Exists ? : Invariant geodesic equation

  • A
  • Thread starter jk22
  • Start date
  • #1
695
21
Does there exist a form of the geodesic equation which is invariant under coordinates change ?
 

Answers and Replies

  • #2
Infrared
Science Advisor
Gold Member
788
408
A curve ##\gamma## is a geodesic if and only if ##\nabla_{\dot{\gamma}}\dot{\gamma}=0.## Is this what you mean?
 
  • Like
Likes Ibix
  • #3
695
21
I don't know this notation.
 
  • #4
Ibix
Science Advisor
Insights Author
6,920
5,824
##\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}##.
 
  • #5
samalkhaiat
Science Advisor
Insights Author
1,713
984
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:

[tex]\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).[/tex]
 
  • #6
21
14
@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.
 
  • #7
695
21
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:

[tex]\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).[/tex]
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.
 
  • #8
Ibix
Science Advisor
Insights Author
6,920
5,824
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.
 
  • #9
695
21
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.
 
  • #10
Ibix
Science Advisor
Insights Author
6,920
5,824
$$\ddot{\phi}=2\cot\theta\dot{\theta}\dot{\phi}$$
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 */
load(ctensor);
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:
  • Like
Likes jk22
  • #11
695
21
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.
 
  • #12
695
21
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.
 

Related Threads on Exists ? : Invariant geodesic equation

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
932
  • Last Post
Replies
1
Views
4K
Replies
5
Views
928
  • Last Post
Replies
1
Views
2K
Replies
5
Views
2K
Replies
14
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
8
Views
8K
Top