Question on Effective Potential

[yuiop]

Hi Mysearch,

Had a look at your spreadsheets. Look OK to me and agree with Jorrie's comments. Only thing I really noticed was that the secondary peaks are an artifact of the software's spline interpolation. You need to set the resolution to 1 in the graph type if you want the graphs to show the correct peak height and remove the artificial secondary peaks that appear even in the Newton graph.

EDIT: I just noticed the secondary peaks only appear in the spreadsheet you sent me and not in the graphs you posted, so maybe you have that sorted already ;)

 

[Jorrie]

I disagree: use \frac {L}{mc}=a=constant and you get the limit to be:

V_{eff}^2 = m^2c^4

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

 

[yuiop]

I think 1effect is 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.

-J

YEp, I've mentioned several times already that the angular momentum term goes to zero while 1effect was insisting it goes to infinty as r\rightarrow \infty showing that I changed my position on that quite a while ago and before 1effect. I slipped up in my last post by lapsing back into substituting L = mvr for the angular momentum term which does not appear to work. I'm still trying to figure out why that is. I guess it is partly due to v not being independent of the radius. For example, when the radius of a rotating object is doubled its tangential velocity halves to conserve angular momentum. (assuming the mass stays constant). Oddly enough, mysearch's spreadsheet seems to get the right curve by assuming L=mvr as that is the equation he is using in spreadsheet formulas.

There are a couple of paradoxical things I can not quite get my head round at the moment which is why I have not posted a response to 1effect yet.

If an observer infinitely far from a gravitational body fired a particle horizontally at 0.99c it would have linear momentum m.v.y as far as he concerned but angular momentum of zero relative to the massive body. So while the angular momentum is zero can it still have horizontal linear momentum?

The effective potential equation seems to implying that nothing can have horizontal momentum at infinity. The other paradoxical part is that if all objects have zero angular momentum at infinity how do they acquire horizontal motion as they fall?

If the term L/m is constant then the particle still has angular momentum L/m at infinty by definition while the energy term L/(mcr) goes to zero. Associating zero energy with non zero L/m seems paradoxical too.

Maybe the resolution to all this, is to take into account that the momentum term uses proper time, suggesting the momentum is measured by an observer co-moving horizontally with the particle. In that case the particle has zero momentum relative to him when it has the same radius and it is the observer that has angular momentum/velocity. As the particle falls its angular velocity changes and it appears to acquire horizontal motion relative to the observer mantaining constant radius.

Any ideas?

 

[Jorrie]

If the term L/m is constant then the particle still has angular momentum L/m at infinty by definition while the energy term L/(mcr) goes to zero. Associating zero energy with non zero L/m seems paradoxical too.

Any ideas?

I did edit my last post just after posting it, so you may not have seen that addition.

As you have pointed sometime before, one must be careful in not always blindly trusting the math once it has been normalized and simplified by cancellation of parameters.

Secondly, one must accept that a particle cannot be at infinite distance from a mass, just approaching infinity. In that sense, there will always be some d\phi/dt for a given L. In any case, the idea of just increasing r arbitrarily without conserving energy is full of dangers for misunderstanding, unless you add energy.

Lastly, yes, one must use propertime even in the equation for L:

L/m = r^2 \frac{d\phi}{d\tau} = r^2 \frac{d\phi}{dt}\frac{dt}{d\tau}

This can be reworked into an equivalent equation, which may shed some light upon your problem:

L/m = \frac{r v_t}{\sqrt{g_{tt}-g_{rr}v_r^2/c^2-v_t^2/c^2}}

where g_{tt} = 1-2GM/(rc^2), g_{rr} = 1/g_{tt} and v_t, v_r are the transverse and radial velocity components in Schwarzschild coordinates.

Note that since v_r influences dtau, it also plays a role in determining L for an arbitrary orbital position.

Edit2: I've removed what I've written here (expression for L_max), because it was wrong. L can approach infinity for any r, when the value below the line (inside the square root) approaches zero.

-J

 

[mysearch]

Jorrie,
Can you clarify your previous statement for me? The classical definition of effective potential (Veff) comes from:

Et = 1/2m\left(\frac{dr}{dt}^2\right)+ \left[1/2mv_o^2-GMm/r\right]

Veff = \left[1/2mv_o^2-GMm/r\right]

As explained, this can be seen to align with the Wikipedia definition:

Veff = \left[1/2mv_o^2-\frac{GMm}{r} \left(1+\frac{vo^2}{c^2}\right)\right]

Veff=\frac{L^2}{2mr^2}-\frac{GMm}{r}-\frac{GML^2}{mc^2r^3}

The units of each term is energy, i.e. joules= kg.m^2/s^2, but you seem to assert that effective potential is joules^2. I presume that this is based on the assumption that:

\left(\frac{dr}{d\tau}\right)^2 = \left(\frac{E}{m}\right)^2-\left(\frac{V}{m}\right)^2 (?)

I am not trying to be pedantic, I am genuinely interested in understanding the root of this difference. Yes, my 3rd graph is identical to the graph you posted in #15 and I appreciate all the help, contributed in this thread, to my understanding. One of things that I didn’t initially appreciate was the sensitivity of this plot to the value of [L>3.4642GM/c] provided by you or the fact that the max/min is not that obvious unless you zoom to a much smaller range on the vertical y-axis. The difference in max/min values is only 6%. I have attached another plot to highlight this fact for any reader who might want to replicate the earlier plot in #15.

 

[Jorrie]

Jorrie,
Can you clarify your previous statement for me? The classical definition of effective potential (Veff) comes from:

Et = 1/2m\left(\frac{dr}{dt}^2\right)+ \left[1/2mv_o^2-GMm/r\right]

Veff = \left[1/2mv_o^2-GMm/r\right]

As explained, this can be seen to align with the Wikipedia definition:

Veff = \left[1/2mv_o^2-\frac{GMm}{r} \left(1+\frac{vo^2}{c^2}\right)\right]

Veff=\frac{L^2}{2mr^2}-\frac{GMm}{r}-\frac{GML^2}{mc^2r^3}

The units of each term is energy, i.e. joules= kg.m^2/s^2, but you seem to assert that effective potential is joules^2.

Hmm..., you are right, but that Wikipedia definition looks like a Newtonian one with a corrective term added, not quite relativistic. I first thought it is just the relativistic one squared, but it surely is not. My bad!

But, it surely does not give the results of MTW and Fourmilab, which are equivalent and correct. Maybe the Wikipedia equation is a good approximation, though.

-J

 

[mysearch]

Jorrie,
Thanks for the considered reply. Personally, I think all equations converge to the same results but do so in different ways. My only point in highlighting this issue was to try to recognise where and why the two approaches differ.

 

[mysearch]

Some general thoughts on Angular Momentum [L]

Some thoughts on the secondary discussion going on about angular momentum. In some of the work I have been doing on the implications of the Schwarzschild metric, I found it useful to separate the derivations into 2 basic classes:

1. Free-falling
2. Circular Orbits

Of course, all practical trajectories are a combination of both types being described. By definition, the free-falling paths has no orbital velocity v_o component and therefore no angular momentum [L]. The radial velocity, with respect to the infinite [dt] and onboard d\tau observers is defined by the following equation:

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

dr/d\tau = -c \sqrt{\frac{Rs}{r}}

These equations are conceptual in the sense that they assume an object free-falling under gravity from zero velocity at an infinite distance from the central mass [M]. In practice, it is quite difficult to go to infinity, but can be matched by setting the initially velocity at radius [r] to that defined by the above equations. It is interesting to note that the classical free-fall velocity corresponds to the on-board observer, not the distanced observer. I raise this point because Kev has made several references to this sort of issue and it questions whether which is more meaningful, but this is another matter all together.

By definition, a circular orbit has no radial velocity and under the specific assumption of the orbit being circular, then mv_or. It is my assumption that the issue of the conservation of angular momentum only applies to trajectories, where any change in v_o or v_r has to be balanced in accordance to this conservation law. Therefore, when only considering circular orbits at difference radii [r], in isolation, the value of [L] changes for each value of [r]. This is why the effective potential is so useful because the minimum corresponds to the stable orbit radius for that specific value of [L]. The actual relativistic orbital velocity is one of the things I now want to double check.

One final generalisation that I always try to keep in mind is the validity of classical physics when relativistic factors don’t exist or are minimal, i.e. gravity and velocity. In part, I have used this argument on several occasions within our discussions of effective potential.

 

[yuiop]

Some thoughts on the secondary discussion going on about angular momentum. In some of the work I have been doing on the implications of the Schwarzschild metric, I found it useful to separate the derivations into 2 basic classes:

1. Free-falling
2. Circular Orbits

Of course, all practical trajectories are a combination of both types being described. By definition, the free-falling paths has no orbital velocity v_o component and therefore no angular momentum [L]. The radial velocity, with respect to the infinite [dt] and onboard d\tau observers is defined by the following equation:

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

dr/d\tau = -c \sqrt{\frac{Rs}{r}}

These equations are conceptual in the sense that they assume an object free-falling under gravity from zero velocity at an infinite distance from the central mass [M]. In practice, it is quite difficult to go to infinity, but can be matched by setting the initially velocity at radius [r] to that defined by the above equations.

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) }


Just for info when investigating orbital velocities,

the vertical coordinate speed of light is

\frac{dr}{dt} = c \left(1-\frac{Rs}{r}\right)

and the horizontal coordinate speed of light is

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

as can be checked by setting the proper time in the Scharzchild metric to zero.

 

[mysearch]

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.

 

[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