0
$\begingroup$

My question is as it sounds. We know that in a vacuum, materials radiate away their heat energy. To my understanding, this is the result of charged particles losing their energy to produce photons. However, I think it is also true that a lower energy state implies that particles on average, will be closer together. Wouldn't this mean there is also a loss of gravitational potential energy? It would be a relatively negligible amount of energy in comparison to the photons emitted, but wouldn't this necessitate that gravitational energy is lost producing a gravitational wave?

$\endgroup$
1
$\begingroup$

Thermal expansion changes the volume of objects, but this has nothing to do with gravity.

In principle vibrating atoms do emit gravitational waves, so there would be a loss. However, this loss is absurdly small. The energy loss is $$L\sim \frac{c^5}{G}\left(\frac{R_s}{r}\right)^2 \left(\frac{v}{c}\right)^6$$ where $R_s=2GM/c^2$. The time it takes to lose a kinetic energy of $K=(1/2)Mv^2$ is $\tau=K/L$. So if we look at a hydrogen molecule at room temperature, the rotational speed is $v\approx 740$ m/s and I get $L=9.2586\times 10^{-69}$ W and $\tau=4.9491\times 10^{46}$ s. This is about $10^{39}$ years, somewhat longer than the expected time to proton decay - gravitational wave cooling is very inefficient.

[Appendix: I noticed a slight error above. The energy loss formula implicitly assumes gravitationally bound orbits, while the hydrogen molecule is electromagnetically bound and will hence move faster and emit more energy. Actually redoing the calculation using the derivation in appendix A of this paper gives an estimate of $5.32\times 10^{36}$ years cooling from 300K - qualitatively the same kind of result, but a bit faster. ]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.