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Maybe another stupid question, but what's the energy of a graviton? Is it $\hbar \omega$?

Does it emit gravitons when an apple falls onto the ground, like photons be emitted when an electron transits from a higher energy level to a lower one?

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up vote 8 down vote accepted

It is assumed, but not measured that e =hw. See this reference about the difficulty of measuring a single graviton.

It is an interesting exercise to compute the amount and wavelength of gravitons from a falling apple. As a first approximation, one can compute the gravitational waves emitted by an apple orbiting the earth and use the formulas developed for binary black holes and neutron stars by the LIGO VIRGO etc experiments. In low earth orbit, the apple will radiate at a wavelength of approximately 90 light minutes, give or take a factor of two. This is very roughly 10^12 meters. Each graviton will then carry about 10^-30 ergs, a very small amount. According to the Wikipedia gravitational wave article the sun earth system emits 200 watts of gravitational radiation, but this would typically emit 10^-34 erg gravitons. 200 watts is 2 10^9 erg seconds, so the sun earth system is emitting 10^43 gravitons per second. Using the formula from the Wikipedia article, the earth-apple system with a one tenth kilogram apple would emit 10^46 times less gravitational wave power, or 10^-42 watts or 10^-35 ergs/second. This implies an average of one graviton every 10^5 seconds, or about once every twenty 5400 second orbits. If your falling apple falls for about one second, it should emit one graviton once out of every one hundred thousand tries. To be more analagous to the orbital picture, your apple should be thrown horizotally like a baseball, rather than falling vertically

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If the exchange of (virtual) gravitons is the underlying QM process corresponding to classical gravitational attraction, then it's hard to see how the apple, or any of its constituent fermions, could be falling at all without gravitons being involved? – Nigel Seel May 30 '11 at 15:21
I think the key here is the distinction between real and virtual gravitons. The virtual gravitons do not result in the emission of gravitational radiation. If you assume all gravitational action is due to virtual gravitons and also assume that the typical wavelength is given by some elementary composition of the size and speed of the system, you could also calculate the number of virtual gravitons exchanged per second. – Jim Graber May 30 '11 at 15:35

This is the case with a caveat: It only works for a weak linearized theory. If gravitons are a small field perturbation on a flat background in a linear Einstein field equation this is the case. In greater generality this is less certain. Energy is not localizable in general relativity. So what I am about to outline below is a linear quantum gravity for long wavelength gravitons in the IR limit.

A gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric $$ g_{ab}~=~\eta_{ab}~+~h_{ab}. $$ The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion $$ R_{ab}~=~\delta R_{ab}, $$ which enters into the Einstein field equation $R_{ab}~-~1/2Rg_{ab}~=~\kappa T_{ab}$, where $\kappa~=~8\pi G/c^4$ is the very small coupling constant between the momentum-energy source and the spacetime configuration or field. The Ricci curvature to first order is then $$ R_{ab}~=~{1\over 2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh~-~\partial_c\partial^ch_{ab}\Big). $$ To within first order the harmonic gauge $\partial_c{h^c}_a~=~1/2\partial_mu h$ the Einstein field equation gives $$ \partial^c\partial_ch_{ab}~-~\frac{1}{2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over c^4}T_{ab}, $$ which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~{1\over 2}\eta_{ab}h$ with the simple wave equation $$ \partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over c^4}T_{ab}. $$

For the wave in vacuum the momentum energy source is set to zero, and the wave equation is $\partial^c\partial_c{\bar h}_{ab}~=~0$. This is a bi-vector analogue to the simple wave equation for an electromagnetic wave in free space. The wave is a transverse traceless wave $h_{ab}~=~A^{TT}_{ab}exp(ik_cx^c)$ with $$ A^{TT}_{ab}~=~\left(\matrix{0 & 0 & 0 & 0\cr 0 & A_{xx} & A_{yx} & 0\cr 0 & A_{xy} & -A_{xx} & 0\cr 0 & 0 & 0 & 0 }\right). $$ The terms $A_{xx}$ and $A_{xy}$ represent a polarization direction. The linearized gravity wave then has a helicity of two, which has its quantum analogue in the di-photon state in quantum optics.

The analogue with a photon in this linearized approximation is for the metric perturbation $h_{ab}~=~\phi^c_a\phi_{bc}$ expanded according to a field. To take this forwards let us expand the field $\phi^a_b$ according to harmonic oscillator operators $b,~b^\dagger$. The fields are then expanded as $$ \phi^a_b~=~\frac{1}{\sqrt{2}}\sum_k E^a_b (b(k)e^{i\theta(k)} + b^\dagger e^{-i\theta(k)}) $$ where $E^a_b$ tetrad, which is discussed more below. The metric perturbation is then $$ \phi^a_c\phi_{cb}~=~\frac{1}{2}\sum_{kk’}\eta^a_b(b(k)b^\dagger(k’)e^{i\theta(k) – i\theta(k’)}~+~b^\dagger(k)b(k’) e^{-i\theta(k’) – i\theta(k)}) $$ $$ +~\frac{1}{2}\sum_{kk’}\eta^a_b(b(k)b^(k’)e^{i\theta(k)+i\theta(k’)}~+~b^\dagger(k)b^\dagger(k’)e^{-\theta(k)-i\theta(k’)}), $$ The first of these terms is a rotating wave, while the second is counter rotating. For the sake of simplicity we ignore this for now. Laplacian operator on the first term gives a wave equation in terms of these fields, and the Hamiltonian is $H~=~\partial_d\phi^c_a\partial^d\phi_{bc}$ is a sum over $\sim~\hbar\omega$ terms. The reader may fill in the details.

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