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Einstein's original 'thought experiment'

  1. Nov 11, 2011 #1
    I'm in the middle of reading 'The Great Equations' by Robert P. Crease (good book) and I've just got to the chapter on E=mc^2. The chapter covers Einstein's original thought experiment. You know, the one with the mirrors and the stationery / moving observers. I've worked through the maths associated with the experiment and as you'd expect the suggested formula does fall out of the end.

    Many, many years ago, at school, we were always taught to feed the results of any calculations back into the starting point of the calculation to make sure that they make sense. Here the starting point is the diagram used to introduce the starting values. Feeding V=0 back into the mirror's experiment is straight forward enough, but what happens when you plug V=c back in ?

    The default presentation of the diagram associated with this experiment already appears to represent the situation where V=c. ie. twice the distance between the mirrors, matches the distance travelled by the moving observer. This being the case, the original diagram represents the limits of the experiment, where V=c, and the distance travelled by the moving observer can never be more than the distance shown on the diagram.

    If this is the case, then it seems to me that the limit of the time dilation factor, that can be derived from this thought experiment, is d/sqrt(2d^2), d/sqrt(2)d, 1/sqrt(2), or approx. 0.71. A smaller value would necessitate half the distance travelled by the moving observer to be greater than the distance between the mirrors. For this, the traveller would have to travel faster than c.

    This obviously doesn't agree with the final formula. Where am I going wrong ? How would V=c be represented in the diagram ?
     
    Last edited: Nov 11, 2011
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  3. Nov 11, 2011 #2

    ghwellsjr

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    It sounds to me like the diagram is depicting a speed of 0.707c which what you get if the distance the mirrors travel is double the stationary distance between the mirrors. In that case the distance the light travels going from one mirror to the other and back to the first is 2d√2 where d is the distance between the mirrors. Gamma in this case will be √2 and the time dilaiton factor will be its reciprocal or .707 which is the answer you got.

    I think you might be confusing the diagonal distance the light travels with the distance the mirrors travel
     
  4. Nov 11, 2011 #3
    No. The diagram depicts the paths of light as observed by a stationery and a moving observer. The base of the triangle represents the distance the mirror travels and the height of the triangle, the distance between the mirrors. The diagonal is the distance that the light travels in the stationery observer's frame of reference. The height over the diagonal is the time dilation factor.

    For the base to equal twice the height, the velocity of the mirror must be c.
     
  5. Nov 11, 2011 #4

    ghwellsjr

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    Is the diagonal at a 45 degree angle? And didn't you calculate 0.707 as the time dilation factor? And isn't the distance the mirror travels equal to the distance between the mirrors and isn't the distance the light travels the length of the diagonal? And isn't the speed of the mirrors equal to the distance the mirror travels divided by the distance the light traveled in the same period of time and isn't that speed 0.707c?
     
  6. Nov 11, 2011 #5
    No ?

    The angle is 45 degrees, if the distance the mirrors travel is (twice) the distance between the mirrors.
    The time dilation factor is 0.707, if the distance the mirrors travel is (twice) the distance between the mirrors.
    The distance the mirror travels is (twice) the distance between the mirrors, only if V=c.
    The distance the stationery observer measures the light travel is the sum of the diagonals.
    The distance the moving observer measures the light travel is twice the height.

    but the speed of the mirrors isn't the distance the mirror travels divided by the distance the light travelled. It's the distance the mirrors travelled over time (t). Given as V in the experiment. The speed of the mirrors is entirely represented by the baseline of the traingle, nothing to do with the diagonals ?


    Mike.
     
  7. Nov 11, 2011 #6

    ghwellsjr

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    In the rest frame of the mirrors, they are not traveling at all. The only speed you can consider is the speed of the mirrors in the stationary frame. Although speed is distance divided by time, we're trying to find out what time to use. That's the whole point of the exercise. So instead, we calculate the speed of the mirrors as a fraction of the speed of light in the stationary frame. And this will also be the same ratio as the distance the mirrors move divided by the distance the light travels, since the time is the same in both cases.

    So, in the stationary frame:
    "The distance the mirror travels is (twice) the distance between the mirrors"​

    And:
    "The distance the stationery observer measures the light travel is the sum of the diagonals."​

    Therefore the speed (as a fraction of the speed of light) that the mirrors moves is the ratio of these two distances which is 0.707c, correct?
     
  8. Nov 12, 2011 #7
    Ah. Yes. Sorry. I see what you're saying and that is the original explanation. However, if you now consider the position from the point of view of the stationery observer, he can measure the speed of the moving observer as a maximum of c. If the moving observer is observed as travelling at c (or close to c), the base of the triangle now represents the speed of the moving observer and the height, the distance between the two mirrors. We can add two stationery mirrors, the same distance apart as the moving mirrors ( no distance dilation perpendicular to the direction of motion ) and send a signal between them. This signal, travelling at the speed of light should return in the time it takes the moving observer to travel twice the distance between the mirrors, both travelling at c, all as observed by the stationery observer.

    The stationery observer, must now measure the speed of the moving observer as c (given), the speed of the signal between the mirrors as c (it's light) and the distance both have travelled as identical. This gives us a maximum time dilation for the moving observer, as measured by the stationery observer, of d/sqrt(2d^2), or 1/sqrt(2), or 0.707. The stationery observer, using this particular experiment, cannot measure the moving observer's time dilation as any lower than 0.707, but the derived formula suggests otherwise.



    Mike.
     
  9. Nov 12, 2011 #8

    ghwellsjr

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    But if you're going to have the moving mirrors travel at close to c, then in the time that the light has made the round trip between the stationary mirrors, the moving mirrors will be close to double that distance away from the stationary mirrors and the light that needs to go between the two moving mirrors will have made practically no progress in the direction of the second moving mirrror but will rather be essentially traveling along with the first mirror, in the same direction as the motion of the first moving mirror. It will take a very, very, very long time for the light to make its way up to the second mirror. The triangle that depicts the path of the light that bounces between the two moving mirrors will look almost like a flat line, having a base that is soooooo long and a height by comparison that is so small (it's still the same height as the distance between the stationary mirrors) that the three angles making up the triangle are something like 0.000000000000001, 179.999999999999998, and 0.000000000000001.

    In the original situation, you had a triangle with a base of 2, a height of 1, and a pair of diagonals of length 1.414. The light traveled a total distance of 2.828 and the mirrors traveled a total distance of 2 so the speed was 2/2.828 or 0.707c, correct? And the time dilation factor was the time for the light to make the trip between the stationary mirrors, 2, divided by the time it takes for the light to make the trip between the moving mirrors, 2.828, making the time dilation will be 0.707, correct?

    Why don't you next think about a triangle with a base of 4 and a height of 1. Each diagonal will be 2.236 [√(22+1)=√5], correct? This makes the total path for the light to travel be 4.472, correct? And the mirrors will have traveled 4, correct? And their speed will be 4/4.472 = 0.894c, correct? And since the time dilation factor is the time for the light to make the trip between the stationary mirrors, always 2, divided by the time it takes for the light to make the trip between the moving mirrors, 4.472, the time dilation will be 0.447, correct?

    You could keep doing this for longer and longer bases for the triangle but you will never get a situation where the limit of the time dilation reaches zero or the speed of the mirrors ever reaches exactly c, correct?
     
  10. Nov 12, 2011 #9
    None of this alters the fact that if the stationery observer bounces a light signal between two mirrors whose distance is the height of the triangle, he can't measure the moving observer as travelling further than twice the distance between the mirrors. This still gives a minimum time dilation factor of 0.707 ?

    Is there a reasonable version of the triangle that represents a situation where the distance along the base is more than twice the height ? ... without exceeding the speed of light ?

    In fact, as you could have two stationery mirrors the same distance apart as the two moving mirrors, the stationery observer could bounce a signal between the two stationery mirrors and the moving observer would see the signal travel the diagonals of a triangle and calculate that the stationery observer's (moving in the moving observer's frame of reference) time was dilated.

    If both of the observers are carrying out the same experiment, at the same time, whose time is dilated ?

    The question is who's moving and who's clock is running slower ?
     
    Last edited: Nov 12, 2011
  11. Nov 12, 2011 #10

    ghwellsjr

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    What's wrong with the version I just gave you?

    You get to decide who's moving and that will be the one who's clock is running slower.
     
  12. Nov 12, 2011 #11
    I've read Einstein's 1905 paper and this thought experiment is not mentioned in there. Einstein was contemplating Maxwell's equations.
     
  13. Nov 14, 2011 #12
    Ok. Going back to the original experiment, all results are viewed from the stationery observer and his view of what is happening. The longer path for the light as measured by the stationery observer is generated purely by the fact that the source of the light is moving.

    If we take it in stages, if the 'moving' observer were stationery, the two paths would be identical, no problem there.

    If the observer were moving at 1/2 c, he'd travel (along the base) 1/2 the distance between the mirrors before the signal hits the mirror. This generates a longer path for the light, giving the time dilation as height/hypotenuse.

    At 3/4c, the moving observer would travel 3/4 distance between the mirrors before the signal hits the mirror.

    At c (or as close as we can get), the moving observer will travel the full distance between the mirrors, before the signal hits the mirror.

    Do you agree ?
     
  14. Nov 14, 2011 #13

    ghwellsjr

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    I'm not sure which mirror you're talking about. Here's how I would say it:

    If the observer were moving at 1/2 c, he'd travel (along the base) 1/2 the distance between the stationary mirrors when the signal hits the stationary mirror at which point the light traveling toward the moving mirror is not quite there yet.

    At 3/4c, the moving observer would travel 3/4 distance between the stationary mirrors when the signal hits the stationary mirror at which point the light traveling toward the moving mirror is a little more than half way there.

    At close to c, the moving observer will travel almost the full distance between the stationary mirrors, when the signal hits the stationary mirror at which point the light traveling toward the moving mirror has barely made any progress at all because the angle at which it is approaching the moving mirror is just a hair over zero and practically all of its progress is in the horizontal direction with essentially none of it in the vertical direction.
     
  15. Nov 14, 2011 #14
    Sorry. No. I've confused the issue ...

    Taking the original thought experiment, there are only the moving mirrors. The original argument goes, that the moving observer would see a signal travel directly back and forth between the mirrors, but the stationery observer would interpret the path travelled by the signal as longer than the moving observer.

    The way the experiment is described, the signal always returns fully to the final mirror in both frames of reference. It has to, so that the stationery observer knows what it is he's comparing. It's just the difference in length of the stationery observer's observed route for the signal that affects the results. It travels further than the moving observer would measure it travelling, so it takes longer. Both must measure it's speed as c, so time is different.

    The experiment then skips straight onto deriving the Lorenz factor, pulling values directly from the experiment to plug into Pythagoras' theorum. The result provides possible ranges of values for V of 0 thru c.

    However, plugging a V=c into the original setup, gives us a lateral movement, in the stationery observer's frame, equivalent to the distance between the mirrors. By the time the signal reaches the mirror, the moving observer would have travelled for the same amount of time at V=c. The signal's travelling at c, so's the moving observer, so you end up with a 45 degree, right-angled triangle.

    Applying the original logic for calculating the time dilation. ie. the distance between the mirrors, over the distance travelled in the stationery observer's frame, you get a minimum time dilation factor of 0.707.

    The original experiment, as originally described, can't result in a value less that this 0.707. To do so, the signal sent between the mirrors, perpendicular to the direction of motion, would have to travel further laterally, than the distance between the mirrors. To do this, the moving observer would need to travel faster than c.


    Where's the flaw in my logic ? ... and I use the term 'logic' loosely :-)
     
  16. Nov 17, 2011 #15
    I was hoping that someone would post a reply confirming that my question was nonsense, because ...

    http://en.wikipedia.org/wiki/File:Time-dilation-002.svg

    No matter how many times I look at this experiment, I can see no way that the time dilation calculated can be a value less than 0.707 (1/sqrt(2)).

    Please can someone provide an example of how this graph would look for time dilation values below this figure ?

    Thanks.
     
  17. Nov 17, 2011 #16

    ghwellsjr

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    Are you are thinking that if the light follows two diagonal paths that it couldn't also follow a straight path to the right? You've already considered the light going straight up and down for the stationary observer and his stationary mirrors at the same time that more light is following the diagonal paths. Now suppose you think of some more light going in the same direction as the moving observer. So pretend those two diagonal lines are sticks with a hinge at the top and push them down so they are flat. Now the distance is the same, 2D, but it is longer than the distance the mirror and the observer have moved in the same period of time, correct? So in a given time, the observer has moved the distance shown in the picture but the light going in the same direction is way ahead of mirror C, correct? That's how we calculate the speed of the observer with respect to the speed of light. And no matter how far we move C away from A (with B half way between), the distance the light travels following the two diagonals will always be the same as the distance another light pulse would travel if it went in the direction of the observer and that will always be farther than the observer will travel in the same period of time.
     
  18. Nov 21, 2011 #17
    Thanks for the response ghwellsjr.

    I agree that if you squash the two diagonals along the x-axis, they'd be 2D and longer than the 2L that the stationery observer measures. That's the basis for the experiment and the results.

    My concern is that as the mirrors can only be moving at a maximum of c in the stationery observers frame, the distance between A and C can be no more than 2L, unless the moving observer can travel faster than c.

    If the base of the diagram can be no more than 2L, D is limited to sqrt( L^2 + L^2 ), which gives a maximum dilation factor of 1/sqrt(2). This doesn't agree with the result of the experiment.

    The 1/2 v t1 at the base can't be greater than 1/2 c t1, which is the same as L ?
     
  19. Nov 21, 2011 #18
    I also would like to know what M1keh is talking about; is there an online reference perhaps?
     
  20. Nov 21, 2011 #19

    ghwellsjr

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    You keep comparing 2D and A-C to 2L which I agree they will be longer than 2L but that's not what you need to compare. You need to compare the length of 2D to A-C. 2D is always longer than A-C. 2D is the distance the light travels when the mirrors travel A-C. When you have a flattened triangle, that is, when 2D and A-C are many times greater than 2L, the ratio of A-C to 2D can approach 1 which means the speed of the mirrors can approach the speed of the light. Nothing happens at 0.707c to prevent the mirrors from going any faster.

    When you are considering what happens with the light for the stationary mirrors, the light goes directly up and down and the distance the light travels, 2D is equal to 2L while the A-C distance is zero (since the mirrors are not moving). When you are considering what happens to the light for the moving mirrors, the distance the light travels, 2D is now longer than 2L, we don't care about 2L anymore except to use the Pythagorean theorem to calculate the distance 2D as a function of the distance A-C. And no matter how long A-C is compared to 2L, the distance the light travels, 2C will always be more than the distance the mirrors travel, A-C.

    So please go back and look at my examples of A-C being longer than 2L and see if they now make sense to you.
     
  21. Nov 21, 2011 #20

    ghwellsjr

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    M1keh is thinking that since the distance the light travels for the stationary case is 2L, that that is the limit of how far the mirrors can move in order to not exceed the speed of light.
     
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