Deriving geodesic equation using variational principle

[dwellexity]

I am trying to derive the geodesic equation using variational principle.
My Lagrangian is $$ L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$
Using the Euler-Lagrange equation, I have got this.
$$ \frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$

How do I prove the right hand side to be zero to get the geodesic eqaution?

I know that the derivation might be simpler using a different Lagrangian but I want to do it using this one.

 

[Orodruin]

In general, you will not get the geodesic equations from the variational principle. The reason for this is that there is an inherent ambiguity in the parametrisation of a curve, which does not affect its length. If you just look at the EL equations based on curve length, you would end up with something like
$$
\nabla_{\dot \gamma} \dot\gamma = \alpha(t) \dot \gamma,
$$
where $\dot\gamma$ is the tangent vector of the curve (or in other words, the change in the tangent vector is proportional to the tangent vector itself). In order to ensure that you obtain the geodesic equations, you need to require that the parametrisation is affine, which corresponds to a reparametrisation such that $\alpha(t) = 0$. This is mathematically equivalent to letting $\mathcal L = g(\dot\gamma,\dot\gamma)$ rather than $\mathcal L = \sqrt{g(\dot\gamma,\dot\gamma)}$.

 

[dwellexity]

$$
\nabla_{\dot \gamma} \dot\gamma = \alpha(t) \dot \gamma,
$$
This is mathematically equivalent to letting $\mathcal L = g(\dot\gamma,\dot\gamma)$ rather than $\mathcal L = \sqrt{g(\dot\gamma,\dot\gamma)}$.

I didn't understand it. You are differentiating $\dot \gamma$ wrt $\dot \gamma$? Also $g(\dot\gamma,\dot \gamma)$?

Having found the equation I have given above, how do I proceed to do this reparametrisation?

 

[Orodruin]

I didn't understand it. You are differentiating $\dot \gamma$ wrt $\dot \gamma$? Also $g(\dot\gamma,\dot \gamma)$?

Yes, the definition of a geodesic is that its tangent vector is parallel transported along the curve. The parallel transport equation for a general vector $X$ is given by $\nabla_{\dot\gamma} X = 0$, the only difference here is that the geodesic tangent vector replaces $X$. The $g(\dot\gamma,\dot\gamma)$ is the value of the metric tensor (which is a linear map from the tensor product of the tangent space with itself to real numbers) evaluated when both its arguments is the tangent vector $\dot\gamma$.

 

[dwellexity]

In order to ensure that you obtain the geodesic equations, you need to require that the parametrisation is affine, which corresponds to a reparametrisation such that $\alpha(t) = 0$.

I am just a beginner in these things. Can you guide me on how to proceed from the equation I have to get the geodesic equation?

 

[Orodruin]

It is a bit unclear to me exactly how you arrived at your formula. Can you give more details? In particular, how did you get the Christoffel symbols in there? They should result from the computation, not be there from the beginning.

 

[dwellexity]

It is a bit unclear to me exactly how you arrived at your formula. Can you give more details? In particular, how did you get the Christoffel symbols in there? They should result from the computation, not be there from the beginning.

I put the above mentioned Lagrangian $$
L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$
in the EL equation $$\frac{d}{dt}\frac{\partial L}{\partial \dot x^l} = \frac{\partial L}{\partial x^l }$$
and solved to get this.

Please see the attached images. I hope it is legible.

 

[Orodruin]

The point is that you have not really solved for $\ddot x^\mu$. There is a term in the RHS which contains time derivatives of $\dot x^\mu$, which then becomes $\ddot x^\mu$. If you solve for $\ddot x^\mu$ correctly, you should get something on the RHS which is proportional to $\dot x^\mu$ as well as the Christoffel symbols.

 

[stevendaryl]

The point is that you have not really solved for $\ddot x^\mu$. There is a term in the RHS which contains time derivatives of $\dot x^\mu$, which then becomes $\ddot x^\mu$. If you solve for $\ddot x^\mu$ correctly, you should get something on the RHS which is proportional to $\dot x^\mu$ as well as the Christoffel symbols.

I don't quite understand that comment. His right-hand side IS proportional to $\dot{x^\mu}$

 

[stevendaryl]

I am just a beginner in these things. Can you guide me on how to proceed from the equation I have to get the geodesic equation?

Your original equation, in the first post can be written as:

$\frac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{dt} \frac{dx^\lambda}{dt} = \frac{dx^\mu}{dt} \frac{d\ log(L)}{dt}$

where $L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}}$. (Remember: $\frac{d log(X)}{dt} = \frac{dX}{dt}/X$)

If you make the change of parameters from $t$ to $s$, you have to use the chain rule:

$\frac{d x^\mu}{dt} = \frac{ds}{dt} \frac{d x^\mu}{ds}$
$\frac{d^2 x^\mu}{dt^2} = (\frac{ds}{dt})^2 \frac{d^2 x^\mu}{ds^2} + \frac{d^2s}{dt^2} \frac{d x^\mu}{ds}$

If you make those sorts of changes, then in terms of $s$, your equation becomes (after rearranging):

$\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\lambda}{ds} = \frac{dx^\mu}{ds} [ \frac{d\ log(L)}{ds} - \dfrac{\frac{d^2 s}{dt^2}}{(\frac{ds}{dt})^2}]$

So to get the geodesic equation, you have to make the right-hand side equal to zero, which means that $s$ has to solve the equation:

$\frac{d\ log(L)}{ds} - \dfrac{\frac{d^2 s}{dt^2}}{(\frac{ds}{dt})^2} = 0$

That can be rewritten (going back to derivatives with respect to $t$) as:

$\frac{d\ log(L)}{dt} - \dfrac{\frac{d^2 s}{dt^2}}{(\frac{ds}{dt})} = 0$

which looks complicated, but implies the simple equation:

$\frac{ds}{dt} = K L$ where $K$ is some constant.

 

[lavinia]

The formula

$\frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = 0$ is the equation for a geodesic.

So setting the variation equal to zero says that a geodesic is a critical point of the variation of your Lagrangian. BTW: The square root in the Lagangian is unnecessary. Try your equations without it.

To see that this is the geodesic equation, write the the definition of a geodesic,

$\nabla_{\dot \gamma} \dot\gamma = 0$ in your local coordinate system $x^{j}$

One thing that is not determined in this formula is the speed of your geodesic. It may not be pararameterized by arc length. But it is of constant speed. This follows because the connection is a Levi-Civita connection, in particular the connection is compatible with the metric. So

$\dot\gamma<\dot\gamma,\dot\gamma> = <\nabla_{\dot \gamma} \dot\gamma,\dot\gamma> + <\dot\gamma,\nabla_{\dot \gamma} \dot\gamma> = 0$ which says that the length of the tangent vector,$\dot\gamma$ ,is constant.

 

[Orodruin]

I don't quite understand that comment. His right-hand side IS proportional to $\dot{x^\mu}$

You are right of course. For some reason it did not look familiar to me, probably because I would generally write it as $\dot x^\mu d\ln\mathcal L/dt$ instead.

 

[Orodruin]

So setting the variation equal to zero says that a geodesic is a critical point of the variation of your Lagrangian. BTW: The square root in the Lagangian is unnecessary. Try your equations without it.

The square root allows him to express the minmal distance between two points. As I already mentioned in #2, this does not fix the parametrisation of the curve and this is solved by using the squared integrand. Technically these are two different problems, but of course several curves are equivalent wrt curve length (forming an equivalence class consisting of the different possible parametrisations of the curve). The option of using the squared version of the integrand just fixes the parametrisation to be the curve with an affine parameter rather than an arbitrary one. The affine parametrisation also happens to be just the geoedesic condition that the tangent vector is parallel transported along the curve.

 

[pervect]

I am trying to derive the geodesic equation using variational principle.
My Lagrangian is $$ L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$
Using the Euler-Lagrange equation, I have got this.
$$ \frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$

How do I prove the right hand side to be zero to get the geodesic eqaution?

I know that the derivation might be simpler using a different Lagrangian but I want to do it using this one.

Other people have answered this, but I'm going to try and make a shorter answer.

Look at $$\frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$

The term inside the brackets is just the square of the length of the tangent vector $u$ where $u^i = \frac{dx^i}{dt}$. We can write $||u||^2 = g_{rs} u^r u^s$

If the curve is parameterized such that this length (and hence the square of the length) of it's tangent vector is constant, then you have your proof, because d/dt of a constant is zero. As other posters have implied, you need a particular parameterization of the curve to make this happen. The correct parameterization that makes the length of the tangent vector constant is usually called an affine parameterization. For a time-like curve, the affine parameterization is one where the curve ((the $x^i$)) are parameterized by proper time $\tau$ so that $u^i$ is the four-velocity.

The geodesic equation will only give you a geodesic if you have an affine parameterization, if you have a general parameterization of the curve, you need to re-parameterize the curve affinely first.

I'm assuming that the rest of the math here is correct, I haven't double checked it. I vaguely recall seeing ||u|| appear in the denominator (in MTW), not the numerator, but I haven't gone to the length of tracking this down.

 

[Orodruin]

Just to add to the previous post: The affine parameter for timelike geodesics (and geodesics in Riemann spaces) can be taken as the proper time (curve length). However, for null geodesics the curve parameter is also affine, but not equal to the proper time (the proper time is zero since it is a null curve).

 

[lavinia]

The square root allows him to express the minmal distance between two points. As I already mentioned in #2, this does not fix the parametrisation of the curve and this is solved by using the squared integrand. Technically these are two different problems, but of course several curves are equivalent wrt curve length (forming an equivalence class consisting of the different possible parametrisations of the curve). The option of using the squared version of the integrand just fixes the parametrisation to be the curve with an affine parameter rather than an arbitrary one. The affine parametrisation also happens to be just the geoedesic condition that the tangent vector is parallel transported along the curve.

Let's see if I understand your point.

For the Riemannian case at least, if one uses the Lagrangian without the square root then it follows directly that the tangent to the critical curve is parallel along the curve. So the solution is already an affinely parameterized geodesic.

If one uses the square root Lagrangian on the other hand, the solution equation merely says that the geodesic curvature of the critical curve is zero. This means that if it were to be reparameterized by arc length, then its tangent would be parallel along the curve. or what is equivalent if it were reparameterized to have constant speed.

In each case one derives the equation for the geodesic from the condition that the geodesic curvature is zero. In the second case, the equation holds for the curve after reparameterization.

 

[Orodruin]

That seems to summarise it, yes.

 

[stevendaryl]

If one uses the square root Lagrangian on the other hand, the solution equation merely says that the geodesic curvature of the critical curve is zero. This means that if it were to be reparameterized by arc length, then its tangent would be parallel along the curve. or what is equivalent if it were reparameterized to have constant speed.

I don't understand that paragraph. If you are using "parallel" in the sense that two vectors are parallel if one is a positive multiple of the other, then reparametrizing the path doesn't change whether its tangent is parallel or not. Does it?

 

[Orodruin]

I don't understand that paragraph. If you are using "parallel" in the sense that two vectors are parallel if one is a positive multiple of the other, then reparametrizing the path doesn't change whether its tangent is parallel or not. Does it?

Parallel has a different meaning in differential geometry. A tensor field $T$ is parallel along a curve $\gamma$ if $\nabla_{\dot\gamma} T = 0$.

The defining property of a geodesic is that its tangent vector is parallel along the geodesic, i.e., $\nabla_{\dot\gamma} \dot\gamma = 0$.

 

[PAllen]

The standard definition of parallel transport (as given e.g. by Orodruin, above), forces an affine parameter (not just in the geodesic case), because it forces preservation of the vector norm as well as direction, and this is only possible with affine parameters. You need to use a more general definition of parallel transport to allow for arbitrary parameter for the curve. Specifically, you need to allow the RHS of the covariant derivative be a scalar function of the parameter times the vector. This allows the magnitude to vary, with direction being locally preserved. Starting from curve affinely paremetrized with a vector parallel transported on it, if you change parameter in a non-linear way, the simple equation for parallel transport will no longer be satisfied.

 

[Orodruin]

because it forces preservation of the vector norm as well as direction

This is slightly misleading. In a space where there affine connection is metric compatible, sure. In a space where it is not, parallel transport does not necessarily preserve inner products of parallel transported vectors and in a space without a metric the notion of a vector norm is not defined. Or are you by "standard" referring to the Levi-Civita connection? The definition I gave is then not the "standard" case but the more general one where only the connection itself matters.

For the general parallel transport, it does not matter how the curve is parametrised. You can have two different parametrisations of the same curve and parallel transport the same vector along it and you will end up with the same vector in the end. This follows directly from the relation $\nabla_{fX}Y = f\nabla_X Y$. The parametrisation becomes important only when we require parallel transport of the tangent vector.

 

[lavinia]

OK. This has led me into a corner. Here is the confusion.

This will be for a Riemannian metric:

The first variation of the Energy Intergral,$1/2∫<\dot\gamma,\dot\gamma>dt$, equals $-∫<V_{t},\nabla_{\dot\gamma} \dot\gamma>dt$ where $V_{t}$ is the variation vector field and the variation is chosen to be smooth. If $V_{t}$ is chosen to be $f(t)\nabla_{\dot\gamma} \dot\gamma$ where $f$ is zero at the end points of the parameter interval and strictly positive on its interior, the first variation is strictly positive unless $\nabla_{\dot\gamma} \dot\gamma$ equals zero. So the only possible critical path is an affinely parameterized geodesic. On the other hand, an affinely parameterized geodesic satisfies the first variation formula for all variations.

Repeating this argument for $∫<\dot\gamma,\dot\gamma>^{1/2}dt$ ,one gets a more complicated formula which, up to mistakes, is

$∫<\dot\gamma,\dot\gamma>^{-3/2}<\dot\gamma,\nabla_{\dot\gamma} \dot\gamma>dt$ - $∫<\dot\gamma,\dot\gamma>^{-1/2}<V_{t},\nabla_{\dot\gamma}\dot\gamma>dt$

This formula again shows that an affinely parameterized geodesic solves the equation for all variations.

Once again, choose the variation vector field to be $f(t)\nabla_{\dot\gamma} \dot\gamma$. The first variational formula becomes

$0 = ∫<\dot\gamma,\dot\gamma>^{-3/2}<\dot\gamma,\nabla_{\dot\gamma} \dot\gamma>dt$ - $∫f(t)<\dot\gamma,\dot\gamma>^{-1/2}<\nabla_{\dot\gamma}\dot\gamma,\nabla_{\dot\gamma}\dot\gamma>dt$

Since one has complete freedom in the choice of $f$,except that it is zero at the endpoints, this formula can only be zero for all $f$ if $\nabla_{\dot\gamma} \dot\gamma = 0$.

So how does this method differ from Euler-Lagrange?
- Euler-Lagrange does not allow all variations?
- Euler-lagrange makes a weaker statement about critical paths?
-?

 

[Orodruin]

So how does this method differ from Euler-Lagrange?
- Euler-Lagrange does not allow all variations?
- Euler-lagrange makes a weaker statement about critical paths?

It does not differ from Euler-Lagrange. It is just that the length of the path is parametrisation independent and so you have a degenerate minimum in the path length. it is like looking for the minimum of $(\vec x^2)^2 - \vec x^2$ in $\mathbb R^n$. It only depends on how far you are from the origin and the rest is rotational freedom. In the same sense here, the path length is invariant under path reparametrisations. You can get rid of this ambiguity by using one of the coordinates as the curve parameter or by parametrising with the path length. In a way, this is similar to gauge fixing.

 

[PAllen]

This is slightly misleading. In a space where there affine connection is metric compatible, sure. In a space where it is not, parallel transport does not necessarily preserve inner products of parallel transported vectors and in a space without a metric the notion of a vector norm is not defined. Or are you by "standard" referring to the Levi-Civita connection? The definition I gave is then not the "standard" case but the more general one where only the connection itself matters.

Yes, I was thinking of a metric compatible connection, because we were talking about norms and intervals, but of course what I said does not apply to a general connection.

For the general parallel transport, it does not matter how the curve is parametrised. You can have two different parametrisations of the same curve and parallel transport the same vector along it and you will end up with the same vector in the end. This follows directly from the relation $\nabla_{fX}Y = f\nabla_X Y$. The parametrisation becomes important only when we require parallel transport of the tangent vector.

I see, I hadn't thought of that.

 

[lavinia]

The standard definition of parallel transport (as given e.g. by Orodruin, above), forces an affine parameter (not just in the geodesic case), because it forces preservation of the vector norm as well as direction, and this is only possible with affine parameters.

- I don't understand what you mean by parallel translation forces preservation of direction. Parallel translation around a closed curve can end up with a different vector than what you started with.

 

[Orodruin]

- I don't understand what you mean by parallel translation forces preservation of direction. If parallel translation around a closed curve can end up with a different vector than what you started with.

The point is that it is the connection which specifies what is meant by "preservation of direction". Of course there is no general way of defining this without a connection, the vectors along the curve belong to different tangent spaces. Parallel transport around a closed curve will give you back the same vector only if the curvature tensor is equal to zero (or if you just happen to "get lucky" with your choice of curve).

 

[lavinia]

The point is that it is the connection which specifies what is meant by "preservation of direction". Of course there is no general way of defining this without a connection, the vectors along the curve belong to different tangent spaces.

So non-trivial holonomy doesn't enter into the idea of preservation of direction. If one parallel translates the same vector along two different curves to the same point and gets different vectors, they are both translated without change of direction. Direction is relative to the the curve.

Parallel transport around a closed curve will give you back the same vector only if the curvature tensor is equal to zero (or if you just happen to "get lucky" with your choice of curve).

Even if the curvature tensor is zero you might have to get lucky.

 

[Orodruin]

So non-trivial holonomy doesn't enter into the idea of preservation of direction. If one parallel translates the same vector along two different curves to the same point and gets different vectors, they are both translated without change of direction. Direction is relative to the the curve.

Well this is the point why we are talking about parallelism along a curve and not as a general concept. Even if some manifolds allow parallel vector fields (in the general sense such that $\nabla_Y X = 0$ for all $Y$) to exist, most of them do not.

Even if the curvature tensor is zero you might have to get lucky.

Fine, let us add the condition of having a curve which is continuously deformable to a point such that the curvature tensor is zero along the deformation. It is obviously not (necessarily) true in manifolds with non-trivial homotopy groups.

 

[samalkhaiat]

$$ \frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$
How do I prove the right hand side to be zero to get the geodesic eqaution?

1) This is the geodesic equation in disguise. You simply wrote it using the arbitrary parameter “$t$”. So, you don’t need to make the right-hand-side zero. You only need to go back to the (affine) proper-time: You used
$$d\tau = L dt , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
where
$$L = \sqrt{g_{ab}\frac{dx^{a}}{dt} \frac{dx^{b}}{dt}} . \ \ \ \ \ \ \ \ \ \ \ (2)$$
Now, if we do what you did and substitute (2) in the Euler-Lagrange equation, we obtain
$$\frac{1}{2L} \frac{\partial g_{ab}}{\partial x^{c}} \frac{dx^{a}}{dt} \frac{dx^{b}}{dt} - \frac{d}{dt} \left( \frac{1}{L} g_{cb} \frac{dx^{b}}{dt} \right) = 0 . \ \ \ \ (3)$$
I am sure you recognise this equation. With a little bit of algebra, we can easily show that Eq(3) is identical to your equation. But we are not going to do that. Instead, we multiply Eq(3) by $2L^{-1}$ and go back to the proper-time $\tau$ using Eq(1). This gives us
$$\frac{\partial g_{ab}}{\partial x^{c}} \frac{dx^{a}}{d\tau} \frac{dx^{b}}{d\tau} - \frac{d}{d\tau} \left( g_{cb} \frac{dx^{b}}{d\tau} \right) = 0 . \ \ \ \ \ \ (4)$$
Again, with a bit of algebra, we can show that Eq(4) is indeed the geodesic equation
$$\frac{d^{2}x^{a}}{d\tau^{2}} + \Gamma^{a}_{bc} \frac{dx^{b}}{d\tau} \frac{dx^{c}}{d\tau} = 0 . \ \ \ \ \ \ \ \ \ \ (5)$$
2) Either by direct calculation or using the condition of parallel transport, we can show that the geodesic equation (5) implies that
$$\frac{d}{d\tau} \left( g_{ab} \frac{dx^{a}}{d\tau} \frac{dx^{b}}{d\tau}\right) = 0 . \ \ \ \ \ \ \ \ (6)$$
This means that the quantity (Polyakov Lagrangian)
$$\mathcal{L} = \frac{1}{2} g_{ab} \frac{dx^{a}}{d\tau} \frac{dx^{b}}{d\tau} , \ \ \ \ \ \ \ \ \ (7)$$
is constant along a geodesic path.
So, if you take the parameter $t$ in your equation to be an affine parameter , say you set $t = \tau$, then Eq(6) tells you that the right-hand-side of your equation is zero.

3) When a geodesic path satisfies
$$\frac{d^{2}x^{a}}{d\lambda^{2}} + \Gamma^{a}_{bc} \frac{dx^{b}}{d\lambda} \frac{dx^{c}}{d\lambda} = k (\lambda) \frac{dx^{a}}{d\lambda} , \ \ \ \ \ \ \ \ \ (8)$$
for some function $k (\lambda)$, we conclude that $\lambda$ is a “wrong” parameter for the geodesic path $x^{a}$. The meaning of the word “wrong” will become clear soon. The point is this: the (proper-time) action
$$S = - m \int d\tau = - m \int \sqrt{g_{ab} dx^{a} dx^{b}} ,$$
is invariant under an arbitrary change of parametrisation
$$\tau \to \lambda = \lambda (\tau) .$$
This is clear because $d\tau = (d\tau / d\lambda) d\lambda$ is independent of $\lambda$. So, you can rewrite the action as
$$S = - m \int \ d\lambda \ \sqrt{g_{ab} \frac{dx^{a}}{d\lambda} \frac{dx^{b}}{d\lambda}} . \ \ \ \ \ (9)$$
Parametrisation-invariance means that the action is independent of what you choose to parameterise the path $x^{a}$. This is, however, not the case for the geodesic equation. Indeed, if you change the proper-time according to $\tau \to \lambda (\tau)$, the geodesic equation transforms into
$$\frac{d^{2}x^{a}}{d\lambda^{2}} + \Gamma^{a}_{bc} \frac{dx^{b}}{d\lambda} \frac{dx^{c}}{d\lambda} = - \frac{d^{2}\lambda / d\tau^{2}}{(d\lambda / d\tau)^{2}} \frac{dx^{a}}{d\lambda} . \ \ \ \ \ \ \ \ (10)$$
Thus, the geodesic equation remains invariant only under a class of parametrisation defined by
$$\frac{d^{2}\lambda}{d\tau^{2}} = 0 \ \ \Rightarrow \ \ \lambda = a \tau + b .$$
This is the class of affine parameters: parameters related to the proper-time $\tau$ by an affine transformation $\tau \to \sigma = a \tau + b$ are called affine parameters. So, the arbitrary, i.e., “wrong” parameter $\lambda$ in Eq(8) is the reason for the appearance of $k(\lambda)$ in the equation of the geodesic path $x^{a}$.
Like the geodesic equation, the “gauge fixed” Polyakov action
$$S’ = \frac{1}{2} \int d\tau \ g_{ab}(x) \frac{dx^{a}}{d\tau} \frac{dx^{b}}{d\tau} ,$$ is also invariant under affine transformation. This is why the Euler-Lagrange equation of the Polyakov Lagrangian leads directly to the geodesic equation.
Comparing Eq(8) with Eq(10), we see that, given $k(\lambda)$, an affine parameter $\tau$ can be introduced by integrating
$$k (\lambda) = - \frac{d^{2}\lambda / d\tau^{2}}{(d\lambda / d\tau)^{2}} = \frac{d}{d\lambda} \ln \left(\frac{d\tau}{d\lambda} \right) ,$$
or
$$\frac{d\tau}{d\lambda} = e^{\int^{\lambda} ds \ k(s)} .$$

4) Finally, a piece of advice for you: To avoid head ach, always derive the equations of motion from variation principle especially when the symmetry of the system is the group of diffeomorphism. Let’s see how this work for the square-root action: We write
$$S = \int \ d\tau = \int d\sigma \ L ,$$
with
$$L = \sqrt{g_{ab} \frac{dx^{a}}{d\sigma}\frac{dx^{b}}{d\sigma}} , \ \ \mbox{and} \ \ d\tau = d\sigma \ L .$$
As usual, we vary the path $x \to x + \delta x$, keeping the end-points fixed and set the variation of the action $\delta S = 0$
$$\int d\sigma \ \delta \left(\sqrt{g_{ab} \frac{dx^{a}}{d\sigma}\frac{dx^{b}}{d\sigma}} \right) = 0.$$
Carrying out the variation, we get
$$\int d\sigma \frac{1}{2 L} \Big \{\frac{\partial g_{ab}}{\partial x^{c}} \frac{dx^{a}}{d\sigma} \frac{dx^{b}}{d\sigma} \delta x^{c} + 2 g_{cb}\frac{dx^{b}}{d\sigma}\frac{d}{d\sigma} \delta x^{c}\Big \} = 0 .$$
Going back to the affine parameter $\tau$, using $d\tau = d\sigma L$ and denoting $dx/d\tau = \dot{x}$, we find
$$\int d\tau \ \Big \{ g_{ab,c} \ \dot{x}^{a} \ \dot{x}^{b} \ \delta x^{c} + 2 g_{cb} \ \dot{x}^{b} \frac{d}{d\tau}(\delta x^{c}) \Big \} = 0 .$$
Integrating the second term by parts and dropping the total derivative term, we obtain
$$\int d\tau \Big \{ g_{ab,c} \ \dot{x}^{a} \ \dot{x}^{b} – 2 \frac{d}{d\tau}\left( g_{cb} \ \dot{x}^{b}\right)\Big \} \delta x^{c} = 0 .$$
Since $\delta x^{c}$ is an arbitrary variation, then the fundamental theorem of variation says
$$2 \frac{d}{d\tau} \left( g_{cb} \ \dot{x}^{b}\right) - \frac{\partial g_{ab}}{\partial x^{c}} \ \dot{x}^{a} \ \dot{x}^{b} = 0.$$
Well, this is nothing but the geodesic equation, and it is not hard to see that:
$$2g_{cb} \ \ddot{x}^{b} + 2 \frac{\partial g_{cb}}{\partial x^{a}} \ \dot{x}^{a} \ \dot{x}^{b} -\frac{\partial g_{ab}}{\partial x^{c}} \ \dot{x}^{a}\ \dot{x}^{b} = 0 .$$
Finally, write the second term as a sum of two terms and let $(a \leftrightarrow b)$ in one of them: So, we arrive at
$$g_{cb} \ \ddot{x}^{b} + \frac{1}{2} \ \left( \frac{\partial g_{cb}}{\partial x^{a}} + \frac{\partial g_{ca}}{\partial x^{b}} - \frac{\partial g_{ab}}{\partial x^{c}}\right) \ \dot{x}^{a} \ \dot{x}^{b} = 0$$
From this, the geodesic equation follow by contracting with $g^{ce}$
$$\ddot{x}^{e} + \Gamma^{e}_{ab} \ \dot{x}^{a} \ \dot{x}^{b} = 0 .$$
Good Luck

 

[stevendaryl]

I think this question about geodesics has come up before, but I don't think it had a succinct memorable answer:

The geodesic equation $\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\lambda}{ds} = 0$ can be derived by extremizing the proper time: $\tau = \int ds \sqrt{g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\lambda}{ds}}$ (and choosing an appropriate parameter $s$).

The mathematical manipulations to get the geodesic equation from the proper time integral are only valid when the proper time is nonzero. (At one point, you have $L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\lambda}{ds}}$ in the denominator, which doesn't make sense if $L = 0$). However, the resulting geodesic equation is perfectly able to handle null geodesics.

Is there some way to see how a geodesic equation that applies to both null and timelike geodesics can come from a variational principle that only makes sense for non-null geodesics? Is there some simple way (for example) to see a null geodesic as some sort of limit of timelike geodesics?