Order of events in a train traveling at v = 0.76c

[techsingularity2042]

Homework Statement

A train with a proper length of 500 m is moving at a speed v = 0.76c with respect to an external stationary observer X.

When exactly half the train has passed next to the observer, two light bulbs placed at opposite ends of the train are turned on simultaneously, according to the frame reference of the train.

According to observer X, which light bulb is turned on first and what is the time interval, Δt between the two lights turning on?

Relevant Equations

(Δs)^2 = (c*Δt)^2 - (Δx)^2

I tried to come up with answers using the spacetime interval equation.

(Δs)^2 = (c*Δt)^2 - (Δx)^2
Let train's frame of reference be S.
Δt = 0
Δx = 500 m (since proper length is measured in object's rest frame)
Then I get
(Δs)^2 = -250,000

Since L = L0 / lorentz factor, where L0 is the proper length and lorentz factor = 1.54

L = 325 m = Δx'

Because spacetime interval for every inertial frame is equal:
(Δs)^2 = (c*Δt')^2 - (Δx')^2
(c*Δt')^2 - (325)^2 = -250000
c*Δt' = sqr(144375)
Δt' = 1.27 * 10^-6 s.

My second attempt incorporated setting spacetime interval an absolute value.

(c*Δt')^2 - (325)^2 = 250000
c*Δt' = sqr(355,625)
Δt' = 1.99 * 10^-6 s.

But the mark scheme says I need to use Lorentz transformation equations, and the answers are slightly different.
Answer according to the mark scheme is:

Δt' = 1.95μs

Why does such discrepancy arise? Is using spacetime interval equation in this context wrong?

Does it have to do with the fact that the events are space-like separated?

 

[Orodruin]

Since L = L0 / lorentz factor, where L0 is the proper length and lorentz factor = 1.54

L = 325 m = Δx'

This is an incorrect application of the length contraction formula. While correct that the train’s length is L, it is not correct that the distance between the events is L in the ground frame. The distance between the events would be L if they were simultaneous in the ground frame, but they are not and you have to correct for this.

 

[techsingularity2042]

No. Because the light bulbs move together with the train, the train’s proper length,L0, measured in its own rest frame equals the distance between the events in the train's rest frame. In contrast, L denotes the train’s length as measured by a ground observer, which corresponds to the distance between events used when calculating the spacetime interval in the observer’s inertial frame.

 

[Orodruin]

No. Because the light bulbs move together with the train, the train’s proper length,L0, measured in its own rest frame equals the distance between the events in the train's rest frame. In contrast, L denotes the train’s length as measured by a ground observer, which corresponds to the distance between events used when calculating the spacetime interval in the observer’s inertial frame.

You are simply wrong. The length L is the length of the train, which is defined as the distance between its end-points at a fixed time. Since the train moves, L will not be the distance between the events at the ends if they occur att different times. This is true even in classical mechanics.

 

[Orodruin]

A demonstration I usually make when teaching special relativity is to pick up the pointer stick, hold it horizontally and start walking. I then ask students to first note where the back is. Half a classroom later I ask them to note where the front of the stick is. Then I ask them what the distance between their noted events are (back of stick at early time and front of stick at late time). In decades of doing this, nobody has ever made an argument that that distance is equal to the stick’s length.

 

[techsingularity2042]

Yup, you are right. Thanks!

 

[Orodruin]

Yup, you are right. Thanks!

Of course …

 

[Orodruin]

It should also be mentioned: Lorentz transformation is the easiest way forward, but your approach can be amended to work as well. You just have to correct your $\Delta x’$ to also include the contribution from motion $v \Delta t’$. You should end up with a second order polynomial equation for $\Delta t’$ where only one solution is physical.

 

[techsingularity2042]

It should also be mentioned: Lorentz transformation is the easiest way forward, but your approach can be amended to work as well. You just have to correct your $\Delta x’$ to also include the contribution from motion $v \Delta t’$. You should end up with a second order polynomial equation for $\Delta t’$ where only one solution is physical.

Thanks! I wasn't getting the hang of it because English is not my first language

 

[techsingularity2042]

It should also be mentioned: Lorentz transformation is the easiest way forward, but your approach can be amended to work as well. You just have to correct your $\Delta x’$ to also include the contribution from motion $v \Delta t’$. You should end up with a second order polynomial equation for $\Delta t’$ where only one solution is physical.

I tried with this method, but I only get imaginary numbers because the spacetime interval is a negative value.

 

[Orodruin]

I tried with this method, but I only get imaginary numbers because the spacetime interval is a negative value.

If you show your work we can go from there.

 

[PeroK]

I tried with this method, but I only get imaginary numbers because the spacetime interval is a negative value.

You have to decide which event is the spacetime origin. Why not use the centre of the train passing X?

Then you have lightbulb events at $t'=0$ and $x'=\pm L/2$. Where $L$ is the proper length of the train.

Now can you use the Lorentz Transformation to get the $t$ coordinate of those events?

 

[Orodruin]

You have to decide which event is the spacetime origin.

Not really. The only interesting thing here is the separation vector between the events, which is independent of whatever origin you choose.

 

[PeroK]

Not really. The only interesting thing here is the separation vector between the events, which is independent of whatever origin you choose.

Given the OP's difficulties, I suggest calculating the explicit time of each event. Not least because the question of which event comes first seems to be missing from his calculations.

 

[Orodruin]

Given the OP's difficulties, I suggest calculating the explicit time of each event. Not least because the question of which event comes first seems to be missing from his calculations.

I disagree. The OP’s difficulties stem from trying to apply the invariance of the spacetime line element rather than Lorentz transforming. This is essentially equivalent to using the invariance of the magnitude of the separation vector between the events. It is a perfectly viable route, but Lorentz transformation is more direct.

There is however no need whatsoever (and possibly confusing) to introduce a particular origin because the relevant 4-vector to transform is the very same separation vector OP was attempting to use - which is origin independent. The time ordering will fall out naturally from such a transformation as well.

Introducing unnecessary and arbitrary quantities when not required just risks complicating things for no good reason in my opinion.

 

[techsingularity2042]

I have two questions. The First one is:

I can obtain the time difference between the events in the observer's reference frame easily from the Lorentz transformation equation. It seems that the negative sign implies the order of events. But I do not get how.
Instead, I can draw a Minkowski diagram to determine the order.

Given that light bulb A was on the left end of the train, I could find that A happened earlier than B.
How can I tell from the sign which event occurred earlier?

The second question is:
@Orodruin said the spacetime equation method could work by substituting Δx with vΔt, so I tried directly subbing in vΔt for Δx. Here's what I got:

where v = 0.76 c (I forgot to put the the squared sign, but it was included in the calculation)

Using the calculator, I get
Δt' = ± 2.6i μs

To explain this with my shallow knowledge on relativity, it seems that directly substituting v into the equation may break the invariance (sorry for my terrible English). The expressions involving Δt and Δx come from the Lorentz transformations, and they ensure the invariance because they are taking account of the gamma factor.

When I use Δx' - calculated from the Lorentz transformation equation - for the spacetime equation, I get the correct value for Δt' which is 1.9 μs.

Is my interpretation of the error with the substitution method correct? And please let me know if there were any calculation mistakes.

Thank you.

 

[PeroK]

Note that $\Delta t$ is inherently ambiguous unless you specify the order of events. Also, which frame is primed frame in this case? And what is the direction of the velocity? All these variables lead to an answer of $\pm$ something. And, as you can see, it's impossible to figure out which event occurred first without being more explicit about how you've set the problem up.

It seems to me that you need practice at setting up the Lorentz transformation (and inverse transformation). I would prefer to see you do it that way.

 

[anuttarasammyak]

Not $v \triangle t'$ but $L'+ v \triangle t'$, is it ?
When v=0, it should become L as you have written.

 

[techsingularity2042]

which frame is primed frame in this case?

It is explicitly stated that in the primed frame is X's frame in the picture.

And what is the direction of the velocity?

You can tell from the graph that the velocity of the train is positive because the worldline of X has a negative gradient.

Which part of my working made you think I needed practice at setting up Lorentz transformation equations? Thank you.

 

[techsingularity2042]

Not vδt′ but L′+vδt′, is it ?

L' as in contracted length?

 

[PeroK]

It is explicitly stated that in the primed frame is X's frame in the picture.

You can tell from the graph that the velocity of the train is positive because the worldline of X has a negative gradient.

Which part of my working made you think I needed practice at setting up Lorentz transformation equations? Thank you.

With that set up you need the inverse Lorentz transformation to transform to the X frame.

 

[techsingularity2042]

How could you tell that I had to use the inverse Lorentz transformation equation?
I get different signs from the spacetime and Lorentz transformation equation.

If I got a negative sign, does that mean I assigned a wrong sign to my variables?

Thank you.

 

[PeroK]

How could you tell that I had to use the inverse Lorentz transformation equation?
I get different signs from the spacetime and Lorentz transformation equation.

The velocity of the rod relative to X is $v$. Hence, the velocity of X relative to the rod is $-v$. This implies the inverse transformation going to the X frame.

 

[anuttarasammyak]

L' as in contracted length?

Yes, it is the train length observed in dashed system which is shorter than the proper length L or more clearly L0.

 

[PeroK]

Yes, it is the train length observed in dashed system which is shorter than the proper length L or more clearly L0.

If we use the Lorentz transformation, then we don't have to worry about length contraction.

There are a dozen ways to do this problem. Getting the Lorentz Transformation set up correctly, with all the signs and directions correct is the way forward, IMO.

 

[Orodruin]

There are a dozen ways to do this problem. Getting the Lorentz Transformation set up correctly, with all the signs and directions correct is the way forward, IMO.

This is essentially what I said from the beginning. Using the Lorentz transformation of the separation vector is the easiest way forward. Not much setup is needed, just a proper definition of the spatial separation vs the relative velocity between frames.

The point however is that the OP’s approach will work if corrected appropriately. It is instructive for the understanding of length contraction and relative simultaneity to work through those ideas correctly because many times students will apply these concepts without actually understanding the required assumptions underneath.

 

[PeroK]

You have to decide which event is the spacetime origin. Why not use the centre of the train passing X?

Then you have lightbulb events at $t'=0$ and $x'=\pm L/2$. Where $L$ is the proper length of the train.

Now can you use the Lorentz Transformation to get the $t$ coordinate of those events?

Just to follow up on this. I'll change my idea to have the origin at the rear of the train when the lightbulb goes on. So we have events at $t' =0$ and $x' =0, L$ in that frame.

The event at the rear of the train has $t=0$ in the X frame. And the time of the event at the front of the train is:
$$t =\gamma(t' +\frac v {c^2}x') = \gamma(0 +\frac{vL}{c^2} ) = \frac{\gamma vL}{c^2}$$So, $t >0$ and this event occurs second.

That is essentially using the Lorentz transformation to confirm the "leading clocks lag" rule.

 

[Orodruin]

Just to follow up on this. I'll change my idea to have the origin at the rear of the train when the lightbulb goes on. So we have events at $t' =0$ and $x' =0, L$ in that frame.

The event at the rear of the train has $t=0$ in the X frame. And the time of the event at the front of the train is:
$$t =\gamma(t' +\frac v {c^2}x') = \gamma(0 +\frac{vL}{c^2} ) = \frac{\gamma vL}{c^2}$$So, $t >0$ and this event occurs second.

That is essentially using the Lorentz transformation to confirm the "leading clocks lag" rule.

It is also confirming the fact that the choice of origin is irrelevant to the problem. Ultimately it of course boils down to the Lorentz transformation being linear.

I would just state the fact that the separation vector between the events is $\Delta X = (0,L)$. Lorentz transforming gives $\Delta X’ = L\gamma (-v, 1)$. In the primed frame the train moves to the left and so the event separation is back-front.

 

[anuttarasammyak]

If we use the Lorentz transformation, then we don't have to worry about length contraction.

It might be noteworthy that the way of post #17 with correction #18 does not need Lorentz contraction formula as given. In solving the quadratic equation, it is obvious that the solution $\triangle t$ is unike positive. D=0 gives Lorentz contraction formula.