Going to higher degrees to obtain other solutions?

[jk22]

Suppose the equations of motion coming from GR in Schwarzscild spacetime for constant radius :

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

Could it be that by solving this system of ODE in increasing the degree, hence inducing an initial condition over speed and acceleration, that the symmetry gets broken and that 3d trajectories happen, not planar ones ?

 

[Office_Shredder]

Sorry for the dumb question, but what does going to a higher degree mean? Just taking the derivative of the equations with respect to time and using only those instead?

 

[jk22]

I thought rather for example isolating $\dot{\phi}$ in the first equation and substituting in the second thus giving a jerk of $\theta$ or other substitution in the system of equations.

It's rather my question that seems dumb, since the way should be to simplify hence reduce the degree. Going the other way would be to get further away from the solution.

 

[jk22]

Big error : not the degree but the order of the ODE.

 

[Office_Shredder]

That doesn't seem crazy as an idea. To pick a simple example to show why it might help, suppose you have

x'=y
y'=x

Then you could sub in y' for x in the first equation and get
y''=y

Which turns out to be pretty easy to solve.

I'm not sure if it helps in this specific case. You would just have to try it I think. Probably someone more expert could identify what's useful here.

 

[pasmith]

I don't think that will help. Whether you write a system as a one-dimensional sixth-order ODE or a three-dimensional second-order ODE or a six-dimensional first-order ODE should not change the symmetries of the system. In each case you have six independent constants of integration (initial conditions) and converting between them is a merely a question of algebra (which may or may not be tractable depending on the specific system).

 

[jk22]

That's what I doubt too. In fact this question came to my mind when I tried to prove general relativistic geodesics for Schwarzschild were planar.

In classical mechanics, one simply derive the angular momentum towards time and considering Newton's acceleration law and central forces give that symmetry.

But in GR I thought : The fact that in classical physics it's planar comes from the orthogonality to a vector. But in GR there are 4 dimensions, so an orthogonal to a vector is a space.

So this could not be the argument. That's why I tried to find three dimensional trajectories, but these seem not to exist in two-bodies central problems or one-body in relativity.

Maybe there is a simple argument in spherically symmetric GR solutions to prove geodesics are planar ?