Question on Effective Potential

[1effect]

In this case I think the math is hiding something. When we say r -> infinity, we do not mean r = infinity, which is impossible.

I am very familiar with calculus, I understand very well the limiting process. :-)

Further, in order to have L > 0 initially, we must have a finite orbital velocity v_o. Why would that velocity (and hence momentum) change if we just increase r without limit?

All I have pointed out is that if $$a=\frac{L}{m}=constant$$ then $$lim \frac{a}{r^2}=0$$. No? Something changed in the theory of limits specifically for this thread? :-)

 

[yuiop]

I am very familiar with calculus, I understand very well the limiting process. :-)



All I have pointed out is that if $$a=\frac{L}{m}=constant$$ then $$lim \frac{a}{r^2}=0$$. No? Something changed in the theory of limits specifically for this thread? :-)

Hi 1effect,

I think we have generally agreed that if $$a=\frac{L}{m}=constant$$ then $$lim \frac{a}{r^2}=0$$ but we were just having some issues when the situation was looked at from different viewpoints.

I think we resolved those issues by assuming that, because L contains proper time dtau in the equation, then the "baseline" observer with the same radial coord as the particle (possibly at infinty) must be comoving with the particle. That way the baseline momentum of the particle is zero.

My substituting mvr for L and then assuming mvr/(mcr) = v/c taking out the radial dependence of L possibly caused the confusion, because mvr is a constant (in this case) so the r of mvr (which is a constant) does not cancel out the r of mcr (which is not a constant). The use of geometrical units makes some of this stuff hard to follow. As you pointed out earlier, does $E/(m^2c^2) -c^2 = 0$? It is hard to tell when m is not clearly defined. That raises another question. If $E/(m^2c^2)= (dt/d\tau)^2(1-2GM/(rc^2))$ does that mean $E/(m^2c^2)= (1-2GM/(rc^2))^2$ ?

 

[Jorrie]

I think we resolved those issues by assuming that, because L contains proper time dtau in the equation, then the "baseline" observer with the same radial coord as the particle (possibly at infinty) must be comoving with the particle. That way the baseline momentum of the particle is zero.

But shouldn't one view the particle in the frame of the massive body (Schwarzschild coordinates, which is non-rotating)?

In such a case, a particle that had transverse momentum in the frame will have to retain that, i.e., the energy is $E^2 = (mc^2)^2 + (mvc)^2$ at r -> infinity.

I agree with the math limits exercise, but argue that it does not represent the problem correctly.

-J

 

[mysearch]

Opening new thread

Jorrie, just a quick thanks for the two equations in post #101.
A clarification on the equation L=GM/c, the units of this equation only add up if L=GMm/c. So is the reference implicitly with respect to unit mass?

I wanted to also say that I have opened a new thread `Circular Orbits of a Black Hole` which, in part, extends the discussion of this thread to some specific questions I have concerning the implications of effective potential on circular orbits. Hopefully you might all wish to put me straight on these issues as well

 

[Jorrie]

A clarification on the equation L=GM/c, the units of this equation only add up if L=GMm/c. So is the reference implicitly with respect to unit mass?

I don't know what you mean by L=GM/c, because it is not an equation for angular momentum. You are not perhaps thinking of R_s=GM/c^2?

-J

 

[mysearch]

Quick reply to #110:
To be honest, I was rushing to pull some ideas together and had just picked on on the reference in your post #15. L>3.4642GM/c and just assumed that there was some correlation, but then noticed a discrepancy in the units. Can you save me some time and tell me where this lower limit comes from and how they are related? Thanks

 

[Jorrie]

Quick reply to #110:
To be honest, I was rushing to pull some ideas together and had just picked on on the reference in your post #15. L>3.4642GM/c and just assumed that there was some correlation, but then noticed a discrepancy in the units. Can you save me some time and tell me where this lower limit comes from and how they are related? Thanks

I did state in my post #15 that V_eff and L are per unit mass in that equation. I think we later agreed that we should distinguish between 'real' V_eff and L and specific (per unit mass) use, by writing V_eff/m and L/m explicitly.

The 'lower limit of L' is not really a limit, but simply the smallest angular momentum the will just be unable to cause an stable circular orbit around a black hole, i.e., below that value the particle will spiral into the hole. This 'minimum value' of L/m corresponds to a circular orbit at r=6GM/c^2.

-J