# Do Events Conditional to Simultaneity Occur in Every Reference Frame?

Imagine a train with a table that contains a light on both ends (so that the line between the two sources of light is parallel with the line of motion of the train) and a light sensor in the perfect middle. The lights start out off with the train moving at a high speed. Then, the lights simultaneously (in the reference of the train) turn on. The sensor detects whether both beams of light hit it at the same time or not. If the light hits the sensor from both sides at once, the sensor turns on a bright light that can be seen from out side the train.

Since, in the reference of the train, the lights both go on simultaneously, the light sensor will detect that and turn on its bright light.

However, a camera sits on an embankment near the train waiting for it to pass (let's just say that the lights are triggered by something on the tracks right before the camera. When the lights turn on, the camera standing on the embankment will see... What comes next of the following options?

1. The camera sees both lights turn on (and can calculate that it was perfectly equidistant to both light sources when the rays of light left the light source), but since the photo sensor is moving away from one beam of light and toward the other, the beams of light do not reach the photo sensor at the same time.

1. But the camera sees the bright light turn on anyway.
2. And the camera doesn't see the bright light turn on.
2. The camera sees both sources of light turn on at different times such that the beams reach the photo sensor at the same time.
3. The camera sees both sources of light turn on at different times, but the beams of light still don't reach the photo sensor at the same rate.
1. But the camera still sees the bright light turn on anyway.
2. And the camera doesn't see the bright light turn on even though the camera on board the train saw it turn on.

Would this thought experiment follow option 1.1, 1.2, 2, 3.1, or 3.2?

If it follows option 2, is there a way to set up a thought experiment so that the criteria to turn on the bright light is met in the reference frame of the train but not in the reference frame of the embankment? If so, would the camera see the bright light turn on anyway, or would the camera on the embankment see the light stay off?

In short, if an event is conditional upon the simultaneity of two events, will the conditional event take place even in a frame of reference where the condition is not met, or will the event take place regardless of any frame of reference other than that of whatever is deciding the simultaneity of the two events?

EDIT

This program graphs the thought experiment with the horizontal component being position and the vertical component being time. The frame of reference of the train is given in black, and the frame of reference of the embankment is given in red. The yellow lines represent the positions of photons given a particular time. Note that the red points and the black points are not in the same coordinate system. The red points are the black points after the Lorentz transformation.

Black A and B are the positions (in space and in time) that the lights first turn on giving off their first photons, whose positions versus time are graphed in yellow. Black M is the midpoint of line segment AB and is the location of the light sensor. Note that since M is just a location and not a time, it should technically be represented as a vertical line. Both yellow lines intersect with this should-be-line-segment M and each other at black point C (a time and location). Since they intersect with M at the same point, the events are recorded as simultaneous.

Red A' and B' are black A and B after the Lorentz transformation (aka, they're the equivalents of A and B in the reference frame of the embankment). While the black points A and B are at the same height in the screen (they both posses the same time component and are thus simultaneous), red A' and B' are not graphed at the same height and thus are not simultaneous.

So that eliminates option 1 because A and B are not simultaneous in the reference frame of the camera on the embankment.

Notice now the yellow lines protruding from red A' and B'. Those are the photons that are emitted by both lights. Now, as black M should have been graphed as a vertical line, red M' should have been graphed as a line slanted toward the left (the line that passes through red M' and red C'). Thus, it is clear that the light from B' intersects M' long before (below) the light from A' intersects M'. Now, to put the speed of light more to scale, all the yellow lines should have been much more horizontal. Thus, putting the whole thing to scale would exaggerate the fact that the light from B' intersected 'M before the light from 'A did (you can change the speed of light in this simulation by changing the number on the first line of code: "var c = 3;". You can also change the speed of the train on the second line to put things into scale. Everything is in SI units).

Correct me if I'm wrong, but using the math of the Lorentz transformation seems to indicate option 3. Both red A' and B' are not simultaneous because they have different time components. Also, the beams of light do not intersect the photo sensor simultaneously (in the perspective of the embankment) because the rays of light do not intersect line M' -> C' at the same height or time component.

So it seems that option 3 is correct, in which case I still don't know if sub-option 1 or 2 is correct.

UPDATE 2

I was being inconsistent in my diagram with representing time as seconds of meters. I fixed it, and it's now working as option 2.

• did you consider that you have to plot it in a minkowski diagram, not on a cartesian one? – Wolphram jonny Jan 30 at 22:11
• @Wolphramjonny I don't think so (as I learned about them just now). I graphed the diagram so that, if each variable at the start of the program is to scale, each pixel corresponds to a meter of space or time. To convert from seconds to meters, I divided by the speed of light. The horizontal component of the Minkowski diagram is $ct$, and I just divided by $c$ so that the vertical component of my graph was $t$. Does it make any difference if I made sure to convert seconds to meters? – ElliotThomas Jan 30 at 22:43
• the issue is that the axes of the two reference frames are not paralell to each other, so a standard graph will draw the points in an incorrect place – Wolphram jonny Jan 30 at 22:57
• @Wolphramjonny I just found out that I was not consistent with my units (sometimes, I treated time as seconds, and sometimes as meters). I fixed it, and now it works as is expected. The two reference frames not being parallel to each other shouldn't make a difference because I mathematically transform the old points to their new position. How I translate the output of the Lorentz transformation into positions to graph shouldn't matter as long as I am consistent within each coordinate system (and realize that there may be some relationship between the two that was lost) – ElliotThomas Jan 30 at 23:01
• yes, it is fundamental that the axes are not perpendicular to each other. in any case, think it this way: an event that happens at the same time and space in a give reference frame will transform also into a single point in any other reference frame, just using the lorentz transformations. If your graph indicates something different is because you did something wrong. – Wolphram jonny Jan 30 at 23:05