# How can you have $\frac{DA^\mu}{d\tau}$?

If a covariant derivative is given by: $$D_\nu A^\mu=\partial_\nu A^\mu +\Gamma^\mu_{\nu \lambda} A^{\lambda}$$ Then how does $\frac{DA^\mu}{d\tau}$ make any sense? Since there are no 'differentials' in $D_\nu$ for $d\tau$ to act on.

Clarification

The parameter $\tau$ can be seen as an arbitrary parameter, although $\tau$ is often used for proper time (interpreting as either arbitrary or proper time does not change the question). I came across the expression $\frac{D A^\mu}{d\tau}$ when looking at parallel transports, but it is also used in the geodesic equation in general relativity and probably a lot of other places to.

• While clear to most people familiar with the topic, it would improve the question to state in which context $\frac{DA^\mu}{\mathrm{d}\tau}$ arises, and that $\tau$ denotes. – ACuriousMind Sep 9 '15 at 16:50
• A way to think about this: $\partial_\mu:=\partial/\partial x^\mu$ and $\mathrm{D}_\mu:=\mathrm{D}/\partial x^\mu$. Then consider the relationship $\mathrm{d}/\mathrm{d}\tau=\dot x^\mu\partial_\mu$. Try to find the analogy with $\mathrm{D}/\mathrm{d}\tau$. – Ryan Unger Sep 9 '15 at 19:17
• While we're on the topic of notation, you should be aware that 99% of GR literature uses $\nabla$ for the covariant derivative in your first equation, not $D$. – user10851 Sep 12 '15 at 6:43

This is a covariant derivative along a world line (if you would not consider a world line the proper time $\tau$ would not make any sense).

So you consider a curve in space time parametrized in dependence of the proper time $x^\mu(\tau)$. Then you have:

$$\frac{DA^\mu}{d\tau} = \frac{\partial A^{\mu}\big(x(\tau)\big)}{\partial \tau} + \Gamma^\mu_{\nu\lambda} A^\nu \dot x^\lambda = \dot x^\lambda A^\mu_{,\lambda} + \Gamma^{\mu}_{\nu\lambda} A^\nu \dot x^\lambda = \dot x^\lambda A^\mu_{;\lambda}.$$

For more details see Wikipedia on Covariant Derivative Along a Curve and on the Levi-Civita connection along the curve. (These articles use an index free notation, not the Ricci calculus usually employed by physicists).

When you are doing things rigorously (i.e. if you are doing mathematics) the connection along a curve only requires the values a vector field along the curve and then therefore is more general than the expression $\dot x^\nu A^\mu_{;\nu}$ (which requires the vector field on an open set which contains the curve).

On a side note, this notion also occurs in the well known gedesic equation: $$\frac{D \dot x^\mu}{D\tau} = 0 = \ddot x^\mu + \Gamma^\mu_{\lambda\nu} \dot x^\lambda \dot x^\nu.$$ (With the difference, that $\dot x^\mu$ is a function of $\tau$ not of $x^\mu(\tau)$).

• Is there any way to determine the expression for $\frac{DA^{\mu}}{d\tau}$ from the expression for $D_\nu A^{\mu}$ or are do they have no link. (sorry if you explain this in your answer, and I have misunderstood) – Quantum spaghettification Sep 9 '15 at 16:58
• @Joseph: If you look closely at the expression in the answer, you'll see that $\frac{\mathrm{D}A^\mu}{\mathrm{d}\tau} = \dot{x}^\nu\mathrm{D}_\nu A^\mu$ (SebastianRiese wrote $A^\mu_{;\lambda}$ for $D_\lambda A^\mu$). – ACuriousMind Sep 9 '15 at 17:13
• @ACuriousMind Ok, I think I have got it now, is this correct: $\frac{DA^{\mu}}{d\tau}\equiv \nabla_{\vec V} \vec A$ where $\vec V$ is the tangent to some curve $\vec r(\tau)$ – Quantum spaghettification Sep 9 '15 at 17:15
• @Joseph Yes, that is correct. – Ryan Unger Sep 9 '15 at 19:15