# Why is $g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$ a constant for geodesics in GR?

In Sean Carroll's spacetime and geometry chapter 5 Carroll states the following

In addition we always have another constant of the motion for geodesics: the geodesic equation (together with metric compatibility) implies that the quantity$$\epsilon=-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\tag{5.55}$$ is constant along the path. (For any trajectory we can choose the parameter $$\lambda$$ such that $$\epsilon$$ is a constnat; we are simply noting that this is compatible with affine parameterization along a geodesic.)

I feel like I understand this now but my understanding of formal tensors is still a bit shaky so could you tell if my reasoning is valid and correct any misunderstandings?

1. From the metric components $$g_{\mu\nu}$$ we can construct the metric tensor given by $$g(x)=g_{\mu\nu}(x)dx^\mu\otimes dx^\nu$$. Since it is a tensor it is reparametrization invariant but can still vary over space.
2. Metric compatibility tells us that actually it doesn't vary over space. Since $$\nabla_\alpha g_{\mu\nu}=0$$ the metric can be parallel transported to any point in space so it is constant.
3. If $$x(\lambda)$$ is a geodesic then from the geodesic equation we have $$\nabla_{\dot x}\dot x=0$$. This means the direction of the tangent vector $$\frac{d}{d\lambda}$$ is conserved. By a suitable (reparametrization) of $$\lambda$$ we get that $$\frac{d}{d\lambda}$$ is conserved.
4. Since $$g_{\mu\nu}dx^\mu\otimes dx^\nu$$ and $$\frac{d}{d\lambda}$$ are both conserved we have that $$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=g\left(\frac{d}{d\lambda}\otimes \frac{d}{d\lambda}\right)=\left(g_{\mu\nu}dx^\mu\otimes dx^\nu\right)\left(\frac{d}{d\lambda}\otimes \frac{d}{d\lambda}\right)$$ is also conserved.
• I think you're a bit confused with what a tensor is. The metric g takes two vectors as arguments, not a (2,0)-tensor. Dec 13, 2020 at 14:42
• @Krup'a So would you write 4. as $g\left(\frac{d}{d\lambda},\frac{d}{d\lambda}\right)=\left(g_{\mu\nu}dx^\mu\otimes dx^\nu\right)\left(\frac{d}{d\lambda},\frac{d}{d\lambda}\right)$? So still with a tensor product between the differentials? Dec 13, 2020 at 14:46
• As a note, the metric tensor $\textbf{g}$ should be written as $\textbf{g}=g_{\mu\nu}(\textbf{e}^\mu \otimes \textbf{e}^\nu)$. Dec 13, 2020 at 15:51

You don't need any knowledge of tensor calculus to understand this.

What Does $$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^ \nu}{d\lambda}$$ Represent?

For a curve $$x^\mu(\lambda)$$ in a manifold, the quantity $$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^ \nu}{d\lambda}$$ is the square of the length of the tangent vector $$\textbf{t}$$ to the curve $$x^\mu(\lambda)$$ at any point $$P$$.

To see this, note that at any point $$P$$ on the curve, the tangent vector $$\textbf{t}$$ is defined as $$\textbf{t}=\frac{d\textbf{s}}{d\lambda},$$ where $$d\textbf{s}$$ is the infinitesimal separation vector between point $$P$$ and a nearby point $$Q$$ on the curve corresponding to the parameter value $$\lambda+d\lambda$$.

In a given coordinate system with basis vectors $$\textbf{e}_\mu$$, we can write $$d\textbf{s}=\textbf{e}_\mu dx^\mu$$ so that the tangent vector is now $$\textbf{t}=\frac{dx^\mu}{d\lambda}\textbf{e}_\mu.$$ (Note: More explicitly, $$dx^\mu \equiv dx^\mu(\lambda)$$.)

The square of the length of the tangent vector $$\textbf{t}$$ is then $$|\textbf{t}|^2=g_{\mu\nu}t^\mu t^\nu=g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}.$$

So the quantity $$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^ \nu}{d\lambda}$$ being constant throughout the curve $$x^\mu(\lambda)$$ means that the tangent vector $$\textbf{t}$$ has a constant length throughout the curve.

Parameterising A Curve

For any curve $$x^\mu(\lambda)$$, we can paramaterise the curve such that the length of the tangent vector is constant. Note that if we choose the parameter $$\lambda$$ to be $$\lambda=as+b,$$ where $$s$$ is the distance measured along the curve and $$a$$, $$b$$ are constants, the length of the tangent vector will be constant. This can be shown through $$|\textbf{t}|=\frac{d|\textbf{s}|}{d\lambda}=\frac{ds}{d\lambda}={1\over a},$$ where $$1\over a$$ is a constant.

Summary: For any curve $$x^\mu(\lambda)$$, we can always parameterise it such that the length of the tangent vector is constant throughout the curve. The length of the tangent vector is given by $$|\textbf{t}|=(g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^ \nu}{d\lambda})^{1\over2}$$.

References:

1. Hobson, Efstathiou & Lasenby General Relativity: An Introduction for Physicists pg. 75

This is simply a choice of parametrization of the geodesic $$x^\mu(\lambda)$$. If we were in a Euclidean-signature manifold, $$\lambda$$ would be proportional to the arc length along the curve. Here, for a timelike geodesic, it would be proportional to the proper time along the curve.

• So would you say 3. is correct? Is the direction of $d/d\lambda$ conservered for a general geodesic? Dec 13, 2020 at 14:06
• It depends on what you mean by $d/d\lambda$. Do you mean $(d x^\mu/d \lambda) \partial_\mu$ with $\partial_\mu$ considered as a tangent space basis vector? Certainly geodesics are autoparallels. I don't like saying that "direction is conserved" as directions at different points are not comparable. Dec 13, 2020 at 14:14
• To me $d/d\lambda$ and $(dx^\mu/\lambda)\partial_\mu$ are two expressions of the same thing. I don't know what autoparallels are. Dec 13, 2020 at 14:17
• But I have to agree that saying directions are conserved is a bit of a vague statement. Dec 13, 2020 at 14:19
• I mean that parallel trasnporting the tangent along the geodesic takes you to the tangent at the new point. If you parallel transport to the same point along a different curve you will get a different vector. So directions at different points are not comparable. The attempted comparison depends on how you try to do it. Dec 13, 2020 at 14:19