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Clocks in gravitational field

  1. Sep 19, 2007 #1
    It is often said that a clock will run slower in a stronger gravitational field. This is not
    always true. Give a case where a clock will run slower in a WEAKER gravitational field.
    Bob
     
  2. jcsd
  3. Sep 19, 2007 #2
    Curious question.

    Firstly it is posed like a homework question, which is disappointing in this forum.

    It doesn't clarify the meaning of "in a stronger gravitational field", which is a concept of dubious reality in GR.

    Nor does it properly explain what "slower" should be taken relative to, which suggests a trivial answer (relative to a distant spaceship that uses engine thrust to rapidly turn in small circles)?

    Does the question really have a more satisfying answer?
     
  4. Sep 19, 2007 #3
    I guess not.
     
  5. Sep 19, 2007 #4
    There are two factors thet control the clock rate: the gravitational potential and the clock speed (relative to the observer). A fast moving clock in a weaker gravitational potential may run slower than the clock at rest in a strong potential.

    Eugene.
     
  6. Sep 20, 2007 #5

    pervect

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    Science Advisor

    Also note that the rate at which a clock ticks does not depend on the field (the Newtonian field), but it depends on the _potential_.

    Consider a clock at the center of the Earth. What is the field there? What is the potential? Does a stationary clock at the center of the Earth tick slower or faster than one on the Earth's surface?
     
  7. Sep 20, 2007 #6
    Yes, pervect, you've answered my question.
    Indeed, a clock at the center of the earth will tick more slowly than one on
    the surface although the FIELD at the center is zero.

    Interesting that Martin Gardner explaines the Pound-Rebka result as the FIELD
    lower down is stronger making the bottom clock run slower but doesn't consider
    that if the bottom clock was buried in the earth the FIELD would be weaker but
    it would still run slower.

    Thanks for your reply.

    Bob.
     
  8. Sep 20, 2007 #7
    You can find a derivation I did on this web page. Its pretty much a standard derivation.
    http://www.geocities.com/physics_world/gr/grav_red_shift.htm

    Pete
     
  9. Sep 21, 2007 #8
    There is, actually, a simpler derivation, which doesn't involve general relativity at all.

    Fundamentally, any clock is a non-stationary physical system that
    exhibits a periodic process with a stable frequency. In quantum
    mechanics, a state of any non-stationary system is represented by a
    superposition of two or more stationary states. For simplicity, we
    will assume that these states have distinct discrete energy levels
    [itex]E_i[/itex] and [itex]E_f[/itex] with masses [itex]m_i = E_i/c^2[/itex] and [itex]m_f = E_f/c^2[/itex],
    respectively. Then the rate of the corresponding "clock" is
    proportional to the energy difference [itex]E_i - E_f[/itex]. For example, the
    frequency of the electromagnetic radiation emitted in an
    atomic transition is


    [tex]
    \nu(0) = \frac{2 \pi}{\hbar} (E_i - E_f) = \frac{2 \pi
    c^2}{\hbar} (m_i - m_f)
    [/tex]

    Total energies of the system "stationary clock + Earth" in the
    ground and excited states are

    [tex]
    \mathcal{E}_f \approx Mc^2 + m_fc^2 + \phi m_f
    [/tex]
    [tex]
    \mathcal{E}_i \approx Mc^2 + m_ic^2 + \phi m_i [/tex]


    where [itex]M[/itex] is the Earth's mass, [itex]\phi = - GMR^{-1}[/itex] is the
    value of the gravitational potential on the Earth's surface,
    and [itex]R[/itex] is the Earth's radius. Then the frequency of the clock on
    Earth can be found as

    [tex]
    \nu(\phi) = \frac{2 \pi}{\hbar} (\mathcal{E}_i - \mathcal{E}_f)
    \approx \nu(0)(1 + \frac{\phi}{c^2})[/tex]

    This means that a clock on the Earth surface is running
    slower than an identical clock at a higher elevation (higher
    gravitational potential [itex]\phi[/itex]).

    Eugene.
     
  10. Aug 22, 2008 #9
    Sorry to revive such an old thread but I have never heard of gravitational potential not equaling the gravitational field in GR. It seems that the same subject was covered in an even older thread:

    https://www.physicsforums.com/showthread.php?t=40391

    It seems to me that Pervect would argue that if you have a massive hollow spherical shell in outer space falling freely, that a clock inside the shell will tick at a slower rate due to the shell than if there were no shell. Even though the gravitational field in the shell is zero at all points Pervect's idea of gravitational potential would not be zero and importantly would not be uniform throughout the interior, it would be greater as you approached the shell walls and least in the center. This would have startling consequences for a test particle placed in the shell, stationary with respect to the shell. The test particle would follow a geodesic path of least time and move toward the shell wall. I've never heard of GR predicting anything like that.
     
  11. Aug 23, 2008 #10

    Jonathan Scott

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    Gold Member

    The gravitational field is the gradient of the potential, and is what gives rise to accelerations and forces. The potential determines potential energy and relative clock rates. In a uniform field, or a near-uniform field such as that at the surface of the earth, the potential varies with the height. Inside a hollow sphere, there is no field anywhere, the potential is the same everywhere and so is the clock rate. The potential and clock rate are lower than they are outside the sphere or within the spherical layer of matter.
     
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