Question on Effective Potential

[Jorrie]

There appears to be a small error with your two equations because they imply

d\tau = dt\left(1-\frac{Rs}{r}\right)

when the gravitational time dilation factor should be

d\tau = dt \sqrt{\left(1-\frac{Rs}{r}\right) }

I think the two equations of 'mysearch':

a) <br /> dr/dt = -c\left(1-\frac{Rs}{r}\right) \sqrt{\frac{Rs}{r}} <br />

b) <br /> dr/d\tau = -c \sqrt{\frac{Rs}{r}} <br />

are correct, where dr/dtau is the (negative of the) radial escape velocity measured locally. For the 'infinity' observer, there are two equal factors \sqrt{1-Rs/r} that multiply: one from gravitational time dilation and one from spatial curvature.

While we're at it, the circular orbital velocities that 'mysearch' is looking for are probably:

b) <br /> rd\phi/dt = c\sqrt{\frac{Rs}{2r}} <br />

c) <br /> rd\phi/d\tau = c{\sqrt{\frac{Rs}{2r (1-Rs/r)}} <br />

where b) is identical to Newton's and c) is measured locally and is b) divided only by the gravitational time dilation factor \sqrt{1-Rs/r}, since there is no spatial curvature along a circular orbit.

It is one of those interesting cases where the locally observed escape velocity remains "Newtonian", while it is the 'observed from infinity' orbital velocity that remains "Newtonian". I hope I've got all this right, so please check.

[Edit] 'measured locally' is supposed to mean by an observer static in the coordinate system at radial parameter r.
-J

 

[yuiop]

Kev,
Just a quick reply. I believe the two equations are standard results for the free-falling case, but I will double check. Time dilation for this case has to account for gravity & velocity. For the special case of a fre-falling object, the velocity is proportional to radius giving a second factor that equals the effect of gravity. Therefore, the total effect is (1-Rs/r) not just the square root. Again, I will try to detail more in a subsequent post as time permits.

Thanks for the info about the relative velocities, I want to work through some of the details and currently have another problem balancing a derivation of the Schwarzschild metric to standard text.

Ok, I'll buy into the additional time dilation due to falling velocity. Forgot about that :(

I am assuming that the Sharzchild metric includes all time dilation effects due to gravity as well as motion.

The metric <br /> c^2 {d \tau}^{2} = <br /> \left( 1 - \frac{r_{s}}{r} \right) c^{2} dt^{2} - \frac{dr^{2}}{1 - \frac{r_{s}}{r}} - r^{2} d\theta^{2} - r^{2} \sin^{2} \theta \, d\varphi^{2}

... with g used to symbolise 1/ \sqrt{1-r_s/r} and with zero angular motion, simplifies to:

c^2 {d \tau}^{2} = g^{-2} c^{2} dt^{2} - dr^{2} g^2

It is easy to see that there is additional time dilation factor due to the vertical radial motion in the above equation, but I am not sure how to get to mysearch's equations from there.

================================================================

On the subject of derivations, I did some calculations on orbital periods in strongly curved spacetime a long time ago here: https://www.physicsforums.com/showpost.php?p=1526798&postcount=25 but never got any feedback. Do they seem reasonable to you or jorrie ? (or anyone else who knows this stuff)

 

[Jorrie]

... with g used to symbolise 1/ \sqrt{1-r_s/r} and with zero angular motion, simplifies to:

c^2 {d \tau}^{2} = g^{-2} c^{2} dt^{2} - dr^{2} g^2

It is easy to see that there is additional time dilation factor due to the vertical radial motion in the above equation, but I am not sure how to get to mysearch's equations from there.

It's easy: just separate dr/dt out and solve! Just joking, because it's not that easy. You also need the total energy equation:

E/m = \frac{1}{g^2}\frac{dt}{d\tau} = 1

since E/m is constant and equals unity in geometric units at infinity. Solve the two together and you can get the radial free-fall velocity equation.

On the subject of derivations, I did some calculations on orbital periods in strongly curved spacetime a long time ago here: https://www.physicsforums.com/showpost.php?p=1526798&postcount=25 but never got any feedback. Do they seem reasonable to you or jorrie ? (or anyone else who knows this stuff)

Will have a look.

-J

 

[1effect]

I think 1effect might be right here. With constant L, d\phi/dt \rightarrow 0 as r\rightarrow \infty; with radial velocity also zero, only the rest energy mc^2 is left. On the other hand, Kev might argue that the original transverse velocity v = L/mr is still present and hence he is right.

Again, I think Kev takes the more physical route and 1effect the more mathematical one and the two routes do not always agree.(?)

-J

It is really simple, look at post no.87.
Unless the defintion of limit has changed lately :-)

 

[Jorrie]

It is really simple, look at post no.87.
Unless the defintion of limit has changed lately :-)

In this case I think the math is hiding something. When we say r -> infinity, we do not mean r = infinity, which is impossible. 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? We should have r -> infinity and v_o = constant. In the weak field, low velocity limit, we should still have the relativistic energy as E^2 = (mc^2)^2 + (pc)^2

Maybe you are falling into the same trap that I probably did in my argument with https://www.physicsforums.com/showpost.php?p=1679102&postcount=32" - trusting the equations a but too much. :-)

-J

 

[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