Timeline for Is there a simple formula for the gravitational self-force (due to emission of gravitational waves) in the classical limit?
Current License: CC BY-SA 4.0
20 events
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| Feb 26, 2020 at 4:14 | answer | added | Paul T. | timeline score: 1 | |
| Feb 24, 2020 at 14:45 | answer | added | TimRias | timeline score: 1 | |
| Feb 24, 2020 at 13:18 | history | edited | Mitsuko | CC BY-SA 4.0 |
deleted 508 characters in body
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| Feb 24, 2020 at 13:11 | history | edited | Mitsuko | CC BY-SA 4.0 |
clarfified things in response to the comments
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| Feb 24, 2020 at 9:34 | history | edited | Frederic Thomas | CC BY-SA 4.0 |
put the a type-writer written formula to MathJax
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| Feb 24, 2020 at 7:54 | comment | added | G. Smith | Just to be clear... it is extremely unlikely that your physics teacher knew the force formula and used it to determine the power. He almost certainly knew the power formula but described what was happening in terms of a force. Good luck. | |
| Feb 24, 2020 at 7:49 | comment | added | G. Smith | But you can do that yourself based on the power formulas. | |
| Feb 24, 2020 at 7:46 | comment | added | G. Smith | It would be possible to express that power in terms of the position, velocity, acceleration, and jerk vectors of the Earth relative to the Earth-Sun center of mass, although I have never seen that done. (It is much easier to express it simply in terms of the first three derivatives of the separation distance.) If that is what you want, I might have time to work it out in the next few days. | |
| Feb 24, 2020 at 7:40 | comment | added | G. Smith | That power is exactly what the links I have provided compute. | |
| Feb 24, 2020 at 7:39 | comment | added | Mitsuko | @G.Smith : My physics teacher talked about the power of that force, i.e., the scalar product of the force and velocity. This is the work done by the stopping force per unit of time. By the stopping power, I meant the power of the stopping force. | |
| Feb 24, 2020 at 7:34 | comment | added | G. Smith | If you are interested in the force, I can’t help you. I have never seen a simple formula for it. Hopefully someone else can help you. | |
| Feb 24, 2020 at 7:33 | comment | added | G. Smith | this gravitational stopping force acting on our planet is so small that the corresponding stopping power is comparable to the power of an ordinary electric lamp You cannot compare a force to a power. They don’t even have the same dimensions. If your physics teacher compared it to the power of a lamp, they could not have been talking about the stopping force. | |
| Feb 24, 2020 at 7:18 | comment | added | Mitsuko | @G.Smith : No, I am interested in the momentary stopping force rather than the energy radiated in the entire period. Imagine a single object of mass M, and imagine you know the momentary value of the derivative of its acceleration of any order. What will be the gravitational stopping force on that object, in the classical limit? Or do you mean to say that no such formula can in principle be derived? | |
| Feb 24, 2020 at 6:56 | comment | added | G. Smith | If this is really what you are interested in, edit your question to ask for the power formula rather than the force formula. | |
| Feb 24, 2020 at 6:55 | comment | added | G. Smith | Applied to the Earth-Sun system, this formula gives the results here: en.wikipedia.org/wiki/… | |
| Feb 24, 2020 at 6:54 | comment | added | G. Smith | It depends on the square of the third time derivative of the traceless mass quadrupole tensor. | |
| Feb 24, 2020 at 6:53 | comment | added | G. Smith | The general formula for the power radiated in gravitational waves by any system is on page 6 here: physics.usu.edu/Wheeler/GenRel2013/Notes/GravitationalWaves.pdf | |
| Feb 24, 2020 at 6:47 | comment | added | G. Smith | The Earth orbiting the Sun radiates about 200 watts of gravitational waves. This is almost certainly what you heard. See en.wikipedia.org/wiki/Gravitational_wave#Binaries | |
| Feb 24, 2020 at 6:45 | comment | added | G. Smith | This is a very complicated topic. See physics.stackexchange.com/q/220886. It is much simpler to consider the energy loss than the radiation-reaction force. | |
| Feb 24, 2020 at 6:36 | history | asked | Mitsuko | CC BY-SA 4.0 |