$$ \frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$
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
