# How to incorporate effects of gravity in a many-electron system?

In the weak interaction limit, behaviors of electrons can still be described by a 1-partitcle equation with a modified mass, where the change in mass can be understood as effects of other neighboring electrons as a single electron drags along. Since the cluster of electrons couples to gravity as usual, then the effective mass of a quasi-particle couples to gravity in the same way as that of a genuine electron. Moreover, the effective mass can be negative which is the case of holes which are “bubbles” in the electron liquid.

In graphene, there are massless Dirac electrons. Do they couple to the gravity?

I think what I intended to ask should be "How to incorporate effects of gravity in a many-electron system?"

(Originally I asked "do effective masses couple to gravity?")

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## 3 Answers

I doubt that the quasi-particle mass couples to gravity like normal mass. Interactions can cause a significant change in the quasi-particle mass while the energy due to the interaction is way below the energy equivalent of the electron mass. This means that looking at the problem as single electrons interaction and calculating the total mass as the individual masses plus some interaction energy, this interaction energy will be negligible compared to the electron mass and to a good approximation the total mass is the same as that for non-interaction electrons.

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The answer is "no", you shouldn't hedge, but +1 anyway. The effective mass of an electron is the hopping parameter on the lattice, and has nothing to do with the stress-energy coupling to gravity. – Ron Maimon Sep 14 '12 at 2:40

The 00 component of the stres tensor in kinetic theory is $$T_{00}=\int p_0p_0 f(x,p,t) \sqrt{g}\frac{d^3p}{p_0}.$$ For a slowly moving particle $$p_0=\sqrt{g_{00}} \left( m-\frac{g_{ij}p^ip^j}{2m}+\ldots\right).$$ When we talk about effective masses in many-body systems we mean that the second term is renormalized, $1/m\to1/m^*$. This will have some effect on gravitational interactions, but the main term is obviously unchanged.

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A possible answer:

A single electron couples to gravity in the usual way. When a large number of electrons are put together, coupling to gravity shouldn't disappear. Gravity, gravitational force actually, can be thought of an additional "electric field" ${\bf E} = \frac{m}{q_e}{\bf g}$ or $q_e{\bf E}=m{\bf g}$.

Then, for whatever dispersion relations, the excitations couple to gravity through $\bf E$ as if an electric field is applied. This is opposed to my previous statements about Bloch electrons.

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