Finding geodesics using the squared Lagrangian

In summary, The geodesic equation can be extremized using either the functional ##\displaystyle \int_{\lambda_1}^{\lambda_2}\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d\lambda ## or ## \displaystyle\int_{\lambda_1}^{\lambda_2}g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} d\lambda ##, and they are equivalent. However, using an affine parameter as the scalar world-line parameter makes the quadratic form more convenient. This is achieved by introducing an
  • #1
ShayanJ
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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
 
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  • #2
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.
 
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  • #3
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.
 
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  • #4
pervect said:
The way to get rid of the square root is to use a mathematical trick called an einbein.
Is this related to vielbeins in any way?
 
  • #5
ShayanJ said:
Is this related to vielbeins in any way?

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

Einbein elds were introduced to get rid of square roots which enter the Lagrangians of relativistic systems, though at the price of introducing extra dynamical variables. For example ...

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

"Vielbein" = "many legs"
"Vierbein" = "four legs"
"Einbein" = "one leg"
 
  • #7
Orodruin said:
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.
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.
 
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  • #8
I have two questions about your post @vanhees71.

vanhees71 said:
##\displaystyle A=\int d\lambda \left[ \sqrt k-\frac \Lambda 2 k+\frac \Lambda 2 g_{\mu \nu}\dot x^\mu \dot x^\nu \right]=0##

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
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.
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=k=\text{const}.$$
So the EoM finally reads
$$\left (\frac{1}{\sqrt{k}}+\Lambda \right) \frac{\mathrm{d}}{\mathrm{d} \lambda} [g_{\mu \nu} \dot{x}^{\nu}] =\frac{1}{2} \partial_{\mu} g_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta} \left (\frac{1}{\sqrt{k}}+\frac{\Lambda}{2} \right).$$
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}##.
 
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1. What is the squared Lagrangian method?

The squared Lagrangian method is a mathematical approach used to find geodesics, which are the shortest paths between two points on curved surfaces. It involves squaring the Lagrangian function, which is a mathematical expression that describes the dynamics of a system, and using this to solve for the geodesic equations.

2. How does the squared Lagrangian method work?

The squared Lagrangian method works by taking the Lagrangian function, which is a function of the position and velocity of a particle, and squaring it. This results in a new function that can be used to solve for the geodesic equations, which describe the path of a particle moving under the influence of gravity or other forces.

3. What are the advantages of using the squared Lagrangian method?

One advantage of using the squared Lagrangian method is that it allows for the efficient calculation of geodesics on curved surfaces. It also provides a systematic way to solve for geodesics, making it a useful tool in many areas of physics and engineering.

4. Are there any limitations to the squared Lagrangian method?

One limitation of the squared Lagrangian method is that it can only be applied to systems that can be described by a Lagrangian function. Additionally, it may not work for highly complex systems or systems with non-conservative forces.

5. How is the squared Lagrangian method used in real-world applications?

The squared Lagrangian method has many real-world applications, including in the fields of physics, engineering, and computer graphics. It is used to model the motion of particles in space, calculate optimal paths for spacecraft, and simulate the behavior of materials under different conditions.

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