# Why is there no permittivity-type constant for gravitation?

When I look at electric or magnetic fields, each of them has a constant that defines how a field affects or is affected by a medium. For example, electric fields in vacuum have a permittivity constant $ϵ_0$ embedded in the electric field expression of a point charge: $E = q/4π ϵ_0r^2$. However, if I put this point charge in some dielectric that has a different permittivity constant $ϵ$, the value of the electric field changes. On a similar note, magnetic fields behave very similar but have the permeability constant $μ_0$ instead.

From my understanding, I believe that this is not the case for gravitational fields since the universal gravitational constant $G$ is consider to be a fundamental constant. So I am assuming that even though gravitational fields do operate in different types of mediums, this somehow doesn’t affect the gravitational field value. My question is why is this the case, that is, why isn’t there a permittivity-type constant for gravitation?

• – jinawee Nov 12 '13 at 23:01

## 3 Answers

Permittivity $\varepsilon$ is what characterizes the amount of polarization $\mathbf{P}$ which occurs when an external electric field $\mathbf{E}$ is applied to a certain dielectric medium. The relation of the three quantities is given by

$$\mathbf{P}=\varepsilon\mathbf{E},$$

where permittivity can also be a (rank-two) tensor: this is the case in an anisotropic material.

But what does it mean for a medium to be polarized? It means that there are electric dipoles, that units of both negative and positive charge exist. But this already gives us an answer to the original question:

There are no opposite charges in gravitation, there is only one kind, namely mass, which can only be positive. Therefore there are no dipoles and no concept of polarizability. Thus, there is also no permittivity in gravitation.

• There is a notion of polarization in gravity. In canonical quantization of GR, we expand the metric as $f_{\mu \nu} \sim \int \mathrm{d}^4 p \, \epsilon_{\mu \nu} a^{\dagger}_p ...$, and $\epsilon_{\mu \nu}$ is interpreted as a traceless symmetric rank 2 tensor specifying polarization. In addition, in string theory, the vertex operator of the graviton contains a polarization tensor, which is also contracted with creation operators to produce states with gravitons. – JamalS Apr 6 '14 at 16:41
• @JamalS: You are right, yet this is different from the subject of this question. The polarization tensor of the graviton/gravitational waves does not refer to the polarization due to some charge, but to the polarization of a wave, comparable to that of an electromagnetic wave. For comparison: in electrodynamics, there are two notions of polarization: that of charges in a medium, and that of electromagnetic waves. – Frederic Brünner Apr 6 '14 at 16:46
• @JamalS does (in e.g. string theory) vacuum polarization exist in the sense of the QED vacuum (for instance) i.e. photon going to electron positron pair hence screening the bare electric charge? My guess would be no since gravity as far as I'm aware only has one "charge". – Your Majesty Apr 14 '14 at 19:54

This is speculation, but I would guess that it's for the same reason gravity never repels, only attracts. The electric permittivity tells you how the electrons in a material rearrange themselves to oppose an applied field. I'm not sure what the analogue for gravity would be there.

• I think that you are correct after reading the link provided above by jinawee. This is what I think: if I put a charge inside a dielectric, it will form a ring of negative charges around itself and this in turn reduces the field. However, if I put a solid dense mass-sphere inside the some other mass, I will also attract a thin ring of mass that instead increasing the gravitational field of the dense mass-sphere. – Jesus Nov 12 '13 at 23:27
• That is, because there are two kinds of charges and one type of mass, this is ultimately the reason for why there is no permittivity constant for gravitation. I am sure someone else can summarize this better than I, so please do. – Jesus Nov 12 '13 at 23:27
• It's just a rewriting of the existing constant. It's rare to solve Gauss's Law-type problems for gravity, so the equation is left in the simpler form involving $G$ – Jerry Schirmer Nov 27 '13 at 13:52

You have significantly over complicated the situation.

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N/C ~ N/Kg

So there must be an analogous quantity for ε. Where

k=(4πϵ0)^-1 Therefore there must be an analog G=(4π@)^-1.

As for the positive negative issue it really isn't the permativity has no direction. The direction is determined by the charge, and the mass. It just happens there is only one possibility for mass... or is there? Why given all the parallels does electricity have an anti particle.

• Welcome on Physics SE :) We have a help section e.g. for typesetting formulas - maybe that will be helpful :) – Sanya Nov 28 '16 at 21:42

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