# How does this derivation of the proper time derivative of a covariant vector work?

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.

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 $$B_{\lambda}$$:

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

• 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! Jan 30 at 22:31
• 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? Jan 30 at 22:33
• 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! Jan 31 at 0:07

This essentially follows from that $$\frac{D}{D\tau}=\nabla_{\dot{x}}$$ and from how a connection $$\nabla$$ acts on tensors.

• Thank you, this is also enlightening. Jan 31 at 9:38