Can anyone show me some maths to explain the train problem with relative simultaneity

PhysicsLaura

Hi, I feel really out of my depth and this is just to satisfy a personal frustration. I can see that the light from both flashes should reach a passenger on board the train at the same time and also that it should hit from the front first, but I just can't seem to make it work in my head. Are there equations I could try and work through to show what point the light has physically reached at each point in actual time? Each line of reasoning makes sense but I am still struggling to figure it out as a whole.

Sorry if it's too simplistic.

Muphrid

We have three objects. Sensor A starts at the origin, Sensor B at [itex]2d[/itex] from the origin in the x-direction, and Light Source C at [itex]d[/itex] from the origin in the same direction.

All three object have some four-velocity [itex]u[/itex] that is a linear combination of [itex]e_t[/itex] (the timelike direction) and [itex]e_x[/itex]. Specifically, [itex]u = \gamma (e_t + \beta e_x)[/itex] with, let's say, [itex]e_t \cdot e_t = -1[/itex].

When the light source C creates a pulse, photons go along [itex]n_\pm = e_t \pm e_x[/itex]. Now, from here, we can calculate the spacetime points where these null lines intersect the worldlines of our sensors.

For sensor A, only [itex]n_-[/itex] intersects its trajectory. Let [itex]\alpha[/itex] be some scalar, and...

[tex]x_A = \tau_A \gamma (e_t + \beta e_x) = \alpha_A (e_t - e_x) + d e_x[/tex]

Applying equalities on each component gives us

[tex]\alpha_A = \tau_A \gamma, \quad \tau_A = \frac{d}{\gamma (1 + \beta)}[/tex]

Doing the same thing for sensor B gives

[tex]\alpha_B = \tau_B \gamma, \quad \tau_B = \frac{d}{\gamma (1 - \beta)}[/tex]

To measure the coordinate time elapses between events, all me must do is calculate take the vector from the source C to each of A and B and dot it with [itex]e^t = -e_t[/itex] to extract the time component. The result is that, from our perspective, the beam reaches A faster (A is at the back of the car). This is the correct result when we impose the condition that A, B, and C all be synchronized with respect to our frame, which is exactly what we've done here. We ensured that their positions all had no component in our [itex]e_t[/itex] direction at some initial time.

When instead we synchronize with respect to the train, then we must enforce the notion that each of the "initial" position vectors of A, B, and C will have no component in the direction of the train's four-velocity. In that framework, we'll get the result that both sensors detect the pulse at the same time. You can try that if you want; the only difference is in setting up the initial position vectors or, if you feel confident in the math, you can set it up as a stationary problem and Lorentz boost it at the end.

PhysicsLaura

Nope, guess I'm no where near your level I lost it here.

Specifically, [itex]u = \gamma (e_t + \beta e_x)[/itex] with, let's say, [itex]e_t \cdot e_t = -1[/itex].

It would be great if someone could explain this to me, and I would LOVE to have an image of this graph as a visual aid! It's partly that I don't know what gamma and beta are representing here!

Laura

Muphrid

[itex]\beta[/itex] denotes some speed relative to the speed of light. So if you're traveling at half of lightspeed, [itex]\beta = 0.5[/itex]. [itex]\gamma = 1/\sqrt{1-\beta^2}[/itex] is a normalization factor to ensure that [itex]u \cdot u = -1[/itex]. This is in [itex]c=1[/itex] units, so what that really means is that the four-velocity's magnitude is always the speed of light.

See this:


Flat spacetime is a Minkowski or hyperbolic space. Where a unit vector in a Euclidean plane would be confined to the unit circle, unit vectors in spacetime are confined to a unit hyperbola instead. The asymptotes of the hyperbola are what light rays follow from the origin. This is why no real particle can ever travel like a light ray--no matter how it goes along the hyperbola, it will never have the same four-velocity as a light ray.

jartsa

Here is a very simple way:

Here is a table of positions of train cabin front wall, relative to the ground, measured at times 0,1,2,3,4,5,6,7,8,9,10:

Table1: 50 51 52 53 54 55 56 57 58 59 60

Here is a table of positions of light, relative to the ground, measured at times 0,1,2,3,4,5,6,7,8,9,10:

Table2: 40 42 44 46 48 50 52 54 56 58 60

This is a table of the difference between the light position table and the cabin wall position table:

Table3: 10 9 8 7 6 5 4 3 2 1 0


At time 10 the difference becomes 0, light hits the cabin front wall, and this happens at position 60


(we have here a train that moves at half the speed of light)

ghwellsjr

I hope you realize that you are talking about two different scenarios here because the two flashes of light cannot both hit the passenger at the same time and at different times.

If we set up the scenario such that the two flashes are emitted at the same time in the passenger's rest frame, then they will hit the passenger at the same time. That's how we define "emitted at the same time". Another way of saying this is that define "emitted at the same time" when the two flashes arrive at the passenger at the same time.

If, on the other hand, we set up the scenario so that the two flashes are emitted at the same time in the ground's rest frame, then they will hit the passenger at different times and that means that they were emitted at different times in the passenger's rest frame. Again, this is how we define what "at the same time" means for remote events.

Just remember, they are two different scenarios and both cannot happen in a single scenario.

solarflare

i said that the light from both flashes hits the passenger of the train at the same time.

Accordng to relativity light travels at the same speed ( the speed of light ) relative to EVERYTHING. Therefore it must travel at the speed of light relative to the train in all directions. so the light must hit her at the same time or light does not travel at the speed of light relative to everything.

im assuming that the train behind the passenger would go through length contraction to make it work.

TheEtherWind

Imagine a light cone diagram, place two points on the x axis at ct=0. These events are space-like. Now consider a moving frame's coordinates. They're slightly angled in towards the light line. Now you can see that in that frame one event happens at negative ct and another at ct=0.

Doc Al

No, it was you who were wrong, not Einstein.
Well, that's not true. In that scenario (that you commented on), lightning strikes the ends of the train simultaneously according to an observer on the platform.
Right!
Wrong!

solarflare

in the video it states the observer is actually the exact distance and that it is a fact that the strikes happen at the same time.

Doc Al

(1) Which observer? The observer on the platform is exactly at the center of the train when he sees the lightning strikes.
(2) The strikes happen at the same time for the platform observer, not for the observer on the train.

solarflare

anyway the observer on the platform has nothing to do with what the video says happens. it says that the light from the rear passes her at a different time even tho they strike at the same time.

what if there was no observer on the platform - and there were two lightening strikes that did actually hit at the same time - what would the passenger see?

Jimmy

The train is moving toward the light from one flash and away from the flash at the rear. The light from the flash at the rear has to travel further to reach the passenger.

Doc Al

Watch it again.
Even though they strike at the same time according to the platform observer.

At the same time according to what frame of reference?

solarflare

Relativity states that light travels at the speed of light RELATIVE to everything. so it would still take the same time no matter how fast the train was moving. that is the whole reason we have time dilation and length contraction.

Doc Al

Right. Every observer will measure the speed of light with respect to themselves to be the same speed c.
For train observers it does take the same amount of time for the light to travel from each end of the train to the middle. And since the light does not arrive at the middle of the train at the same time, the train observer concludes that the lightning strikes did not happen at the same time.

Of course, for platform observers the light from the front of the train has a shorter distance to travel so it takes less time to reach the middle. (Since the middle of the train is moving towards the oncoming light.) Similarly, it takes longer for the light from the rear of the train to reach the middle, according to platform observers.

Muphrid

This is fairly easy to demonstrate with math, solarflare.

Let some person S be standing stationary at [itex]x=0[/itex] on a train platform watching a train go by. Another person, T, is riding this train and, at time [itex]t=0[/itex], is also at [itex]x=0[/itex]. There are two light sources F and B at the front and back of our train, respectively; F is at [itex]x=+d[/itex] and B at [itex]x=-d[/itex] at time [itex]t=0[/itex]. Let the four-velocity of T (and also of the sources F and B) be [itex]e_t' = u = \gamma (e_t + \beta e_x)[/itex].

Case 1: Person S perceives two light pulses simultaneously from sources F and B.

Person T will then measure the time at which F and B emitted these pulses relative to his own four-velocity. As with all vectors, you can measure the component of a vector along another with a dot product. The vectors involved are [itex]s_F = d e_x[/itex], the position of the front source, [itex]s_B = -d e_x[/itex], the position of the back source, and [itex]{e^t}' = \gamma (e^t - \beta e^x)[/itex].

[tex]t'_F = s_F \cdot {e^t}' = -d \gamma \beta, \quad t'_B = s_B \cdot {e^t}' = +d \gamma \beta[/tex]

In the limit that [itex]\beta \to 0[/itex], the person on the train perceives the flashes simultaneously also, but he clearly does not for any nonzero x velocity.


Case 2: Person T on the train perceives two flashes simultaneously.

Change [itex]\beta[/itex] to [itex]-\beta[/itex], and you can now work the problem in reverse. No additional logic is required because both frames are inertial; we need only say that the train platform is moving backwards relative to the inertial train.