Question on Effective Potential

[Jorrie]

This last derivation makes clear the assumption of constant radius in the familiar equation for effective potential.

Nice summary, Kev!

MTW's effective potential is not quite defined for constant radius: it just ignores kinetic energy of the radial component of the orbital velocity. In other words, it defines it as the sum of potential energy and angular velocity energy. That's why it has that peculiar humped shape if plotted. It happens to be the total orbital energy at the turning points (peri- and apo-apsis) of elliptical orbits, because there the radial velocity component is zero. The total orbital energy is obviously constant for $r > 4GM/rc^2$.

-J

 

[mysearch]

Part 1 of ?: Summary

Hi, many thanks for all the work you have submitted in response to my original question and apologises for some minimal contributions this week, but I had some other commitments. Hopefully, I will have some more time over the upcoming weekend. I have tried to look at the issue of effective potential Veff from two perspectives, i.e. its classical derivation based on Newtonian concepts and its relativistic derivation stemming from the Schwarzschild metric. Starting with the classical definition:

$$Et = 1/2m\left( v_r^2 + v_o^2\right)-GMm/r$$

$$Et = 1/2m\left(\frac{dr}{dt}^2\right)+ 1/2m\left(\frac{d\phi}{dt}^2\right)-GMm/r$$

$$Et = 1/2m\left(\frac{dr}{dt}^2\right)-Veff$$

$$Veff = 1/2m\left(\frac{d\phi}{dt}^2\right)-GMm/r$$

This appears to suggest that Veff is defined as the addition of the positive orbital kinetic energy $$1/2mv_o^2$$ and the negative gravitational potential energy. As shown earlier, the total energy [Et=Ep/2] with [Ek=-Ep/2]. Note, this assumption is based on the radial velocity being zero. The purpose of raising this point again is that the classical derivation has no component related to rest energy, e.g. $$mc^2$$. If we start from a form of the Schwarzschild metric rationalise to an equatorial path, we can see both the radial and orbital velocities

$$\left( \frac{dr}{d\tau} \right)^{2} = c^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac{Rs}{r}\right) \left(r^2\left(\frac{d\phi}{d\tau}\right)^2 + c^2 \right)$$

If we proceed with the following assumptions:

$$dr/d\tau = v_r = 0$$
$$d\phi/d\tau = v_o$$

$$0 = c^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac{Rs}{r}\right) \left(v_o^2 + c^2 \right)$$

To normalise to kinetic energy, multiply through by m/2 and expand [Rs]:

$$Rs = \frac{2GM}{c^2}$$

$$1/2mc^2\left(1-\frac{2GM}{rc^2}\right)^2 \left(\frac{dt}{d\tau}\right)^2 = 1/2m\left(1-\frac{2GM}{rc^2}\right) \left(v_o^2 + c^2 \right)$$

If we now just expand the term on the right (RHS)

$$RHS = 1/2m\left(1-\frac{2GM}{r}\right) \left(v_o^2 + c^2 \right)$$

$$RHS = 1/2mv_o^2 + 1/2mc^2 -\frac{GMm}{r}\left(1+\frac{v_o^2}{c^2}\right)$$

Goto Part-2:

 

[mysearch]

Part 2 of 2

Now the parallels with the earlier classical derivation of Veff are obvious, except for the $$1/2mc^2$$ term.

$$Veff = 1/2mv_o^2-GMm/r$$

If we were to transfer this term to the left hand side of our Schwarzschild derivation, we might be attempt to equate the relativistic form of Veff to:

$$RHS-1/2mc^2 = Veff = 1/2mv_o^2 -\frac{GMm}{r} - \frac{GMm}{r}\left(\frac{v_o^2}{c^2}\right)$$

Of course, at this point, we might wish to correlate this result with angular momentum [L=mvr] for a circular orbit:

$$Vr = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GML^2}{c^2mr^3}$$

At which point we are back to the Wikipedia definition, which unfortunately didn't quite seem to produce the max/min curve, although I still have to double check this assumption in light of the minimum value of L=3.4642GM/c. Now another source of the Fourmilab equation is Taylor & Wheeler: Exploring Black Holes (p.4-14). This starts with the equation:

$$\left( \frac{dr}{d\tau} \right)^{2} = \left(\frac{E}{m}\right)^2 - \left(1-\frac{2M}{r}\right) \left(1 + \frac{(L/m)^2}{r^2}\right)$$

Now this reference work is almost entirely in geometric units, which sometime makes it difficult to follow exactly what is actually implied. The following form is comparable to my earlier derivation from the Schwarzschild metric.

$$\left( \frac{dr}{d\tau} \right)^{2} = c^2\left(1-\frac{2GM}{rc^2}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac{2GM}{rc^2}\right) \left(\frac{L^2}{m^2r^2} + c^2 \right)$$

If you set G=c=m=1, as per geometric units you get back to the T&W starting point:

$$\left( \frac{dr}{d\tau} \right)^{2} = \left(1-\frac{2M}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac{2M}{r}\right) \left(1 + \frac{(L/m)^2}{r^2}\right)$$

With $$\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}$$

Taylor & Wheeler proceed from their starting point to:

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

From which a form of the Veff equation appears to originate:

$$\left(\frac{V}{m}\right)^2 = \left(1-\frac{2M}{r}\right) \left(1 + \frac{(L/m)^2}{r^2}\right)$$

I still need to double check whether this equation is still appropriate to draw the max/min curve, but I would still argue that this equation appears inconsistence with the Schwarzschild and Wikipedia derivations because it has not separate out the implicit $$mc^2$$ term that is still embedded in the right hand side of the equation.

Thoughts?

 

[Jorrie]

...
From which a form of the Veff equation appears to originate:

$$\left(\frac{V}{m}\right)^2 = \left(1-\frac{2M}{r}\right) \left(1 + \frac{(L/m)^2}{r^2}\right)$$

... but I would still argue that this equation appears inconsistence with the Schwarzschild and Wikipedia derivations because it has not separate out the implicit $$mc^2$$ term that is still embedded in the right hand side of the equation.

Thoughts?

But you have explicitly set c=G=1 above, so how can you expect a c^2 in the equation?!

Look at my posts 64 amd 67, where the mc^2 is obvious in standard units. Your result is perfectly correct in geometric units. Just set L/m somewhat larger than 3.465, plot it and you will see the bulge.

-J

 

[1effect]

$$v=r \frac{d \phi}{dt}$$

$$d\phi =v\frac{dt }{r} = \frac{dx}{dt} . \frac{dt}{ r} = \frac{dx}{r}$$

.

I don't think so. You are trying to justify your mistakes again. All you can write is :

$$v=r \frac{d \phi}{dt}$$

Here is another way of showing that you made a mistake, you wrote initially :

$$L=mr^2 \frac{d \phi}{dt}$$ which is correct. Based on this, it is obvious that :

$$\frac{L}{mcr}->infinity$$

 

[yuiop]

I don't think so. You are trying to justify your mistakes again. All you can write is :

$$v=r \frac{d \phi}{dt}$$

Here is another way of showing that you made a mistake, you wrote initially :

$$L=mr^2 \frac{d \phi}{dt}$$ which is correct. Based on this, it is obvious that :

$$\frac{L}{mcr}->infinity$$

Your argument that

$$L=r\frac{d \phi}{cdt}$$

goes to infinity as r goes to infinity is incorrect because you are making the false assumption that ${d \phi}/{dt}$ remains constant as r varies.

By looking at the original expression $$L=mr^2 \frac{d \phi}{dt}$$

in Newtonian terms, then considerations of conservation of angular momentum clearly indicate that $d\phi/dt$ is inversely proportional to r^2.

If L is conserved then the term $$\lim_{r\to\infty} \frac{L}{mcr}$$ goes to zero (rather than to infinity as you suggest) as r goes to infinity.
However L is not necessarily conserved (independently of energy) in relativity.

Infinity is a tricky subject and before this becomes a discussion of whether $\infty/\infty = 1$ or what $\infty*0$ is equivalent to, I would like to suggest this alternative expression for gravitational gamma factor that is less ambiguous:


$$\left({1-2GM / (r c^2 ) \over 1-2GM / (r 'c^2 )}\right)$$

where r' is the radial displacement of the observer from the massive body while r is the radial displacement of the test point particle.

The potential can then be expressed as


$$V_{eff}^2 = m^2c^4 \left({1-2GM / (r c^2 ) \over 1-2GM / (r 'c^2 )}\right)\left(1+{L^2 \over m^2c^2r^2}\right)$$

As r' tends to infinity the expression aproximates to

$$V_{eff}^2 = m^2c^4 \left(1-2GM / (r c^2 )\right) \left(1+{L^2 \over m^2c^2r^2}\right)$$

and when r = r' the expression becomes:

$$V_{eff}^2 = m^2c^4 \left(1+{L^2 \over m^2c^2r^2}\right) = m^2c^4 + {L^2 c^2 \over r^2} = m^2c^4 + m^2 v^2 c^2$$

which can be interpreted as rest energy + momentum energy $M^2 + P^2$

 

[1effect]

Your argument that

$$L=r\frac{d \phi}{cdt}$$

Huh? I never said that. At least quote me correctly.

goes to infinity as r goes to infinity is incorrect because you are making the false assumption that ${d \phi}/{dt}$ remains constant as r varies.

$$\phi$$ and $$t$$ are primary variables so, contrary to what you claim, ${d \phi}/{dt}$ is not dependent of r. When will you stop making up phony proofs :-)

 

[yuiop]

Your argument that

$$L=r\frac{d \phi}{cdt}$$

goes to infinity as r goes to infinity is incorrect because you are making the false assumption that ${d \phi}/{dt}$ remains constant as r varies.

Huh? I never said that. At least quote me correctly.

I was paraphrasing what you said. What you actually said is:

" $$\frac{L}{mcr}->infinity$$"

and it obvious to me that that term goes to zero (not infinty) as r goes to infinity.

 

[1effect]

I was paraphrasing what you said. What you actually said is:

" $$\frac{L}{mcr}->infinity$$"

and it obvious to me that that term goes to zero (not infinty) as r goes to infinity.

I don't understand why you keep piling errors on top of other errors when your proofs are found to be phony. This doesn't make any sense.

I also pointed out to you that $$\phi$$ and $$t$$ are primary, independent variables so, contrary to what you claim, ${d \phi}/{dt}$ is not dependent of r.

 

[yuiop]

I don't think so. You are trying to justify your mistakes again. All you can write is :

$$v=r \frac{d \phi}{dt}$$

Here is another way of showing that you made a mistake, you wrote initially :

$$L=mr^2 \frac{d \phi}{dt}$$ which is correct. Based on this, it is obvious that :

$$\frac{L}{mcr}->infinity$$

Wikpedia defines $$\frac{L}{m} = r^2 \frac{d \phi}{d\tau}$$ as a constant of motion.

On that basis, your statement that $$\frac{L}{mcr}->infinity$$ as r goes to infinity is wrong.

Assuming L/m and c are constants then the term goes to zero as r goes to infinty.

 

[1effect]

Wikpedia defines $$\frac{L}{m} = r^2 \frac{d \phi}{d\tau}$$ as a constant of motion.

This only means that $$\frac{L}{m} = r^2 \frac{d \phi}{d\tau}$$ is conserved. It doesn't mean that the quantity is finite when r becomes infinite. You are making up your own definitions, like when you said that $$\frac{d \phi}{d\tau}$$ depended on $$r$$.

On that basis, your statement that $$\frac{L}{mcr}->infinity$$ as r goes to infinity is wrong.

Assuming L/m and c are constants then the term goes to zero as r goes to infinty.

$$\frac{L}{mcr}=\frac{r}{c} \frac{d \phi}{d\tau}$$ and this goes to infinity when $$r->infinity$$.

Why do you have such a hard time admitting to error?

 

[yuiop]

$$\frac{L}{mcr}=\frac{r}{c} \frac{d \phi}{d\tau}$$ and this goes to infinity when $$r->infinity$$.

Why do you have such a hard time admitting to error?

When an ice skater pulls her hands in during a spin her angular velocity increases. Angular velocity is defined as $d\phi/dt$. From that it is obvious that $d\phi/dt$ is NOT independent of radius as you seem to think it is. In normal physics angular momentum is conserved and and as radius increases, angular velocity reduces to compensate. If she could stretch her hands out towards infinity her spin rate would become almost zero, NOT almost infinite. Another example is very dense collapsed stars. Their spin rate $d\phi/dt$ increases dramatically as their radius gets smaller, not the other way round. If $d\phi/dt$ is independent of radius why do things spin faster with reduced radius?


In the particular case here involving the Scharzchild metric a lot depends on how L and $d\phi$ is defined. $d\phi$ and all other variables in the metric are as measured by an observer at infinity except for d\tau which is explicitly a proper (local) measurement. L is not a variable in the metric and so we have to be clear how it is defined and who measures it. Wikipedia clears that up by stating L/m is a constant.

Maybe it is time to admit you are wrong for a change?

 

[Jorrie]

This only means that $$\frac{L}{m} = r^2 \frac{d \phi}{d\tau}$$ is conserved. It doesn't mean that the quantity is finite when r becomes infinite. You are making up your own definitions, like when you said that $$\frac{d \phi}{d\tau}$$ depended on $$r$$.

$$\frac{L}{mcr}=\frac{r}{c} \frac{d \phi}{d\tau}$$ and this goes to infinity when $$r->infinity$$.

I think you two are arguing somewhat in circles, due to misunderstanding each other, with Kev perhaps slight more correct. L/m is the conserved (constant) specific angular momentum of an orbiting particle and is given (as you agreed) by

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

It means that for a given L/m, if you take r to infinity, $d \phi/d\tau$ goes to zero, and visa versa. If you somehow keep $d \phi/d\tau$ constant and increase r without limit, you will soon run into the speed of light for the particle and $d \phi/d\tau$ will have to drop. In fact you will run into infinite energy required as well.

So the bottom line is that one must use L/m as intended, a conserved quantity in all Schwarzschild orbits. Practically, you can choose a distance and a desired tangential (orbital) speed and then calculate L/m. This value then remains constant throughout the orbit, be it circular, elliptical or hyperbolic.


-J

 

[1effect]

I think you two are arguing somewhat in circles, due to misunderstanding each other, with Kev perhaps slight more correct. L/m is the conserved (constant) specific angular momentum of an orbiting particle and is given (as you agreed) by

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

It means that for a given L/m, if you take r to infinity, $d \phi/d\tau$ goes to zero, and visa versa. If you somehow keep $d \phi/d\tau$ constant and increase r without limit, you will soon run into the speed of light for the particle and $d \phi/d\tau$ will have to drop. In fact you will run into infinite energy required as well.

So the bottom line is that one must use L/m as intended, a conserved quantity in all Schwarzschild orbits. Practically, you can choose a distance and a desired tangential (orbital) speed and then calculate L/m. This value then remains constant throughout the orbit, be it circular, elliptical or hyperbolic.-J

Ok, this explanation made sense.
From the conservation of $$\frac{L}{m}$$
you get the conservation of
$$r^2 \frac{d \phi}{d\tau}$$
So far we are all in agreement. Where the disagreement starts is the use of the above (see kev's classical example of the skater) to prove that $$mr^2 \frac{d \phi}{dt}$$ does not tend to infinity when $$r->infinity$$
The above is indeed true for a system with radius variable in time $$r=r(t)$$, like the skater but it is not true for the case where one let's $$r$$ go to infinity just by increasing $$r$$ as an independent variable, not as a function of time. This is exactly the case in discussion, where r is taken to infinity as a result of using an observer at infinity, to paraphrase kev's own words in his last post. In the latter case ware not dealing with any conservation of momentum, we are dealing with taking a limit when the independent variable r goes to infinity. There is no reason to surmise the "compensatory" effect of $$\frac{d \phi}{dt}$$ decreasing in this case.

 

[Jorrie]

In the latter case ware not dealing with any conservation of momentum, we are dealing with taking a limit when the independent variable r goes to infinity. There is no reason to surmise the "compensatory" effect of $$\frac{d \phi}{dt}$$ decreasing in this case.

From a pure math point of view, you are right, but what you describe is unphysical.

No matter how small you choose your constant $d \phi/dt$, as r increases without limit, velocity $v=rd \phi/dt$ will eventually approach the speed of light and $d \phi/dt$ will be forced downwards.

-J

 

[mysearch]

The 3 attached graphs correspond to the different perspectives of effective potential (Veff) that have been under discussion. I will email the actual spreadsheet to Kev so that he can double-check the figures.

1. Newton Veff
2. Wikipedia Veff
3. Veff based on posts #64 & #67.

The Newtonian plot is just for reference, as it does not include the additional gravitation terms associated with relativity. However, there is a direct correlation between the style of these first two graphs, which will be expanded in my next post. The final plot is intended to correspond to the equation forwarded in post #64 and #67. I believe it is based on a different premise, again to be discussed in my next post, which I believe is at the root of the initial confusion, at least mine.

 

[1effect]

In the opening post mysearch wanted to have a better understanding of the physical concepts behind the effective potential so I would like to post some conclusions that might help and clear up some misconceptions (of mine).

The derivation given by Jorrie appears to be correct and and he is right that $L = m( d\phi/dt)r^2$ and not $m(d\phi/dt)r$ as I suggested.

Looking at the equation for effective potential given by Jorrie

$$V_{eff}^2 = m^2c^4 \left(1-{2GM \over rc^2}\right)\left(1+{L^2 \over m^2c^2r^2}\right)$$

it is easy to see that at infinity the potential^2 is

$$V_{eff}^2 = m^2c^4\left(1+{L^2 \over m^2c^2r^2}\right)$$

The L term does not go to zero because L includes a hidden r^2 within its definition that cancels out the visible r^2 in the equation

.

Ok,

Then the clean way of calculating the limit is to start with:

$$V_{eff} = mc^2 \sqrt{\left(1-{2GM \over rc^2}\right)\left(1+{L^2 \over m^2c^2r^2}\right)}$$

(we all agree on this) and to use the fact that $$\frac{L}{mc}=a=constant$$

Then,

$$V_{eff} = mc^2 \sqrt{\left(1-{2GM \over rc^2}\right)\left(1+{a^2 \over r^2}\right)}$$

which, for $$r->infinity$$ is clearly:

$$V_{eff} = mc^2$$

 

[1effect]

Your argument that

$$L=r\frac{d \phi}{cdt}$$

goes to infinity as r goes to infinity is incorrect because you are making the false assumption that ${d \phi}/{dt}$ remains constant as r varies.

By looking at the original expression $$L=mr^2 \frac{d \phi}{dt}$$

in Newtonian terms, then considerations of conservation of angular momentum clearly indicate that $d\phi/dt$ is inversely proportional to r^2.

If L is conserved then the term $$\lim_{r\to\infty} \frac{L}{mcr}$$ goes to zero (rather than to infinity as you suggest) as r goes to infinity.
However L is not necessarily conserved (independently of energy) in relativity.

Infinity is a tricky subject and before this becomes a discussion of whether $\infty/\infty = 1$ or what $\infty*0$ is equivalent to, I would like to suggest this alternative expression for gravitational gamma factor that is less ambiguous:


$$\left({1-2GM / (r c^2 ) \over 1-2GM / (r 'c^2 )}\right)$$

where r' is the radial displacement of the observer from the massive body while r is the radial displacement of the test point particle.

The potential can then be expressed as


$$V_{eff}^2 = m^2c^4 \left({1-2GM / (r c^2 ) \over 1-2GM / (r 'c^2 )}\right)\left(1+{L^2 \over m^2c^2r^2}\right)$$

As r' tends to infinity the expression aproximates to

$$V_{eff}^2 = m^2c^4 \left(1-2GM / (r c^2 )\right) \left(1+{L^2 \over m^2c^2r^2}\right)$$

and when r = r' the expression becomes:

$$V_{eff}^2 = m^2c^4 \left(1+{L^2 \over m^2c^2r^2}\right) = m^2c^4 + {L^2 c^2 \over r^2} = m^2c^4 + m^2 v^2 c^2$$

which can be interpreted as rest energy + momentum energy $M^2 + P^2$

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

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

 

[mysearch]

Explanation of graphs: See post#86

First, a response to Jorrie #74. Either you missed the point I was trying to make in my post #72 & #73 or I missed yours. Hopefully this post will clarify the issue I was trying to highlight. What the two graphs associated with the Wikipedia and posts #64/67 highlight is that both forms are essentially equivalent in that they both produce the same max/min values. However, I will now argue that the Wikipedia is actually the more precise definition of effective potential (Veff) in that it correlates to the classical definition of effective potential. However, the second definition adopted by Fourmilabs, posts #64/67 plus Taylor & Wheeler possibly produces the most useful graph. What I was trying to do in my post #72/73 was to show the definition of classical effective potential and then how the relativistic variant is derived from the Schwarzschild metric starting from:

$$\left(\frac{dr}{d\tau}\right)^2 = c^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac{Rs}{r}\right) \left(v_o^2 + c^2 \right)$$

Now it is clear that in this form (Rs/r) is just a ratio without unit, as is $$dt/d\tau$$. In fact, every term reduces to $$velocity^2$$ with unit $$metres^2/sec^2$$. Only if we multiply by [1/2m] does this equation take on the energy form analogous to the kinetic energy associated with radial and orbital velocity:

$$1/2m\left(\frac{dr}{d\tau}\right)^2 = 1/2mc^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left[1/2m\left(1-\frac{Rs}{r}\right) \left(v_o^2 + c^2 \right)\right]$$

Expansion of the term on the right leads to:

$$1/2m\left(\frac{dr}{d\tau}\right)^2 = 1/2mc^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left[1/2mv_o^2+1/2mc^2-\left(\frac{GMm}{r}\right) \left(1+\frac{v_o^2}{c^2} \right)\right]$$

In essence, this equation parallels:

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

Now in classical physics it is the bracketed [] term on the right that is defined as [Veff]. So the question is whether the following component relates to total energy [Et]:

$$Et = 1/2mc^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - 1/2mc^2 (?)$$

Now if this is the case:

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

Substituting [v] for [L]:

$$v_o^2=\frac{m^2v_o^2r^2 }{m^2r^2} = \frac{L^2}{m^2r^2}$$

$$Veff=\frac{L^2}{2mr^2}-\frac{GMm}{r}\left(1+\frac{L^2}{m^2r^2c^2} \right)$$

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

This is the Wikipedia form that I contend is more reflective of the classical definition of effective potential. The alternative form forwarded by posts #64/67, Fourmilab and Taylor & Wheeler are simply a normalised variant of this form, which can also be started at the same point in the Schwarzschild metric, i.e.

$$\left(\frac{dr}{d\tau}\right)^2 = c^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac{Rs}{r}\right) \left(v_o^2 + c^2 \right)$$

But the next step defines the point of separation between the two derivations:

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

This is based on the assumption that
$$\left(\frac{E}{m}\right) = \left(1-\frac{Rs}{r}\right) \left(\frac{dt}{d\tau}\right)$$

As such, this leads to the basic form in geometric units:

$$\left(\frac{V}{m}\right) = \sqrt{\left(1-\frac{Rs}{r}\right) \left(1+\frac{(L/m)^2}{r^2}\right))$$

Post #64/67 seem to align to this derivation, but clarify the issue of geometric units. What the graphs in my previous post show is that while the form of the Wikipedia and Fourmilab plots look different, they are essentially equivalent in that they do, in principle, define the same max/min point – I think.

Thoughts?

 

[Jorrie]

What the two graphs associated with the Wikipedia and posts #64/67 highlight is that both forms are essentially equivalent in that they both produce the same max/min values. However, I will now argue that the Wikipedia is actually the more precise definition of effective potential (Veff) in that it correlates to the classical definition of effective potential. However, the second definition adopted by Fourmilabs, posts #64/67 plus Taylor & Wheeler possibly produces the most useful graph.

Your graphs look OK to me. Just note that the Wikipedia equation you used:

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

is in units of the square of the effective potential (i.e., Joules^2) and to me looks like an approximation of the full relativistic one, because relativistic energy is normally taken as only positive. This may however only be an arbitrary constant offset. For accuracy in Joules, use

$$V_{eff} = c^2\sqrt{\left(1-\frac{Rs}{r}\right) \left(1+\frac{(L/m)^2}{r^2c^2}\right)$$

Your third graph is identical to the one I posted https://www.physicsforums.com/showpost.php?p=1670815&postcount=15,", which was just normalized and my spatial axis was in units of $rc^2/GM$, which doubles my spatial parameter relative to yours.

-J

 

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

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

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) $$rd\phi/dt = c\sqrt{\frac{Rs}{2r}}$$

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

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 $$c^2 {d \tau}^{2} = \left( 1 - \frac{r_{s}}{r} \right) c^{2} dt^{2} - \frac{dr^{2}}{1 - \frac{r_{s}}{r}} - r^{2} d\theta^{2} - r^{2} \sin^{2} \theta \, d\varphi^{2}$$

... 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