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.