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Time dilation in uniform field?

  1. Aug 28, 2009 #1
    Imagine two spaceships a distance z apart both moving with the same constant acceleration a. The trailing spaceship shoots a beem of light which will be redshifted by an amount

    [tex]\frac{\Delta \lambda}{\lambda_0} = \frac{az}{c}[/tex]

    (assuming that [tex]\frac{\Delta v}{c}[/tex] is very small). Due to the equivalence principle the same redshift will happen in a uniform gravitational field - this is the famous gravitational redshift. In a non-uniform gravitational field a is not constant, and

    [tex]\frac{\Delta \lambda}{\lambda_0} = \Delta \Phi[/tex]

    I have a pretty good understanding of the latter type of redshift, I think. Basically it expresses how time moves at different speeds in areas with different potentials. The deeper you are in a potential well, the slower time goes.

    But what about the equation for the uniform field? Does it imply as well that time moves slower for the front spaceship? I hardly think that is 'fair' - I mean, the two spaceships are moving at the exact same speed. The same goes for the uniform gravitational field - why should time move faster for an observer in a uniform field, just because he is higher up?
     
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  3. Aug 28, 2009 #2

    A.T.

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    Yes
    Well, nature is pretty 'unfair'.
     
  4. Aug 28, 2009 #3
    They are moving at the exact same instantaneous speed as measured in some inertial frame. But neither of them are in that frame. Because of the relativity of simultaneity, they will disagree with each other about how long each has been accelerating, and therefore about how fast each is going in terms of their own "instantaneous" (i.e. simultaneous) speeds.
     
  5. Sep 3, 2009 #4
    Originally Posted by dianaj
    But what about the equation for the uniform field? Does it imply as well that time moves slower for the front spaceship?

    I thought that the gravitational dilation in accelerating systems theorem predicted that the dilation was greater in the rear???
    or is there some other applicable principle involved wrt two different ships?
    I certainly agree that nature is unfair. Or at least fairness doesn't seem to be an operative principle in either physics or life.
     
  6. Sep 3, 2009 #5
    time will run slower behind the accelerating rocket and eventually at some point will even stop completely. a photon emitted from that point will never reach the rocket. from the rockets point of view it is a kind of black hole.

    beyond that point time will run in reverse.
     
  7. Sep 3, 2009 #6

    A.T.

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    Yes I misread, what dianaj wrote.
    Yeah, just watch those animal documentaries on discovery chanel.
     
  8. Sep 3, 2009 #7

    Cleonis

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    How exactly do you take 'deeper in a potential well' to mean?

    Take a spaceship that is using thrusters to accelerate. A Pound-Rebka experiment conducted onboard that spaceship will find a frequency shift. If we choose to phrase that result in terms of a gravitational field then we will say that the redshift has occurred due to the difference in gravitational potential between ceiling and floor.

    In the case of the Earth clocks at different altitudes count different amount of proper time. Also we have that gravitational acceleration is a function of the distance to the center of gravitational attraction. I suppose it's _tempting_ to attribute that difference in clock rate to the fact that at each altitude there is a different gravitational acceleration.
    But whether or not there is different gravitational acceleration at different altitudes, the sole reason for different rates of clocks is different gravitational _potential_.

    Phrased in a different way:
    Let me call a gravitational field a 'slope'.
    In the case of a spaceship accelerating itself with thrusters there is a uniform 'slope' from nose to rear of the ship.

    In the case of the Earth's gravitational field the 'slope' is not uniform; there is a 'gradient' in the 'slope'. However, the local gravitational redshift as measured by a Pound-Rebka experiment is due solely to the local 'slope', not due to 'gradient of the slope'.

    I don't know if I actually addressed your problem here. I mentioned a 'tempting supposition', but I'm not sure whether your question actually comes from that particular supposition.

    Cleonis
     
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