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Variable speed of light

  1. Nov 11, 2003 #1
    [SOLVED] variable speed of light

    OK people, I'm sure you'll find this one easy. For myself, I have been working on it for a while now and am not getting far enough.

    My work colleague has posed me this problem:

    how can you fit a 5m(eter) car into a 3m garage?

    Now, I know this has something to do with the variable speed of light. I know that the speed of light varies in a gravitational field. And I know that the physical dimensions of bodies vary when these are plunged into a gravitational field.

    However, that's about all I know so far.

    And my work colleague is soooo smug.

    Then I found you...

    Please help me challenge his smugness.
     
  2. jcsd
  3. Nov 11, 2003 #2

    chroot

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    Light actually does not vary in speed. All observers will always measure light as moving through their local frame at c.

    What your friend's question involves is actually a phenomenon called length contraction. You see, the length of an object (like the car) is not a fixed quantity. Different observers moving at different relative speeds to the car will actually measure its length differently.

    Let's say you're standing still beside the 3 m garage. A physicist would say you are using the garage as your 'rest frame,' and you are at rest in it. The garage's length, 3 m, is called its 'proper length.' The car's proper length is 5 m -- this means that a person standing still beside it would measure it as being 5 m in length.

    If you propel the car to a high enough velocity, it is possible that you, standing beside your garage, will measure the car's length as being shorter than its proper length of 5 m.

    The equation governing this effect (called length contraction) is this:

    l = l0 / γ

    where l0 is the proper length of the car, and l is the measured length of the car.

    γ (lowercase Greek gamma) is a unitless number that is always greater than or equal to one. When you are at rest with respect to the object you're measuring, γ = 1, and you will measure its length as its proper length.

    γ can be found by plugging in v into the following definition:

    γ = 1 / sqrt(1 - v2 / c2)

    To find the velocity the car must have relative to the garage to cause an observer standing beside the garage to measure its length to be 3 m, you must solve the equation:

    3 = 5 / γ

    3 = 5 * sqrt(1 - v2 / c2)

    v = (4/5) c, or four-fifths the speed of light.

    If you drive your 5 m (proper) car at four-fifths the speed of light with respect to your 3 m (proper) garage, it will fit inside it -- before promptly crashing through the back wall!

    - Warren
     
  4. Nov 11, 2003 #3

    jcsd

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    Nothing to do with VSL theories (which are a far from conventional) or gravitational fields even, it's special relativty and length contraction, which is given by the following equation:

    L = L0√(1 - v2/c2)

    Solving for L = 3 and L0 = 5, we find that v = 0.8c. So in other words: to fit a 5m car into a 3m garage the car must be travelling at 4/5 the speed of light in a vacuum (relative to the garage).

    edited to add: looks like chroot beat me to it!
     
    Last edited: Nov 11, 2003
  5. Nov 12, 2003 #4
    Hey, Thanx people! Now that's what I call service.

    Remind me to visit this board more often.
     
  6. Nov 12, 2003 #5
    Warren,

    If you get the car going fast enough to measure its length as 3m instead of 5m,and don't cra**** into the garage, how long is the car after you slow it back down to a halt in your rest frame?

    -zoob
     
  7. Nov 12, 2003 #6

    jcsd

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    5m of course, length contraction doesn't deform the object. Interestingly, due to the failure of simultaneity at distance, though an observber in the rest frame of the garage will see both the front and back end of the car in the garage at the same time, an observer in the rest frame of the car will not.
     
  8. Nov 15, 2003 #7
    JCSD,

    This being the case (which makes perfect sence to me) why is it that people claim a clock that is measured to be running slow when moving past an observer, will actually be found to have lost time when slowed down and returned to the observer's rest frame (which doesn't make sence to me)?
     
  9. Nov 15, 2003 #8

    Ivan Seeking

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    Hey Zooby, I thought I would jump in since I'm here.
    I suspect that you don't really believe the statements above. From the frame of the garage, the car really is shorter. Next, if observers in both frames of reference must observe light to travel with speed C, and since we know the length contraction is given by

    L = L0√(1 - v2/c2)

    and we know that for any observer, C = L0/t0 = L/t,

    it can easily be shown that

    t = t0√(1 - v2/c2)

    Therefore just as length contracts, observers in the frame of the garage will see clocks running slowly in the car. Again, this is not just an illusion; this is real. Now, it gets interesting since for an observer in the car, the garage is in motion and the car is at rest. So, an observer in the car sees clocks in the length contracted garage running slowly. Whose clocks are right? Both. When someone accelerates, in this case when the car comes to a stop and comes back, we have chosen a preferred observer - the garage - and we find that the clocks in the car have lost time. It we speed up to catch the car, assuming we started out in motion with the car and synchronized our clocks, we would find that the clocks in the frame of the garage have lost time. Again, the one who changes their state of motion is the one whose clocks have lost time. Note that in reality, the presence of gravity complicates this situation.
     
  10. Nov 15, 2003 #9
    If the clock in the car comes back into the rest frame of the garage having lost time, why hasn't the car lost length?
     
  11. Nov 15, 2003 #10

    Ivan Seeking

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    When the car is in motion, it is shorter in the frame of the garage. Likewise, from the frame of the garage, the car's clocks are running more slowly. When the car stops, that is, when the frame of the car coincides with the frame of the garage, the two lengths L and L0 agree. Likewise, it we compare the ticks of the clocks, we find that again they agree – they occur at the same rate. However, and this was a key test of relativity, we find that while the frames of the car and garage did not coincide, ie, while the car is in motion as viewed from the garage, the clocks in the car really were running more slowly...just as observed and predicted.

    This was finally verified I think in the early sixties using two atomic clocks; one on the ground, and one in a jet. After flying one of the clocks around for a while, and after accounting for the effects of gravity, the clock on the plane had indeed lost time as predicted to within the accepted margins of error. This has since be replicated in many other ways. Also, we see the lifespan of subatomic particles increase according to Relativity and their relative speed – since their clocks run more slowly, we see them live longer. This is seen in particle accelerators as well as in nature.
     
    Last edited: Nov 15, 2003
  12. Nov 15, 2003 #11
    Both parties see the other as in motion according to their own frame. Both parties see the other's clock as running slow.
    When both parites come to be in the same frame again neither can say if it was he or the other who decelerated. Both must find the the other's clock has lost some time, in which case they will both find both clocks to agree both as to the time and the rate of timekeeping. The illusion of slow clocks only exists while the relative motion is occuring. When the relative motion stops the illusion stops.
     
  13. Nov 15, 2003 #12

    Ivan Seeking

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    We can tell who accelerates [decelerates] - the one who experiences a force. If the relative motion of observer B changes wrt observer A, and if observer A experiences no forces, then A knows that B has changed his state of motion. Likewise, B feels a force and is also aware of whose frame of reference has changed. This indicates who is at rest -the preferred observer. This is no illusion; it is a 98 year old, well tested theory. The clocks do not agree when we compare the results in the same frame of reference; and the two clocks vary by the amount predicted by Special Relativity [or General Relativity if required].

    Edit: A key concept here is that until someone experiences a force, indicating a change in their state of motion, there is no preferred observer. We can define either [or any] frame of reference to be at rest as long as the state of motion remains constant.
     
    Last edited: Nov 15, 2003
  14. Nov 15, 2003 #13
    So, it seems that acceleration is the critical point.

    If we accelerate the car away from the garage, what does each observe about the other's clock?
     
  15. Nov 15, 2003 #14

    Ivan Seeking

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    The clocks in the car would be seen to run slowly as before; and due to the acceleration.

    From General Relativity: Clocks run more slowly in gravity fields.
     
  16. Nov 15, 2003 #15
    No, acceleration isn't the critical point. You can construct variants of the twin paradox in which nobody accelerates, but the twins come back with different ages. (e.g., going around a closed universe, or a gravitational slingshot back home). All that is required for different elapsed times is that, by whatever means (acceleration or not), the two twins take spacetime paths of different lengths.
     
  17. Nov 15, 2003 #16
    In which case the one who takes the longer path ages less?

    This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.
     
  18. Nov 15, 2003 #17

    Ivan Seeking

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    How do we do this without any acceleration?
     
  19. Nov 15, 2003 #18

    Ivan Seeking

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    SR is only valid in the inertial frame - no forces.
    We need GR to calculate the effects of acceleration [gravity].
     
  20. Nov 15, 2003 #19

    chroot

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    A path in spacetime is called an interval.

    The length of an interval is called its proper time. If you have a clock follow some path, the clock will measure that much time having been elapsed when moved along that path.
    This is not really true -- a distant observer will measure a clock as running slowly when it is moving at high relative velocity to the observer. According to the clock, however, everything is just fine.

    Imagine Picard is flying along in the Enterprise at 0.9c with respect to the Earth. An observer on the Earth will measure Picard's clock as running slow compared to an identical Earth-bound clock. Picard, however, will see everything on the bridge of the Enterprise as running completely normally, but will measure the Earth-bound clock as running slowly.

    If you think about it, it has to be that way... if it weren't, then some cosmic ray particle moving at 0.9c with respect to you somewhere in the depths of space would somehow affect YOUR clock!

    - Warren
     
  21. Nov 16, 2003 #20
    The longer path ages more, if you measure "length" using the spacetime interval.

    Special relativity too.

    Consider the case where the Earth twin stays at home for 10 years according to his own clock. The other twin travels at 80% of light speed, travelling 4 lightyears in 5 years according to the Earth twin, then returns the same way (after an instantaneous deceleration and acceleration back home).

    The Earth twin's worldline is a line from (t,x) = (0,0) to (10,0). The travelling twin's worldline consists of two line segments, one from (0,0) to (5,4), the other from (5,4) to (10,0).

    The proper time measured by the Earth twin is the spacetime length of his worldline,

    [tex]
    \tau = \sqrt{{\Delta t}^2-{\Delta x}^2} = \sqrt{(10-0)^2-(0-0)^2} = 10
    [/tex]

    The proper time measured by the travelling twin is the spacetime length of his worldline,

    [tex]
    \begin{equation*}
    \begin{split}
    \tau &= \tau_1+\tau_2 = \sqrt{{\Delta t_1}^2-{\Delta x_1}^2}+
    \sqrt{{\Delta t_2}^2-{\Delta x_2}^2} \\
    &= \sqrt{(5-0)^2-(4-0)^2}+\sqrt{(10-5)^2-(0-4)^2}\\
    &= 3+3 = 6
    \end{split}
    \end{equation*}
    [/tex]

    The travelling twin ages 6 years to the Earth twin's 10 years.
     
    Last edited: Nov 16, 2003
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