Why is 4-velocity not defined as the covariant derivative of position instead of the regular time derivative?

Question

The geodesic equation is usually written as

\begin{equation} D_\tau u^\mu = 0 \end{equation}

where $D_\tau= u^\mu \nabla_\mu$ is the covariant proper time derivative and $u^\mu=\frac{dx^\mu}{d\tau}$ is the 4-velocity. I understand why the equations of motion for a free particle in curved spacetime have the covariant derivative $D_\tau$ and not the usual $\frac{d}{d\tau}$: As we move in curved spacetime we need to account for the effect that curvature has on vectors. Usually this term is called the acceleration

\begin{equation} a^\mu=D_\tau u^\mu \end{equation}

My question is: If velocity is defined in this way... Why is 4-velocity not defined in this way too? In other words, why is the formula for 4-velocity this

\begin{equation} u^\mu =\frac{dx^\mu}{d\tau} \end{equation} and not this

\begin{equation} u^\mu =D_\tau x^\mu \end{equation}

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user P. C. Spaniel

Comments

One problem is that this would be defining the four-velocity in terms of itself, since $D_\tau$ involves $u^\mu$.

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user G. Smith

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I think no answer here can be a good substitute for a good book on differential geometry, especially one applied to relativity. Try for example Burke: Applied Differential Geometry, it is very helpful for building a geometric and visual intuition of these concepts.

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user pglpm

Answer 1

I think you're being somewhat misled by your terminology and by the ambiguity of the standard, compact notation. Note that some texts use only the symbol "$\mathrm{d}/\mathrm{d}\tau$", or only "$\mathrm{D}_\tau$" (or similar symbols), for both operations you mention. Because its meaning is determined by what it is applied to. A general warning here is not to take mathematical notation at face value: as usual it is there to remind us, or give us a summary or signpost, of what we are doing; but its syntax cannot be taken too strictly (as it can in some formal-logic contexts instead). For example, "$\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}$" is already inaccurate: if $x^\mu$ is a coordinate, it is a function from spacetime to the real numbers, $x^\mu \colon M \to \mathbf{R}$, and therefore any derivative with respect to some parameter $\tau$ makes no sense at all.

Geometrically we are taking the following steps:

1. We consider a curve into spacetime, that is, a map from the reals (or an interval thereof) into spacetime, $C \colon \mathbf{R}\to M$.

2. We consider the field of tangent vectors $\pmb{u}$ to the curve; this is what is denoted by "$\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}$". These vectors embody the effect of deriving a function defined over spacetime, $f \colon M \to \mathbf{R}$, along the curve, which we can write as $\pmb{u}(f)$. This is an operation that we can do because the composite function $f\circ C \colon \mathbf{R} \to \mathbf{R}$ is from the reals to the reals. I personally prefer to avoid the inconsistent notation "$\mathrm{d}x^\mu/\mathrm{d}\tau$", and denote the function mapping the parameter $\tau$ to the coordinate chart by, say, $C^\mu(\tau)$, which is just $x^\mu[c(\tau)]$. It now makes sense to take the derivative of $C^\mu$, and we can denote it with a dot: $\dot{C}{}^\mu := \mathrm{d}C^\mu/\mathrm{d}\tau$. These are the components of the tangent vector $\pmb{u}$, which we can also write in invariant notation: $\pmb{u} = \dot{C}{}^\mu\ \partial_\mu$. The action of this vector on a function is then $\pmb{u}(f) = \dot{C}{}^\mu\ \partial_\mu f$.

3. We cannot derive any other tensor field along the curve, though, because we would not know how to "move" the tensor from one point on the curve to another nearby, to take their difference, which is necessary if we are considering a derivative. This is also true of the vector field $\pmb{u}$ defined above. In flat space this is done by moving the tensor keeping it parallel to itself, but in a generic manifold there is no notion of parallelism. To move the tensor we must therefore introduce a notion of "nearby parallelism", which is not unique. This is embodied in the choice of a connection. The directional covariant derivative $\nabla_{\pmb{u}}$ expresses the result of infinitesimally moving a tensor along the direction of the vector $\pmb{u}$, keeping the tensor parallel to itself according to the connection chosen, and then taking the difference with the tensor at the end place. We can in particular apply this to $\pmb{u}$ itself: $\nabla_{\pmb{u}}\pmb{u}$.

This shows that we do not use $\nabla_{\pmb{u}}$ because "we need to account for the effect of curvature". Rather, we need to introduce curvature – or properly speaking a connection – if we want to be able to take a derivative of $\pmb{u}$ or of other tensors along the curve.

But I think no answer here can really be a substitute for a good book on differential geometry, especially one applied to relativity. I recommend:

• Burke: Applied Differential Geometry (also at Archive): it is very helpful for building a geometric and visual intuition of these concepts;

• Bossavit: Differential Geometry: for the student of numerical methods in electromagnetism;

• Choquet-Bruhat & al: Analysis, Manifolds and Physics. Part I: Basics;

• Misner, Thorne, Wheeler: Gravitation.

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user pglpm

Comments

Thanks for your reply! You mention that some books use only one symbol, does that mean that the same symbol acting on any vector would be the covariant derivative but acting on position it would be just a time derivative? For example, If I have to integrate $x^\mu D_\tau u_\mu$ by parts, would I get $- u_\mu D_\tau x^\mu$ or $- u_\mu \frac{d}{d\tau} x^\mu$

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user P. C. Spaniel

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@P.C.Spaniel Yes, its meaning would depend on what it acts upon. But you seem to be confused about the time derivative. There is no time derivative in relativity theory. There are derivatives along wordlines, derivatives along fluxes of vector fields (Lie derivative), covariant derivatives, exterior differentials, and some other derivatives more or less constructed from these. But no time derivative.

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user pglpm

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@P.C.Spaniel I fail to make sense of the quantity $x^\mu u_\mu$, especially within general relativity. It is not covariant. What do you use it for?

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user pglpm

Answer 2

Because position is not a vector.

This post imported from StackExchange Physics at 2025-01-21 21:54 (UTC), posted by SE-user TimRias