I Conserved Quantity Along Affine Parameter

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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##.
 
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Conservation of the squared 4-velocity. Essentially the affine parameter requirement.
 
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##.
 
Dale said:
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).
 
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Dale said:
But what is ?
##Q## is just the Lagrangian itself. Would we ever expect the parameter ##\lambda## to appear in the Lagrangian?
 
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Orodruin said:
Note that ##Q = \mathcal L##
PeterDonis said:
##Q## is just the Lagrangian itself
Embarrassingly, I didn't even notice that.

PeterDonis said:
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. :doh:
 
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Orodruin said:
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##.
 
This is from my GR lecture notes:
1658328140871.png

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 ...
1658328341267.png

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##.
 
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Dale said:
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.
 
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