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In the context of killing vectors, authors like to talk about some quantity being constant wrt. the proper time $\tau$ as it moves along a geodesic. They express this condition as

$ U^{\alpha} \nabla_{\alpha}p^{\beta} = 0, $

$U^{\alpha}$ represents 4-velocity and $p^{\beta}$ is one of the components of the 4-momentum which is constant. Also, $\nabla_{\alpha}p^{\beta}$ stands for $\partial_{\alpha} p^{\beta} + \Gamma^{\beta}_{\alpha \lambda} p^{\lambda}$ (the covariant derivative). I don't see why this should be. Let me work out an alternate expression:

$ \frac{d p^{\beta}}{d\tau} = 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~p^{\beta}\mathrm{~is~constant} \\ \implies \frac{\partial p^{\beta}}{\partial x^{\alpha}}\frac{d x^{\alpha }}{d \lambda } =0 \mathrm{~~~~~~~~~~~~~~~using~the~usual~chain~rule}\\ \implies U^{\alpha} \frac{\partial p^{\beta}}{\partial x^{\alpha}} =0, $

which is not the same as the first equation (the one found in textbooks), in that I have a partial derivative instead of a covariant derivative. What has gone wrong?

More queries: If we are thinking about retaining the tensorial character of the derivative, why consider covariant derivative only? Why not Lie derivative or exterior derivative? All these three derivatives are valid tensors after all.

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In your equation, it is not $p$ that is constant, but the component(s) of $p$. This is not a coordinate-invariant statement because the coordinates themselves are changing as you move through the manifold, so whether the components of a vector/tensor field change or not (and how they do) does not give any information on whther or not the actual vector/tensor field changes or not.

In some cases (when your manifold is not endowed with a metric or connection) it is not even possible to separate a "true" rate of change from the "apparent" rate of change.

If it is theoretically possible to do so, that means you have a connection $\nabla$.

In this case the total change of a quantity can be written symbolically as $$ dY=\nabla Y+\delta Y, $$ for some quantity $Y$ where $\nabla Y$ is the "true" or "physically/geometrically" meaningful rate of change and $\delta Y$ is an "apparent" rate of change due to how the coordinate system shifts around.

By analizing the algebraic properties of first-order differential operators, one may realize that $\delta$ must be a zeroth order differential operator, so essentially a pointwise-linear transformation, which we write as $-\Gamma$. So we have $$ \nabla Y=dY+\Gamma Y. $$

I have essentially now introduced the covariant derivative. The point is that it is pretty much always $\nabla$ that we are interested in, the fact that in some coordinate system $p^\beta$ might be constant carries no meaning at all. If you switch to another coordinate system it won't be constant. What you are interested in is the coordinate-invariant derivative which in this case is $$ u^\alpha\nabla_\alpha p^\beta . $$

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  • $\begingroup$ So you should actually consider $\tfrac{d}{d\tau}(p^\alpha e_\alpha)$ rather than $\tfrac{d}{d\tau}p^\alpha$, right? $\endgroup$ – Photon Nov 24 '17 at 11:54
  • $\begingroup$ @Uldreth I am still confused. Why consider covariant derivative only? Why not Lie derivative or exterior derivative? All three derivatives are tensors. $\endgroup$ – Sashwat Tanay Nov 24 '17 at 12:12
  • $\begingroup$ The reason is the following. What were trying to say is that a certain quantity (p) is constant along a geodesic. But to define a geodesic you need a metric. The Lie derivative and the exterior derivative do not need a metric to be defined. Therefore they cannot be the right kind of derivative to express this idea of "constant along a geodesic" $\endgroup$ – John Donne Nov 24 '17 at 13:09
  • $\begingroup$ @JohnDonne Please see my reply below. $\endgroup$ – Sashwat Tanay Nov 24 '17 at 23:18
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I think Uldreth's answer has already given the key points.

Unfortunately, Bingo's answer to his own question is wrong. $p^\beta$ is not a scalar, it is the component of a vector. This is because if I make a coordinate transformation, $p^\beta$ will change with the usual formula, but a scalar doesn't. Therefore the covariant derivative does not reduce to the partial derivative in this case.

The essential mistake in Bingo's derivation is to adopt the "usual" chain rule. This is an understandable mistake which is due to subtle notation. In this context, $\frac{d}{d\tau}$ means "directional derivative along a geodesic" (as stated in the reference quoted by the OP, Carroll page 136). If $U^\mu$ is the four velocity of a geodesic, then by definition: $$\frac{d}{d\tau}=U^\mu \nabla_\mu$$ This is the same as what the OP has written if the object on which it acts is a scalar; it is not in general however.

The confusion can be avoided by avoiding that notation. As in Carroll, page 136, eq. (3.174) one can simply state that: $$\nabla_{(\mu}K_{\nu)} = 0 \implies U^\mu \nabla_\mu (K_\nu U^\nu)=0$$ (Carroll actually uses $p^\mu=mU^\mu$, but it doesn't matter) That is, if $K_{\nu}$ is Killing, then $K_\nu U^\nu$ is constant along a geodesic.

It is to be noted however that $K_\nu U^\nu$ is indeed a scalar. So in this case it is true that $$U^\mu \partial_\mu (K_\nu U^\nu)=U^\mu \nabla_\mu (K_\nu U^\nu)=0$$ This may help clear the OP's confusion.

Now the OP also asks why we are using the covariant derivative and not other notions of derivative. In this case, since we're concerned with a scalar being constant, we can express the same notion with the Lie derivative for instance: $$\mathcal{L}_U (K_\nu U^\nu) = U^\mu \partial_\mu (K_\nu U^\nu)=U^\mu \nabla_\mu (K_\nu U^\nu)=0$$

In general however, the notion of an object remaining constant along a curve is given by parallel transport, which depends on the metric. Carroll discusses the issue at page 105. This is ultimately the reason why in general we need to use the covariant derivative. However, if the object you're considering is truly a scalar, then the two agree and there's no issue.

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The alternate derivation given seems to have nothing wrong. After all, what you employed were standard techniques of calculus (e.g. the chain rule). The conflict can be resolved easily by noting that we define the action of covariant derivative on scalars (functions) to be the same as partial derivative (pg 96 of Sean Carroll or any other text).

A more subtle point is that here, $p^{\beta}$ is treated as a scalar (again see Sean Carroll: pg 136) despite its tensorial index appearance. These two above points help us reach the conclusion $\nabla_{\alpha} p^{\beta}= \partial_\alpha p^{\beta}$.

Carroll's notation is a bit different, in that he uses lower indices instead of upper and $\sigma_{\star}$ instead of $\beta$. Basically, he is working with $p_{\sigma_\star}$ instead of $p^{\beta}$.

How can $p^{\beta}$ be treated like a scalar? Here the meaning of $p^{\beta}$ is a subtle one. It means the following. Choose a coordinate system which has at least one coordinate $x_\beta$ such that $\partial_\beta g_{\mu \nu} = 0$. Now let's define a scalar $S$ whose value exactly equals $p^{\beta}$. So the value of $S$ does not change wrt. a change in coordinates. When Sean Carroll uses $p_{\sigma_{\star}}$ (or $p^{\beta}$ in the question above), he means '$S$' as defined above. That is why he introduces a weird $\sigma_{\star}$ index (just to mean that coordinate system and the coordinate is fixed: the one wrt. which the metric does not change).

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