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Local gravity calculation

  1. Jul 16, 2011 #1

    Pengwuino

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    One term I fully understand yet I have never seen how one actually does the calculation is the local gravity a particle feels in a gravitational field.

    Now, I honestly feel this is as stupid of a question as they come, but intuitively I'd say, if I wanted a(r), the acceleration as a function of the radius (useful for things like Hawking temperature), you simply take your geodesic equation

    [tex]{{d^2 x^{\mu}}\over{d\lambda^2}} + \Gamma^{\mu}_{\alpha \beta} {{dx^{\alpha}}\over{d\lambda}} {{dx^{\beta}}\over{d\lambda}} = 0[/tex]

    to find the acceleration as a function of the radial distance, multiply by the redshift, and do your usual simple radially-infalling particle and wala, local acceleration.

    And what is the physical interpretation? My assumption is that it's the acceleration felt in the test particle's frame. Correct? Naive? :)
     
    Last edited: Jul 16, 2011
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  3. Jul 16, 2011 #2

    atyy

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  4. Jul 16, 2011 #3

    Mentz114

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    Solving the geodesic equation will give geodesic worldlines which won't experience any proper acceleration by definition.

    The proper acceleration of a particle on a worldline [itex]u^\mu,\ \ u^\mu u_\mu=-1[/itex] is [itex]\dot{u}_\mu=u_{\mu;\nu}u^\nu \equiv \nabla_\nu u_\mu u^\nu[/itex].

    This is the covariant derivative of [itex]u_\mu[/itex] projected in the direction [itex]u^\mu[/itex]. To get the local proper acceleration it has to be expressed in the frame basis.

    The calculation for the Schwarzschild vacuum is discussed in this article http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity
     
  5. Jul 17, 2011 #4

    Bill_K

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    The 'local gravity' is the coordinate acceleration of a freely falling particle, in your example d2xi/dλ2. And in turn 'coordinate acceleration' means with respect to a distinguished system of coordinates, such as the stationary coordinates in a stationary solution.
     
  6. Jul 17, 2011 #5

    bcrowell

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    Sorry to be pedantic, but my wife is a French teacher, and I want to make sure she doesn't choke on a cherry pit and die if she sees this. It's "voilà," not "wala." Some French speakers sometimes pronounce it in certain speciific contexts without the "v" sound, but it's not normal: http://forum.wordreference.com/showthread.php?t=465332 It would be sort of like an American saying "I'm a," instead of "I'm gonna."
     
  7. Jul 17, 2011 #6

    Pengwuino

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    Well apparently there was all sorts of wrong going on in my thread so might as well butcher some languages as well.
     
  8. Jul 17, 2011 #7

    pervect

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    For starters, do you know that 'r' is a radial coordinate, and that you have to process it to convert it into a "radial distance"?

    I could give details, if you're interested and if you don't already know this. The short version though is that it's the job of the metric to convert changes in coordinates to actual distances.
     
  9. Jul 17, 2011 #8

    Pengwuino

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    Yes I'm glad to say I've gotten that far into my GR education :P I was just hoping there was a simpler way to do this... le sigh.
     
  10. Jul 17, 2011 #9

    pervect

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    Have you been exposed to "frame fields" yet? That might be just the tool you're looking for.
     
  11. Jul 17, 2011 #10

    Pengwuino

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    Nope :( One of my texts seems to have a lot on it, i'll have to check it out.
     
  12. Jul 18, 2011 #11

    Mentz114

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    See post #3.

    Frame field is more than a tool. It is necessary to find out what is experienced by actual observers on worldlines.
     
    Last edited: Jul 18, 2011
  13. Jul 19, 2011 #12

    pervect

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    I'll try to describe, briefly, the super-simple method for frame fields. In a rather non-rigorous way.

    Let's start with polar coordinates, in flat space time, which should be familiar. You have some metric

    ds^2 = -dt^2 + r^2 + r^ d theta^2

    Because of the coefficient of r^2, [itex]d\theta[/itex] doesn't represent a constant distance.

    What you do is you introduce locally, some vectors [itex]\hat{r}[/itex] = dr, and [itex]\hat{\theta} = r d\theta[/itex] that are unit length. At this point, I'm not worring about whether the "vector" dr is covariant or contravariant, though it turns out to be the former, when you think of it as a vector at all, that is.

    More formal treatments will distinguish the coframes , dr, from the frames [itex]\partial / \partial r[/itex], I'm beeing very lax by glossing over this. It's eventually important to understand the fine distinctions here, but it's not good if it distracts you from understanding the basic idea of what's going on. It's the whole covariant/ contravariant mess...

    You are essentially introducing new coordinates nearby any local event, the new coordinates have hats.

    You can think of it as doing a coordinate transformation. There are lots of ways you coulc do a coordinate transformation, for instance going back to the definition of the metric as a tensor, but the easiest one is algebra:

    [itex]\hat{r} = r, \hat{\theta} = r \left( \theta - \theta_0\right) = r d\theta [/itex]

    THen you can just write, in these local coordinates

    [itex] dr^2 + r d\theta^2 = \hat{r}^2 + \hat{\theta}^2 [/itex]

    You visualize [itex]\hat{r}, \hat{\theta}[/itex] as unit vectors, little unit arrows, that form a local coordinate system .

    And you just do the same thing in Schwarzschild coordinates. You've got non-unity metric coefficeints

    gtt dt^2 + grr dr^2

    so you just define local "vectors" [itex]\hat{t}[/itex] = sqrt(|g11|) dt and [itex]\hat{r}[/itex] = sqrt(|grr|) dr

    and you proceed on just as you did in polar coordinates, and you interpret the symbols the same way, the little hats are local vectors, that form their own little local coordinate system, which is orthonormal and very familiar and easy to deal with.
     
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