# A Finding geodesics using the squared Lagrangian

1. Mar 3, 2017

### ShayanJ

To find the geodesics of a space(time), what we need to do is extremizing the functional $\displaystyle \int_{\lambda_1}^{\lambda_2}\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d\lambda$. But sometimes the presence of the square root makes the equation of motion too complicated and it would be good if we could use the functional $\displaystyle\int_{\lambda_1}^{\lambda_2}g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} d\lambda$ instead.
It turns out that if we use an affine parameter (s) to parameterize the geodesic, then the two Lagrangians become equivalent:
$\frac{d}{ds}\frac{\partial L^2}{\partial \dot x^\mu}=\frac{\partial L^2}{\partial x^\mu} \rightarrow \frac{d}{ds}\left(2L\frac{\partial L}{\partial \dot x^\mu}\right)=2L\frac{\partial L}{\partial x^\mu} \rightarrow \frac{dL}{ds}\frac{\partial L}{\partial \dot x^\mu}+L \frac {d}{ds}\frac{\partial L}{\partial \dot x^\mu}=L\frac{\partial L}{\partial x^\mu} \xrightarrow{\frac{dL}{ds}=0} \frac {d}{ds}\frac{\partial L}{\partial \dot x^\mu}=\frac{\partial L}{\partial x^\mu}$.

My problem is, $\frac{dL}{ds}=0$ means that the Lagrangian is a constant and its possible to use a parametrization in which $L=1$. In this case, what does it mean to extremize the functional $\int L^2 ds$? Also a constant Lagrangian gives 0=0 as the EL equations of motion. What's wrong here?

Thanks

2. Mar 3, 2017

### Orodruin

Staff Emeritus
It should be clear that the path length is invariant under reparametrisations of a curve. When you extremise the first integral, the extrema are therefore degenerate and you need to add an additional condition. Note that you are doing this on-shell and not off-shell. It might be possible to introduce this additional requirement as a holonomic constraint on the first functional but I have never tried that explicitly.

3. Mar 3, 2017

### pervect

Staff Emeritus
The way to get rid of the square root is to use a mathematical trick called an einbein.

https://cds.cern.ch/record/331888/files/9708319.pdf and http://www.ellipsix.net/blog/2010/8/the-origin-of-the-einbein.html [Broken] may be of some help.

Last edited by a moderator: May 8, 2017
4. Mar 3, 2017

### ShayanJ

Is this related to vielbeins in any way?

5. Mar 3, 2017

### pervect

Staff Emeritus
Well, other than both being German terms, I'm not aware of any relationships.

A quote from https://cds.cern.ch/record/331888/files/9708319.pdf

Vielbeins are basically tetrads, AFAIK, a description of space-time via four basis vectors at every point in the space-time, allong with a connection that I would loosely describe as "connecting" the different tangent spaces at each point together. (I'm a bit hazy on the specifics, in particular Wiki mentions Cartan connections vs affine connections.)

The words do look very similar, perhaps there is a linguistic connection in the word roots, but I really don't know.

6. Mar 3, 2017

### Ben Niehoff

"Vielbein" = "many legs"
"Vierbein" = "four legs"
"Einbein" = "one leg"

7. Mar 4, 2017

### vanhees71

This is the hint to how to derive the equivalence between the "quadratic" and the "square-root" form of the geodesics equation! Since the choice of the scalar world-line parameter (I call it $\lambda$) is arbitrary, we can as well choose it to be an affine parameter, which means to introduce the constraint
$$\frac{\mathrm{d} s}{\mathrm{d} \lambda}=k=\text{const}.$$
In the physics context usually you either have $k=1$ for time-like curves (possible trajectories of particles with mass) or $k=0$ (possible trajectories of ficitious particles with zero mass or "photons", i.e., light-rays in the sense of the eikonal approximation of electromagnetic waves).

Now it's easy to see, why one can use the much more convenient quadratic form. You simply include the constraint using a Lagrange parameter $\Lambda$ leading to the action
$$A=\int \mathrm{d} \lambda [\sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}} + \frac{\Lambda}{2}(g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}-k).$$
Now the corresponding Lagrangian is not explicity dependent on $\lambda$, and thus
$$H=\dot{x}^{\mu} \frac{\partial L}{\partial \dot{x}^{\mu}}-L=\text{const},$$
but
$$H=\frac{\Lambda}{2}(g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}+k).$$
Due to the constraint the expression in the brackets is constant, and from $\mathrm{d} H/\mathrm{d} \lambda=0$ we thus have $\Lambda=\text{const}$. In the action we can use the constraint, and thus the action finally reads
$$A=\int \mathrm{d} \lambda [\sqrt{k}-\frac{\Lambda}{2} k + \frac{\Lambda}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=0.$$
Since the value of $\Lambda=\text{const}$ is arbitrary and we can cancel any constant additive term in the Lagrangian, we finally have as an equivalent action
$$\tilde{A}=\int \mathrm{d} \lambda \frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\int \mathrm{d} \lambda \tilde{L}(x,\dot{x}).$$
Because it's a quadratic form in $\dot{x}^{\mu}$ the constraint defining $\lambda$ to be an affine parameter is always compatible with the equations of motion: Since $\tilde{L}$ is not explicitly depenent on $\lambda$ again
$$\tilde{H}=\dot{x}^{\mu} \frac{\partial \tilde{L}}{\partial \dot{x}^{\mu}}-\tilde{L}=\tilde{L}=\text{const}=\frac{k}{2}.$$
This can be even generalized to non-free particles, at least for a large class of interactions with external fields. E.g., for the motion of a massive particle in an electromagnetic field instead of the original Lagrangian
$$L=-m \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}-q A_{\mu} \dot{x}^{\mu}$$
one can use
$$\tilde{L}=\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}+q A_{\mu} \dot{x}^{\mu},$$
since the interaction part is linear in $\dot{x}^{\mu}$ and thus the constraint still holds true, because now
$$\tilde{H}=\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}$$
implies that again the parameter $\lambda$ is again automatically affine for the solutions of the equations of motion, and for an affine parameter again the equations of motion from $L$ and $\tilde{L}$ are equivalent as for the free particle.

8. Mar 5, 2017

### ShayanJ

1) Here you're selectively replacing only the square-root part of the action with k. If you do it for the whole Lagrangian, what you get is $\displaystyle A=\int d\lambda \sqrt k$, which is the on-shell value of the action, not zero. So it seems that can't be the way to get to the squared Lagrangian.

2) The equations of motion of the action $\displaystyle A=\int d\lambda \left[ \sqrt{g_{\mu \nu}\dot x^\mu \dot x^\nu}+\frac \Lambda 2 \left( g_{\mu \nu}\dot x^\mu \dot x^\nu-k \right) \right]$ are:

$\Lambda:g_{\mu \nu}\dot x^\mu \dot x^\nu=k$
$\dot x^\sigma: g_{\mu\sigma}\dot x^\mu \left( \frac 1 {\sqrt{g_{\mu \nu}\dot x^\mu \dot x^\nu}}+\Lambda \right)=const$

Which together give $g_{\mu\sigma}\dot x^\mu=const$. This doesn't look like the geodesic equation at all!

9. Mar 5, 2017

### vanhees71

Start from
$$L=\sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}+\frac{\Lambda}{2} \left (g_{\mu \u} \dot{x}^{\mu} \dot{x}^{\nu}-k \right).$$
Then you have
$$p_{\mu}=\frac{\partial L}{\partial \dot{x}^{\mu}}=\frac{g_{\mu \nu} \dot{x}^{\nu}}{\sqrt{...}} + \Lambda g_{\mu \nu} \dot{x}^{\nu}.$$
The Euler Lagrange equation then is
$$\dot{p}_{\mu} = \frac{\partial L}{\partial x^{\mu}}=\frac{1}{2} \frac{\partial g_{\alpha \beta}}{\partial x^{\mu}} \frac{\dot{x}^{\alpha} \dot{x}^{\beta}}{\sqrt{...}} + \frac{\Lambda}{2} \frac{\partial g_{\alpha \beta}}{\partial x^{\mu}} \dot{x}^{\alpha} \dot{x}^{\beta}.$$
For $\Lambda$ we get the constraint.
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=k=\text{const}.$$
$$\left (\frac{1}{\sqrt{k}}+\Lambda \right) \frac{\mathrm{d}}{\mathrm{d} \lambda} [g_{\mu \nu} \dot{x}^{\nu}] =\frac{1}{2} \partial_{\mu} g_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta} \left (\frac{1}{\sqrt{k}}+\frac{\Lambda}{2} \right).$$
Sorting a bit and symmetrizing the coefficient of $\dot{x}^{\alpha} \dot{x}^{\beta}$ gives
$$g_{\mu \nu} \ddot{x}^{\nu} + \frac{1}{2} [\partial_{\alpha} g_{\mu \beta} + \partial_{\beta} g_{\mu \alpha} -\partial_{\mu} g_{\alpha \beta} ] \dot{x}^{\alpha} \dot{x}^{\beta}=0$$
or using the definition of the Christoffel symbols
$$\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=0.$$
This is indeed the geodesics equation under the constraint of using an affine parameter.

It's also easy to see by direct calculation that you get the same by simply using the quadratic Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
just varying the $x^{\mu}$ as independent degrees of freedom. The constraint to have an affine parameter follows automatically as shown in my previous posting from $\tilde{H}=p_{\mu} \dot{x}^{\mu}-L=L=\text{const}$.