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Define the operator $\frac{D}{D\tau}$ by its action on an arbitrary contravariant vector $A^\lambda$:

$$\frac{DA^\lambda}{D\tau} = \frac{dA^\lambda}{d\tau} + \Gamma^\lambda_{\mu\nu} \frac{dx^\mu}{d\tau} A^\nu$$

(with the motivation that this allows us to express the geodesic equation in a nice form).

Now, in an attempt to deduce the corresponding action on a covariant vector $\frac{DA_\lambda}{D\tau}$, contract with an arbitrary covariant vector $B_\lambda$:

$$\begin{align} B_\lambda\frac{DA^\lambda}{D\tau} &= B_\lambda\frac{dA^\lambda}{d\tau} + B_\lambda\Gamma^\lambda_{\mu\nu} \frac{dx^\mu}{d\tau} A^\nu \\ &= \frac{d(A^\lambda B_\lambda)}{d\tau} - A^\lambda\frac{dB_\lambda}{d\tau} + B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau} A^\lambda \\ &= \frac{d(A^\lambda B_\lambda)}{d\tau} - A^\lambda\left(\frac{dB_\lambda}{d\tau} - B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau}\right) \\ \end{align}$$

The following is the bit which I don't understand - we claim that the term in brackets must be the derivative of this covariant vector, i.e.

$$ \frac{DB_\lambda}{D\tau} = \frac{dB_\lambda}{d\tau} - B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau} $$

I agree that it is definitely a vector, but I don't see how we can make the leap to saying that it is certainly the form that this operator takes when acting on a covariant vector. Is there something I'm missing?

I considered that it might have something to do with the fact that if you substitute this definition and rearrange you obtain

$$ B_\lambda\frac{DA^\lambda}{D\tau} + A^\lambda\frac{DB_\lambda}{D\tau} = \frac{d(A^\lambda B_\lambda)}{d\tau} $$

but I can't see exactly where this leads.

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2 Answers 2

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We can start with the following reasonable assumptions: (1) the action of the operator $\frac{D}{D\tau}$ reduces to that of $\frac{d}{d\tau}$ on scalar functions i.e. $$\frac{D\phi}{D\tau}=\frac{d\phi}{d\tau}$$ for any scalar function $\phi$ , and (2) the operator $\frac{D}{D\tau}$ satisfies the Leibniz product rule.

These assumptions fix the action of $\frac{D}{D\tau}$ on a covector ($B_{\lambda}$) as follows.

From assumption (1), keeping in mind that $A^{\lambda}B_{\lambda}$ is a scalar, we have:

$$\frac{D(A^{\lambda}B_{\lambda})}{D\tau} = \frac{d(A^{\lambda}B_{\lambda})}{d\tau}$$

From assumption (2), we apply the Leibniz product rule on the left side (and the Leibniz product rule on the right as with any derivative):

$$ B_{\lambda} \frac{DA^{\lambda}}{D\tau}+A^{\lambda} \frac{DB_{\lambda}}{D\tau} = B_{\lambda}\frac{dA^{\lambda}}{d\tau} + A^{\lambda}\frac{dB_{\lambda}}{d\tau} $$

Apply the definition of the action of $\frac{D}{D\tau}$ on the vector $A^{\lambda}$:

$$ B_{\lambda} \left(\frac{dA^{\lambda}}{d\tau} + \Gamma_{\mu \nu}^{\lambda} \frac{dx^{\mu}}{d\tau}A^{\nu} \right)+A^{\lambda} \frac{DB_{\lambda}}{D\tau} = B_{\lambda}\frac{dA^{\lambda}}{d\tau} + A^{\lambda}\frac{dB_{\lambda}}{d\tau} $$

$$ B_{\lambda} \frac{dA^{\lambda}}{d\tau} + \Gamma_{\mu \nu}^{\lambda} \frac{dx^{\mu}}{d\tau}B_{\lambda}A^{\nu} +A^{\lambda} \frac{DB_{\lambda}}{D\tau} = B_{\lambda}\frac{dA^{\lambda}}{d\tau} + A^{\lambda}\frac{dB_{\lambda}}{d\tau} $$

Rearrange:

$$ A^{\lambda} \frac{DB_{\lambda}}{D\tau} = A^{\lambda}\frac{dB_{\lambda}}{d\tau} - \Gamma_{\mu \nu}^{\lambda} \frac{dx^{\mu}}{d\tau}B_{\lambda} A^{\nu} $$

Rename the indices on the last term on the right:

$$ A^{\lambda} \frac{DB_{\lambda}}{D\tau} = A^{\lambda}\frac{dB_{\lambda}}{d\tau} - \Gamma_{\mu \lambda}^{\nu} \frac{dx^{\mu}}{d\tau} B_{\nu} A^{\lambda} $$

As this equation must be true for any vector $A^{\lambda}$, we can remove it from all the terms and conclude with the action of $\frac{D}{D\tau}$ on $B_{\lambda}$:

$$ \frac{DB_{\lambda}}{D\tau} = \frac{dB_{\lambda}}{d\tau} - \Gamma_{\mu \lambda}^{\nu} \frac{dx^{\mu}}{d\tau} B_{\nu} $$

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  • $\begingroup$ Nice - so the derivation I give just does things backwards, and just needs to bring in the assumptions at the end to give the necessary conclusion. Thank you! $\endgroup$ Commented Jan 30, 2023 at 22:31
  • $\begingroup$ Actually, why can we make assumption (1)? I assume it has something to do with the fact we can't contract a scalar with $\Gamma$ since it has no components? $\endgroup$ Commented Jan 30, 2023 at 22:33
  • $\begingroup$ One of the aims of defining a covariant derivative is to have a tensorial derivative of tensors. You can check that the partial derivative of a vector, for instance, is not tensorial; this is why a connection is added. On the other hand, the partial derivative of a scalar function is tensorial and does not need any extra term. So we define the covariant derivative of a scalar as its partial derivative. An equivalent way of saying all this is that a scalar remains invariant under parallel transport regardless of curvature! $\endgroup$
    – user319197
    Commented Jan 31, 2023 at 0:07
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This essentially follows from that $\frac{D}{D\tau}=\nabla_{\dot{x}}$ and from how a connection $\nabla$ acts on tensors.

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  • $\begingroup$ Thank you, this is also enlightening. $\endgroup$ Commented Jan 31, 2023 at 9:38

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