# Error in special relativity albert einstein assumed

JesseM

no in both cases they dont....
as u have quoted it will return to its old position(which it started propagation) which is equal to the new position of the flash light- v*t
@fredrik:
ok here are the 3 events:
1- light emitted moving up.
2- light reflected from mirror
3- light is moving diagnolly down to the florr of the train(not back to the flash light)
in the first phase: flash light moved v*0.5t
int the third phase flash light moved v*0.5t
so what happned according to observer 1 (in the car) he sees light moving up and reflected back no to the flash light postion but flash light postion- v*t
OK, but you're just wrong about that. observer 1 will see it reflected back to the flashlight position because a vertical flashlight will always shoot light straight up in the flashlight's own rest frame. Observer 2 will see the light come back to (position on tracks that flashlight was when it was first turned on ) + v*t.

If you disagree, can you explain why you think this is wrong? Give some type of argument as to why you don't think this is a perfectly coherent possibility? Or is it just a matter of gut feeling?

Dale
Mentor

Hi thetrice,

Thanks for the drawings they are very helpful for understanding what you are thinking. There are a couple of points to notice here:

1) If both observer 1 and observer 2 see that the flash light and the mirror moving to the right at $v_t$ then you are describing the situation from one single frame of reference. In other words, observer 1 and observer 2 are essentially the same observer, they are not moving relative to each other. At most they differ in position, not velocity.

2) It is not logically possible for the reflected flash to both miss the light (top row, last diagram) and hit the light (bottom row, last diagram). Both observers are observing the same scenario, and are merely assigning different coordinates to the events, but the same events must occur.

Janus
Staff Emeritus
Gold Member

Hi thetrice
You can begin by telling what one of the observer sees. For, eg. You can begin by telling Obs 1.(moving one) sees light bounce vertically up and down or You can begin by saying Obs 2. Sees light moving vertically up and down.

Once you describe what one of the observer sees, then Everyone here are happy and everyone here are consistent in describing what other observer would see. We derive what other observer will see logically based on what the first observer sees.

You can't describe what both of them should see (by using your own fancy rules) and shout out! Hey Einstein was wrong!. Thats crazy, you know!

I think i understand your question. Correct me if I'm wrong, but here's what I'm getting:

The observer emits a "pulse of light" from his flashlight to the center of the mirror ("straight upward"). The light goes up. Since the mirror is moving with a velocity (let's say v), the light pulse should hit at a spot vt away from (or "behind") the center of the mirror, where t is the time it takes for light to travel from floor to ceiling.

But, a contradiction! According to Observer 1 (who is moving with a constant velocity so that he can't even tell that he's moving), he emitted a pulse "straight up" and it should also come "straight down," right back at his flashlight.

So, the question is, what does it mean to emit light in a "straight line upwards"?

Observer 1 aimed his flashlight right in the center of the mirror, and it bounced right back at him (remember, Observer 1 and the mirror are moving at the same speed, in the same frame).

But, Observer 2 was watching this, and here's what he saw:
-he saw Observer 1 aim his flashlight at the center of the mirror.
-he saw the "pulse of light" leave the flashlight.

Now, the equivalence principle. Since Observer 1 saw the pulse of light go up and bounce right back at him, it must be that Observer 2 sees the pulse of light bounce back to Observer 1 as well. The only way this could be is that Observer 2 doesn't see the light go "straight up" horizontally, but on a diagonal.

Pause and think about that for a sec...

...do you agree? ok, continue:

That's all well and good, but what about the other postulate of relativity: the speed of light is the same in all frames.

Why does this matter? Well, say the height of the train car is h. Say that the distance travelled in the time for the pulse to go up and come back down is d.

For Observer 1, the time that it took for the light to go up and then down is: 2h/c (distance travelled divided by speed of light).

For Observer 2, recall that he saw the light go up, hit the center of the mirror, and bounce back to hit Observer 1. But for that, the light had to be moving diagonally with respect to him.

He see's the light move over a larger distance: 2 * (h^2 + d^2)^(1/2)

But the thing is, he also measures the speed of light to be c! So, the time it took for the light to travel that distance is [2 * (h^2 + d^2)^(1/2)]/c, which is a longer time than Observer 1 measured.

Observer 2 measured more time than Observer 1. This is time dilation.

They say "moving clocks run slow" because Observer 1 (the moving clock) measured less time than Observer 2 (the non-moving clock).

You're thoughts?
Does this shed any "light" on the matter? :P

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I think i understand your question. Correct me if I'm wrong, but here's what I'm getting:

The observer emits a "pulse of light" from his flashlight to the center of the mirror ("straight upward"). The light goes up. Since the mirror is moving with a velocity (let's say v), the light pulse should hit at a spot vt away from (or "behind") the center of the mirror, where t is the time it takes for light to travel from floor to ceiling.

But, a contradiction! According to Observer 1 (who is moving with a constant velocity so that he can't even tell that he's moving), he emitted a pulse "straight up" and it should also come "straight down," right back at his flashlight.
no i dont agree ..
here what is wrong and i said about 200 times but seems everybody ignore it YOU said:
he emitted a pulse "straight up" and it should also come "straight down," right back at his flashlight.
yes it will come straight down but not back to his flashlight
this observer emitted a flashlight straight upward moving with velocity C
observer 1 is in the train.
observer 2 is outside.
[PLAIN]http://img686.imageshack.us/img686/8618/imgwkv.jpg [Broken]

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yeah another way of explaining it(AFTER READING ABOVE POST):
if u say that according to observer1 the mirror is not moving away from him and so is the light
he sees the light moving upward in c and downward again and that what really happens as he cant see that light returned to flash light while it didn't
but since the frame was moving with velocity vtrain then light should be moving upward with velocity c and moving diagonally with velocity v train and that s impossible

Dale
Mentor

If the pulse of light is aimed vertically in the frame where the flashlight is moving to the right then it is aimed to the left in the frame where the flashlight is stationary. You cannot have it be vertical in both frames.

yeah another way of explaining it(AFTER READING ABOVE POST):
if u say that according to observer1 the mirror is not moving away from him and so is the light
he sees the light moving upward in c and downward again and that what really happens as he cant see that light returned to flash light while it didn't
but since the frame was moving with velocity vtrain then light should be moving upward with velocity c and moving diagonally with velocity v train and that s impossible
Observer1 sees the light moving straight up and down with speed "c".
Observer2 sees the light "aberrated", i.e. moving at an angle in the direction of vtrain. Light in the frame of Observer2 moves at "c" just the same.

jtbell
Mentor

no i dont agree ..
here what is wrong and i said about 200 times but seems everybody ignore it
No, you are either ignoring or misunderstanding what everybody else has said in this thread. I don't see any sign that either "side" is going to budge in this discussion, so there's no point in continuing it.