# How many DOF does gravity actually have

• I
Commonly it is said that the gravitational field has 6 degrees of freedom (a symmetric tensor, that is 10, minus 4 conservation laws). But is this accepted without any dispute?
The reason for my question are so called "indirectly coupling" models, where the primary gravitational field is something, and this something creates the metric of spacetime which couples in the usual way. In the 1970s, Wei-Tou-Ni had a scalar as the primary field. So this distinguished expert deemed possible that the gravitational field only has a single dof.
Ni`s theory was not successful, but meanwhile there are considerations to have a primary vector field, what would mean 4 dof at most (if there are no further restrictions on this vector field).

Thank you very much in advance!

PeterDonis
Mentor
Commonly it is said that the gravitational field has 6 degrees of freedom

Actually, it's not. There are also 4 degrees of gauge freedom that are unphysical; when you eliminate those via a choice of gauge, you are left with only 2 degrees of freedom. These correspond to the two possible polarizations of a gravitational wave.

Demystifier
Peter Donis, thank you very much.

Your answer hit a sore point, since I never have fully comprehended this dof-count. I would like to focus on electrodynamics first, and come back to gravity later.

I can understand that ##A^0## and ##div A## are c-numbers, so they do not correspond to dynamical degrees of freedom in second quantization.

But for a „classical“ (which actually means first quantization) configuration emerging from the solution of Maxwell’s equations, are these two dof superfluous as well? I know that the source only has 3 dof because of charge conservation. I have always thought that this corresponds to the gauge degree of freedom of the field, because charge conservation is a consequence of the gauge invariance of the action. But I understand you in the way that the gauge degree of freedom comes in additionally. So ##A## has 4 dof minus one from charge conservation minus another one from gauge invariance?

The field having only two degrees of, does this mean that I can make such gauge- and Lorentz transformations that ##A^0## and, say, ##A^1##, are zero everywhere for any solution of Maxwell’s equations?

There are two Lorentz invariants of the electromagnetic field, which are gauge invariant by their very construction, namely ##E^2 - B^2## and ##E \cdot B##. Are these the (only) two dof of the electromagnetic field?

I have to apologize if these questions are dumb, but for me this is among the most confusing aspects of field theory.

PeterDonis
Mentor
I never have fully comprehended this dof-count.

I'm not an expert on it either. I agree that focusing on the EM case first is probably helpful.

Another thing to keep in mind is that when we talk about dof of the field by itself, we are talking about dof of the source-free field. In the presence of a source, there are more dof in the system because the source has its own dof. I'm not entirely clear about how the count works in the presence of a source so let's just focus for now on source-free EM.

„classical“ (which actually means first quantization)

I'm not sure I understand. Are we talking about classical EM? (And by extension, classical gravity?) That's what I had assumed since you posted this question in the relativity forum, not the QM forum. If we're just talking about classical fields, then there's no such thing as quantization (first, second, or otherwise). If you are really interested in the quantum case, I would suggest starting another thread in the QM forum for that case, since it is more complicated and I don't feel confident enough of my understanding of it to say anything useful.

The field having only two degrees of, does this mean that I can make such gauge- and Lorentz transformations that ##A^0## and, say, ##A^1##, are zero everywhere for any solution of Maxwell’s equations?

I think so, but I'm not as confident of this as of the next answer below.

There are two Lorentz invariants of the electromagnetic field, which are gauge invariant by their very construction, namely E##E^2 - B^2## and ##E \cdot B##. Are these the (only) two dof of the electromagnetic field?

To the best of my understanding, yes, that's one way of looking at it, and probably the clearest way of seeing why there are only two dof, since it's straightforward to show that those are the only two possible Lorentz scalars that are consistent with Maxwell's equations. (It's even easier in the 4-vector formalism.)

samalkhaiat
Consider the Maxwell’s equation,

$$\partial^{ \sigma }\partial_{\sigma} A^{ \mu } = \partial^{ \mu } \partial_{ \nu } A^{ \nu } , \ \ \ \ \ \ \ \ \ \ (1a)$$

and the gauge transformation

$$A^{ \mu } \rightarrow A^{ \mu } + \partial^{ \mu } f ( x ) . \ \ \ \ \ \ \ \ (2a)$$

Now, if you substitute

$$A^{ \mu } ( x ) = \epsilon^{ \mu } ( p ) \ e^{ i p \cdot x } ,$$

in Eq(1a) and Eq(2a), you get (the momentum space versions of Maxwell’s equation and the gauge transformation)

$$p^{ 2 } \ \epsilon^{ \mu }(p) = \left( p \cdot \epsilon (p) \right) p^{ \mu } , \ \ \ \ \ \ \ \ (1b)$$
$$\epsilon^{ \mu } ( p ) \rightarrow \epsilon^{ \mu } ( p ) + i f ( p ) \ p^{ \mu } , \ \ \ \ \ (2b)$$

where (we have defined) $$f ( p ) \equiv f ( x ) \ e^{ - i p \cdot x } .$$

Now, for “massive” electromagnetic field $p^{2} \neq 0$, Eq(1b) implies that the polarization vector $\epsilon^{\mu}(p)$ is proportional to $p^{\mu}$,

$$\epsilon^{\mu}( p ) = \frac{ p \cdot \epsilon (p)}{ p^{ 2 } } \ p^{\mu} \ . \ \ \ \ \ \ \ \ \ \ (3)$$

However, this “massive” solution (to Maxwell equation) is not physical because it can be gauged away completely by the following choice for the gauge function
$$f ( p ) = i \ \frac{ p \cdot \epsilon (p) }{ p^{ 2 } } . \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$
To see that, just substitute Eq(3) and Eq(4) in the RHS of Eq(2b). This explains why the photon cannot have non-zero physical (i.e., invariant) mass.
For massless mode (i.e., $p^{2}= 0$ solution), Eq(1b) leads to the (momentum space) Lorenz condition $p_{\mu} \epsilon^{\mu} (p) = 0$. So, in the frame $p^{\mu}= (\omega , 0 , 0 , \omega)$, the polarization vector has only 3 independent components $\epsilon^{\mu} = (\epsilon^{0} , \epsilon^{1} , \epsilon^{2} , \epsilon^{0})$. Again, the $\epsilon^{0}$ component is not physical, for we can get rid of it by the gauge choice $$f ( p ) = i \frac{ \epsilon^{ 0 } }{ \omega } .$$ So for electromagnetic wave moving in the z-direction, the physical polarization vector is confined in the xy-plane and has only 2 degrees of freedom.

Similar analysis can be applied to the Einstein field equations. But, for simpler reasoning, see the following post
https://www.physicsforums.com/posts/5916275/

PeterDonis
haushofer
Also, Zwiebach's string theory book covers this extensively. Recommended.

Many thanks to all.

I first have to say that my reference to first quantization w.r.t.\ Maxwell’s equations was only meant in the sense that these are wave equations like the Klein-Gordon equation or so. But, of course, Planck’s constant does not show up because of the masslessness. So, what I am interested in are the ’’classical’’ (non-quantum) equations of electrodynamics and gravitation.

Having looked through the recommended references, I conclude that the following simple relations hold for the degrees of freedom in the entire region of validity (not only for the description of waves) of Maxwell’s and Einstein’s equations:

Electrodynamics:

4 (vector field) - 1 (charge conservation) - 1 (gauge invariance) = 2

Gravitation:

10 (symmetric tensor) - 4 (energymomentum conservation) - 4 (gauge invariance) = 2

I hope this conclusion is right, else please correct me.

Demystifier