Vinaur, I understand what you're saying. It's been too long since I've taken SR and I'm too lazy to do the math, so I didn't want to do it out explicity. I was trying to reformulate your question in such a way that I didn't have to do any math.
Anyway, maybe I am misunderstanding you, but:
Quote:
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My reasoning is that, according to the law of propagation of light the speed of the light beam relative to the moviong object has to be c, therefore the distance between the object and either of the two light beams has to be decreasing at the same rate c. If the object is heading towars one of the beams, then it "slows down" the one it's heading towards, and "speeds up" the one it moving away from. So basically relative to both beams (taken at the same time) it is not moving.
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Let's say that L1 is light 1, and L2 is light 2 and M' is a train, M the platform. Arrows show travel direction. If I'm standing on a platform, I see the following
Code:
M'
*--> * <--^ <--*
L1 -d1- M -d2- L2
where d1 and d2 are distances, and the ^ is the train. When I'm sitting on the platform, reading my newspaper, I look at L1 and L2, and the distance between me and L1 and L2 is decreasing at c. The train is moving toward me at .9c.
If I'm on the train, I see the following:
Code:
M
*--> *--> ^ <---*
L1 -d1'- M' -d2'- L2
If I'm on the train, I'm not moving anymore, so M' doesn't have an arrow. The platform is now moving toward me, and the two lights are still moving toward me. I've redefined some distances. d1' and d2' are the distance between me (on the train now), and the two lights. (If it helps, you can imagine that at this particular time, the train ^ and the platform M are at the same place in space. I can't draw it in ASCII like that, so this diagram will have to do.)
In Newtonian physics, the rate of chance of d1' is c+.9c, and the rate of change of d2' is c-.9c. However, this is false. Special relativity tells us that the rate of change of d1' and d2' are the same, the are both c.
Here's another illustration (from the point of view of sitting on the platform):
At this point in time (t=0), M' is at the midpoint of AB. And at this point in time, light from A and B hit M' at the same time, From The Point of View of The Platform. (This is important)
However, 1) M' is moving. 2) It takes light some time to get from either A or B to the midpoint of AB. Therefore, at the time that the light is actually emited, the picture looks something like this:
Code:
* <--^ *
A -d1- M' -d2- B
d1' > d2'
At this point, two lights are emitted, so the picture looks like this:
Code:
*--> <--^ <--*
L1 -d1- M' -d2- L2
Remember now, from the frame of M', things look like this:
Code:
*--> ^ <--*
L1 -d1'- M' -d2'- L2
In the frame of M', M' isn't moving, and L1 and L2 are moving. But remember, d1' and d2' decrease
at the same rate, c. Also, d1' and d2' is the distance between the
light and the train now.(we can now ignore the points A and B) This is what makes SR kooky. In the frame of M', since d1'>d2', and the
distance is deceasing at the same rate, from the point of view of M',
L2 will reach M' sooner than L1.
In the frame of the platform, since the platform is not moving with respect to A and B, the lights hit each other and the moving train simultaneously. From the frame of the train, L2 hits the train before L1.
There is nothing wrong with this. The idea of simulataneity no longer exists in special relativity. Something which is simultaneous in one frame, is not simulateous in another frame.
Why my earlier reply talked about order and causality:
In the example above, we can see that in special relativity, order is not preserved! L1 and L2 hit at the same time in the frame of M, but L2 hit before L1 in the frame of M'. These events do not have an absolute order in which they occur!
However, what if I do something at point A which
causes something else to occur at B? The philosophical implication is that no matter what frame I'm in, in order for A to cause B, A must always occur
before B. But our example shows us that in general, order is not preserved in SR. Is there any way that order can be preserved?
To preserve Order (A before B) in SR, the condition is the following. Given events A and B and the distance between them D, A will always occur before B if and only if t, the time between A and B, is t>=D/c. Thus, this says that for A to cause B (A to occur before B) the time between evens A and B must be greater than or equal to the time it takes light to travel between A and B. This is the
only case in SR where order is preserved. In all other cases, order is not preserved.
In the example with the train:
Since on the platform, A and B are seen to occur at the same time, then the events at A and B do not meet the criterea stated above for causation. (the time between A and B is not greater than or equal to D/c). Because A and B are not causally related, I can always find a frame where A occurs before B, and viceversa.
phew. that was a long post. If anyone finds any errors, please correct me.