ls at that moment, then x(t) for the front is 0.6t + 8 while x(t) for the light beam is 1t, so they'll meet when 1t = 0.6t + 8, meaning 0.4t = 8, so t = 20, yes.

and strikes the back of the train at 1/1+.6(.8)(10) =5s

x(t) for the back of the train is 0.6t - 8, x(t) for the light beam is -1t, so they meet when 0.6t - 8 = -1t, so 1.6t = 8, so t=5, agree again.

As observed from inside the trains reference frame the light strikes the front at t-vx/LZ
20-.6(12)/.8 =16s

I don't quite follow the basis for the formula you're using, but if the clock at the front reads 0 when the flash occurs, then after 20 seconds in the platform frame, the clock at front has advanced forward by 20*0.8 = 16 seconds due to time dilation, so your answer for what it reads when the light hits it is right.

And strikes the back at 5-.6(3)/.8 = 4s
Also both train clocks as observed from platform read t16

Well, you're correct that the clock at the back will also read t=16 seconds at the moment the light hits it, since in the platform frame it read 12 seconds when the flash was set off, and it took 5 seconds in the platform frame for the light from the flash to reach it, during which time the clock at the back only advanced forward by 5*0.8 = 4 seconds, so it read 12 + 4 = 16 seconds when the light hit it. But the first part of your statement appears wrong--how can the time on the back clock when the light hits it be different from the time-coordinate of the light reaching the back in the train's frame? The coordinates of events in a given frame are defined in terms of local readings on synchronized clocks at rest in that frame. Again, I don't know what the basis is for the formula you're using, but it seems to give the wrong answer in this case--the two flashes hit the front and back of the train simultaneously at t' = 16 seconds in the train's rest frame. And note that a clock at the center of the train would have read 6 seconds when the flash was set off there (halfway between the readings of the clocks at the front and back of the train at that moment, as observed in the platform frame), so in the train's frame the light from the flash going in both directions took 16 - 6 = 10 seconds to get from the center to either end, and the distance from the center to either end in the train's rest frame is 10 light-seconds--so the speed of the light beams going in either direction was 1 light/second per second in the train's frame, as required by second postulate of special relativity.

So is the above correct. that is from inside both train and platform reference frames the
light strike event at back of train is first ?

No, as I said above all your figures are correct except for the one about the time-coordinate of the light hitting the back in the train's frame. Again, time-coordinates in a given frame are simply defined in terms of local readings on synchronized clocks at rest in that frame, so if you agree that both clocks read 16 seconds when the light hits them you should agree both events happen simultaneously at a time-coordinate of t' = 16 seconds in the train rest frame. And if light from a single flash at the center took different amounts of time to reach the front and back in the train's rest frame, this would clearly violate the rule that each frame must measure the speed of light to be the same, so that's another way of seeing that your figure for the time-coordinate in the train's rest frame can't be right.

 

[pervect]

It's a bit hard to follow what you wrote without a diagram, does the following attached scanned space-time diagram match what you want?

In the lab frame, the front of the train starts out at x=8,t=0 while the back of the train starts out at x=-8, t=0.

The length of the train is thus 16 light seconds in the lab frame, which corresponds to a proper length of 20 light seconds.

Solving the equations for the light beam emitted at t=0 in the lab frame we find

point F: t=20, x=20
point B: t=5, x=-5

where point F is where the light beam intersects the front of the train, and point B is the point where the light beam intersects the back of the train.

I think this matches your lab-frame numbers now, so hopefully this is what you had in mind.

We can use the Lorentz transforms to convert these lab-frame coordinates into the train frame.

Let t' be the time in the train frame. Then using the standard Lorentz formula with c=1 (1 light second / second) we have:

$$t' = \gamma \left(t - vx \right)$$

where v=.6 and $\gamma = 1/.8 = 1.25$

This gives

point F: t' = 1.25 ( 20- .6*20) = 10
point B: t' = 1.25 (5 - .6* -5) = 10

thus t' is equal for point F and point B in the frame T', as we expect, and it's also equal to 1/2 the proper length of the train, as we would expect.

I didn't bother to compute x', but you can compute it if you like (it doesn't matter for the problem, though).

 

[JesseM]

pervect, I think part of the problem is that morrobay wants to have the clock at the front read a time of 0 s and the clock at the back read a time of 12 s at the moment the flash is set off, as observed in the platform frame. This would mean the clock at the center of the train would read 6 s at the moment the flash is set off at coordinates x=0, t=0 in the platform frame...and this in turn means that if you use the Lorentz trasform, which assumes that x=0, t=0 in one coordinate system matches up with x'=0, t'=0 in the other, then the time coordinates of events happening at the front and back of the train in the train's coordinate system won't actually match up with the local readings of the clocks on the train when the events happen. In particular, while it's true that the Lorentz transform says that in the train's frame the events of the light hitting the front and back will both happen at time-coordinate t'=10 seconds, the clocks will actually read 16 seconds when the light strikes them. Alternately, you could have the train-observer use a coordinate system whose time-coordinates did match up with the clocks (which is what I suggested seemed simplest to me), but then you could no longer use the Lorentz transformation formulas, I suppose you'd need to use a Poincaré transformation instead.

The problem would be a lot simpler if we assumed that at the moment the flash was set off, the clock at the front would read -6 seconds and the clock at the back would read 6 seconds in the platform frame, so the clock at the middle would read 0 seconds at this moment (at coordinates x=0 and t=0 in the platform frame), and a coordinate system based on this system of clocks would relate to the platform-observer's coordinate system by the simple Lorentz transformation equations.

 

[morrobay]

If the below algebra is correct then a light flash originating from the center of a train with .6c will be observed inside both train and platform reference frames as striking the back in less time.
Given : train length= 20ls
v = .6c
lorentz factor .8

As observed from the platform the front train clock=t0 and back train clock = vx/c sq
=t12
As observed from platform clock the light strikes the front of the train at 1/1-.6 (.8)(10)
=20s and strikes the back of the train at 1/1+.6(.8)(10) =5s
this is in accord with (t)front- (t)back= vL/lorentz = .6(20)/.8 =15s
As observed from inside the trains reference frame the light strikes the front at t-vx/LZ
20-.6(12)/.8 =16s
And strikes the back at 5-.6(3)/.8 = 4s
Also both train clocks as observed from platform read t16
So is the above correct. that is from inside both train and platform reference frames the
light strike event at back of train is first ?
morrobay B.S. biology, chemistry minor San Diego State

Thanks for the clarifications in posts 2,3,4
regarding post # 2 question on basis for formula used , the (x) used in t' = t-vx/Lz
was obtained from x=v/c(t) So x front = .6(20) and back =.6(5) I realized that the results for t' = t-vx/Lz were equal to (Lz)(platform clocke times)
I agree that there is only one train clock time at light strikes,t16.
The question I am attempting to formulate is: Could the elapsed time for the light in flight, the travel time, for the light going from center of train at t6 to the back be less
than the elapsed time going to the front- independent of clock readings in the train reference frame .

 

[JesseM]

Thanks for the clarifications in posts 2,3,4
regarding post # 2 question on basis for formula used , the (x) used in t' = t-vx/Lz
was obtained from x=v/c(t)

I still don't understand, what is c(t), and why are you dividing the velocity by it? If x is supposed to be a position, what is it the position of? If x has units of position and v has units of position/time, it seems like the only way the units of this equation would make sense is if c(t) has units of 1/time.

So x front = .6(20) and back =.6(5)

Here you seem to be multiplying the velocity by the times the light reaches each ends, whereas in your formula you were dividing by c(t), whatever that is.

realized that the results for t' = t-vx/Lz

Where does this formula come from? What is Lz? Are you just rewriting the Lorentz transformation formula t' = (t - vx/c^2)*gamma, where gamma = $$\frac{1}{\sqrt{1 - v^2 / c^2}}$$?

I agree that there is only one train clock time at light strikes,t16.
The question I am attempting to formulate is: Could the elapsed time for the light in flight, the travel time, for the light going from center of train at t6 to the back be less
than the elapsed time going to the front- independent of clock readings in the train reference frame .

In what frame? Obviously in the platform frame it's less, since the light only takes 5 seconds to go from center to back but 20 seconds to go from center to front in this frame.

 

[morrobay]

post # 6 answer
x=v/c (t) is intended to = x=vt/c , as a pt. on t'
the misinterpretation is due to typography.
Yes t'=t-vx/lz is intended to equal t' =t-vx/c sq ( gamma)

 

[JesseM]

post # 6 answer
x=v/c (t) is intended to = x=vt/c , as a pt. on t'
the misinterpretation is due to typography.

But where did you get this equation from? How can x be equal to vt/c, when x has units of distance and v*t/c has units of time? And physically, what event is x supposed to be the position of?

Yes t'=t-vx/lz is intended to equal t' =t-vx/c sq ( gamma)

The equation is (t - vx/c^2) * gamma, not (t - vx/c)*sqrt(gamma). Anyway, as I said in post #4, this formula only works if you assume that x=0, t=0 in the first frame is equivalent to x'=0, t'=0 in the other frame...but didn't you assume the flash happened at x=0, t=0 in the platform frame, yet the clock on the train there (if it was synchronized with the other two clocks in the train's frame) read 6 seconds at the time the flash was set off? In this case the time-coordinates in the train frame wouldn't actually match up with readings on the synchronized clocks at rest in the train frame, which is permissible but it makes things a bit confusing (it would mean that even though both clocks read a time of 16 seconds when the light hit them, you're using a coordinate system where both these events happened at t'=10 seconds). It would be simpler if the clock at the front of the train read -6 seconds when the flash was set off (in the platform frame), the clock at the middle read 0 seconds, and the clock at the back read 6 seconds (again in the platform frame), that way their readings would match up with the time-coordinate assigned to them in the train's coordinate system.

 

[morrobay]

Re post #8
I got x=vt/c from htt://panda.unm.edu/Courses/Finley/p262/spacetimeI/SpacetimeI.html
I have also referred to htt:galileo.phys.Virginia.edu/classes252/
as well to google.
For whatever the interpretation of the formulas I have been using with a regular keyboard
t' =(gamma) (t-vx) If this application is so far off why is it giving the correct times : t16 and t4

 

[JesseM]

Re post #8
I got x=vt/c from htt://panda.unm.edu/Courses/Finley/p262/spacetimeI/SpacetimeI.html
I have also referred to htt:galileo.phys.Virginia.edu/classes252/
as well to google.

The second link doesn't work (did you mean http://galileo.phys.virginia.edu/classes/252/? If so, which sub-page?), and in the first link, are you talking about this paragraph?

Let us be more precise. We first look at the event where S is 2 hours older than he was when the origin was established, i.e., the event on the ct-axis marked by the symbol 2. (For convenience we will label this event as A.) Then we draw a line parallel to his x-axis through this point, i.e., a horizontal line through the point (x,ct) = (0,2). This line constitutes a line of simultaneity for S, of all events that he measures as happening at exactly the same time as his being 2 hours older. This cuts the worldline for S' at a point satisfying x=(v/c)t, the equation for the worldline belonging to S'. We label this event as B. On our diagram this means that x = (0.6)2 = 1.2 light-hours. At this event, on the worldline for S', our observer S measures the values (x,t) = (1.2,2). This is an event that S believes to be simultaneous with his being 2 hours older.

This appears to be a mistake, since like I said, the units just don't work out in that equation. (have you ever studied http://www.physics.uoguelph.ca/tutorials/dimanaly/?) What they should have wrote was x = vt, but since they were using units where c = 1, their equation luckily gave the right numbers even though it was incorrect.

In any case, the context of this equation is that they are calculating the position at time t of an object S' moving at velocity v in the frame of S, assuming that S and S' were at the same position x=0 at time t=0. But that doesn't work for calculating the position of the front and back of the train at time t in the platform-observer frame, since they did not start at the position x=0 at time t=0, instead at time t=0 in the platform frame the back was at position x=-8 and the front was at position x=8. So, after a time t the back of the train would be at position x = vt - 8, while the front would be at position x = vt + 8 (again, working in the platform frame).

t' =(gamma) (t-vx) If this application is so far off why is it giving the correct times : t16 and t4

Those are not the correct times in either frame. In the platform frame, the coordinates of the light hitting the back end are x=-5, t=5, while the coordinates of the light hitting the front are x=20, t=20. So if we use the formula t' = gamma * (t - vx/c^2), with v = 0.6 and gamma = 1/0.8 = 1.25, then for the event of the light hitting the back in the train's coordinate system we get:

t' = 1.25 * (5 - 0.6*-5) = 1.25*(8) = 10

And for the event of the light hitting the front in the train's coordinate system we get:

t' = 1.25 * (20 - 0.6*20) = 1.25*(8) = 10

So, these events both happen at t = 10 seconds in the coordinate system of the train when you use the Lorentz transform (again, the Lorentz transform assumes that the event with coordinates x=0, t=0 in the first coordinate system corresponds to x'=0, t'=0 in the second coordinate system, so the time-coordinate of a reading on a given clock on the train won't be the same as the reading itself, since in your example a clock placed at the middle of the train and synchronized with the other two would actually read 6 seconds rather than 0 seconds at the moment the flash was set off, which happens at x=0 and t=0 in the platform frame).

 

[morrobay]

Yes the second link ( Galileo) you listed is correct and the relativity of simultaneity pages.
The first link (Panda). yes that is the paragraph I was referring to.
Well I am going to be looking at all this for awhile- thanks for the information.
I would like to see a good spacetime diagram of this situation and with the train frame
at t0 front t6 center and t12 back

 

[morrobay]

diagram for" simultaneity check" post 12/04/07

Diagram for Simultaneity Check post of 12/04/07