In Physics we say we have a conserved entity $\vec{P}$ (to use a common example) if we can write:

$$\frac{\partial}{\partial x^{\mu}}P^{\mu}=P^{\mu},_{\mu}=0\tag{1}$$

Where "," denotes the partial derivative. Such a formulation allows us to apply Gauss's law and say that the change of our entity within some volume is equal to the rate of flux of that entity (momentum here) through a bounding surface. This can be simply written:


The situation is changed in General Relativity, as we take covariant derivatives such that the above expression becomes:


Many claim that this essentially rules out a general integral conservation law in General Relativity (Moller, theory of relativity pg 337; Wheeler, gravitation and inertia pg 44; Fock space time and gravitation 163). However; a vector can be written in terms of arbitrary basis $\hat{e}$ as:


And the covariant derivative can be simply written as:

$$\vec{P};_{\alpha}=\frac{\partial P^{\mu}}{\partial x^{\alpha}}\hat{e}_{\mu}+P^{\mu}\frac{\partial\hat{e}_{\mu}}{\partial x^{\alpha}}\tag{5}$$

Where the latter term (involving derivatives of the basis) is most commonly re-written in terms of Christoffel symbols. This expression can clearly be written in the form:


The same is true of of the standard covariant expression for the stress energy tensor in General Relativity:


Which can be written as:


I don't see why this isn't a conservation law, as it only deals with the partial derivative, could someone please point me in the right direction? much appreciated

  • 2
    $\begingroup$ Do you have a citation for the claim that "this essentially rules out a general conservation law in General Relativity"? I've never heard anyone say that, and it's an unclear-enough statement that I think you'd need more context for the original claim in order to evaluate its correctness. $\endgroup$
    – tparker
    Commented Jun 10, 2017 at 0:50
  • 1
    $\begingroup$ @tparker I added some references, but most textbooks on the subject, will claim that in general there is no global (integral form) of the conservation laws, such is a well-known argument concerning general relativity. The argument is usually based on not being able to apply Gauss's theorem to the vanishing covariant divergence of the stress-energy tensor. $\endgroup$
    – R. Rankin
    Commented Jun 10, 2017 at 1:03
  • $\begingroup$ @tparker This is typically where the gravitational stress energy pseudotensor comes into play, yet many claim this is valid only in an asymptotically flat spacetime. $\endgroup$
    – R. Rankin
    Commented Jun 10, 2017 at 1:09
  • $\begingroup$ @tparker FYI I'm not a fan of this view of non-conservation, hence my question. I've asked a related question here: physics.stackexchange.com/q/335210 $\endgroup$
    – R. Rankin
    Commented Jun 10, 2017 at 1:23

3 Answers 3

  1. The problem seems to be OP's eq. (5), which incorrectly uses partial instead of covariant derivatives on the right-hand side.

  2. If we write a $(r,s)$ tensor $$T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}$$ in index-free notation $$T~=~ \frac{\partial}{\partial x^{\mu_1}}\otimes \cdots \otimes \frac{\partial}{\partial x^{\mu_r}} ~T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s} ~ dx^{\nu_1}\otimes \cdots \otimes dx^{\nu_s},$$ it doesn't mean that covariant derivatives may be replaced with partial derivatives.

  3. Concerning how covariant derivatives act on basis elements, the reader may find my Phys.SE answer here interesting.

  • $\begingroup$ Thank you, I found that your link cleared this up for me. @Uldreth also provided a good answer, but I wanted equations. Thank you. $\endgroup$
    – R. Rankin
    Commented Jun 10, 2017 at 19:42

Switching to invariant language solves nothing. What you write as $(\vec{P})_{,\alpha}$ is still a covariant derivative.

On a general manifold, we have no canonical way of differentiating, we need a connection. The metric provides one and we use that.

In flat space, we have a natural way of differentiating, but if we use a basis that itself changes throughout space, the component of the derivative is not the derivative of the components, because we need to factor in the rate of change of the basis fields too, hence why we talk about "covariant derivative" in flat-space context. But that is a coordinate-based phenomenon, in the invariant sense, you still use the same derivative you'd use otherwise. In particular, if in curves spacetime, you'd calculate something like $\vec{P}_{[,\mu,\nu]}$, it would give nonzero result (provided you do it properly and contract the derivative-index with a basis vector one you do the first derivative). (Mathematicians would call this $\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z=R(X,Y)Z$.)

This doesn't change the fact that the main problem is that a vector $v_x\in T_xM$ at $x$ and a vector $v_y\in T_yM$ live in different vector spaces, operations between them don't make mathematical sense.

So if you'd calculate the integral of a vectorial (or tensorial) object, the operation $$ \int_M\vec{P}\ d\mu $$ would correspond to $$ \lim_{\Delta x^\mu\rightarrow 0}\sum_{i}\vec{P}(x_i)\sqrt{-g}(x_i)\Delta x^0\cdot...\cdot\Delta x^3, $$ where the index $i$ labels a partition of the manifold $M$ into small coordinate-rectangles, and it is impossible to add up $\vec{P}(x_i)$ and $\vec{P}(x_j)$ for $i\neq j$, since they are located in different spaces. No vector-valued integrals in differential geometry, sorry.


This isn't the root reason why you don't, in general get integral forms of conservation equations in general relativity. The root reason why is that integral conservation laws set a time derivative of some quantity equal to an integral of a field.

But, in a fully covariant spacetime, you can mess this up by choosing a crazy time coordinate.

So, if you're going to use the Gauss's law trick, and set the time derivative of some charge equal to the integral over some timelike or null stack of two-surfaces that bound a stack of 3-spaces, the timelike or null derivative of the charge will have to be an isometry of the 3-space, so that you don't get any geometrically induced weirdness.

At asymptotic infinity, any timelike vector will do, because you're asymptoting to Minkowski space. But this isn't the only case where this is true. The event horizon of an isolated black hole also has these properties, with the null tangent providing the role of "time."

  • $\begingroup$ @Jerry_Schirmer I appreciate that; however it makes more sense in general relativity to speak of four volumes and bounding three surfaces in the spirit of covariance. See this question physics.stackexchange.com/questions/335252/… if you can show that integral over bounding three surface is a constant where different foliations/3-surfaces (space like) are at different times, is this not the same as time independence of the charge quantity $\endgroup$
    – R. Rankin
    Commented Jun 10, 2017 at 2:44
  • $\begingroup$ @R.Rankin: a stack of 2-surfaces is a bounding 3-surface, and a stack of 3-surfaces is a 4-surface. I'm just making the generalization explicit, because we care about spacelike/timelike/etc $\endgroup$ Commented Jun 10, 2017 at 2:54

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