Question on Effective Potential

[yuiop]

I guess the answer hinges on whether an angle mesured in radians is dimensionless or not. The perimeter of a circle with a radius of 5 metres is 10 Pi metres. In terms of radians the perimeter is 5 radians. Does the perimeter cease to have dimemsions of length? While the total displacement is zero for a complet circunavigation of a circle, this is not always true for a partial rotation. Now, radians, is the length of a segment divided by the radius and so Si units suggest it is dimensionless. Wikpedia suggests this dilemna is solved by using extended SI units and talking of the perimeter of the circle as 5 radian metres to make clear we do not mean 5 metres in a straight line. Anyway, velocity in a straight line is measured in metres per second, and angular velocity still essentially has dimensions distance over time but maybe we should call it radian-metres per second.

 

[Jorrie]

Anyway, velocity in a straight line is measured in metres per second, and angular velocity still essentially has dimensions distance over time but maybe we should call it radian-metres per second.

In geometric units, velocity will then be radian-meters per meter, which is dimensionless, just like light-years per year is dimensionless. I like the radian-meters for clarity!

 

[yuiop]

In geometric units, velocity will then be radian-meters per meter, which is dimensionless, just like light-years per year is dimensionless. I like the radian-meters for clarity!

If we roll a wheel along the ground so that it completes one full rotation, can we agree that the ratio of the distance rolled to the circumferance of the wheel is dimensionless whether we choose to measure the circumferance in radian-meters or meters? In other words radian-meters and just plain meters can both be considered as measures of (path) length and cancel out?

I mention path length, because the total displacement of a particle on the rim of a wheel after completing one full rotation is zero, but the path length is not.

 

[Jorrie]

It is not clear to me that mc^2 is still not being included in Veff. However, it would be a start to get the curve and then work backwards towards the validity of its derivation.

I think the last equation I wrote in reply to Kev:

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

may solve your concern, but now you must remember that V_eff is in Joule (kg^2 m^2/s^2), so V_eff^2 is in Joule^2. L is in units kg m^2/s.

Note that your minimum L for a bound, stable orbit is now not just 3.4641, but 3.4641GM/c ~ 7.7E-19 M kg m^2/s. Below this value, you will not get the "trough and the bulge" in the V_eff-curve and it may confuse the issue.

-J

 

[Jorrie]

In other words radian-meters and just plain meters can both be considered as measures of (path) length and cancel out?

Yep, I would think so. Radians are always considered to be dimensionless ratios, AFAIK.

In geometric units, time is also measured in meters - one unit is actually the amount of time it takes light to travel one meter distance in vacuum and could be called light-meters; hence velocity is dimensionless, acceleration is in meter^{-1} and so on. It is much simpler to keep track of the units in geometric units than it is in conventional units and it helps that G=c=1!

-J

 

[yuiop]

Yep, I would think so. Radians are always considered to be dimensionless ratios, AFAIK.

In geometric units, time is also measured in meters - one unit is actually the amount of time it takes light to travel one meter distance in vacuum and could be called light-meters; hence velocity is dimensionless, acceleration is in meter^{-1} and so on. It is much simpler to keep track of the units in geometric units than it is in conventional units and it helps that G=c=1!

-J

OK, I've checked out the Wikpedia reference on geometric units and I am a bit more comfortable with then now. Takes a bit of getting used to, measuring time and mass in centimetres :P

So where are we now?
What is the final agreed form of the potential in conventional units? The one quoted in post #64 presumably?

 

[Jorrie]

So where are we now?
What is the final agreed form of the potential in conventional units? The one quoted in post #64 presumably?

My background is engineering, so I also favor the conventional units. My vote is for the form in conventional units of Joules and always indicating that it is effective potential V_eff (not to be confused with normal potential energy):

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

That mc^2 is a nice way of indicating that it is actually an energy equation, because the rest are dimensionless.

-J

 

[yuiop]

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

If we expand L to base units we get

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

which simplifies to

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


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

This can now be expressed as the well known conserved energy-momentum expression of relativity E^2 = M^2 + P^2

For radii less than infinity the effective potential is simply the potential at infinity reduced by the gravitational gamma factor.



The above equations assumes constant orbital radius and glosses over proper time issues.

A better derivation can be obtained from this equation given in Wikpedia that makes the issues clear. http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity#Geodesic_equation

<br /> \left( \frac{dr}{d\tau} \right)^{2} = \frac{E^{2}}{m^{2}c^{2}} - \left( 1 - \frac{r_{s}}{r} \right) \left( c^{2} + \frac{L^{2}}{m^{2} r^{2}} \right)<br />

Divide both sides by c^2 and rearrange:

E^{2}-<br /> \left( \frac{mcdr}{d\tau} \right)^{2} = m^2c^4\left( 1 - \frac{r_{s}}{r} \right) \left( 1 + \frac{L^{2}}{m^{2} r^{2}c^2} \right)<br />

For constant radius this becomes the familiar

E^{2} = m^2c^4\left( 1 - \frac{r_{s}}{r} \right) \left( 1 + \frac{L^{2}}{m^{2} r^{2}c^2} \right)<br />

where L is defined by Wikpedia in terms of proper time as is the expression for radial velocity that has just been eliminated.

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

 

[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

If we expand L to base units we get

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

which simplifies to

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

I'm afraid that you can't do any of the stuff that you are showing above.
You are using the fact that v=r \frac{d \phi}{dt} and you are taking the limit for V_{eff} for r-&gt;infinity . Obviously, this makes v-&gt;infinity. In other words, you forgot that \frac{L}{mcr}-&gt;infinity. Not good :-)

 

[yuiop]

I'm afraid that you can't do any of the stuff that you are showing above.
You are using the fact that v=r \frac{d \phi}{dt} and you are taking the limit for V_{eff} for r-&gt;infinity . Obviously, this makes v-&gt;infinity. In other words, you forgot that \frac{L}{mcr}-&gt;infinity. Not good :-)

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

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

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

The radius terms cancel out, so no v going to infinity to worry about.

 

[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 &gt; 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}-&gt;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}-&gt;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 &#039;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 &#039;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) = <br /> 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}-&gt;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}-&gt;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}-&gt;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}-&gt;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}-&gt;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-&gt;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-&gt;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-&gt;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

<br /> \frac{L}{m} = r^2 \frac{d \phi}{d\tau}<br />

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

<br /> \frac{L}{m} = r^2 \frac{d \phi}{d\tau}<br />

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-&gt;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:

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

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

Then,

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

which, for r-&gt;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 &#039;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 &#039;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) = <br /> 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:

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

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

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

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