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Power radiated

  1. Nov 29, 2012 #1

    nos

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    Hi all,

    If an electron is being accelerated, it will emit electromagnetic radiation.
    Will it radiate gravitational radiation? If yes, how much power does it radiate?

    Im talking about linear acceleration.

    I am not familiar with general relativity.
    I guess you cant have a gravitational larmor formula?
     
  2. jcsd
  3. Nov 29, 2012 #2

    phyzguy

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    There is a gravitational enalog to the Larmor formula. The energy radiated by a changing mass distribution is given by:
    [tex] \frac{dE}{dt} = \frac{G}{5 c^5} (Q''')^2[/tex]

    where Q''' is the third time derivative of the quadrupole moment of the mass distribution. In most ordinary situations this radiation is ridiculously small. For an electron, for example, it is at least 10^40 times weaker than the EM radiation and hence completely unobservable. However, it can become significant in astrophysical situations like merging neutron stars or black holes.
     
  4. Nov 29, 2012 #3

    PeterDonis

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    For the case of linear acceleration, no gravitational radiation is emitted. phyzguy's formula is correct, but for the case of linear acceleration, the third time derivative of the quadrupole moment is zero.
     
  5. Nov 30, 2012 #4

    nos

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    Oh thanks for replying! Why is there no gravitational radiation for linear acceleration? Is it because the field is not a vector field? Or does it have to do with conservation of mass energy?
     
  6. Nov 30, 2012 #5

    PeterDonis

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    Because, as phyzguy said, gravitational radiation requires a time-varying quadrupole moment. That means there has to be a mass distribution that not only changes with time, but changes with time in a more complicated way than just linear acceleration. The Wikipedia page on gravitational radiation has some examples:

    http://en.wikipedia.org/wiki/Gravitational_wave
     
  7. Nov 30, 2012 #6

    Bill_K

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    Don't believe it. For uniform acceleration, x ~ cosh(aτ/c), t ~ sinh(aτ/c) and Q ~ mx2. No way (d/dt)3(x2) vanishes!
     
  8. Nov 30, 2012 #7

    PeterDonis

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    But this solution isn't complete; you are only including the accelerated object, not what is making it accelerate. For a self-consistent solution, you have to include the "rocket exhaust" as well, and its effect on the total quadrupole moment.

    For example, in the Kinnersley photon rocket, there is no gravitational radiation, but to arrive at that conclusion, you have to include the photons (the rocket exhaust) as well as the rocket itself. This paper discusses that:

    http://arxiv.org/abs/gr-qc/9412063
     
  9. Nov 30, 2012 #8

    Bill_K

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    OMG I take it back, it does vanish! Thanks.
     
  10. Nov 30, 2012 #9

    PeterDonis

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    It vanishes if m is constant, correct? I get

    [tex]Q = mx^2 = mc^2 \left( \frac{c^2}{a^2} + t^2 \right)[/tex]

    so if m is constant,

    [tex]\frac{d^3Q}{dt^3} = m c^2 \frac{d^3}{dt^3} \left( t^2 \right) = 0[/tex]

    But if m is not constant (as in the rocket case), things get more complicated, and one has to include the rocket exhaust in the analysis to verify that no gravitational radiation is emitted. At least, that's my understanding.
     
  11. Dec 4, 2012 #10

    Bill_K

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    One more try. For small t, Newton tells us that x ≈ ½ at2. Then the quadrupole moment is Q = mx2 = (m/4) a2t4. Clearly the third time derivative of this does not vanish. :confused:
     
  12. Dec 4, 2012 #11

    PeterDonis

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    Whoa, you're right, that is confusing. One quick note: your "x" is defined using a different spatial origin than my "x" was; you put the spatial origin at the point where the mass starts accelerating, but I put it at the Rindler horizon (more precisely, at the origin of the Minkowski frame in which the lines x = +/-t are the Rindler horizon, and in which the mass is at rest at the instant it starts accelerating). I don't see how that would clear up the confusion any, though, since adding or subtracting a constant to x doesn't change the time derivatives. :uhh:
     
  13. Dec 4, 2012 #12

    mfb

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    If this acceleration is caused by a second object with mass M, we have another contribution to the dipole moment, as M accelerates with ##a'=-\frac{m}{M}a##.
    This gives ##Q_M=-\frac{M}{4}{m^2}{M^2}a^2t^4 = -\frac{m^2}{4M} a^2 t^4 \neq -Q_m## for m≠M. A displacement adds a constant to the quadrupole moment, this vanishes in the time-derivative.

    Linear acceleration of two objects with different mass towards (or away from) each other gives a non-zero third derivative of the total dipole moment. At least if we can ignore the finite speed of the interaction between both objects.
     
  14. Dec 4, 2012 #13

    PAllen

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    I think Kinnersley's photon rocket is the special case. As the paper Peter linked to shows, the absence of GW is due to:

    - all nonstatic character of the scenario (motion) being due to the null fluid + conservation.
    - the null fluid only has anisotropy of monopole and dipole character

    Mfb's example suffers from non-conservation of momentum, so it is not a physically plausible example.

    However, I have seen, in the literature, that two non-rotating, uncharged black holes radially falling toward each other produce gravitational radiation. This seems the closest analog of the Newtonian two mutually attracting point particles. If I find a link for this, I'll post it, but I think it is pretty well known radially attracting BH radiate power.

    [edit: Here is a link for radiation from radial infall of particle into a BH: http://arxiv.org/abs/1012.2028 ]
     
    Last edited: Dec 4, 2012
  15. Dec 4, 2012 #14

    PeterDonis

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    I agree, but it would still be nice to see something similar in the non-relativistic (i.e., small t) limit. I have been trying to add in the contribution of the rocket exhaust to the quadrupole moment that Bill_K wrote down for the rocket in the small t limit, to see if that extra contribution makes the third time derivative of the complete quadrupole moment (rocket + exhaust) vanish, as it does in the relativistic case, but so far all I've got is a mess.
     
  16. Dec 4, 2012 #15

    mfb

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    Where do you see a violation?

    With a common force F between two objects, the accelerations are a=F/m and a'=F/M, which gives the relation am=a'M or a'=m/M a.
     
  17. Dec 4, 2012 #16

    PeterDonis

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    I assume PAllen was referring to the fact that, as you said at the end of your post, you were not taking into account the finite speed of interaction between objects. What you wrote assumes instantaneous interaction, which implies non-conservation of momentum if Lorentz transformed into any frame other than the one in which you wrote your post.
     
  18. Dec 4, 2012 #17

    PAllen

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    Action at a distance cannot be made compatible with SR. Third law can only work for collisions in SR. There's a theorem on this going back, I think, to Sudarshan.

    I should have clarified, because obviously you are right for pre-relativity physics, and you did mention the action at a distance issue.
     
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