Conserved Quantity Along Affine Parameter

[Dale]

TL;DR Summary

What is the meaning/interpretation of the conserved quantity along the affine parameter in the Schwarzschild spacetime

In the usual Schwarzschild coordinates the Lagrangian can be written: $$\mathcal{L}= \frac{\dot r^2}{1-\frac{2M}{r}} - \left( 1- \frac{2M}{r} \right) \dot t^2 + r^2 \dot \phi^2$$ where all derivatives are with respect to a (affine) parameter $\lambda$, and where for convenience I have considered units such that $c=G=1$ and coordinates such that $\theta = \pi/2$ so everything is in the equatorial plane.

Inspecting the Lagrangian we see that $t$, $\phi$, and $\lambda$ do not appear. So we have three easy conserved quantities: $$ E=\left( 1-\frac{2M}{r} \right) \dot t^2$$ $$ L=r^2 \dot \phi $$ $$Q = \frac{\dot r^2}{1-\frac{2M}{r}}-\left( 1-\frac{2M}{r} \right) \dot t^2 + r^2 \dot \phi^2$$

I understand that $E$ is interpreted as a conserved energy and $L$ is interpreted as a conserved angular momentum. But what is $Q$?

It is conserved, but it isn't apparent to me what it is. I also am not sure if it is useful. I can solve for $\dot t$ in terms of $E$ and for $\dot \phi$ in terms of $L$ and use those to simplify my geodesics. But I don't see anything similar to be done for $Q$.

 

[Orodruin]

Conservation of the squared 4-velocity. Essentially the affine parameter requirement.

 

[Dale]

Oh, so it just is the condition in these coordinates that makes $\lambda$ into an affine parameter instead of some generic parameter.

I guess then that different values for $Q$ correspond to different scalings for the affine parameter. So the value of $Q$ can be arbitrarily and freely selected and is not physically meaningful like $E$ or $L$.

 

[Orodruin]

I guess then that different values for $Q$ correspond to different scalings for the affine parameter. So the value of $Q$ can be arbitrarily and freely selected and is not physically meaningful like $E$ or $L$.

The standard choice would of course be $Q =-1$ or 0 (the latter for light-like geodesics). Note that $Q = \mathcal L$.

Inserting the other constants of motion into $Q$ gives the typical effective equation of motion for $r$ which is very reminiscent of the classical Newtonian one (with some additions).

 

[PeterDonis]

But what is ?

$Q$ is just the Lagrangian itself. Would we ever expect the parameter $\lambda$ to appear in the Lagrangian?

 

[Dale]

Note that $Q = \mathcal L$

$Q$ is just the Lagrangian itself

Embarrassingly, I didn't even notice that.

Would we ever expect the parameter $\lambda$ to appear in the Lagrangian?

No. This is part of what confused me. I can easily think of Lagrangians without conserved energy or angular momentum, but not $Q$. I guess it is unsurprising that I could not think of a Lagrangian without a Lagrangian.

 

[Dale]

Inserting the other constants of motion into $Q$ gives the typical effective equation of motion for $r$ which is very reminiscent of the classical Newtonian one (with some additions).

Interesting. So I get $$ \frac{r^3 \left(\dot r^2-E^2\right)-2 L^2 M+L^2 r}{r^2 (r-2 M)}=-1 $$ I could solve that for $\dot r$ to get a first-order differential equation that seems useful. Otherwise, even with the conserved quantities, the original Euler equations gives me a second-order differential equation for $r$.

 

[Orodruin]

This is from my GR lecture notes:

In essence, you find the Newtonian equation of motion with time replaced by proper time and the additional term caused by $\alpha$ (your $Q$, essentially a constant term) and the additional term caused by the cross term of the angular momentum barrier with the Newtonian potential.

Edit: Note that I also defined $E$ differently ...

This choice is obviously just to reproduce the similarity to the Newtonian equation.

Edit 2: You should find the same if you replace $E^2 \to 2E$ and solve for $\dot r^2/2$.

 

[vanhees71]

Oh, so it just is the condition in these coordinates that makes $\lambda$ into an affine parameter instead of some generic parameter.

I guess then that different values for $Q$ correspond to different scalings for the affine parameter. So the value of $Q$ can be arbitrarily and freely selected and is not physically meaningful like $E$ or $L$.

Yes, that's the great thing when choosing the "squared form" of the Lagrangian. Your parameter is automatically affine since $Q=\text{const}$ means that $g_{\mu \nu} \dot{q}^{\mu} \dot{a}^{\nu}=\text{const}$. If you have massive particle, you can choose this to be $c^2$, and your affine parameter is the proper time of the particle. You can of course also choose $Q=0$ for a massless particle.