# Noether's theorem in general relativity

Noether's theorem yields a conservation law for every symmetry. Is that independent of the Lagrangian i.e. when $\mathcal{L}\neq T-V$? In general relativity the integral that is minimised will be the geodesic: $$S=\int ds$$ What form would Noether's theorem take? I am also looking for a proof of this. All the proofs I've seen assume $\mathcal{L}=T-V$.

• If you're considering an arbitrary geodesic in a fixed spacetime background, then the charges you get out of Nother's theorem will depend on the fixed metric and its symmetries. – Jerry Schirmer Apr 22 '15 at 21:02
• Continuing @Jerry Schirmer's comment, it seems that OP is considering a point particle in a fixed curved spacetime background rather than dynamical GR. Then the trajectories will generically not be geodesics unless the external potential $V=0$. Moreover, it seems that OP is implicitly assuming that $L=T=\sqrt{\dot{x}^{\mu} g_{\mu\nu}\dot{x}^{\nu}}$ and $V=0$. (Note that the kinetic term $T$ is not the kinetic energy.) – Qmechanic Apr 22 '15 at 21:42

Assuming that your primary question is about Noether's theorem when $\mathcal{L} \ne T - V$, we can proceed as follows. Denoting the generalized coordinates by $q_i$, we consider a transformation $q_i \rightarrow q_i + \delta q_i$ which leaves the Lagrangian $\mathcal{L}(q_i, \dot{q}_i)$ invariant: $$\delta \mathcal{L} = 0$$ i.e. $$\sum_i \left( \frac {\partial \mathcal{L}}{\partial q_i}\delta q_i + \frac {\partial \mathcal{L}}{\partial \dot{q}_i}\delta \dot{q}_i\right) = 0$$ Now, the equations of motion are $$\frac {\partial \mathcal{L}}{\partial q_i} = \frac{d}{dt}\frac {\partial \mathcal{L}}{\partial \dot{q}_i}$$ We can substitute this in the expression for the variation in $\mathcal{L}$ to get $$\sum_i \left(\left(\frac{d}{dt}\frac {\partial \mathcal{L}}{\partial \dot{q}_i}\right)\delta q_i + \frac {\partial \mathcal{L}}{\partial \dot{q}_i}\left(\frac{d}{dt} \delta q_i\right)\right) = 0$$ We can now combine the terms: $$\frac{d}{dt}\left(\sum_i\frac {\partial \mathcal{L}}{\partial \dot{q}_i}\delta q_i\right) = 0$$ Thus, the quantity $Q = \sum_i \frac {\partial \mathcal{L}}{\partial \dot{q}_i}\delta q_i$ is conserved, upto a constant multiplier.

Now, considering the action: $$S = m\int ds = m\int \sqrt{g_{ab}\dot{x}^a\dot{x}^b} dt$$ the Lagrangian is $\mathcal{L} = m\sqrt{g_{ab}\dot{x}^a\dot{x}^b}$. As such, what transformation leaves this invariant depends on $g_{ab}$. If, for example, we assume that $g_{ab} = \eta_{ab}$, the Minkowski metric, then the Lagrangian is invariant under displacements $x^a \rightarrow x^a + c^a$ and we get the conserved quantity: $$Q = \eta_{ab}c^a\frac{m\dot{x}^b}{\sqrt{\eta_{ab}\dot{x}^a\dot{x}^b}} = p_a c^a$$ Now, this holds for any and all $c^a$, and therefore, the coefficient of each component $c^a$ must also be conserved (as can be seen by setting the other components to zero). Therefore, the conserved quantities are: $$p_a = \eta_{ab} \frac{m\dot{x}^b}{\sqrt{\eta_{ab}\dot{x}^a\dot{x}^b}}$$ This is essentially the conservation of the 4-momentum of a free particle in special relativity.

The formulation of Noether's Theorem in General Relativity requires the use of something called a Killing Vector Field. It's a really fascinating subject, but understanding it does require having a fairly strong grasp of Tensor calculus. I'll explain the concept here, and if you need any clarification on the mathematics I'll be happy to update my explanation.

So, as you hopefully know, the general relativity assumes that spacetime is an ordered pair $(M, g_{ab})$, where $M$ is a 4-dimensional, smooth, real manifold, and $g_{ab}$ is a symmetric, bilinear, non-degenerate rank $(0, 2)$ tensor (known as the metric of spacetime). We define differentiation of tensors on spacetime such that the derivative of the metric is zero -- that is, $\nabla_a g_{bc} = 0.$

We need one more definition. A one-parameter group of diffeomorphisms is a smooth group action of the additive group $\mathbb{R}$ on the manifold $M$ -- equivalently, it is a map

$$\phi : \mathbb{R} \times M \rightarrow M$$

such that for any $p \in M$ we have $\phi(0, p) = p$ and for any $t, s \in \mathbb{R}$ we have $\phi(t, \phi(s, p)) = \phi(t + s, p).$ We also require that for any fixed $t$, $\phi_t : M \rightarrow M$ is a diffeomorphism.

There's a lot to think about there, but you don't need to think about it too hard. The only important thing is to think about what happens when we fix a point $p$ on the manifold. Let $\phi_p : \mathbb{R} \rightarrow M$ be the map given by $\phi_p(t) = \phi(t, p).$ This is a smooth map from the reals into $M$, otherwise known as a curve on the manifold, which passes through $p$ at $t = 0.$

There is a general notion in the study of smooth manifolds of being able to "push forward" a tensor along a curve. That is, if we have a tensor at the point $p$ and a curve passing through $p$, there is a natural way of pushing it along the curve and obtaining a new tensor at a different point on the manifold. We shall be interested in doing this for $g_{ab}.$ I'll omit a more general discussion of push-forwards (also known as differentials of diffeomorphisms), but if you want a better explanation, see Lee's Introduction to Smooth Manifolds.

There's just ONE more thing you need to think about: if I have a one-parameter group of diffeomorphisms, I can use it to fill up the entire spacetime $M$ with similar curves. If I take the tangent vectors to these curves at all points, I can get a smooth vector field on the manifold $M$. Likewise, if you give me a smooth vector field, I can find you a corresponding one-parameter group of diffeomorphisms (by taking the set of integral curves that has that vector field as its set of tangent vectors). So from now on I'll only talk about pushing tensors along using smooth vector fields, not one-parameter groups of diffeomorphisms.

Okay, so that was a lot of formalism. Where does it get us? Why did I waste your time with it? Well, think about how Noether's theorem is phrased in classical mechanics: "Every continuous symmetry gives rise to a conserved quantity." If we want to study the conserved quantities in general relativity, we need to interpret a notion of "continuous symmetry."

Well, we believe that the metric $g_{ab}$ governs nearly all of the physics of spacetime. So the only symmetries we can really consider are the symmetries of the metric. What does that mean? Well, we know that every smooth vector field gives us a way to take the metric at one point on the manifold and move it along to get a tensor at a different point on the manifold. So the most natural way to think of symmetries of the metric is to look at the vector fields that leave the metric unchanged. That is, if I take the metric at a point $p$ and push it along to a point $q$ using a given vector field, I get precisely the metric that we expect to have at the point $q$.

This vector field is called a Killing Vector Field. The associated diffeomorphisms are called isometries of the metric.

So to complete this formalism you'd need to fully understand the notion of Lie derivatives. I've already spent a lot of time on formalism, so I'll glaze over this part. The idea, though, is that if I am given a vector field I can define a derivative with respect to that vector field by taking the value of a tensor field some distance along an integral curve, pulling it back along the curve in the way defined by push-forwards, subtracting whatever value I get from the actual value of the tensor field at point $p$, dividing by the distance along the curve, and taking the limit.

The only thing you really need to know from that discussion is that since symmetries of the metric give rise to curves that leave the metric unchanged, we conclude that a vector field is a symmetry of the metric if and only if the Lie derivative of the metric with respect to that vector field is zero.

So we said that a Killing Vector Field is a vector field that corresponds to a symmetry of the metric. We conclude: a vector field is a Killing Vector Field if and only if its Lie derivative is zero.

We're almost there, I promise. This is where it gets good.

Suppose $v^a$ is a Killing Vector Field on a spacetime $(M, g_{ab})$. We denote the Lie derivative with respect to $v^a$ as $\mathscr{L}_v$. There is a formula for taking the Lie derivative of a general tensor, which I will omit. The point is that we know that the Lie derivative of the metric is zero. So we have

$$0 = \mathscr{L}_v g_{ab} = v^c \nabla_c g_{ab} + g_{cb} \nabla_a v^c + g_{ac} \nabla_b v^c = v^c \nabla_c g_{ab} + \nabla_a v_b + \nabla_b v_a.$$

Since we stated earlier that $\nabla_c g_ab = 0$, this reduces to

$$\nabla_a v_b + \nabla_b v_a = 0.$$

So any Killing Vector Field satisfies this property.

Here I finally claim the final from of Noether's Theorem:

Given any Killing Vector Field $v^a$ (i.e. a smooth symmetry of the metric), and a geodesic with tangent $u^a$, the inner product $u^a v_a$ is conserved along the path of the geodesic.

What does it mean for a quantity $C$ to be conserved along the path of a geodesic? It means that $u^a \nabla_a C = 0.$ So we evaluate:

$$u^a \nabla_a (u^b v_b) = (u^a \nabla_a u^b) v_b + u^b (u^a \nabla_a v_b).$$

Since $u^a$ is the tangent vector to a geodesic, $u^a \nabla_a u^b = 0$ by definition of an affinely-parameterized geodesic. So we have

$$u^a \nabla_a (u^b v_b) = u^b (u^a \nabla_a v_b).$$

Since $a, b$ are dummy variables we may interchange them. But we also have $\nabla_a v_b = - \nabla_b v_a$, which we proved earlier to be a consequence of the fact that $v^a$ is a Killing Vector Field. So ultimately, we have

$$u^b (u^a \nabla_a v_b) = u^a (u^b \nabla_b v_a) = - u^a (u^b \nabla_a v_b) = - u^b (u^a \nabla_a v_b).$$

This quantity is equal to its own negative, so we have

$$u^a \nabla_a (u^b v_b) = u^b (u^a \nabla_a v_b) = 0.$$

That's quite a mouthful, but there it is: Noether's Theorem. If we have a vector field $v^a$ such that pushing the metric $g_{ab}$ along $v^a$ leaves the metric unchanged, then any observed with a tangent vector $u^a$ will observe that the quantity $u^a v_a$ remains unchanged over the course of their motion.

You're presumably familiar with the idea that "Energy is the conserved quantity corresponding to time-translation symmetry." Well, if the metric $g_{ab}$ is not explicitly dependent on time, then the vector $\left( \frac{\partial}{\partial t} \right)^a$ is a Killing Vector Field. We define the energy of a system as measured by an observer with tangent vector $u^a$ to be

$$E = - u_a \left( \frac{\partial}{\partial t} \right)^a$$.

So the fact that a metric is time-independent implies that $\left( \frac{\partial}{\partial t} \right)^a$ is a Killing Vector Field, which corresponds to the statement that energy is conserved.

Although users35736's answer is certainly correct, and the question is old, I think it should be noted that each Killing vector, $\xi^i$, also gives rise to a Noether current: $J^i = T_j{}^i\xi^j$. First note that $$J^i{}_{;i} = T^{ji}\xi_{i;j} + \xi_jT^{ji}{}_{;i} = 0,$$ by the Killing equation $\xi_{(i;j)} = 0$, the symmetry of the stress-energy tensor $T^{ij} = T^{(ij)}$, and the vanishing divergence of the stress-energy tensor $T^{ji}{}_{;i} = 0$.

Then consider the Hodge star operator $*: \Omega^k(M) \to \Omega^{n-k}(M)$ which we can take to be defined by $$*\alpha = \sqrt{|\det[g_{ij}]|}\alpha^{\vec{J}}\epsilon_{\vec{J}\vec{I}}\omega^{\vec{I}},$$ where $\omega^i$ is the local frame field on $T^*M$, $\epsilon_{i_1\ldots i_n}$ is the $n$-dimensional Levi-Civita symbol, and $\vec{I},\vec{J},\ldots$ denote strictly increasing multi indices of appropriate lengths (above $\vec{J}$ is of length $k$ and $\vec{I}$ is of length $(n-k)$). To simplify notation we take $\varepsilon_{I} := \sqrt{|\det[g_{ij}]|}\epsilon_{I}$ to be the Levi-Civita tensor.

In the case of a 1-form the Hodge dual is essentially a $(n-1)$-form completely ortogonal to the original 1-form. For let $\alpha_i$ be the 1-form, then contraction on some index with $*\alpha$ yields $$\alpha^i\alpha^j\varepsilon_{ik_1\ldots k_{r-1} j k_{r+1} \ldots k_n} = 0.$$

It therefore should come as no surprise that the Hodge star operator can be used when considering flows across a surface. To make matters more precise consider a hypersurface, $\Sigma$, defined by the normal $\eta_i$, which we take at first to be non-null and normalized. Then the surface element of $\Sigma$ is given by $*\eta$, and by abuse of notation we may designate the directed surface element by $d\Sigma_i := \eta_i{*\eta}$. However, we can also write $$d\Sigma_i = \varepsilon_{i\vec{J}}\omega^{\vec{J}}|_\Sigma,$$ where $|_\Sigma$ denotes projection onto the surface $\Sigma$. To see this note that contraction with $\eta_i$ produces the same result for both expressions of $d\Sigma_i$, as does contraction with any 1-form or vector orthogonal to $\eta_i$. The latter expression is however also well defined for null hypersurfaces, and by continuity also gives the directed surface element in these cases.

For the flux described by the Noether current $J^i$, i.e. the flux of the $\xi^i$ component of energy-momentum, the total flux over $\Sigma$ is given by $\int_\Sigma J^id\Sigma_i$. Suppose now that $\Sigma$ is a closed hypersurface, enclosing a sufficiently well behaved 4-volume $V$ then by Stokes' theorem on smooth manifolds $$\oint_\Sigma J^id\Sigma_i = \int_V d\left(J^id\Sigma_i\right) = \int_V J^i{}_{;i}dV = 0.$$

Depending on your preferences, the final expression can be confirmed in a coordinate frame from the formula $$X^j{}_{;j} = \frac{1}{\sqrt{|\det g|}}\left(\sqrt{|\det g|}X^j\right)_{,j},$$ which is valid in coordinate frames and follows from the Jacobi formula for the derivative of a determinant; in a rigid frame from the first Cartan equation $$d\omega^i = \omega^j\wedge\gamma^i{}_j,$$ where $\gamma^i{}_j$ are the connection forms; and in a mixed frame from both.