This question is actually 2 parts

  1. Just like binary star system or the 2 black wholes (which generated the recently discovered gravitational waves) shouldn't the Earth also radiate giving off energy. Reading the Wikipedia article confirmed my belief that Earth should radiate and give off gravitational waves. Now, since the Earth is 'radiating' shouldn't it lose energy (howsoever small it may be) resulting in shrinking of its orbit (howsoever ever small the shrink may be)? Radiating means losing energy after all.

  2. All mass (except few like neutrinos) is made up of charges and accelerating charges radiate. So normal matter, though neutral, must radiate in the sense that each of its individual charge should radiate. Shouldn't it? And there should be a net loss of energy?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 15 '17 at 16:39

Not all masses, only those that have an asymmetric mass generate gravitational waves. A perfect sphere will not.

Unlike charge, which exists in two polarities, masses always come with the same sign. This is why the lowest order asymmetry producing electromagnetic radiation is the dipole moment of the charge distribution, whereas for gravitational waves it is a change in the quadrupole moment of the mass distribution. Hence those gravitational effects that are spherically symmetric will not give rise to gravitational radiation. A perfectly symmetrical collapse of a supernova will produce no waves, while a non-spherical one will emit gravitational radiation. A binary system will always radiate. Gravitational waves distort spacetime: in other words, they change the distances between free macroscopic bodies.

Italics mine.

So for 1: the radiation from the Earth , which is almost spherical, will be very small in any case.

For 2: The elementary particles are point particles, so symmetric and they will not radiate gravitational waves in acceleration. Molecules and atoms accelerated may, if the outside orbitals are not S orbitals. Even in the last case as can be seen here it will be very small, as there is G and divisions by power of c involved.

The energy has to be provided by the source that is accelerating the objects.

The amplitude of the gravitational wave, formula 2.34 :


where εE_kin(with 0≤ε≤1), is the fraction of kinetic energy of the source that is able to produce gravitational waves. The factor ε is a measure of the asymmetry of the source and implies that only a time varying quadrupole moment will emit gravitational waves. For example, even if a huge amount of kinetic energy is involved in a given explosion and/or implosion, if the event takes place in a spherically symmetric manner, there will be no gravitational radiation

  • $\begingroup$ Ok I understand that spherically symmetric objects won't radiate G waves .Since the earth is not perfectly spherical and for that matter " any heavenly body will not be spherical due its own gravitational effect " will give off G waves and the energy the loose by this would be very less. But it would be there and would cause a shrink in the orbit howsoever may small. I got that electromagnetic radiation is primarily due to the dipole. But why is Gravitational .R because of change in quadrupole moment . Why is it due to change in and what does the quadrupole moment of mass mean Why not dipole $\endgroup$ – Shashaank Feb 15 '17 at 18:50
  • $\begingroup$ It is all in the mathematics,the dipole radiation is zero. see this en.wikipedia.org/wiki/… it is a tensor dependence that gravitational waves have. I think that the quadrupole moment in the quote for the mass is in an expansion of the shape in moments, but it could possibly be a mistaken usage. It is the asymmetry that is important. $\endgroup$ – anna v Feb 15 '17 at 19:16
  • $\begingroup$ Ok thanks ! Perhaps my maths is not that good right now that I will able to understand tensors. I just wanted to know that it's just because of maths. And just one thing more . About the 2 question is this not correct : The electron and proton are bound in an atom . The both have a common COM. In such a bound condition there cannot be a radiation from the atom. Is this thinking correct ? $\endgroup$ – Shashaank Feb 15 '17 at 19:20
  • $\begingroup$ It is in the problem of quantum mechanics, they are not "rotating" but are in orbitals, and to find the gravitational interaction between them would again need quantum mechanics so at that level one needs to wait for research. One knows though that the numbers would be very very small from the constants in the classical two body formula. Quantum mechanics makes fuzzy solutions of classical problems. For a classical two body it depends on the trajectories, but some asymmetry exists because of difference in masses. $\endgroup$ – anna v Feb 16 '17 at 5:04
  • $\begingroup$ Thank you ! I am learning Schroedinger's equation now and perhaps after the equation and this question I have started understanding the essence and the true necessity of quantum mechanics $\endgroup$ – Shashaank Feb 16 '17 at 7:41

The earth-sun system absolutely emits gravitational waves, and they will tend to reduce the radius of the earth's orbit. But the effect will be very small (I think negligible) because the accelerations involved are small in relativistic terms.

  • $\begingroup$ The effect of the gravitational wave IS negligible. What do you mean by accelerations involved are small in relativistic terms? An accelerating atom produces EM waves which cancel out? $\endgroup$ – Yashas Feb 15 '17 at 18:28
  • $\begingroup$ This answer has a lot of technical inaccuracies. $\endgroup$ – Yashas Feb 15 '17 at 18:28
  • $\begingroup$ EM waves cancel out . What does that mean . There is no destructive interference ! Energy can 't cancel out. Its not charge that it will cancel out ! $\endgroup$ – Shashaank Feb 15 '17 at 18:52
  • $\begingroup$ I decided to just delete that part, @Shashaank $\endgroup$ – Mark Foskey Feb 16 '17 at 14:08
  • $\begingroup$ (continued) I'm not confident enough about it. Two charges at opposite ends of a very short stick are a poor model of an atom. (However, if you accelerate such an assembly I think you will get destructive interference.) $\endgroup$ – Mark Foskey Feb 16 '17 at 14:24

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