Effective potential and curved spacetime

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I was looking at this chart and I didn't understand how increased angular momentum of the test particle curves the spacetime around the center mass. If that is how it's interpreted. Now the way it looks like is that the curvature is dependent on the angular momentum of the test particle.
 

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  • #2
Ibix
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If that is how it's interpreted.
It isn't. That's a graph of the effective potential, not the potential. The kinetic energy of the orbiting particle splits neatly into a radial and a tangential component, and the tangential component can be moved to the other side of the equals sign and treated as part of the potential controlling the radial behaviour. It's an extremely useful mathematical device, but it doesn't mean that there's any curvature associated with anything to do with the orbiting test particle.
 
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But we got this from the metric of the spacetime. And this chart with the energy level would allow you to draw an orbit around the center mass.
 
  • #4
A.T.
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Now the way it looks like is that the curvature is dependent on the angular momentum of the test particle.
Because the rotating frame is defined to have the same angular velocity as the particle. The effective potential depends on the rotation of the reference frame, just like in Newtonian mechanics.
 
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Ibix
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But we got this from the metric of the spacetime.
...and a bunch of terms that depend on the orbit of the test particle.
 
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That is what setting phi = pi/2 means right?
 
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Ibix
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That is what setting phi = pi/2 means right?
It's theta (##\theta##) that's set to ##\pi/2##, not phi (##\phi##). And that's just choosing coordinates so that the particle orbits in the coordinates' equatorial plane. It simplifies the maths with no loss of generality in this spherically symmetric case.
 
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Well bare with me for a minute, I know that there is a kinetic term and also a "potential term" that is the slope of the curved line I posted up and the total energy should be fixed. I'm just trying to understand if increasing the total energy of the system, right, does anything to the curvature of the spacetime. I know that if I have somekind of angular momentum T my orbit will be different.

I don't understand why I can't equate this curved line with the "curved spacetime". Even if I take the example of two separate systems, with the same center mass, but where the initial angular momenta of the test particles was different, my intuition based on everything I've read so far would tell me that the space times would be curved differently.


But ok....
 
  • #9
PeterDonis
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I know that there is a kinetic term and also a "potential term" that is the slope of the curved line I posted up and the total energy should be fixed.
The effective potential is not the same as the total energy. The effective potential is derived by treating the total energy as a constant of geodesic motion, but that does not mean that the sum of all the terms in the effective potential equation is the total energy. It isn't.
 
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PeterDonis
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I don't understand why I can't equate this curved line with the "curved spacetime".
Because the effective potential is not a part of the spacetime. It's an abstraction constructed to help simplify the analysis of geodesic motion. It is not the same thing as the spacetime geometry or any part of it.
 
  • #11
Ibix
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I'm just trying to understand if increasing the total energy of the system, right, does anything to the curvature of the spacetime
No. These are test particles - by definition they have no effect on curvature. If you want to consider orbits of bodies that contribute to curvature then you'll need a different metric from the Schwarzschild metric.
I don't understand why I can't equate this curved line with the "curved spacetime".
Because you've included terms that come from curvature and terms that have nothing to do with curvature. So the result cannot be described solely in terms of curvature.

Even if I take the example of two separate systems, with the same center mass, but where the initial angular momenta of the test particles was different, my intuition based on everything I've read so far would tell me that the space times would be curved differently.
Then you are either reading bad sources or misunderstanding good ones. Curvature is the Riemann tensor. Full stop.

It's true that you can change the components of the Riemann tensor by picking different coordinate systems, but that isn't changing the Riemann tensor, merely picking a different expression of it.
 
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is that what changing the angular momentum means in this instance? picking a different coordinate system?
 
  • #13
Ibix
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is that what changing the angular momentum means in this instance? picking a different coordinate system?
Depends what you mean by "changing the angular momentum". If you mean "imagining dropping another test object with a different angular momentum" then yes, that's one way of interpreting it. As @A.T. noted some posts back, you're picking a coordinate system that rotates with the particle and if it has a different angular momentum you need a different coordinate system.
 
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pervect
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It might be helpful to look up "effective potential" in a Newtonian context. I can't recall where I saw this - probably Goldstein. Possibly MTW, but while they used it, I don't think they explained it.

Anyway, you write the differential equations of motion for the orbit, and you eliminate the angular part, solving only for the radial motion. Then you can interpret that differential equation of motion as if a mass were moving in an "effective" 1 dimensional potential. But basically, it's just a way of getting some intuition about the differential equation of motion.

From https://users.physics.ox.ac.uk/~harnew/lectures/lecture17-mechanics-handout.pdf with changes in notation.

We can write the conserved Newtonian energy as

$$E = \frac{1}{2}m \left( \dot{r}^2 + r^2 \, \dot{\theta}^2 \right) + U(r)$$

Here a dot is used to represent differentiation with time, and U(r) is the potential of the central force, usually -GMm/r, and E is the energy, a constant of motion.

Now the conserved angular momentum L is L = ##m r^2 \dot{\theta}##. So we can re-write the energy equation as

$$E= \frac{1}{2}m \left( \dot{r}^2 + r^2 \left( \frac{L}{m r^2} \right)^2 \right) + U(r) = \frac{1}{2}m \dot{r}^2 + \frac{1}{2} \frac{L^2}{m r^2} + U(r)$$

So, this is a differntial equation in r and ##\dot{r}##. This is the same equation of motion that one would get for the mass m, moving in a different potential of ##U(r) + L^2 / 2 m r^2##. As I said earlier, this is just a way to help visualize the differential equation for r - rather than an abstract equation, one visualizes it as the motion of a mass m in the "effective" potential.

The analysis in GR is similar in concept and different in detail. E and L exist as conserved quantities in GR for a test mass orbiting a massive body, but the equations are slightly different.

I don't want to go into the GR case in this post. "Orbits in Strongly Curved Space-time", https://www.fourmilab.ch/gravitation/orbits/, has a theoretical discussion based on MTW's textbook description in "Gravitation", plus some interactive java elements.
 
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