Error in special relativity albert einstein assumed

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thetrice
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Error in special relativity
albert einstein assumed (in the time dilation) a box-car moving with velocity V with a mirror in its ceiling and observer 1 sitting in the box car holding flash light vertically so when he sends a flash pulse , this pulse will go vertically , hit the mirror , and return back to the observer 1 so delta time = 2d/c.
while according to observer 2 who is outside the box car what he have seen is that the flash light starts from observer 1 old position , and ends in observer 1 new position . (where difference in the distance is (delta time)*V) so according to observer 1 the light pulse has moved a longer distance than 2d.
where delta time can be calculated afterward using pythogeran therom.
and here lies the error
that pulse went vertically , hit the mirror , and returned back , it didnt return back to observer 1 but it returned back to its same starting position as y vector of velocity of this light pulse is equal to c , while the x-vector is equal to zero(not affected by V of the box-car, and it has to be un-affected because if x-vector of the light pulse velocity has a value then the total V of the light will be more than c and its a fact for observer 1 and 2 that light velocity is c)
so since x-vector is zero the light didnt change its x-coordinate so when light pulse returned back to its position (which was observer 1 oldpostion) the distance between the light pulse and observer 1 after delta time is (delta time * V) since the box car moved.
while according to observer 2 the light pulse starts from observer 1 old position , and ends in observer 1 old position so simply delta time according to observer 2 the light pulse moved 2d
so to both observer the light pulse moved the same distance. with the same velocity so equal delta time.
This not an argue , i just wantto know whats wrong in the above...so i am not gonna argue just asking to understand.
 
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Answers and Replies

  • #2
Doc Al
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Why do you think the light doesn't return back to observer 1? Pretend observer 2 isn't even there. Someone in the box car is just reflecting the light off the (very tall) ceiling; to him, the light just moves vertically. Everything works just like you'd think, regardless of the box car's motion.

Now add observer 2; Clearly observer 2 sees the light move sideways, since the boxcar is moving.
 
  • #3
Fredrik
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I think you need to keep in mind that no matter whose point of view you're using to describe what happens, the description must include the same events. In this scenario there are three events: 1. The flashlight emits light. 2. The light gets reflected. 3. The reflected light hits the flashlight.

When you describe these three events from observer 1's point of view, the light is going straight up and then straight down, and when you describe the same three events from observer 2's point of view, it isn't.

The "up" component of the velocity of light in observer 2's coordinates is less than c. If it had been c, the speed of light would have been greater than c.
 
  • #4
thetrice
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as i said , the light being reflected by the moving mirror(due to the car motion) doesnt affect the position of the x-cordinate of the light it will just reflect it perpendicular .
and the velocity of light is equal to c according to observer 1 and 2 since the direction of the light in both cases is downward BTW i am not talking in the case where light particles moving with the box-car (as it shouldnt) and that what i stated above that why do u think light particles move with the boxing car and when it get reflected back it comes to a new position(delta time * V) , untrue because from where did the light particles gain horizontal velocity to move relative to the boxing car?
 
  • #5
JesseM
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as i said , the light being reflected by the moving mirror(due to the car motion) doesnt affect the position of the x-cordinate of the light it will just reflect it perpendicular .
and the velocity of light is equal to c according to observer 1 and 2 since the direction of the light in both cases is downward BTW i am not talking in the case where light particles moving with the box-car (as it shouldnt) and that what i stated above that why do u think light particles move with the boxing car and when it get reflected back it comes to a new position(delta time * V) , untrue because from where did the light particles gain horizontal velocity to move relative to the boxing car?
Light always has the same speed regardless of the motion of the emitter, but the velocity (which includes direction) does depend on the motion of the emitter. This follows from the first postulate of relativity, which says the laws of physics work the same way in all frames. So, if you have a laser emitter standing vertically on the floor of a train car, and when the train is at rest relative to the ground the beam goes straight up and hits a point on the ceiling directly above the laser, then the beam still has to hit the same point on the ceiling if the train car is moving at constant velocity relative to the ground. If you're in a windowless train car that's moving inertially (not accelerating), any experiment you do inside the car has to give the same results regardless of how the train car is moving relative to outside landmarks; that's just an application of the first postulate.
 
  • #6
Matterwave
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as i said , the light being reflected by the moving mirror(due to the car motion) doesnt affect the position of the x-cordinate of the light it will just reflect it perpendicular .
and the velocity of light is equal to c according to observer 1 and 2 since the direction of the light in both cases is downward BTW i am not talking in the case where light particles moving with the box-car (as it shouldnt) and that what i stated above that why do u think light particles move with the boxing car and when it get reflected back it comes to a new position(delta time * V) , untrue because from where did the light particles gain horizontal velocity to move relative to the boxing car?

If you assume this, then there IS a preferred reference frame wherein there is absolute rest. This frame can be found as the frame in which a light shone straight up, will come straight down. So, if I were on the ground, i could tell that I was preferred, because if I were in the boxcar, the light would look like it's drifting. This is directly contrary to the postulate of relativity.
 
  • #7
Doc Al
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as i said , the light being reflected by the moving mirror(due to the car motion) doesnt affect the position of the x-cordinate of the light it will just reflect it perpendicular .
From the view of observer 1, the mirror doesn't move.

and the velocity of light is equal to c according to observer 1 and 2 since the direction of the light in both cases is downward BTW i am not talking in the case where light particles moving with the box-car (as it shouldnt)
From the view of observer 1, the light just bounces vertically up and down. Of course the light moves with the boxcar.

and that what i stated above that why do u think light particles move with the boxing car and when it get reflected back it comes to a new position(delta time * V) , untrue because from where did the light particles gain horizontal velocity to move relative to the boxing car?
The light gained horizontal velocity due the motion of the car. Since the speed of light is constant from any frame, that implies that the vertical component of the light's velocity is less than c from observer 2's frame. A beam of light that is vertical in one frame, appears slanted from a horizontally moving frame.
 
  • #8
thetrice
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Light always has the same speed regardless of the motion of the emitter, but the velocity (which includes direction) does depend on the motion of the emitter. This follows from the first postulate of relativity, which says the laws of physics work the same way in all frames. So, if you have a laser emitter standing vertically on the floor of a train car, and when the train is at rest relative to the ground the beam goes straight up and hits a point on the ceiling directly above the laser, then the beam still has to hit the same point on the ceiling if the train car is moving at constant velocity relative to the ground.
well, that cant be because if the train moving with constant velocity , and light or laser moving vertically with velocity c so to hit the same point as if the train wasnt moving it has to have also velocity in the x direction (let it V) so total velocity of the light will be sqrt(v^2 + c^2) so greater than c . and thats impossible
the light cannot move vertically with speed c and move horizontally relative to the train because its velocity then will be greater than c.
@ dr al said "that implies that the vertical component of the light's velocity is less than c from observer 2's frame. A beam of light that is vertical in one frame, appears slanted from a horizontally moving frame."
yes observer 2 will see the vertical component of light is less than c, BUT THAT IF THE LIGHT IN FIRST PLACE WAS REFLECTED BACK IN THE NEW POSITION(V*DELTA TIME) but that what i am talking about that light not be reflected to new position but same position( as stated above light cannot gain velocity in x direction while moving vertically with speed of c)
 
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  • #9
JesseM
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well, that cant be because if the train moving with constant velocity , and light or laser moving vertically with velocity c so to hit the same point as if the train wasnt moving it has to have also velocity in the x direction (let it V) so total velocity of the light will be sqrt(v^2 * c^2) so greater than c . and thats impossible
But that's the point, in the frame where the light is moving diagonally, it moves slower than c in the vertical direction, which is how you can derive the time dilation equation via the idea of a light clock. In the frame where the train is moving, the vertical speed v1 will be less than c, and the horizontal speed v2 (which is just the velocity of the train) is also less than c, such that sqrt(v12 + v22) = c.
 
  • #10
thetrice
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plz read the post above :) , as i said that the light cannot move diagonally in the first place as when it will reflect back it will reflect in the same position it started, so it will appear to observer1 as moving vertically not diagonally
 
  • #11
JesseM
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plz read the post above :) , as i said that the light cannot move diagonally in the first place as when it will reflect back it will reflect in the same position it started, so it will appear to observer1 as moving vertically not diagonally
In the frame of observer1 on the train it moves vertically, in the frame of observer2 on the ground who sees the train in motion it moves diagonally. What's the problem with this? It would also be true that if observer1 tossed a ball vertically in the air it would move diagonally in the frame of observer2, although in that case the speed would not be the same in both frames.

Is your argument that if its vertical speed is less than than c in observer2's frame, then it must also be less than c in observer1's frame? If so then you're forgetting about time dilation, which says the time that observer1 measures for the light to go from flashlight to ceiling is less than the time in observer2's frame, because in observer2's frame the clock of observer1 is running slow. Please read the light clock link I gave above if this isn't clear.
 
  • #12
Matterwave
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I think you are still positing a notion of absolute rest, or an aether. Both ideas are unnecessary in the context of special relativity.

If there is no absolute rest, there is no "it must be bounced back vertically". This is because "vertical" is different for the two observers.
 
  • #13
Dale
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this pulse will go vertically , hit the mirror , and return back to the observer 1 ...
that pulse went vertically , hit the mirror , and returned back , it didnt return back to observer 1 but it returned back to its same starting position
Think about this logically. How can the light pulse logically both return back to observer 1 and not return back to observer 1? This is a clear contradiction, your logic is unsound. Suppose observer 1 is a photodiode which triggers a bomb. Does the bomb explode or not? It cannot possibly both explode and not explode.
 
  • #14
thetrice
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thats because the light will not be reflected back to the flash light but it will be reflected back to a postion = postion of the flashlight- (V*delta time)
 
  • #15
thetrice
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it seems you dont read my posts as i explained why should light start at pos (x,y) and return back to pos(x,y) while when the car and the observer 1 will be in pos(x+(t*V) ,y)
and observer2 didnt even move so basically
observer 2 and the light didn't move from their position and observer 1 will see light bouncing up and reflecting back but not in the same position.
even if you assume that light reflected back from the mirror again to the flash light(which is impossible) observer 2 will not assume then that light moved diagonally , he will assume that light changed it's position as it was moving with train and observer1(which is again impossible ) and both will have the same delta time!
 
  • #16
thetrice
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jeesm , the train will move horizontally or even diagonally according to observer2 but not the light , as the light have nothing to do with moving train and mirror as it will just start from postion1 and back to it(as i said million times above) and will not return to a new position(which was assumed x +v*t) as light moved vertically and back again(no velocity in x direction) so:
to observer1: light moved vertically upward and returned back not in the same position as observer 1 has moved a little bit(which equal to v*t) while observer 2 will see light bouncing up and down normally.
 
  • #17
Fredrik
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thats because the light will not be reflected back to the flash light but it will be reflected back to a postion = postion of the flashlight- (V*delta time)
I don't know why you think so, but it's wrong. You're supposed to describe the same sequence of events from two different points of view. If observer 1's description includes the three events I listed in my previous post, then so does observer 2's description.

even if you assume that light reflected back from the mirror again to the flash light(which is impossible)
Again, both descriptions must include the same events, so if (and only if) this is impossible in observer 2's description, then it's impossible in observer 1's description.

to observer1: light moved vertically upward and returned back not in the same position as observer 1 has moved a little bit(which equal to v*t) while observer 2 will see light bouncing up and down normally.
Now you seem to be describing a different sequence of events than you did at first. That's fine, as long as you make sure that both descriptions include the same events. So for example, you can consider the scenario where observer 2 (on the ground outside the train) sees light go straight up and straight down. This scenario includes the following events:

1. Light is emitted from a flashlight on the floor near the front of the train car.
2. Light is reflected by a mirror in the ceiling near the middle of the train car.
3. Light reaches the floor near the rear of the train car.

So in this scenario observer 1 would have to describe the light as going "diagonally".
 
  • #18
Dale
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thats because the light will not be reflected back to the flash light but it will be reflected back to a postion = postion of the flashlight- (V*delta time)
it seems you dont read my posts as i explained why should light start at pos (x,y) and return back to pos(x,y) while when the car and the observer 1 will be in pos(x+(t*V) ,y)
I read and understood your posts. The problem is that your posts are illogical (self contradictory). You have specifically proposed that in one frame the light pulse returns to the flashlight, and in another frame it does not. Remember that the two different frames are simply two different and equivalent ways of describing the same situation. It is simply not logically possible for something to both happen and not happen. Just think about the implications of what you have proposed. It should be clear that if the pulse returns to the flashlight in one frame it must do so in all frames.
 
  • #19
Ich
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and it has to be un-affected because if x-vector of the light pulse velocity has a value then the total V of the light will be more than c and its a fact for observer 1 and 2 that light velocity is c
You didn't understand the whole point of the example. As there clearly must be an x-component to the light speed, it follows logically that the y-component must be smaller acordingly, so that the speed remains c. That's a valid derivation of time dilatation.

If I'm allowed a personal remark: If you, as a layman, get to clearly nonsensical results by trying to follow a popular account of time dilatation, it would be more appropriate to title your post "An error in my understanding?".
 
  • #20
Matterwave
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@thetrice: In your scenario, observer 2 is obviously favored. You remark again and again that the light will "come back to the same position". But "the same position" is only from observer 2's point of view! Special relativity says that if you are traveling with uniform velocity relative to another observer, you couldn't tell whether it was you moving towards them, or them moving towards you; both views are equally valid. If you take this postulate as fact, there is no reason to think that observer 2 is favored, and the light must work according to his perspective. The light must actually go back to the flashlight because it was emitted by observer 1's flashlight!
 
  • #21
Janus
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as i said , the light being reflected by the moving mirror(due to the car motion) doesnt affect the position of the x-cordinate of the light it will just reflect it perpendicular .
and the velocity of light is equal to c according to observer 1 and 2 since the direction of the light in both cases is downward BTW i am not talking in the case where light particles moving with the box-car (as it shouldnt) and that what i stated above that why do u think light particles move with the boxing car and when it get reflected back it comes to a new position(delta time * V) , untrue because from where did the light particles gain horizontal velocity to move relative to the boxing car?

The biggest problem with this viewpoint is that experiment has shown it to be wrong!.

An equivalent experiment has actually been done and the result was that the light does return to the flashlight regardless of any "motion" it has. You can't argue the light "should" behave one way when the actual evidence says it behaves another.
 
  • #22
thetrice
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@thetrice: In your scenario, observer 2 is obviously favored. You remark again and again that the light will "come back to the same position". But "the same position" is only from observer 2's point of view! Special relativity says that if you are traveling with uniform velocity relative to another observer, you couldn't tell whether it was you moving towards them, or them moving towards you; both views are equally valid. If you take this postulate as fact, there is no reason to think that observer 2 is favored, and the light must work according to his perspective. The light must actually go back to the flashlight because it was emitted by observer 1's flashlight!
yes the light will return to the same position according to observer 2 as both are not moving(x-axis axis of motion where y-axis is along gravity) while observer1 (in the train)is the one moving so he will not see light returning to the same position.
@janus: what experiment?? (any link or name??)
 
  • #23
thetrice
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You didn't understand the whole point of the example. As there clearly must be an x-component to the light speed, it follows logically that the y-component must be smaller accordingly, so that the speed remains c. That's a valid derivation of time dilatation.

If I'm allowed a personal remark: If you, as a layman, get to clearly nonsensical results by trying to follow a popular account of time dilatation, it would be more appropriate to title your post "An error in my understanding?".
PLEASE READ THE POSTS:
no the speed in y direction wont be smaller as when calculated we substitute vY with c and vx with zero because if we made a train move with a v nearly to c then oobserver1(in the train) will surely realize that light didn't reflect and return back to the same position but rather was left back by v(of train)*delta time
and thats the problem einstein assumes that observer1 will see the light falling on the flash light(and that wont happen since observer1 with flash light will move distance v*t while light being reflected back)
 
  • #24
thetrice
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I read and understood your posts. The problem is that your posts are illogical (self contradictory). You have specifically proposed that in one frame the light pulse returns to the flashlight, and in another frame it does not. Remember that the two different frames are simply two different and equivalent ways of describing the same situation. It is simply not logically possible for something to both happen and not happen. Just think about the implications of what you have proposed. It should be clear that if the pulse returns to the flashlight in one frame it must do so in all frames.
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
while observer 2 will see light moving up and down
so simply observer 1 and 2 see the same 3 events but 1 see it different as he is moving with v
 
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  • #25
thetrice
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here is what i meant:(SOORY FOR BAD HAND WRITING):http://img693.imageshack.us/img693/9056/imgkqn.jpg [Broken]
 
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  • #26
JesseM
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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?
 
  • #27
Dale
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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 [itex]v_t[/itex] 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.
 
  • #29
I_am_learning
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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!
 
  • #30
Tuomaaca
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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.

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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  • #31
thetrice
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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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  • #32
thetrice
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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
 
  • #33
Dale
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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.
 
  • #34
starthaus
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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

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.
 
  • #35
jtbell
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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.
 

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