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I was revisiting some of my undergraduate notes on general relativity and came across a statement where I mention that if the geodesic equation is satisfied for the parameters $\alpha$ and $\beta$: $$\frac{d^2x^{\mu}}{d\alpha^2}+\Gamma^{\mu}_{\nu\rho}\frac{dx^{\nu}}{d\alpha}\frac{dx^{\rho}}{d\alpha}=0$$ and $$\frac{d^2x^{\mu}}{d\beta^2}+\Gamma^{\mu}_{\nu\rho}\frac{dx^{\nu}}{d\beta}\frac{dx^{\rho}}{d\beta}=0$$ Then $$\alpha=C_1\beta+C_2$$ where $C_1$ and $C_2$ are constants.

Well I feel like this is correct cause the geodesic equation is of second order in the parameter with respect to which we differentiate. But I also feel like I miss something deeper. What is the physical interpretation of this?

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  • $\begingroup$ I removed my answer because it did not concern the physical sense but just the mathematical reason. However a geometric sense is the following. Geodesics are curves with, among other properties, the peculiarity that the tangent vector has constant length along the curve. It happens only for a class of parametrizations. This class is made if a parametrisation with that propery and all possibile affine reparametrisations. This property holds for timelike and spacelike geodesics. $\endgroup$ – Valter Moretti Sep 24 '17 at 13:11
  • $\begingroup$ Hi! Thank you for your comment. I did not see your answer on the mathematical reasoning. Was it something more rigorous than what I mention in my question? I mean just applying the chain rule requires that $\frac{d^2\alpha}{d\beta^2}=0$. $\endgroup$ – Helen Sep 24 '17 at 14:41
  • $\begingroup$ OK I undelete it. $\endgroup$ – Valter Moretti Sep 24 '17 at 14:53
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Actually there is a delicate point that uses the validity of uniqueness theorem for second order ordinary differential equations. It is true that affine re-parametrisations preserve the equation of geodesics. The converse fact is however more articulate.

I assume the metric is $C^2$ to guarantee the validity of the theorem of existence and uniqueness of geodesics for given initial conditions.

The precise statement you are looking for should be the following one.

Proposition. Suppose that the metric is $C^2$ and that $x^\mu = x^\mu(\alpha)$, in local coordinates, is a geodesics defined for $\alpha \in [a,b]$. There are only two mutually exclusive possibilities referring to the class of $C^2$ re-parametrisations $\alpha= \alpha(\beta)$ where $d\alpha/d\beta \neq 0$ for $\beta \in [c,d]$.

(a) $x^\mu= x^\mu(\alpha)$ is the constant geodesic and $x^\mu=x'^\mu(\beta) = x^\mu(\alpha(\beta))$ still solves the geodesic equation for every re-parametrisation.

(b) $x^\mu= x^\mu(\alpha)$ is not the constant geodesic and $x^\mu=x'^\mu(\beta) = x^\mu(\alpha(\beta))$ solves the geodesic equation if and only if $\beta = C_1\alpha + C_2$ where $C_1 \neq 0$.

Proof. First of all observe that if $x=x(\alpha)$ is the constant geodesic, then every reparametrisation will produce the constant geodesic. If $x=x(\alpha)$ is a geodesic, then an affine reparametrisation trivially preserves the equation of geodesics.

To conclude we have to prove that only affine re-parametrisations are allowed unless the geodesic is constant.

Suppose you change the parameter $\alpha= \alpha(\beta)$ (the function being $C^2$ and $\frac{d\alpha}{d\beta} \neq 0$ everywhere to have a true re-parametrisation) and $x=x'(\beta):= x(\alpha(\beta))$ still satisfies the geodesic equation (the latter you wrote). Since $$\frac{d^2}{d\beta^2} = \left(\frac{d\alpha}{d\beta}\right)^2\frac{d^2}{d\alpha^2} + \frac{d^2 \alpha}{d\beta^2}\frac{d}{d\alpha}$$ comparing the two equations of geodesics appearing in the initial question, you see that
$$\frac{d^2 \alpha}{d\beta^2}\frac{d x^\mu}{d\alpha}=0$$ everywhere along the curve. There are two possibilities: (a) $\frac{d x^\mu}{d\alpha}=0$ somewhere or (b) $\frac{d x^\mu}{d\alpha} \neq 0$ everywhere. In the second case $\frac{d^2 \alpha}{d\beta^2}=0$ everywhere so that $\alpha = C_1 \beta + C_2$ (where $C_1\neq 0$ in view of the initial requirement $\frac{d\alpha}{d\beta} \neq 0$). Suppose that, conversely, $\frac{d x^\mu}{d\alpha}|_{\alpha= \alpha_0}=0$. In this case the function $x^\mu(\alpha)= x^\mu(\alpha_0)$ satisfies the Cauchy problem consisting of your first geodesic equation and initial conditions $x^\mu(\alpha_0)$ and $\frac{d x^\mu}{d\alpha}|_{\alpha= \alpha_0}=0$ and this is the unique solution as the hypothesis of uniqueness theorem are satisfied. So $x^\mu(\alpha) = x^\mu(\alpha_0)$ constantly.

QED

Now I can present a geometric meaning of the class of affine re-parametrisations of a geodesic.

NB. I henceforth assume to deal with spacelike or timelike non-constant geodesics.

The equation of geodesics can be written $$\nabla_{\dot{\gamma}}\dot{\gamma} =0$$ This formulation is intrinsic, $\nabla$ is the covariant derivative arising from the metric $g$ and $\dot{\gamma}$ is the tangent vector to the geodesic $\gamma = \gamma(t)$. As consequence of the said equation, $$g(\dot{\gamma},\nabla_{\dot{\gamma}}\dot{\gamma}) =0$$ that can be re-written as $$\frac{1}{2}\nabla_{\dot{\gamma}}g(\dot{\gamma},\dot{\gamma}) =0\tag{0}$$ i.e. $$\frac{d}{dt}g(\dot{\gamma},\dot{\gamma}) =0\:. \tag{1}$$ It is evident that the length of the tangent vector $g(\dot{\gamma},\dot{\gamma})$ is constant along the geodesic. This property cannot be invariant under arbitrary re-parametrisation, since the tangent vector depends on the choice of the parameter.

It is however possible to characterize the class of the re-parametrisations preserving this property finding that it is exactly the already found class re-parametrisations that leaves fixed the form of the geodesic equation.

If $t=t(u)$ is a $C^2$ re-parametrsation with $dt/du \neq 0$ everywhere, defining $\sigma(u) := \gamma(t(u))$, (1) can be re-arranged into the equivalent requirement $$\frac{d}{du}\left(\left(\frac{dt}{du}\right)^2 g(\dot{\sigma}(u), \dot{\sigma}(u)) \right) =0\:.$$ Expanding the left-hand side, using $\frac{dt}{du} \neq 0$ together with $g(\dot{\gamma},\dot{\gamma}) \neq 0$ (since the geodesic is a non-constant spacelike or timelike geodesic) and assuming that also the length of $\dot{\sigma}$ is constant $$\frac{d}{du}g(\dot{\sigma},\dot{\sigma}) =0\:. \tag{1}$$ we find $$\frac{d}{du} \left(\frac{dt}{du}\right)^2 =0\:.$$ Since $\frac{dt}{du} \neq 0$, this is equivalent to $$\frac{d}{du}\frac{dt}{du} =0\:,$$ namely $u= C_1t + C_2$ with $C_1 \neq 0$. We conclude that if the re-parametrisation produces a curve with tangent vector with constant length then the re-parametrisation preserves the structure of the equations of geodesics and vice versa.

In relativity and in differential geometry, one of the parameters of a (non-constant non-lightlike) geodesic has a relevant physical/geometric meaning: the proper time or the arch-length parameter of the geodesics. According to Ben Crowell's answer, we are however free to change our units and we are also free to arbitrarily fix the origin of the parameter and we expect, i.e., to perform an arbitrary (non singular affine transformation of the parameter). As these choices are completely conventional, the physical/geometrical properties of the curve must remain unchanged. That is the case of Eq.(0) itself and the fact that the tangent vector is constant along the curve.

There is another class of curves whose tangent vector has constant length and this property is preserved if re-parametrising the curves by means of affine transformations and these transformations have a physically relevant meaning. I mean the tangent curves to Killing vector fields $K$. These curves are defined as $$\dot{\gamma}= K(\gamma(t))\:.$$ Since the metric is invariant along the orbits of $K$, we trivially have $$\frac{d}{dt}g(\dot{\gamma}, \dot{\gamma})= \frac{d}{dt}g(K(\gamma(t)), K(\gamma(t)))=0\:.$$ On the other hand $K$ is a Killing field if and only if $K' := cK$ is a Killing fields for $c \neq 0$. The integral curves of $K'$ are those of $K$ with rescaled parameter $u= t/c$. The origin of the parameter can be arbitrarily fixed an we therefore discover the affine class of transformations once more.

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A typical example of an affine parameter would be the proper time along a timelike geodesic. The fact that the curve is a geodesic doesn't depend on what units your clock uses or when you start the clock.

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  • $\begingroup$ And why assuming that good notions of time can only be modified with affine transformations? I think that the reasoning can be reversed: we admit only affine re-definitions of our notion of time because of that property of geodesic equation (or a similar property of parameters of the integral curves of Killing vector fields in curved spacetime). However, yes your remark is correct nomatter if it is the cause or a consequence, for timelike geodesics. $\endgroup$ – Valter Moretti Sep 24 '17 at 12:59
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Good question! Obviously, if $a=C_1\beta+C_2$ then $$\text{d}a=C_1\text{d}\beta,$$ and the geodesic equation remains unchanged upon this transformation if one multiplies through by $C_1^2$. To answer your question, the deeper meaning is that the action $$S = \int \text{ds}$$ is always the same for linear transformations of the parameter. An action can also be thought of as

  1. an arc length, or
  2. the total energy over some time since $$S=\int^tL(\dot{x},x)\text{d}t.$$

Putting these two ideas together reveals that the minimum energy of a system over time will remain same for arbitrary linear transformations of the length measurement parameter. This statement encodes a lot actually; one interesting one is the total energy being locally conserved.

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