# Degrees of freedom in the EFE

1. Nov 16, 2015

### Staff: Mentor

Edit: this thread was split off from another where it was off topic https://www.physicsforums.com/threa...gful-to-speak-of-the-density-of-space.843442/ . The question in the original thread was if you could express GR in terms of a density of spacetime. I made a mistake in the following response, and Demystifier explains:

The number of degrees of freedom is too small. A density field has one degree of freedom. The metric has 10. You cannot replace something with 10 degrees of freedom by something with 1.

Last edited: Nov 17, 2015
2. Nov 16, 2015

### Demystifier

I think that nobody suggested that space-time is determined only by its density. After all, a surface of the rubber ball has its density and a curved (2-dimensional) metric tensor.

In addition, in GR, the metric tensor has only 2 physical degrees of freedom. The rest are "gauge artifacts".

3. Nov 16, 2015

### Staff: Mentor

Hmm. I wasn't aware of that. Do you have a reference (preferably accessible)

Last edited: Nov 17, 2015
4. Nov 16, 2015

### Demystifier

For a simple argument see
http://physics.stackexchange.com/qu...-graviton-versus-classical-degrees-of-freedom
or
http://physics.stackexchange.com/qu...out-the-degree-of-freedom-in-general-relatity

In a nutshell, among 10 Einstein equations (for 10 components of the metric tensor), 4 do not contain the second-time derivatives. They only serve to constrain initial conditions. These constraints remove 4 degrees of freedom, which gives 10-4=6 free degrees. But different metric tensors may describe the same geometry described in different coordinates. To remove the non-physical freedom in the choice of 4-dimensional coordinates one needs 4 additional constraints, which finally gives only 6-4=2 physical degrees. For gravitational waves, they correspond to two possible polarizations of the gravitational wave (spin "up" or spin "down").

Last edited: Nov 16, 2015
5. Nov 16, 2015

### Staff: Mentor

That's interesting, thanks.

The stress energy tensor seems to have more degrees of freedom, although I am not sure how to count it exactly. It would seem that you start out with 10 again, and then lose 4 for the continuity equation. So either the stress energy tensor must lose another 4 degrees of freedom somewhere else, or the extra degrees must go to the other components of the Einstein tensor.

6. Nov 16, 2015

### Staff: Mentor

I think the difference is that the degrees of freedom of the metric tensor are "gravity only" degrees of freedom; but the degrees of freedom of the SET include "matter" degrees of freedom.

7. Nov 17, 2015

### Demystifier

The stress-energy tensor describes matter, so its degrees of freedom are described by the equation of motion for the matter. This is not the Einstein equation, but another independent equation coupled with Einstein equation.

Take, for example, a scalar field described by the Klein-Gordon equation (coupled to gravity). It contains only one degree of freedom (per point in space, of course). This means that different components of the energy-momentum tensor are not independent.

As another example, consider all matter fields of the standard model of particles. Their total number of degrees of freedom is much larger than 10. This means that changes of some (combinations of) matter degrees of freedom do not change the total energy-momentum tensor. For instance, the energy-momentum tensor does not see a difference between an electron with spin-up and electron with spin-down. Or between red quark and blue quark. Or between particle and antiparticle.

8. Nov 17, 2015

### Staff: Mentor

I am interested more in the maximum degrees of freedom for the stress energy tensor. I am pretty sure that it is at least four, ie three principal axes of strain, plus energy density.

It seems odd that a quantity with four or more degrees of freedom could be equal to a quantity with two degrees of freedom.

9. Nov 17, 2015

### Demystifier

What is important is that the total number of dynamical equations of motion is equal to the total number of dynamical degrees of freedom. This principle is not violated here.

10. Nov 17, 2015

### Staff: Mentor

What do you mean by that?

11. Nov 17, 2015

### martinbn

This may be just about terminology, but you can think (if I am not wrong) that there are six degrees of freedom, two of which are dynamical.

12. Nov 17, 2015

### Demystifier

Let me simplify. Initially, the "Einstein equation" contains 10 equations for 10 unknown components of the metric tensor, and depends on 10 components of the energy-momentum tensor. But when you eliminate the 4 non-dynamical equations (i.e., those which do not contain second time-derivatives) and fix 4 coordinate conditions, what remains is 2 equations for 2 components of the metric tensor, which depend on 2 components of the energy-momentum tensor. So everything matches.

13. Nov 17, 2015

### martinbn

I take this back. Now I am convinced that this would be a very non-standard use of terminology. The degrees of freedom should be counted as two.

14. Nov 17, 2015

### Staff: Mentor

This is the part I am missing. How can that happen given a general stress energy tensor with, say 6 DOF?

15. Nov 17, 2015

### Roy_1981

"This is the part I am missing. How can that happen given a general stress energy tensor with, say 6 DOF?"

Not sure what is the inconsistency here. Lets take the case of electrodynamics as a more basic example. Here the source i.e. the 4-current has 4 components, with just one constraint, namely, the continuity, That leaves us three independent sources. The EM fields (not the 4-potential) has 6 components to start with, but the Bianchi Identities (no mag. monopole plus Faraday's law) removes 3 of the, thus giving us three components of the EM fields to be fully independent, same as the number of sources. But when you simply focus on the components of the gauge potential A_mu, you see 2 and not three. The reason is that you have naively set A_0 = 0 (so called Coulomb Gauge). You can't do so when charge-density is non-zero. (Sure A_0 is non-dynamical because it does not have a momentum, but doesn't mean it is zero!.) One you choose to keep the A_0, then you see you have 3 degrees of freedom and not 2. This is so transparent in the so called Lorenz gauge where the gauge condition looks exactly like the continuity equation:
http://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node31.html

Another thing I would like to point out is that the term degrees of freedom, does not only refer to coordinates, but to momenta as well especially for 2-derivative lagrangians or equations. Just the fields such as electromagnetic gauge field and metric tensor DOES NOT comprise the entire set of degrees of freedoms, their derivatives or the momenta are there as well. This happened in our EM example, E and B fields are not just made out of A but we need the derivatives as well. So you only count d.o.f correctly if you first start with fields AND their derivatives as d.o.f's and then start eliminating by using constraints and gauge conditions. Then you will see, the number of d.o.fs and the sources are equal.

Now for gravity/metric of course the analysis is not as simple and to be honest I haven't checked it. But as I said, just counting metric tensor components doesn't exhaust the full set of dof's. Cheers!

16. Nov 17, 2015

### Staff: Mentor

I think the point is that there are two dynamical DOF for the metric, but there may be more dynamical DOF for the SET components; so to have a fully determined set of equations, you may need to add equations of motion for the SET components, derived from some knowledge of the matter present.

In other words, in order to solve the EFE, you have to already know what the SET components are; but knowing those might require solving other equations, containing additional DOF not captured in the EFE.

17. Nov 17, 2015

### Roy_1981

Well not necessarily, you should be able to solve Einstein equations for arbitrary Stress tensor. Consider it to be an blackbox. It is true you can always specify matter and then determine its stress tensor but it is not a necessary condition. All Einstein Field equations specify is that all you need is a conserved stress tensor as a source of warping for consistency.

18. Nov 17, 2015

### Staff: Mentor

Sure; I only said you "might" have to solve other equations, not that you definitely would have to. If you don't care about the physical reasonableness of the stress tensor, you can plug in anything you want. But if we're talking about degrees of freedom, we probably want to have at least some degree of physical reasonableness; nobody cares how many degrees of freedom a configuration has if it's never actually going to occur.

19. Nov 17, 2015

### Roy_1981

Yes indeed, but it so happens the definition of physical reasonableness evolves with experience and with newer examples. DeSitter or Anti-desitter spacetimes violate reasonability expecations (negative pressure or negative energy) which are more appropriate for asymptotically flat spacetimes, yet they are now considered physical spacetimes (AdS describes in a dual way a lot of strongly couple matter while our universe is supposed to have had a a desitter era in the past and is asymptoting to one in the future). So it's better to be general and impose as few as possible restrictions needed to maintain internal consistency so that when we come across novel examples we don't have to change our formalism much. It's purely a matter of taste.

20. Nov 18, 2015

### Demystifier

That's in fact much simpler than you think. Consider a toy model for 2 variables $g_1(t)$ and $g_2(t)$, satisfying the set of 2 equations
$$\ddot{g}_1(t)=T_1(t)$$
$$\dot{g}_2(t)=T_2(t)$$
where $T_1(t)$ and $T_2(t)$ are given sources, playing a role similar to energy-momentum in the Einstein equation. Note that the second equation is not dynamical, in the sense that it does not contain a second time-derivative. Only the first equation is dynamical, so the dynamics of the system above reduces to a single equation
$$\ddot{g}_1(t)=T_1(t)$$
It depends on only one function $T_1(t)$, despite the fact that there is also a second function $T_2(t)$. There should be nothing mysterious about that.

Perhaps the only confusing thing is the role of the second equation. Nobody said that this equation is irrelevant. All what is claimed is that this equation is not dynamical, which means that it does not contain second time derivatives.* The resulting non-dynamical variable is not called a degree of freedom. In post #11 martinbn suggested that such a variable could be called a "non-dynamical degree of freedom", but later he realized that such terminology would be very non-standard.

*Actually, it is slightly more complicated than that; the Dirac equation also does not contain second time-derivatives, and yet it is dynamical. But let me not delve into it now.