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Flat uniform and equivalent

  1. Jun 8, 2008 #1

    Dale

    Staff: Mentor

    In the usual "windowless room" thought experiment it is stated that there is no experiment which can be performed which will determine if the room is undergoing "uniform accleration" far from any massive body or is in a "uniform gravitational field". The two situations are therefore said to be "equivalent". Also, since the spacetime in the uniform acceleration case is obviously flat the equivalent spacetime in the uniform gravitation must also be flat.

    So, my question is a pretty basic one. How do we construct the right coordinate systems for doing the "equivalent" windowless room thought experiments? Specifically:

    In the gravitational/accelerated non-inertial frame what is the form of the gravitational field (as a function of position)?

    If the field is a non-constant function of position then how is it considered either uniform or flat?

    How are simultaneity, distance, and time experimentally defined in the uniform field?

    What is the correct transformation between the accelerated and an inertial frame?

    Can you simply use the Lorentz transform to boost between observers with uniform (non-inertial) motion in the gravitational field? (i.e. since spactime is flat do four-vectors still transform like usual, especially the four-momentum)

    Any other hints would be appreciated.
     
  2. jcsd
  3. Jun 9, 2008 #2

    Mentz114

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    Gold Member

    My guess is that g_00 and g_11 ( x^0 = t, x^1=r say) must depend on r linearly, so that d/dr(g_00) = constant, d/dr(g_11) = constant. This means the Christoffel symbols will exist and will not change over space, but the Riemann tensor is zero.

    The problem with this setup is that it diverges at infinity.

    It's a tough question, let's see what the GR experts say.
     
  4. Jun 9, 2008 #3

    Mentz114

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    Gold Member

    Uniform red-shift

    I've done some work and find that this non-physical vacuum solution gives time dilation in the x-direction proportional to x. Otherwise the space is flat, no Christoffels or Riemann components.

    [tex]ds^2 = -B^{-1}x^{-2}dt^2 + Ax^2dx^2 + dy^2 + dz^2[/tex]

    This must be the space outside an infinite slab of matter, and the red-shift depends not at all on any properties the matter may have - probably because there's an infinite mass.

    Will this do for the acceleration comparison ?

    [edit] No it has no acceleration as required.
     
    Last edited: Jun 9, 2008
  5. Jun 9, 2008 #4

    gel

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    For a gravitation acceleration of a in the -ve x direction at the origin, I get the following metric
    [tex]
    d\tau^2 = \left(1+c^{-2}ax\right)^2dt^2-c^{-2}(dx^2+dy^2+dz^2)
    [/tex]
    where [itex]\tau[/itex] is the proper time.
    It is non-uniform, the gravitational acceleration at coordinate x is [itex]\frac{a}{1+c^{-2}ax}[/itex]. Note, it blows up at x=-c2/a.

    And, you can transform to inertial coords by
    [tex]
    x'=(x+a^{-1}c^2)\cosh(c^{-1}at)-a^{-1}c^{2}, t'=c^{-1}(x+a^{-1}c^2)\sinh(c^{-1}at), y'=y,z'=z.
    [/itex]

    (You might want to check the details - I'm not guaranteeing there's no mistakes.)
     
    Last edited: Jun 9, 2008
  6. Jun 9, 2008 #5

    Mentz114

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    Gold Member

    gel,
    I find no non-zero Christoffels for your metric. Does this not mean zero acceleration ?
     
  7. Jun 9, 2008 #6

    gel

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    btw, I think this is what's called Born rigid motion. It's flat because the curvature tensor vanishes, and can be transformed to Minkowski space.

    You can also work out a metric for a uniform gravitational field, but it isn't flat and the Einstein tensor is non-zero, so it wouldn't be empty space.
     
  8. Jun 9, 2008 #7

    gel

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    I got
    [tex]
    \Gamma^x_{tt}=a(1+c^{-2}ax),\ \Gamma^t_{tx}=a(c^2+ax)^{-1}.
    [/tex]
     
  9. Jun 9, 2008 #8

    gel

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    Google search came up with born rigid motion. I haven't read it all, but the diagram is what I had in mind.
     
  10. Jun 9, 2008 #9

    Mentz114

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    Gold Member

    gel,
    [tex]\Gamma^x_{tt}=a(1+c^{-2}ax),\ \Gamma^t_{tx}=a(c^2+ax)^{-1}.[/tex]

    I agree. I entered 'r' instead of 'x' into my calculator and zapped all the derivatives.

    Phew - thought I had a bad problem.

    Correction to my calculation above, Christoffel symbs are

    x-xx: x^-1
    x-tt: -A^-1*B^-1*x^-5
    t-tx: -x^-1
     
    Last edited: Jun 9, 2008
  11. Jun 9, 2008 #10

    gel

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    To correct what I said above. For a uniform gravitational field, but non-flat space, you have the metric [itex]d\tau^2=\exp(2c^{-2}ax)dt^2-c^{-2}(dx^2+dy^2+dz^2)[/itex]. This isn't flat, so not what the OP asked for, but the Einstein tensor vanishes, so it is empty space.
    In fact, the Einstein tensor must vanish for similar reasons that it does in 2d manifolds.
     
  12. Jun 9, 2008 #11
    The metric for a uniform gravitational field is identical to that of a uniformly accelerating frame of reference. This follows as an immediate consequence of the equivalence principle. Also, the spacetime associated with a uniform gravitational field is flat. The Einstein tensor is vanishes for such a spacetime. Such a spacetime must be empty too.

    Pete
     
  13. Jun 9, 2008 #12
    There are two quantities normally associated with a field. One is the potential and the other is given in terms of derivatives of the potential. In the case of gravity the potential is a tensor known as the metric tensor. The 10 independant components of the metric are gravitational potentials. The field quantity in terms of the derivatives are the components of the affine connection or the Christoffel symbols. The Christoffel symbol for a uniform g-field is given above I believe.
    A gravitational field is said to be "uniform" when there are no tidal gradients in the field. By definition this means that the spacetime is flat.

    Pete
     
  14. Jun 9, 2008 #13
    Aside from the "infinite slab of matter" that Mentz mentioned - otherwise I'd assume that the gravitational field were uniformly zero.

    Regards,

    Bill
     
  15. Jun 9, 2008 #14
    There is more than one way to generate a uniform gravitational field. An infinite slab of matter is just one way which occurs in Newtonian mechanics. I don't believe that such a distribution of matter will generate a uniform g-field in the strong field limit in GR.

    Another distribution of matter which generates a uniform gravitational field is a spherical cavity cut out of a sphere of matter whose center is off center of the original sphere. This is easier to draw than describe. A diagram is shown here

    http://www.geocities.com/physics_world/gr/image_gif/gr10-img-01.gif

    Inside such a cavity the field is uniform, at least in the Newtonian approximation. Whether or not this represents a uniform g-field in general relativity is another matter. I know it will hold in the weak field limit though. And yes. I was speaking about outside the distribution of matter which generates such a field. Inside matter the spacetime will always be curved, i.e. where there is matter there is spacetime curvature.

    Pete
     
    Last edited: Jun 9, 2008
  16. Jun 9, 2008 #15
    Yes. The more common version being that the gravitational field about a spherical body of mass of uniform density is uniform (in magnitude) over all theta and phi associated with a constant radius from the center of gravity.


    Hmm. So a closed room affected by a uniform gravitational field cannot exist.

    Regards,

    Bill
     
  17. Jun 9, 2008 #16
    too cool, pmb. If you're sitting on an attractive ball of r1 and (an intersecting) repulsive ball of r2. the field is unform on the surface of r2...

    you found this via electrostatics, right?
     
  18. Jun 9, 2008 #17

    Dale

    Staff: Mentor

    Thanks for all of the great information everyone. I will post follow-up questions soon. At present, I am not interested in whether or not a uniform field can actually be produced by some specific arrangement of matter, I just want to get a better understanding of the mathematical details of the equivalence principle.
     
  19. Jun 9, 2008 #18
    No.
    Well I did have this as an exam question in an EM class but I was reminded of this in a GR text.

    Pete
     
  20. Jun 9, 2008 #19
    If I understand what you're referring to then I don't see how.
    [quote
    So a closed room affected by a uniform gravitational field cannot exist.
    [/QUOTE]No.

    Pete
     
  21. Jun 9, 2008 #20
    Do you feel that the statement is true or false? I cannot tell from your response.

    Regards,

    Bill
     
    Last edited: Jun 9, 2008
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