Gravitational Curvature

Discussion in 'Physics & Math' started by shalayka, Oct 25, 2007.

  1. shalayka Cows are special too. Registered Senior Member

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    Can the gravitational curvature generated by the momentum and energy of a photon to be expressed as something like this:

    \(G_p = \frac {8\pi \rm G}{c^4} \frac{\it h\nu}{c}\)

    \(G_{\rm E} = \frac {8\pi \rm G}{c^4} \it h\nu\)

    If this is correct, I wonder if these following curvatures exist?

    \(G_{\rm ?} = \frac {8\pi \rm G}{c^3} \it h\nu\)

    \(G_{\rm ?} = \frac {8\pi \rm G}{c^2} \it h\nu\)

    \(G_{\rm ?} = \frac {8\pi \rm G}{c} \it h\nu\)
     
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  3. Reiku Banned Banned

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    I can tell that these are deriviations of EFE's. I suppose they should work, but only as guide equations. Potenetially, from Einsteins Field Equation, you can derive any suggestiong by general relativity.As for the other equations, they could be used, i (think). Wouldn't know how to apply them myself really.
     
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  5. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    shalayka---

    Your units are FUBAR.
     
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  7. shalayka Cows are special too. Registered Senior Member

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    The issue with units has been mentioned a few times, always without explanation.

    I've seen this a lot of times:

    \(G_{\mu\nu} = \frac{8\pi\rm G}{c^4} T_{\mu\nu}\)

    And so I thought I could break out the momentum and energy density components like in the first post.

    Can you please explain why this causes the units to be messed up? Is it because the EFE in this form does not use mks units?

    I thought that it was going well, since the following works out for rest mass:

    \(G_{\rm E} = \frac {8\pi \rm G}{c^4}\rm E\)

    Dividing by \(4\pi\) to get the Schwarzschild radius:

    \( \rm R_{\rm s} = \frac {2\rm G\rm E}{c^4}. \)
     
    Last edited: Oct 26, 2007
  8. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Hopefully someone will know more GR than I do here---I don't know ANYthing when it comes to this stuff.

    What I CAN tell you, though is your Gp and GE have two different units--just compare factors of c. Naively, one expects that ``curvature'' has specific units.

    Also, \(G_{\mu\nu}\) is a tensor, and what you have written with Gp and GE are scalars.

    Also, the Schwarzchild radius thing may be a coincidence.
     
  9. shalayka Cows are special too. Registered Senior Member

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    I was assuming that both curvatures were of the same units (work), even though the definitions for a photon's momentum \(p = \frac {h\nu}{c}\) and energy \(E = h\nu\) differ in units. The same would apply to the difference between viscosity (momentum flux) and energy. I suppose this is the part that I need most clarification on by someone with lots of experience in this type of thing.

    Yes, the \(G_{\rm E}\) and \(G_p\) scalars are only 2 of 16 components. I used them as such because I had assumed that the product of a scalar and a rank-2 tensor can be given by multiplying each component by the scalar (like multiplying a scalar by a quaternion). I was ignoring the other components to keep things simple.

    So in my example, the scalar \(h\nu\) is given by \(T_{0 0}\) and the scalar \(\frac{h\nu}{c}\) is given by \(T_{0 1}\) (or \(T_{1 0}\), taking into account the symmetry of the stress-energy tensor). The critical thing I suppose is that I'm assuming that the momentum is entirely along a single axis, so \(T_{0 2}\), \(T_{0 3}\) (and accordingly, \(T_{2 0}\), \(T_{3 0}\)) are equal to \(0\).

    This leads me to also question whether or not it's correct to equate gravitational curvature with the momentum of mass along a single axis by:

    \(G_{p} = \frac{8\pi\rm G}{c^4} \frac{\rm{E_0}v}{c^2\sqrt{1 - v^2/c^2}}\)


    I used this as a reference:
    http://en.wikipedia.org/wiki/Stress-energy_tensor

    I used this as a reference:
    http://www2.kutl.kyushu-u.ac.jp/seminar/MicroWorld1_E/Part3_E/P37_E/momentum_of_photon_E.htm

    Sorry for so many edits.

    Please Register or Log in to view the hidden image!

    I think I'm done now.
     
    Last edited: Oct 26, 2007
  10. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Right. But each component of the stress energy tensor should have the same units, and this is not the case in your example.
     
  11. Reiku Banned Banned

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    He is right you know.
     
  12. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Who's right, me? Yes I know.

    Shalayka is not right, because his units are wrong. He may be on to something, and perhaps this weekend I will sit down and try to remember how to write down the stress tensor for electromagnetism.

    there may be a factor of 1/c somewhere which fixes his calculation, but untill someone reminds me, or I sit down and figure it out, he cannot be right.
     
  13. Reiku Banned Banned

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    Yes you.
     
  14. shalayka Cows are special too. Registered Senior Member

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    MTW's Gravitation seems to define \(T_{00}\) to \(T_{03}\) and \(T_{00}\) to \(T_{30}\) as mirrored copies of the 4-momentum within a 3-volume.

    The Schwarzschild solution that I got was assuming that the 4-momentum of a massive body in its rest frame is:
    \((E, 0, 0, 0)\)

    Perhaps my error is in assuming that the 4-momentum for a photon traveling in the positive x direction is:
    \((h\nu, h\nu / c, 0, 0)\)

    This is where I got the notion that:

    \(G_{\rm E} = \kappa T_{00}\)

    \(G_{p} = \kappa T_{01}\)

    \(\kappa = \frac{8\pi\rm G}{c^4}\)

    I suppose that \(G_p\) would be \(\kappa\) multiplied by the magnitude of the 3-vector \((T_{01}, T_{02}, T_{03})\) in the case of travel not limited to a single dimension.

    Anyway, thank you guys for looking into this. I do appreciate it, because trying to talk to my university is like trying to pull teeth with your bare hands.
     
    Last edited: Oct 30, 2007

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