# Conservation laws with Killing fields

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1. Nov 17, 2015

### Neutrinos02

Hi,

In general relativity we have no general conservation of energy and momentum. But if there exists a Killing-field we can show that this leads to a symmetry in spacetime and so to a conserved quantity. Thats what the mathematic tells us. But I don't understand what's the meaning of an timelike/spacelike/lightlike Killing-field? For conservation of energy is it important to have the field ∂t or must it be an arbitrary timelike Killing-field?

If we look at the Kerr solution we see that for some spacetime region the field ∂t will be timelike and for another it will be spacelike Killing-field. What consequences has this for the conserved quantities, since its still a Killing-field?

Thanks for help
Neutrino

2. Nov 17, 2015

### pervect

Staff Emeritus
Suppose you have a one-parameter group of transformations that leaves the metric unchanged, a one parameter group of isometries. Example: Suppose you have a metric $ds^2 = -(x+1)dt^2 + dx^2 + dy^2 + dz^2$. Then the transformation $t' = t + \delta t$ leaves the metric unchanged, so it's a one-parameter group of isometries, where $\delta t$ is the parameter.

The group is generated by an infinitesimal transformations. I don't think I could be terribly rigorus about the generators of groups even if I wanted to be, the non-rigorous explanation though is that an infinitesimal transformation such as $t' = t + dt$ generates the one-parameter group of transformation $t' = t + \delta t$.

We can interpret dt as a (co)-vector field, and we are lead to the textbook idea (Wald, for definiteness), that every one parameter group of diffeomorphisms is generated by a vector field".

A Killing vector field is not generated by just any one-parameter group of diffeomorphisms, however, the one parameter group of diffeomprohisms must leave the metric unchanged, i.e it must be a one-parameter group of isometries.

When none of the metric coefficients is a function of one of the coordinates, the isometries are trivially obvious, as in the above example. More generally, one has to resort to Killings equation, which can be written in the form that $\mathcal{L}_\xi \, g_{ab} = 0$ or $\nabla_a \xi_b + \nabla_b \xi_a = 0$, where $\xi^a$ is a vector, $\nabla_a$ is the covariant derivative, and $\mathcal{L}$ is the Lie derivative ($\mathcal{L}_\xi$ is the Lie derivative along the vector field $\xi$).

3. Nov 17, 2015

### Roy_1981

It's actually pretty simple, and I don't mean mathematically slick. If you have a timelike Killing vector, K then you can find a frame/coordinate system/observe which/who only sees that the killing vector has the time-like component to be non-zero (and the spatial components all zero, e.g. K ~ (1,0,0,0)). Now as you know the conserved quanttiy is the particles 4-momentum dotted with the Killing vector i.e. K.p. Since in the case the only non-vanishing component is non-zero, the conserved quantity is p_0 aka the zeroth component of 4-momentum aka the energy!

Similarly, if you have a spacelike killing vector , say along the x-direction then you can show the corresponding x-component of momentum is conserved!

4. Nov 17, 2015

### Neutrinos02

But if you say a timelike Killing-field conserves the energy and a spacelike the momentum, what about a lightlike killing field and what about the conservation of energy and momentum (together)?

5. Nov 17, 2015

### Roy_1981

Null Killing vector implies conservation of a momentum tangent to a null coordinate. It is asking the same question as what is conserved for a photon for which there is no rest frame and the answer will be some null momentum 4-vector.

Ofc energy and momentum or energy-momentum 4-vector is conserved in general. If you don't choose of a rest frame or special frames then the general statement is K.p is conserved. For general K i.e. not in special frames each component is non-vanishing and then some particular linear combination of the components of the momentum-4 vector is the conserved charge.

6. Nov 17, 2015

### Staff: Mentor

I'm not sure what you mean by this. In a curved spacetime, "conservation" has to be defined carefully.

In any spacetime, the covariant divergence of the stress-energy tensor is zero at every event. This is the most general local version of what we call "conservation of energy" in ordinary language. However, there is no general way to integrate this over a region of spacetime to get a global conservation law.

If we take a test object, defined as an object whose effect on the curvature of spacetime is negligible, then we can describe it by a 4-momentum vector, yes. However, the norm of this 4-momentum will not, in general, be "conserved"; that is, it won't be the same at every event on the object's worldline.

However, in a spacetime with a Killing vector field, the inner product of a test object's 4-momentum with the Killing vector belonging to that KVF at any event is a constant of the motion, as you say, provided the object is in free fall, i.e., geodesic motion. So, for example, in a spacetime with a timelike KVF, the inner product of a test object's 4-momentum with the KVF is a constant called "energy at infinity". Even in this case, however, this "energy" is not something that will be measured locally; for example, an observer at some finite spatial location measuring the energy of the object as it passes him will not get this value, but something different. So the physical interpretation of the value, and the fact that it is "conserved" for geodesic motion, is still not straightforward.

7. Nov 17, 2015

### Roy_1981

Sure, but in the present case, the statement of constancy or conservation is along geodesic or along the world line. The derivative is a normal derivative wrt world line parameter or affine parameter, not a covariant derivative. This is a diffeomorphism/gauge invariant statement, totally unambiguous. The stress-tensor you are talking about and which is covariantly conserved in general, regardless of a particle moving or not (in some sense the stress-tensor of the whole spacetime plus matter) does not follow from some symmetry or isometry. Regardless of existence of an isometry/Killing vectors, the stress tensor is covariantly conserved.

8. Nov 17, 2015

### Staff: Mentor

Yes, but you said "energy-momentum 4-vector is conserved in general". That's not the case "in general". It's only the case in particular circumstances, and in those circumstances we have to be clear about exactly what is "conserved" and what it means physically.

What you are calling a "normal derivative wrt world line parameter" is a covariant derivative contracted with the 4-velocity. So it does involve the extra terms in the connection coefficients that distinguish a covariant derivative from an ordinary partial derivative.

9. Nov 17, 2015

### Roy_1981

1. Perhaps I should remind you that I was talking about energy-momentum 4-vector, not the stress tensor, Tμν. These two are distinct. In particular, energy-momentum 4 vector of a particle in a geodesic is conserved when there is a Killing vector or an isometry using Noether's theorem. However the in GR stress tensor does not need any Killing vector to be covariantly conserved (You can check this by taking covariant derivative of Einstein's field equation without using any Killing vector). It's easy to conflate these two but the distinction is crucial.

2. I'm afraid that is incorrect. it is not a contraction of a covariant derivative, it is in fact contraction of the partial derivative along the tangent or the ordinary derivative wrt to the worldline parameter, λ

Kμ d xμ /dλ = 0, where d/dλ = tνν .

t is of course the tangent vector along the world-line.

10. Nov 17, 2015

### Staff: Mentor

Yes, I understand the distinction. But the ordinary language words "energy conservation" don't just apply to the 4-vector; they apply to the stress-energy tensor as well (referring to something different mathematically in each case). So I think it's better to use unambiguous language.

No, $d/d\lambda$ is not given by $t^{\nu} \partial_{\nu}$. It is given by $t^{\nu} \nabla_{\nu}$. In other words, for a vector field $X^{\mu}$, $d X^{\mu} / d \lambda = t^{\nu} \nabla_{\nu} X^{\mu} = t^{\nu} \partial_{\nu} X^{\mu} + t^{\nu} \Gamma^{\mu}{}_{\nu \rho} X^{\rho}$. Check any GR textbook. Your version is only correct in flat spacetime, where $\Gamma^{\mu}{}_{\nu \rho} = 0$.

11. Nov 17, 2015

### Roy_1981

1. I am not sure which ordinary language you are talking about, but I am not privy to such colloquial conversations. The fact is if you look up something called Noether's theorem - which states that corresponding to every isometry i.e. a Killing vector there exists a conserved charge and in the case of a timelike Killing vector conserved charge is called Energy. Period. There is no ambiguity here with the term "energy" and let's not deviate the focus of the question. Stress tensor in GR is a different animal, completely unrelated to the question of Killing vectors and conservation laws in GR.

2. Again your derivation is out of context, you failed to realize the distinction between a coordinate x and a vector field X[x]. In the case of coordinate x[λ], along a curve (not necessarily a geodesic) parametrized by λ, the tangent vector is defined the way I have stated it. You can look up any GR book or even a book on multivariable calculus. The tangent is defined even before you define a metric, has nothing to do with metric or connection or covariant differentiation.

12. Nov 17, 2015

### Staff: Mentor

I'm quite familiar with it, thanks. I'm not disputing that Noether's theorem is correct.

Once again, check a GR textbook. You will see that this "conserved charge" you are talking about is often called "energy at infinity", just as I did. And for a good reason, since, as I said before, this "energy" is not something any observer at any finite spatial position will actually measure.

We're not talking about the tangent vector; you didn't even state a definition of a tangent vector. We're talking about the derivative with respect to the curve parameter. That's the definition you gave, and that I'm disputing. You said that $d / d\lambda = t^{\nu} \partial_{\nu}$, where $t$ is the tangent vector (which you gave no other definition of). Show me a GR textbook that says that is true in a curved spacetime, rather than $d / d\lambda = t^{\nu} \nabla_{\nu}$.

So what? We're talking about the case where there is a metric. If there isn't, how do you propose to pick out the Killing vector fields? You do realize that the definition of a KVF is that the Lie derivative with respect to the metric is zero, right?

13. Nov 17, 2015

### Roy_1981

1. Good at least we can assume you are familiar with Noether's theorem.
2. The energy I defined can be "measured" locally, it is constant at all points on the worldline, you can pick ANY point on the world-line and compute Kμ dxμ/dλ, all locally at a point and find it to be conserved. Perhaps it would be instructive for you to see this in action when you look at an in-falling geodesic in the Schwarzschild geometry which has a timelike Killing vector - energy is conserved along the trajectory. Forget infinity. It has absolutely NOTHING to do with the "energy at infinity" i.e. the Schwarzschild mass parameter "M". Please don't confuse these two, no matter how many GR books you have read and which have told you otherwise. You need to do the derivation yourself to see this happening, not gonna help looking up references, as I tell my class often.

3. The derivative of the curve with respect to the curve parameter is a tangent vector to the curve. This is the definition of a tangent vector, you can look at say a easy GR ref like Carroll's notes or the math book by Nakahara. And you will discover, to your surprise,that this tangent vector to a geodesic contracted with the Killing vector is conserved. I am a bit too busy to go over the derivation but again Carroll does a decent job - there is a nice introduction to geodesics, tangents etc.

4. Again don't confuse the person who posed the question by lumping together unrelated things. We DO NOT need a metric to define the tangent to a curve or a geodesic- all we need is a normal/ordinary derivative wrt to world line parameter which is further using rules of multivariate calculus is d/dλ = tν∂ν. Even if you did not have any killing vectors the tangent to a geodesic can be defined. I don't know which GR book you are referring to but you could learn these kind of basic stuff from Carroll's note on relativity (http://xxx.lanl.gov/abs/gr-qc/9712019), in particular eq. 2.10 on page 43.

.

14. Nov 17, 2015

### Staff: Mentor

Last edited: Nov 18, 2015
15. Nov 18, 2015

### Staff: Mentor

Just to be clear, the original question was about Killing vector fields, not tangent vectors or geodesics. Defining Killing vector fields requires a metric. That's why I have been focusing on things that require a metric, rather than things that don't.

That said, let me try to summarize key points that have arisen, to try to get the discussion back on track. For a reference, since you mentioned Carroll's online lecture notes, we can use that.

Agreed. Carroll, in equation (1.24) and surrounding discussion, gives the coordinate-free definition.

Disagree. Defining the tangent vector means we know $t^{\mu} = dx^{\mu} / d \lambda$; but it does not mean that we know how to take another derivative of $t^{\mu}$ with respect to $\lambda$, at least not the kind we need to define a geodesic. In order to define a geodesic, we need to define parallel transport, and as Carroll discusses in Chapter 3, you need additional structure on a manifold in order to do that. The additional structure you need is a connection, and in the discussion leading up to equation (3.30) he defines what that is and how it is used to define parallel transport. Technically, you don't need a metric on a manifold in order to define a connection, but in GR that's how we do it; the connection we use is the one derived from the metric.

This doesn't surprise me at all; I already agreed to it in post #6. But demonstrating it requires a metric, since a metric is required to define a Killing vector field in the first place. (Technically, it is also required to contract two vectors; without a metric, you can contract a vector with a covector, but not with another vector. But that's a minor point.) Carroll discusses this in Chapter 5; equation (5.42) gives the relevant definition. Equation (5.43) demonstrates conservation of the inner product of a tangent vector with a Killling vector field for a geodesic.

M is not the same as "energy at infinity". M refers to the mass associated with the spacetime geometry, not with a test object. "Energy at infinity" refers to the inner product of the tangent vector of a test object undergoing geodesic motion with a timelike Killing vector field. Carroll does not appear to use this terminology, but other sources do; for example, MTW uses it. (IIRC Wald does too.)

To see what "measured locally" actually means here, perhaps it would be instructive to take the example you suggest, a radially infalling geodesic in Schwarzschild spacetime, and analyze it.

I will use Painleve coordinates because they are more convenient mathematically for this case than Schwarzschild coordinates. In these coordinates, the tangent vector to a radially infalling geodesic is given by (we are including only the $t$ and $r$ components since the other two coordinates are irrelevant here):

$$t^{\mu} = \left[ 1, - \sqrt{\frac{2M}{r}} \right]$$

The line element is:

$$ds^2 = - \left( 1 - \frac{2M}{r} \right) dt^2 + 2 \sqrt{\frac{2M}{r}} dt dr + dr^2$$

The Killing vector field is $K^{\mu} = \partial_t$, so the constant of the motion is

$$g_{\mu \nu} t^{\mu} K^{\nu} = g_{tt} t^t + g_{rt} t^r = - \left( 1 - \frac{2M}{r} \right) - \frac{2M}{r} = -1$$

Of course this is the same as the norm of the 4-velocity vector; so an observer who is falling in along with the test object along the geodesic will of course measure its "energy" (actually its energy per unit rest mass) to be the same at every point during the fall. This is obvious because the observer is always at rest relative to the object, so the "energy" it measures is just the object's rest mass.

But now consider an observer who is "hovering" at a constant finite value of $r$. What energy will he measure when the infalling object passes him? This observer's 4-velocity (the tangent vector to his worldline) is

$$u^{\mu} = \frac{1}{\sqrt{1 - 2M / r}} \partial_t = \frac{1}{\sqrt{1 - 2M / r}} K^{\mu}$$

So the contraction of $t^{\mu}$ and $u^{\mu}$ will be the constant of the motion given above, times the factor $1 / \sqrt{1 - 2M / r}$. In other words, it will be larger in magnitude than the constant of the motion. Physically, of course, this just means that the infalling object is passing the "hovering" observer at some nonzero velocity, so its total energy as measured by the hovering observer is larger than its rest energy.

So describing the constant of the motion as "energy", without qualification, is ambiguous; energy measured by whom? Energy is observer-dependent, so using the word to describe an invariant seems less than optimal. The term "energy at infinity" avoids that ambiguity.

Last edited: Nov 18, 2015
16. Nov 18, 2015

### Staff: Mentor

As a further note, the fact that the constant of the motion happens to equal the norm of the 4-velocity in this case does not generalize. For example, consider a test object in a circular geodesic orbit about a central mass. If $r$ is the orbital radius, the tangent vector to this worldline is (now giving the $t$ and $\phi$ components, and assuming we are in the "equatorial plane" $\theta = \pi / 2$)

$$t^{\mu} = \left[ \sqrt{\frac{r}{r - 3M}}, \sqrt{\frac{M}{r^2 \left( r - 3M \right)}} \right]$$

The constant of the motion here has only one term:

$$g_{tt} t^t K^t = - \left( 1 - \frac{2M}{r} \right) \sqrt{\frac{r}{r - 3M}} = \frac{r - 2M}{\sqrt{r \left( r - 3M \right)}}$$

However, this is not the energy that the test object will be observed to have, either by an observer orbiting along with it, who will measure it to have its rest energy, or a "hovering" observer at the same radius, who will measure it to have energy (per unit mass)

$$g_{tt} t^t u^t = - \left( 1 - \frac{2M}{r} \right) \sqrt{\frac{r}{r - 3M}} \sqrt{\frac{r}{r - 2M}} = \sqrt{\frac{r - 2M}{r - 3M}}$$

This again illustrates why "energy", without qualification, is ambiguous; but it further illustrates that the constant of the motion given by contracting the tangent vector of a geodesic with a Killing vector field might not always be "locally measurable", even by an observer moving with the object traveling along the geodesic.

17. Nov 24, 2015

### Neutrinos02

So i can't get a conservation of the energy and the momentum if I only have one KVF?

What would be measured? And if it we not measure the energy alonge the geodesic what is conserved (So what does conserved mean in this context)?

Last edited by a moderator: Nov 24, 2015
18. Nov 24, 2015

### Staff: Mentor

Not in the sense of "energy" and "momentum" used here; but remember that there are different possible senses of those terms, and in the general case they are all distinct:

(1) There is "energy" and "momentum" in the sense of "the energy-momentum 4-vector describing an object". This 4-vector is covariant, and has an invariant magnitude equal to the object's rest mass, regardless of whether the spacetime has any KVFs, and regardless of the state of motion of the object.

(2) There is "energy" and "momentum" in the sense of "components of the stress-energy tensor describing some piece of matter". The SET satisfies a local conservation law--its covariant divergence is zero--regardless of whether the spacetime has any KVFs, and regardless of the state of motion of the matter.

(3) There is "energy" and "momentum" in the sense of "contractions of an object's energy-momentum 4-vector with a time translation KVF (for energy) or space translation KVF (for momentum)". These are "conserved" only if the appropriate KVFs are present and if the object is in free-fall (geodesic) motion. (There is also "angular momentum", which means "contraction of an object's energy-momentum 4-vector with a spatial rotation KVF", and is conserved only if the appropriate KVF is present and if the object is in geodesic motion.)

(4) There is "energy" and "momentum" in the sense of "contractions of an object's energy-momentum 4-vector with some observer's 4-velocity (for energy) or some spatial unit vector describing an observer's measuring apparatus for spatial distance (for momentum)". The concept of "conservation" doesn't really apply here because different observers, in general, have different 4-velocities and different spatial vectors describing their measuring apparatus, and there's no invariant way to compare them.

See posts #15 and #16.

See above.

19. Nov 27, 2015

### Neutrinos02

This should discribe an invariant under coordinate change but not an invariant under a physical process or along a geodesic, dosen't it?

If i contract the killing field with the 4-velocity i get a scalar, but the momentum for example is a 3-vector how must this fact be understood?

20. Nov 27, 2015

### Staff: Mentor

No. The invariant magnitude of an object's energy-momentum 4-vector is always its rest mass, at every event on its worldline. The easiest way to see this is to observe that, at any event, we can always adopt local inertial coordinates in which the metric, at that event, is the Minkowski metric; and in the Minkowski metric the invariant magnitude of the object's energy-momentum 4-vector is obviously its rest mass. But since the magnitude is invariant under any change of coordinates, it must also be the rest mass in any coordinate chart whatsoever.

The 3 components of ordinary momentum are really three contractions, not one; for the 3-momentum to be conserved, there must be three spatial translation KVFs, and they must all be mutually orthogonal, so the contractions of the object's energy-momentum 4-vector with all three KVFs will give the 3 components of the object's ordinary momentum.