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Power radiated 
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#1
Nov2912, 06:06 PM

P: 32

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
Nov2912, 06:49 PM

P: 2,179

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. 


#3
Nov2912, 08:14 PM

Physics
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PF Gold
P: 6,066

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.



#4
Nov3012, 10:41 AM

P: 32

Power radiated
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?



#5
Nov3012, 10:49 AM

Physics
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PF Gold
P: 6,066

http://en.wikipedia.org/wiki/Gravitational_wave 


#6
Nov3012, 11:13 AM

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Thanks
P: 4,160




#7
Nov3012, 11:19 AM

Physics
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PF Gold
P: 6,066

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/grqc/9412063 


#9
Nov3012, 02:24 PM

Physics
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PF Gold
P: 6,066

[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. 


#10
Dec412, 07:02 AM

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Thanks
P: 4,160




#11
Dec412, 10:35 AM

Physics
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PF Gold
P: 6,066




#12
Dec412, 10:48 AM

Mentor
P: 11,631

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 timederivative. Linear acceleration of two objects with different mass towards (or away from) each other gives a nonzero third derivative of the total dipole moment. At least if we can ignore the finite speed of the interaction between both objects. 


#13
Dec412, 11:28 AM

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PF Gold
P: 5,027

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 nonconservation of momentum, so it is not a physically plausible example. However, I have seen, in the literature, that two nonrotating, 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 ] 


#14
Dec412, 03:03 PM

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PF Gold
P: 6,066




#15
Dec412, 04:05 PM

Mentor
P: 11,631

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. 


#16
Dec412, 04:12 PM

Physics
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PF Gold
P: 6,066




#17
Dec412, 04:44 PM

Sci Advisor
PF Gold
P: 5,027

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


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