# Exists ? : Invariant geodesic equation

• A
Does there exist a form of the geodesic equation which is invariant under coordinates change ?

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Infrared
Gold Member
A curve ##\gamma## is a geodesic if and only if ##\nabla_{\dot{\gamma}}\dot{\gamma}=0.## Is this what you mean?

Ibix
I don't know this notation.

Ibix
##\dot{\gamma}## is the tangent vector to a curve. ##\nabla_{\dot{\gamma}}## is the covariant derivative along the curve. Writing ##\dot{\gamma}=\frac{dx^\mu}{d\tau}##, you get ##\nabla_{\dot\gamma}\dot\gamma=\frac{dx^\mu}{d\tau}\nabla_\mu\frac{dx^\nu}{d\tau}##.

samalkhaiat
Does there exist a form of the geodesic equation which is invariant under coordinates change ?
The geodesic equation cannot be invariant because it has one free (vector) index. It is, therefore, covariant (i.e., form invariant) with respect to an arbitrary change of coordinates. Indeed, it is a good exercise to show that the left-hand-side of the geodesic equation transforms as a vector:

$$\left(\frac{d^{2}\bar{x}^{\mu}}{d\tau^{2}} + \bar{\Gamma}^{\mu}_{\rho \sigma}(\bar{x}) \frac{d\bar{x}^{\rho}}{d\tau}\frac{d\bar{x}^{\sigma}}{d\tau}\right) = \frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \left( \frac{d^{2} x^{\nu}}{d\tau^{2}} + \Gamma^{\nu}_{\rho \sigma}(x) \frac{d x^{\rho}}{d\tau}\frac{d x^{\sigma}}{d\tau}\right).$$

@jk22 I think a precise way of asking your question is 'Does their exist a way to write an equation for a geodesic that does not require using coordinates?" @Infrared in post #2 has given it.

The geodesic equation cannot be invariant because it has one free (vector) index. It is, therefore, covariant (i.e., form invariant) with respect to an arbitrary change of coordinates. Indeed, it is a good exercise to show that the left-hand-side of the geodesic equation transforms as a vector:

$$\left(\frac{d^{2}\bar{x}^{\mu}}{d\tau^{2}} + \bar{\Gamma}^{\mu}_{\rho \sigma}(\bar{x}) \frac{d\bar{x}^{\rho}}{d\tau}\frac{d\bar{x}^{\sigma}}{d\tau}\right) = \frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \left( \frac{d^{2} x^{\nu}}{d\tau^{2}} + \Gamma^{\nu}_{\rho \sigma}(x) \frac{d x^{\rho}}{d\tau}\frac{d x^{\sigma}}{d\tau}\right).$$
I wanted to say rotationally invariant in the 3d space. I ask this because of the following : there are circular uniform orbits by fixing theta at a certain value, but if one rotate this trajectory around the origin then it is no more solution of the 4 geodesic equation.

Ibix
I wanted to say rotationally invariant in the 3d space. I ask this because of the following : there are circular uniform orbits by fixing theta at a certain value, but if one rotate this trajectory around the origin then it is no more solution of the 4 geodesic equation.
If you are using the geodesic equation in a space that is rotationally symmetric about a point and you rotate solutions about that point, they will still be solutions.

Are doing something like simplifying the geodesic equations in Schwarzschild spacetime to solve for paths in the equatorial plane, then rotating those solutions out of the plane? The result will be a solution of the geodesic equations but not of the simplified equations.

For a circular motion ##r## is constant and time flows linearly. The geodesic equation reads

$$\ddot{\theta}=\sin\theta\cos\theta\dot{\phi}^2$$
$$\ddot{\phi}=2\cot\theta\dot{\theta}\dot{\phi}$$
$$\dot{\theta}^2+\sin^2\theta\dot{\phi}^2=C(onstant)$$

The second equation is divided by ##\dot{\phi}## giving by integration :

$$\dot{\phi}=A\sin^2\theta$$

Substituting in the first and multiplying by ##\dot{\theta}## gives

$$\dot{\theta}^2/2=A^2/6\sin^6\theta$$

The last equation becomes

$$\dot{\theta}^2+A^2\sin^6\theta=C$$

Thus one can deduce that ##\theta=\sqrt[6]{3C/4A^2}##

However if one considers a circular uniform motion around the y axis then ##\theta=\omega t## increases linearly with time, hence this cannot be a solution and gives a contradiction since it should be one.

Ibix
$$\ddot{\phi}=2\cot\theta\dot{\theta}\dot{\phi}$$
You haven't stated what spacetime you are working in, so once again we need to guess part of your problem. Please try to state a complete problem.

Assuming you are working in Schwarzschild spacetime, I believe the quoted equation should be $$\ddot{\phi}=-2\cot\theta\dot{\theta}\dot{\phi}$$
Code:
/* Maxima batch file to calculate geodesic equations in Schwarzschild  */
/* spacetime, simplifying for the case of a circular orbit (dr/dtau=0, */
/* dt/dtau=const)                                                      */

/* Load ctensor and set up for Schwarzschild spacetime */
ct_coordsys(exteriorschwarzschild);
depends(r,tau);
depends(t,tau);
depends(theta,tau);
depends(phi,tau);

/* Calculate inverse metric and Christoffel symbols */
ug:invert(lg);
christof(mcs);

/* Function to generate the ith geodesic equation. Note that Maxima  */
/* numbers arrays from 1 and the Christoffel symbol \Gamma^i_{jk} is */
/* stored in array element mcs[j,k,i].                               */
geodesic(i):=block(
[j,k,geo],
geo:diff(ct_coords[i],tau,2),
for j:1 thru 4 do block (
for k:1 thru 4 do block (
geo:geo+mcs[j,k,i]*diff(ct_coords[j],tau)*diff(ct_coords[k],tau)
)
),
return(ratsimp(geo)=0)
);

/* Generate the geodesic equations and simplify */
[geodesic(1),geodesic(2),geodesic(3),geodesic(4)];
substitute(0,diff(r,tau,2),%);
substitute(0,diff(r,tau),%);
substitute(R,r,%); /* R is a constant */
substitute(0,diff(t,tau,2),%);
substitute(K,diff(t,tau),%); /* K is a constant */
ratsimp(%);
expand(%);

Last edited:
jk22
Yes, in fact I used wikipedia equations of motion : https://en.m.wikipedia.org/wiki/Schwarzschild_geodesics and I made a sign mistake.

But my argument is anyhow wrong since the above solution is only if ##\dot{\phi}\neq 0##, whereas a rotation around the y axis is exactly that countercase.

Erratum:
For a circular motion ##r## is constant and time flows linearly. The geodesic equation reads

$$\ddot{\theta}=\sin\theta\cos\theta\dot{\phi}^2$$
$$\ddot{\phi}=-2\cot\theta\dot{\theta}\dot{\phi}$$
$$\dot{\theta}^2+\sin^2\theta\dot{\phi}^2=C(onstant)$$

The second equation is divided by ##\dot{\phi}## giving by integration :

$$\dot{\phi}=a/\sin^2\theta$$

Substituting in the first and multiplying by ##\dot{\theta}## gives

$$\dot{\theta}^2=-a^2/\sin^2\theta+d$$

Giving :

$$(\sin(x) \sqrt(d - a \csc^2(x)) (\sqrt(d) \log(\sqrt(2 a + d \cos(2 x) - d) +\sqrt(2) \sqrt(d)\cos(x)) - \sqrt(a) \tanh^(-1)((\sqrt(2) \sqrt(a) \cos(x))/\sqrt(2 a + d \cos(2 x) - d))))/\sqrt(a + 1/2 d \cos(2 x) - d/2)=t$$(##x=\theta##, by wolframalpha)

The last equation becomes

$$D=C$$

So we see it is a solution.