# Einstein Field Equations, how many?

1. Dec 19, 2015

### Igorr

Hello,

Action of general relativity consists of two terms: action of gravitation, dependent on metric tensor and its derivatives; action of matter, say one freely moving point mass particle, dependent on particle coordinates and metric tensor. Stationary principle for this action gives all the eqs: variation w.r.t. metric tensor gives Einstein eqs with stress-energy tensor on r.h.s.; variation w.r.t. particle coordinates gives motion eqs for the particle, resulting that particle moves along a geodesic.

Further, I read (in Dirac's lectures on General Theory of Relativity) that Einstein eqs are not independent. Due to Bianchi identity for Ricci tensor there are certain compatibility conditions on stress-energy tensor. And from these conditions it follows that the particle moves along a geodesic. It's OK, the system is closed and has a solution. Although it would be enough to consider only Einstein eqs to reproduce the whole dynamics.

My first question comes from an attempt to count degrees of freedom. For 10 independent variations contained in metric tensor I would expect 10 independent gravitational eqs (PDEs); for 4 independent variations in coordinates - 4 independent eqs of motion (ODEs). Not just 10 eqs with dependencies from which 4 others follow.

The related question: let me fix coordinate system and fix the world line of the particle, e.g. let it be a straight line in non-flat metric. The particle is now non-free, by intent. I consider variation of action w.r.t. metric tensor only and obtain 10 eqs. They must be compatible since the action must have a stationary point somewhere also for fixed particle's path. The eqs seem to be just the same Einstein eqs. Which are not compatible since the particle does not move along a geodesic.

Do I miss something?

Cheers, Igor.

2. Dec 19, 2015

### bcrowell

Staff Emeritus
3. Dec 19, 2015

### Staff: Mentor

For the case of a single, freely moving point mass particle, yes. But in your second case (non-flat metric), you are implicitly violating that assumption. See below.

Which means you also have to include something more in the matter Lagrangian to explain why the particle is now non-free; otherwise your solution will not be self-consistent. In other words, if you specify that the only "matter" present is a single, freely moving point particle, then you can't dictate its worldline, whether spacetime is flat or curved and regardless of which coordinates you pick: you have to solve the equations of motion to find the worldline (and if the only matter present is a single particle, that solution will always give you a geodesic worldline, because there's nothing else for it to interact with to exert any non-gravitational force on it). If, OTOH, you specify that the particle follows a particular worldline, then you can't assume that the only matter present is a single particle, since that's obviously wrong; you have to include whatever other matter sources are necessary to make the particle's worldline what you specified it was.

The EFE can perfectly well describe matter that does not move on a geodesic. But, as above, the matter Lagrangian (and therefore the stress-energy tensor) needs to include the necessary matter sources to produce the non-geodesic motion. So the "incompatibility" here is not in the EFE; it's in your assumptions, which are not consistent with each other (non-geodesic worldline + only matter present being a single particle).

4. Dec 20, 2015

### Igorr

Thank you, guys!

@PeterDonis: Physically, one cannot fix the worldline without adding external force and corresponding material term into eqs. Mathematically, one can fix the worldline in the action and ask what is the stationary point of this functional. The purpose is to understand the role of compatibility conditions for EFE.

I think I can answer my own question and count degrees of freedom as follows:

EFE are not independent because the action is invariant under diffeomorphisms. The action depends not on all 10 components of metric tensor, but on those minus 4 diffeomorphisms. So we have 6 physical and 4 unphysical degrees of freedom. 4 compatibility conditions for EFE come from variation w.r.t. unphysical degrees of freedom. They are conditions on stress-energy tensor. If one intentionally adds material term violating the symmetry under diffeomorphisms, the whole action functional becomes dependent on unphysical degrees of freedom. If the term is chosen unluckily, there will be no stationary point for such action. To model this situation, one can imagine an action functional as a sum of a function of physical variables and a linear function of unphysical variables. There is no stationary point w.r.t. unphysical variables. Compatibility condition for EFE helps to prevent this issue.

5. Dec 20, 2015

### Staff: Mentor

And saying that those things are zero corresponds to fixing the worldline as being a geodesic, yes.

This is not "fixing the worldline". Finding the stationary point of the action functional is finding a worldline by solving the equation, not fixing it before finding the solution.

Can you give a reference for the compatibility conditions you refer to? I may just not be familiar with that particular terminology, but I'm not sure what conditions these are.

6. Dec 21, 2015

### Igorr

Action depends on the metric (g) and the worldline (x). Normally one finds stationary point w.r.t. both, variation w.r.t. g gives EFE, variation w.r.t. x gives the shape of the worldline. One can also, for the purpose of mathematical exercise, fix g and find the shape of the worldline in the fixed metric. At the first glance, it seems that one can fix x and find the metric for the given shape of the worldline. However, compatibility conditions for EFE can prevent this. The reason seems to be that fixing an explicit formula for x in the action will not only fix the shape of the worldline, but will also break reparametrization symmetry of the action. As a result, the functional will not have a stationary point, in a way as a function f(x)=x does not have a minimum.

These are just continuity conditions on stress-energy tensor, which appear if one applies Bianchi identity for Ricci tensor in l.h.s. of EFE.

P.A.M.Dirac, General Theory Of Relativity, Florida State University 1975, Chap.25

Or one can read this here:

https://en.wikipedia.org/wiki/Solutions_of_the_Einstein_field_equations#Solving_the_equations

"notice that only 10 of the original 14 equations are independent, because the continuity equation is a consequence of Einstein's equations"

7. Dec 21, 2015

### Staff: Mentor

In other words, this is just the condition that the covariant divergence of the SET is zero. That's fine. But I don't understand why you think this is somehow incompatible with non-geodesic motion of a particle. Can you give a specific example where you think there is a problem?

8. Dec 21, 2015

### Igorr

There is a proof in

P.A.M.Dirac, General Theory Of Relativity, Florida State University 1975, Chap.25

that continuity equation on stress-energy tensor is equivalent to (1) conservation of mass and (2) motion of matter along geodesics. Of course, this is for the case when the matter is involved in gravitational interaction only, so that stress-energy tensor does not contain other terms like EM-field, pressure etc.

This looks quite reasonably, my problem was just to understand why continuity equations appear twice, as a consequence of EFE and as motion equations for the matter. The first ones appear in variation of the action w.r.t. metric (g), the second ones - w.r.t. particle coordinates (x). Those variations are independent, so I would expect they must produce different eqs. That's why I'm trying to disable x-variations and still have the eqs of motion, as a consequence of EFE.

I see that this problem is known:

https://en.wikipedia.org/wiki/Geode...llow_from_the_field_equations_for_empty_space

starting from "Albert Einstein believed that the geodesic equation of motion can be derived from the field equations for empty space..." and ending "...this claim remains disputed".

May be it becomes more clear, if we note that the worldline = geodesic is a condition not only on the world line, but also on metric. So even if we fix the worldline in action to a certain curve, the metric will try to adjust to this curve, in a way that it will become a geodesic.

If the metric cannot adjust itself in this way, due to global topology, boundary conditions or other reasons, this will be an indication that the action functional does not have a stationary point. This is also possible, especially if our intervention to the action functional breaks reparametrization symmetry, so that unphysical degrees of freedom can lead to unbounded variations of action.

9. Dec 21, 2015

### Staff: Mentor

Yes, and that is precisely the restriction that, physically, makes the matter move on geodesics. If you put in those "other terms", the continuity equation is still satisfied (because it's always satisfied), but matter no longer moves on geodesics. See below.

They don't "appear twice", they just have two different relationships to two different things. The EFE tells you the relationship of the SET to the curvature of spacetime; the continuity equations are a consequence of that relationship, so they tell you how the SET components have to be related in order to satisfy the EFE. The equations of motion for the matter described by the SET are, in turn, a consequence of the continuity equations; in other words, the equations of motion tell you how the matter has to move in order to satisfy the continuity equations.

No, it isn't. You have things backwards. See below.

No. Again, this is backwards. You can't "fix the worldline in action to a certain curve". You derive the equation for the worldline from a solution to the EFE for a given SET that you have already obtained.

Let's consider specific examples, to illustrate what I mean. First, consider a spacetime containing a spherically symmetric region of "dust", i.e., matter that has a finite energy density but zero pressure. In a local inertial frame, this matter will have an SET with only one nonzero component $T_{00} = \rho$, the energy density. Let us suppose that, at some instant of time in an appropriate coordinate chart, this region of dust is motionless and has some finite radius $R$. Then we can solve the EFE and apply the continuity equation to the SET to derive the worldlines of each infinitesimal piece of matter in the dust region. These worldlines will be geodesics that start out motionless at the initial instant of time (the one we postulated above), but will then start free-falling inward, ultimately meeting at a single point. (There are other interesting aspects of this solution as well--it is in fact the solution describing idealized gravitational collapse that was published by Oppenheimer and Snyder in 1939--but we don't need to go into them here.)

For our second example, consider a region of matter that is similar to the above, but this time with nonzero isotropic pressure. In a local inertial frame, this matter will have an SET which is diagonal, with components $T_{00} = \rho$, $T_{11} = T_{22} = T_{33} = p$. We again postulate that, at some instant of time in an appropriate coordinate chart, the region of matter is motionless and has finite radius $R$. Again, we can solve the EFE and apply the continuity equation to the SET, and thereby derive the worldlines of each infinitesimal piece of matter. But this time, these worldlines will not be geodesics, because of the nonzero pressure. If the pressure as a function of density and radius is just right, each worldline will in fact be motionless--the object will be a static, spherically symmetric mass like an idealized non-rotating planet.

In both cases above, the SET satisfies the continuity equations, and the worldlines of the matter are determined by solving those equations. But in the first case, the worldlines obtained are geodesics, while in the second, they are not. The difference is in the actual specific components of the SET (which in turn can be traced back to differences in the action for the matter). Also, the metric is different in these two cases (to take just the most obvious difference, the metric is static in the second case if the pressure is right, but not in the first), but this is not because the metric is "adjusting" to the worldlines. The worldlines adjust to the metric; the metric is determined by the EFE and the specific SET which is taken as given (or, as above, derived from an action for the matter which is taken as given).

10. Dec 21, 2015

### rubi

The coupled particle-gravity system of differential equations can be obtained from the action
$S[g,x] = \frac{1}{2\kappa}\int\mathrm d^4x \sqrt{-g}R - mc^2\int\mathrm d s$
Since $g$ shows up in both terms of the action, a simultaneous variation of the action with respect to both $g$ and $x$ will produce a backreaction term in the stress-energy tensor. This backreaction is usually assumed to be small and therefore neglected. However, the variation will still yield only the well-known geodesic equation for $x$, since the first term doesn't contain $x$. No modifications arise.

11. Dec 21, 2015

### Igorr

It sounds like you are not permitting me doing this :)

OK, look:

ΔA=∫ dτ √(x'μ gμν x'ν), x=c0+c1τ

I have fixed the worldline to a straight line. The result is still a functional of metric, which I can variate to find stationary point of action (w.r.t. metric).

So I'm puzzled why I cannot do this. May be because it violates reparametrization invariance, because such fixation is not possible physically, because this is non-standard procedure. However, as mathematician I can do whatever I want. After all, it's just a functional which has a stationary point, or not. If not, I am interested why.

We are talking past each other.

You are trying to explain me a procedure to find a solution of a physical problem. By solving EFE together with continuity eqs for SET and state eqs, which can include other interaction or not, so worldlines will be geodesic or not. Also don't forget to fix particular coordinate conditions, choose initial and boundary conditions, solve it analytically or numerically, by sequential approximations or directly as PDE. Thank you, I understand this.

While I am interested by the other question, why conditions of consistency of one set of equations, appearing in variation of the action w.r.t. one set of variables, exactly coincide with the other equations, appearing in variation of the same action w.r.t. the other set of variables. This is may be a technical question, but I'm interested in such issues.

Especially because it's a non-standard situation.

In electrodynamics there are field eqs, found in variation of action w.r.t. vector-potentials, and eqs of motion for the charges, found in variation w.r.t. particle coordinates. However, eqs of motion for the charges do not follow from the field eqs.

In string models there are equations for the worldsheet and equations for particles which live on the worldsheet. These eqs are coupled, but in no way one eq follows from the other.

Here is a little model which illustrates the point.

A=∫ dτ (m√(x'2)+vx)

Just a relativistic particle in Minkowski metric, modified by a term (vx), where v is a constant vector. This term violates reparametrization symmetry as well as Poincare symmetry, however I will add it here, for a moment, if you allow :) The eqs of motion are

(mx'/√(x'2))'=v

The eqs are not independent, due to reparametrization symmetry of original action. Their multiplication to x' gives:

0=vx'

which is a condition on v. Or x.

Now let v be a timelike vector. Then x' is a spacelike vector, so that square root in the above formulae is undefined. This eq has no solutions. This action functional has no stationary point.

If v is a spacelike vector, solutions are straight lines orthogonal to v. In this way solution is trying to adjust itself to a vector v, which I have fixed in the action. If it cannot, because the initial derivative in Cauchy problem does not satisfy this condition, or initial and final data in boundary problem do not allow it, the action functional has no stationary point.

@PeterDonis: sorry to waste your valuable time explaining me the standard procedure. Although thinking more about it, I have some questions, which I better ask in a different thread.

@rubi: can you please explain me how you write a square root over a long formula? Thank you.

12. Dec 21, 2015

### Staff: Mentor

It isn't a question of "permission". You can do anything you want mathematically, but if what you are doing doesn't make physical sense, you will end up getting answers that don't make physical sense. That's not the fault of the physics; it's the fault of your procedure, writing down equations that don't make physical sense.

No, you haven't, because you haven't said what $x^{\mu}$ and $g_{\mu \nu}$ are. The equation you wrote down is not satisfied for all possible $x^{\mu}$ and $g_{\mu \nu}$. It isn't even always satisfied for $x^{\mu}$ and $g_{\mu \nu}$ such that $x^{\mu}$ is a solution to the equations of motion obtained when you take the Einstein tensor of $g_{\mu \nu}$, multiply it by $8 \pi$, call that the "stress-energy tensor", and derive equations of motion from it. So you can't just arbitrarily write down that equation; you have to look at what it means, physically, and under what physical conditions it will be satisfied. That's what you appear to be ignoring.

Sure, but that doesn't mean it will make any sense physically. We are doing physics here, not mathematics.

It looks to me like you're confusing several different things. There is only one variable with respect to which the action is being varied: the metric. Varying the Einstein-Hilbert action with respect to the metric gives you the Einstein tensor; varying the matter action with respect to the metric gives you the stress-energy tensor. Varying the complete action, containing both parts, with respect to the metric gives you the Einstein field equation.

If you are talking about deriving the equations of motion from the continuity equation satisfied by the stress-energy tensor, that doesn't require any variational principle; you just solve the differential equations.

If you are talking about deriving the geodesic equation for a given metric by varying a curve with respect to its affine parameter, that has nothing to do with equations of motion; it is purely geometry. You are finding the geodesics of a particular geometric object. There is no requirement that anything physical has to actually move along those geodesics. And you're not varying the action; you're just varying a curve.

And what, physically, is the meaning of this extra term? Why is it there? What is it telling you? Again, we're doing physics here, not math; you can't just wave your hands and add a term to an equation without explaining why, physically, it should be there.

13. Dec 21, 2015

### dextercioby

It works only in LaTex with the command \sqrt{however long this may be} $\sqrt{x^2 + 6.24x + \sqrt{7\sqrt{5}}}$

14. Dec 21, 2015

### Igorr

@PeterDonis: First, I said what x is. Second, it seems now that you don't understand the matter, saying that there is only one variable with respect to which the action is being varied, the metric. At least, in Dirac's lectures the action is varied w.r.t. all variables entering in it. Third, I treat your post as an offence.

15. Dec 21, 2015

### Igorr

16. Dec 21, 2015

### Staff: Mentor

Last edited by a moderator: Dec 21, 2015
17. Dec 21, 2015

### Staff: Mentor

Yes, you're right, you said $x = c_0 + c_1 \tau$. I was misreading your equation before. (Using the LaTeX feature helps to make equations more readable, btw.)

Note, however, that the above still, in itself, does not make $x$ a geodesic; it's a "straight line" only in the sense that it's a linear function of the affine parameter $\tau$, but you can find coordinates in which $x^\mu$ is a linear function of $\tau$ for any worldline, geodesic or not.

I understand how variational principles work. I'm not sure I understand what issue you are raising. It would really help if you would give a specific example, as I did in an earlier post.

That's what you vary the action with respect to in order to get the Einstein Field Equation. When I said what you refer to above, that is what I was talking about.

I see that rubi mentioned in an earlier post a simultaneous variation of the coupled particle-gravity action with respect to $g$ and $x$. I'll ask him about that in a separate post.

Disagreement with what you are saying, asking for clarification because I'm not sure I understand what you are saying, and saying that you appear to me to be confusing different things, are not "offenses". They are part of discussion. Instead of taking offense, you should respond by addressing the substance of what I am saying.

18. Dec 21, 2015

### Staff: Mentor

Shouldn't there be a $\sqrt{-g}$ in the last term?

Just to be sure I understand: varying this action with respect to $g$ produces the Einstein Field Equation with a RHS that looks like the particle's energy times a delta function that makes the RHS nonzero only on the particle's worldline. Varying $\int \mathrm{d}s$ with respect to $x$ tells you the equation of the particle's worldline, so you know at which points the delta function is nonzero. What is the "back reaction" term? Is it the delta function term? (Since that is usually ignored and the spacetime geometry is computed using the vacuum EFE, and then the particle is treated as a "test particle" moving on a geodesic of that geometry.)

19. Dec 22, 2015

### rubi

Well, in general, it would be $S_{\text{particle}}=-mc^2\int\sqrt{-g(\dot x(t), \dot x(t))}\mathrm d t$, but for brevity, I have written it in affine parametrization with tangent vectors rescaled to length $1$. One can always parametrize the curve this way by choosing the arc length as the parameter.

That's right. By varying this action with respect to $g$ and $x$, one gets a system of coupled differential equations:
1. $G_{\mu\nu}(g) = \kappa T_{\mu\nu}(g,x)$ (Einstein's field equations with some $x$-dependent stress-energy tensor)
2. $\nabla_{\dot x} \dot x = 0$ (the plain old geodesic equation with no modifications. It depends on $g$ through $\nabla$.)

The back reaction term is $T_{\mu\nu}(g,x) = \frac{2}{\sqrt{-g}}\frac{\delta S_{\text{particle}}}{\delta g_{\mu\nu}}$. I haven't worked it out, but it would probably involve a delta function with support on the worldline. Due to the singular nature of point particles, I expect similar mathematical problems to arise as in electrodynamics, where the infinite self-energy of point particles also needs to be treated with a lot of care if one writes down the coupled Maxwell-particle system. I don't know how much this has been studied in the case of GR.

20. Dec 22, 2015

### Staff: Mentor

I'm not referring to the parameterization of the curve; I'm referring to a factor of $\sqrt{-g}$, i.e., the square root of minus the determinant of the metric tensor, that I thought should appear in the matter term in the action as well as in the curvature term $R \sqrt{-g}$. Normally, that is necessary in order to make the integration measure $\mathrm{d}^4 x$ invariant under diffeomorphisms. However, the way you have written the matter action, as $\int \mathrm{d}s$ instead of $\int \mathrm{d}^4 x$, it might not be necessary.

Ok, good. This is just the stress-energy tensor corresponding to the matter action.

Yes, it would. As you say, that really means that a point particle is not a good matter model; you have to take great care in using it.