# Choice of action to geodesic equations

• Mishra
In summary: The two approaches are not exactly the same. If you minimize the first action integral, you obtain the equations of motion:\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} + \frac{dx^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\t
Mishra
Hello,

I am quite new to GR and I have a question regarding the construction of the action to find the geodesic equation.
In pretty much every book, you'll find:

##S=-m ∫ dS##

using: ## dS=dS\frac{d\tau}{d\tau}=\sqrt{g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau## with ##dS=\sqrt{g_{\mu \nu}dx^{\mu}dx^{\nu}}##

one gets: ##dS= -m∫\sqrt{g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau##

Minimisation of this quantity gives rise to the geodesic equations.
I understand that contruction: given a certain metric, the particule moves in a "straight" line, minimizing proper time.

Here's my problem, when dealing with the Schwarzschild metric, my prof. uses:
##S= -m∫g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}d\tau##
Which makes the computation much easier, I've also been able to re-find the geodesic equation starting from this expression.My questions would be:
What is the diffence between ##d\tau## and ##dS## and what is the difference between these to actions ? I feel like I'm confusing proper time, interval ##dS## and the action differential...

It would be great if someone could explain me the meaning of this!
Thank you very much

The second choice is equivalent to the first if you reparameterise your curve such that the curve parameter is an affine parameter (i.e., essentially you are taking the curve length as the parameter). This is of course a choice you can always make and it will not change the curve that results.

Mishra said:
Hello,

I am quite new to GR and I have a question regarding the construction of the action to find the geodesic equation.
In pretty much every book, you'll find:

##S=-m ∫ dS##

using: ## dS=dS\frac{d\tau}{d\tau}=\sqrt{g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau## with ##dS=\sqrt{g_{\mu \nu}dx^{\mu}dx^{\nu}}##

one gets: ##dS= -m∫\sqrt{g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau##

Minimisation of this quantity gives rise to the geodesic equations.
I understand that contruction: given a certain metric, the particule moves in a "straight" line, minimizing proper time.

Here's my problem, when dealing with the Schwarzschild metric, my prof. uses:
##S= -m∫g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}d\tau##
Which makes the computation much easier, I've also been able to re-find the geodesic equation starting from this expression.My questions would be:
What is the diffence between ##d\tau## and ##dS## and what is the difference between these to actions ? I feel like I'm confusing proper time, interval ##dS## and the action differential...

The two approaches are not exactly the same. If you minimize the first action integral, you obtain the equations of motion:

$\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} + \frac{dx^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}}) = 0$

where $\Gamma^\mu_{\nu \lambda}$ is a combination of derivatives of $g_{\mu \nu}$.

$\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} = 0$

They are only the same if
$\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0$

which means

$\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}} = K$

for some constant $K$, which in turn implies:

$d\tau = \frac{1}{K} \sqrt{g_{\nu \lambda} dx^\nu dx^\lambda}$

which means that $\tau$ is linearly related to proper time.

The first integral works for any parameter $\tau$, while the second only works when $\tau$ is proper time.

stevendaryl said:
They are only the same if
$\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0$
Just to nitpick, I suspect you intended to write:
$$\frac d{d\tau}\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0$$

First thank you both for you answers.

It is very strange, I've done the computations yesterday and both gave me the exact same geodesic equations. I've must have made the assumption ##d\tau=dS## during the second one without knowing it. I'll take a look at it again.

So what is the difference between proper time ##d\tau## and space-time interval ##dS## ? I can't figure it out.
Also, does that mean that I can take any action ##dS^{z}##?

Orodruin said:
Just to nitpick, I suspect you intended to write:
$$\frac d{d\tau}\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0$$

Yes.

Mishra said:
First thank you both for you answers.

It is very strange, I've done the computations yesterday and both gave me the exact same geodesic equations. I've must have made the assumption ##d\tau=dS## during the second one without knowing it. I'll take a look at it again.

So what is the difference between proper time ##d\tau## and space-time interval ##dS## ? I can't figure it out.
Also, does that mean that I can take any action ##dS^{z}##?

The relation between proper time and the space-time interval is the same as the relation between curve length and the space interval in Riemann geometry. In Riemann geometry, you could always pick the length along the curve as your curve parameter. In the same way, you can here pick the proper time along the world line as your curve parameter.

Does that mean I can always pick ##dS=d\tau## ?

That would mean ##g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\mu}}{d\tau}=1## right ?

Mishra said:
First thank you both for you answers.

It is very strange, I've done the computations yesterday and both gave me the exact same geodesic equations. I've must have made the assumption ##d\tau=dS## during the second one without knowing it. I'll take a look at it again.

It's the first integral where you have to make the assumption that $d\tau = dS$ in order to get rid of the term

$\frac{d x^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}})$

Using the Lagrangian equations of motion, with $L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}$, you find

$\frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{g_{\mu \nu} \frac{dx^\nu}{d\tau}}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}}$

Now, take $\frac{d}{d\tau}$ and you get two terms:
$\frac{d}{d\tau} \frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{\frac{d}{d\tau} (g_{\mu \nu} \frac{dx^\nu}{d\tau})}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}} + g_{\mu \nu} \frac{dx^\nu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}})$
The first term leads to the usual geodesic equation (once you multply through by $g^{\mu \mu'} \sqrt{g_{\mu'' \nu}\frac{dx^{\mu''}}{d\tau} \frac{dx^\nu}{d\tau}}$). The second term is an additional term not found in the geodesic equation that you derive from the second action integral.

So what is the difference between proper time ##d\tau## and space-time interval ##dS## ?

In the first integral, $\tau$ is not necessarily proper time. It is any parameter whatsoever that increases continuously along the path. The first integral is invariant under parameter changes.

Orodruin said:
The relation between proper time and the space-time interval is the same as the relation between curve length and the space interval in Riemann geometry. In Riemann geometry, you could always pick the length along the curve as your curve parameter. In the same way, you can here pick the proper time along the world line as your curve parameter.

Except that in the first integral, it isn't necessary to pick $\tau$ to be proper time.

Mishra said:
Does that mean I can always pick ##dS=d\tau## ?

That would mean ##g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\mu}}{d\tau}=1## right ?

Yes, and that allows you to throw away any derivatives of that expression (since they'll be zero).

stevendaryl said:
Except that in the first integral, it isn't necessary to pick $\tau$ to be proper time.
Which is why I used "can" and not "must". :)
It is the same in Riemann geometry, you can, but must not, use the curve length as the curve parameter. For example, I could parameterise the line ##x = y## as ##x = t^3## and ##y = t^3##.

Just getting rid of the square root is wrong, as other posters have mentioned. However, there is a cute trick, known as an einbein, that allows one to accomplish the same end legitimately.

See for instance http://arxiv.org/pdf/hep-ph/9708319v1.pdf

Maybe more later if I have time.

stevendaryl said:
It's the first integral where you have to make the assumption that $d\tau = dS$ in order to get rid of the term

$\frac{d x^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}})$

Using the Lagrangian equations of motion, with $L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}$, you find

$\frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{g_{\mu \nu} \frac{dx^\nu}{d\tau}}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}}$

Now, take $\frac{d}{d\tau}$ and you get two terms:
$\frac{d}{d\tau} \frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{\frac{d}{d\tau} (g_{\mu \nu} \frac{dx^\nu}{d\tau})}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}} + g_{\mu \nu} \frac{dx^\nu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}})$
The first term leads to the usual geodesic equation (once you multply through by $g^{\mu \mu'} \sqrt{g_{\mu'' \nu}\frac{dx^{\mu''}}{d\tau} \frac{dx^\nu}{d\tau}}$). The second term is an additional term not found in the geodesic equation that you derive from the second action integral.
In the first integral, $\tau$ is not necessarily proper time. It is any parameter whatsoever that increases continuously along the path. The first integral is invariant under parameter changes.

I meant the first integral you're right.
I checked my notes: I did not used the Lagrange equation to derive the geodesics. I studied the variation of the action ##\delta S## with respect to the coordinates . It turns out I have indeed set ##dS=d\tau## during the computation.
I have to say, using the Euler-Lagrange equations is much more handy!
stevendaryl said:
Yes, and that allows you to throw away any derivatives of that expression (since they'll be zero).
Then that explain how I got the equations of motion in the Schwarzschild metric! Thanks!
stevendaryl said:
In the first integral, τ\tau is not necessarily proper time. It is any parameter whatsoever that increases continuously along the path. The first integral is invariant under parameter changes.

I am not sure what you mean by that. Is it that in the first integral, ##d\tau## is just any variable in the integral ? In other words ##d\tau## is any infenitesimal space-time interval, as long as it follows the integration path ? I might as well write ##dP,dD,dT## etc.? (like ##dx,dy,dz,dt## mean the same thing in a integral, they are just the "integration variables").
And since I am integrating along the trajectory in space time it turns out that ##d\tau## is the proper time during the integration ?

I hope this makes sense and that I am not too far from the actual meaning of ##d\tau, dS##, thank you very much to all

Last edited:
Mishra said:
I am not sure what you mean by that. Is it that in the first integral, ##d\tau## is just any variable in the integral ? (like ##dx,dy,dz,dt## mean the same thing in a integral, they are just the "integration variables").
And since I am integrating along the trajectory in space time it turns out that ##d\tau## is the proper time during the integration ?

I hope this makes sense and that I am not too far from the actual meaning of ##d\tau, dS##, thank you very much to all

##d\tau## in the first integral is a priori only a curve parameter and does not have to be an affine parameter for the proper time. It is only after setting ##d\tau = dS## that you can interpret it as proper time.

Orodruin said:
##d\tau## in the first integral is a priori only a curve parameter and does not have to be an affine parameter for the proper time. It is only after setting ##d\tau = dS## that you can interpret it as proper time.

Hum so:
##dS## is the infinitesimal displacement in spacetime along the true trajectory
##d\tau## is any curve parameter

And I can always choose ##d\tau=dS## ?

Mishra said:
And I can always choose ##d\tau=dS## ?

Yes. You can always make such a reparameterisation. It is a matter of solving the differential equation ##dS/d\tau = 1##.

I think I'm getting the meaning of all this
You've been very patient thank you very much!

Mishra said:
Hum so:
##dS## is the infinitesimal displacement in spacetime along the true trajectory
##d\tau## is any curve parameter

And I can always choose ##d\tau=dS## ?

I'd like to point out though, that usually the notation ##d\tau## is for a proper time. For a non-proper time parametrization one uses usually ##dt##. People will be confused if you use ##d\tau## to mean an arbitrary curve parametrization.

## 1. What are geodesic equations?

Geodesic equations are mathematical equations that describe the shortest path between two points in a curved space. They are used to study the movement of objects in non-Euclidean spaces, such as in the theory of general relativity.

## 2. How are geodesic equations related to choice of action?

The choice of action in geodesic equations refers to the specific mathematical form of the equations used to describe the motion of objects. Different choices of action can lead to different equations, but they all ultimately describe the same geodesic path.

## 3. What is the importance of studying geodesic equations?

Studying geodesic equations is important because they provide a fundamental understanding of how objects move in curved spaces. They have many applications in fields such as physics, astrophysics, and engineering.

## 4. Can geodesic equations be applied to real-world situations?

Yes, geodesic equations can be applied to real-world situations. For example, they are used in the navigation systems of spacecrafts and satellites, and in the study of gravitational lensing in astrophysics.

## 5. Are there any challenges in using geodesic equations to model motion?

One challenge in using geodesic equations is that they can be complex and difficult to solve. They also require a good understanding of differential geometry and tensor calculus. Additionally, in some cases, there may be multiple geodesic paths between two points, making it difficult to determine the specific path an object will take.

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