Question on Effective Potential

[yuiop]

No, the error is mine, not MTWs! There's a typo in my equation for L; should have an r^2 above the line, not just r. You can't take it out inside the square root, because that will screw up the units there. Correct L:

<br /> L = \frac{r^2 d\phi}{dt\sqrt{1-2M/r-d\phi^2/dt^2}}<br />

Sorry for the inconvenience, but well spotted. I should have checked dimensions. L has the dimension of meter in geometric units (or m^2/s in SI units), while the one I gave was dimensionless geometrically.

I took the liberty to correct my post with the offending formula to reduce confusion.

If we assume

L = mvr = m (d\phi / dt) r

and if we assume

\sqrt{1-2M/r-d\phi^2/dt^2}}

is a dimensionless ratio, then the equation should be:


<br /> \frac{L}{m} = \frac{r d\phi}{dt\sqrt{1-2M/r-d\phi^2/dt^2}}<br />

 

[mysearch]

Comments on some Earlier Posting

Just a couple of points on some of the statements made:

“it is worth noting that the Wikipedia solution uses proper time in its derivation and this may account for the large difference between the fourmilab/MTW and Wikipedia solutions, despite their apparent similarities. I am not sure at this point if the proper time refers to time measured by a stationary clock deep in the gravity well or a clock that is orbiting and falling with the test particle. That is one reason I prefer coordinate time as measured at infinity because we do not have a moving target then. “

I think proper time [d\tau] is appropriate. I equate this time to correspond to the wristwatch time of an observer traveling with the frame of reference in question, e.g. free-falling, orbiting etc. It is my understanding that it is the distant observer’s perspective that is distorted by both gravity and velocity. Local time is always 1 sec per sec.

“I accept the criticism of using the virial theorem relationship of PE= -2.KE and I have already stated I was not certain it applied here, so I will reject it. “

I wasn’t sure whether you were rejecting the virial theorem or the fact that Ep=-2KE? I believe the latter is valid; at least, classical circular orbits. It is an interesting point to check whether this relationship still stands in your relativistic derivation. Here is the classical argument:

Et = Ek – Ep

Et = 1/2mv^2 + (-GMm/r)

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

Where v_r=0 and
v_o=\sqrt {\frac{GM}{r}}.
So substituting:

E_t = \frac{GMm}{2r}(E_k)-\frac{GMm}{r} (E_p)

So for classical circular orbit Et=-GMm/2r, while Ek=GMm/2. The implication is that any stable orbit exists within the gravitational potential well, i.e. has net negative energy based on the accounting of potential energy being negative.

Are you coming to any conclusions as to which equation for Veff can be justified by derivation and does this equation produce the max/min curve?

 

[Jorrie]

If we assume

L = mvr = m (d\phi / dt) r

and if we assume

\sqrt{1-2M/r-d\phi^2/dt^2}}

is a dimensionless ratio, then the equation should be: <br /> \frac{L}{m} = \frac{r d\phi}{dt\sqrt{1-2M/r-d\phi^2/dt^2}}<br />

MTW's relativistic equation 25.17 (pp. 657 in my book) for L is:

L/m = \frac{r^2 d\phi}{d\tau} = \frac{r^2 d\phi}{dt\sqrt{1-2M/r -r^2d\phi/dt^2}}

in geometric units and with dr/dt=0. In general orbits, dr/dt <> 0 for most of the time, so dr/dt influences L, because it influences d\tau.

Check you dimensions and you will also see the problem. Inside the square root, all terms must be dimensionless; your d\phi^2/dt^2 is not, but gives seconds^{-2}.

 

[yuiop]

Just a couple of points on some of the statements made:

“it is worth noting that the Wikipedia solution uses proper time in its derivation and this may account for the large difference between the fourmilab/MTW and Wikipedia solutions, despite their apparent similarities. I am not sure at this point if the proper time refers to time measured by a stationary clock deep in the gravity well or a clock that is orbiting and falling with the test particle. That is one reason I prefer coordinate time as measured at infinity because we do not have a moving target then. “

I think proper time [d\tau] is appropriate. I equate this time to correspond to the wristwatch time of an observer traveling with the frame of reference in question, e.g. free-falling, orbiting etc. It is my understanding that it is the distant observer’s perspective that is distorted by both gravity and velocity. Local time is always 1 sec per sec.

“I accept the criticism of using the virial theorem relationship of PE= -2.KE and I have already stated I was not certain it applied here, so I will reject it. “

I wasn’t sure whether you were rejecting the virial theorem or the fact that Ep=-2KE? I believe the latter is valid; at least, classical circular orbits. It is an interesting point to check whether this relationship still stands in your relativistic derivation. Here is the classical argument:

Et = Ek – Ep

Et = 1/2mv^2 + (-GMm/r)

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

Where v_r=0 and
v_o=\sqrt {\frac{GM}{r}}.
So substituting:

E_t = \frac{GMm}{2r}(E_k)-\frac{GMm}{r} (E_p)

So for classical circular orbit Et=-GMm/2r, while Ek=GMm/2. The implication is that any stable orbit exists within the gravitational potential well, i.e. has net negative energy based on the accounting of potential energy being negative.

Are you coming to any conclusions as to which equation for Veff can be justified by derivation and does this equation produce the max/min curve?

I am assuming that the MTW and fourmilab equations

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

where V is the potential energy per unit mass of the orbiting particle, are the probably the most authoritive and in reasonable agreement, ( I have taken the liberty of converting the normilsed formula to normal units) but I do not understand enough about the equations the derivations are based on to give a clear opinion.

The Wikipedia equation is roughly the MTW equation squared with an additional constant. That is quite a big difference. I am not saying the Wiki equation is wrong, but it may be looking at things from a different angle, maybe proper time versus coordianate time.
Also bear in mind that the Wiki derivation is ultimately flawed in the region of a black hole because it is based on the Newtonian formula mv^2/2 for kinetic energy rather than the relativistic equation mc^2/\sqrt{1-v^/c^2}-mc^2 for KE.

As for the virial theorem, I was not rejecting the theorem itself, just the use of PE= -2KE in the context of strong gravitational "field" near a black hole for example, as suggested by Jorrie. The wiki article seems to be using PE = -KE

Any accurate derivation for gravitational potential should probably be based on the invariant relationship E^2 -P^2 = M^2 as this is the quantity that is conserved in relativity.

You might be interested in this derivation I did a year ago of orbital periods in strongly curved spacetime, using a mixture of classical and relativistic equations, that seemed to work (in this instance) https://www.physicsforums.com/showpost.php?p=1526798&postcount=25

As for coordianate measurements being distorted compared to proper measurements, is it fair to say in special relativity that the clock of an observer moving relative to you is distorted, or do you both have equally valid viewpoints of time from your own reference frames?

 

[mysearch]

Thanks for the reply. I will try to sit down today and review all the exchanges and then plot the max/min curve based on all the different equations suggested. 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.

 

[Jorrie]

I am assuming that the MTW and fourmilab equations

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

I guess by now you have spotted the dimensional problem in your foumilab conversion!

Without the offending m^2 it will work perfectly for a normalized V_eff^2 in normal SI units.

-J

Edit: OK, I may be nitpicking, but if you write V without showing it actually means V/m, then you should not write L/m. That's why MTW uses the top-tilde to make clear which convention is followed. If you do not want to use the top-tilde convention, it is better to write:

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

-J

 

[yuiop]

MTW's relativistic equation 25.17 (pp. 657 in my book) for L is:

L/m = \frac{r^2 d\phi}{d\tau} = \frac{r^2 d\phi}{dt\sqrt{1-2M/r -r^2d\phi/dt^2}}

in geometric units and with dr/dt=0.

Check you dimensions and you will also see the problem. Inside the square root, all terms must be dimensionless; your d\phi^2/dt^2 is not, but gives seconds^{-2}.

Sorry, my typo this time.

I meant for a point particle the moment of inertia is I = mr^2 and angular momentum is

L = Iw = (mr^2)w = (m r^2)\frac {v}{r} = mvr = \frac{mr d\phi}{dt}

then

L/m = \frac{rd\phi}{d\tau} = \frac{r d\phi}{dt\sqrt{1-2GM/(rc^2) -d\phi/(dt^2c^2)}}

It is hard to check if units are dimensionless using geometrical units.

G/c^2 , 1 , 1/c^4 , c^5/G^9 , c^5/c^2 all appear to be dimensionless in units of G = c = 1

 

[yuiop]

I guess by now you have spotted the dimensional problem in your foumilab conversion!

I guess it is because by momentum I usually mean p = mv and for angular momentum I usually mean L = mvr whereas the text prefer to use the definitions of momentum per unit mass, p = v and L = vr.


I was just explicity showing the mass which is easier than trying to typset tilde marks :P


To be fair I did explicity state under the equation that V is potential energy per unit mass. The terms (potential energy) and (potential) seem to used interchangeably but they are different things with different units and was trying to bring that out. Potential is a gradient while potential energy is not.

 

[yuiop]

MTW's relativistic equation 25.17 (pp. 657 in my book) for L is:

L/m = \frac{r^2 d\phi}{d\tau} = \frac{r^2 d\phi}{dt\sqrt{1-2M/r -r^2d\phi/dt^2}}

in geometric units and with dr/dt=0. In general orbits, dr/dt <> 0 for most of the time, so dr/dt influences L, because it influences d\tau.

Check you dimensions and you will also see the problem. Inside the square root, all terms must be dimensionless; your d\phi^2/dt^2 is not, but gives seconds^{-2}.

If d\phi^2/dt^2 has units of seconds^{-2}, then your r^2 d\phi^2/dt^2 has units of metres^{2} seconds^{-2} which is not dimensionless either.

 

[Jorrie]

If d\phi^2/dt^2 has units of seconds^{-2}, then your r^2 d\phi^2/dt^2 has units of metres^{2} seconds^{-2} which is not dimensionless either.

Hmm... we were actually both wrong! I should have written that "your d\phi^2/dt^2 has units of meters^{-2} ", because the geometric units of time is either meters or cm, not seconds, unless, like in astronomy, one start to use light-seconds for distance and seconds for time.

In any case d\phi is always dimensionless, but dt never, that's why the r^2 on top makes it dimensionless in geometric units. Checking dimensions in either geometrical or conventional systems is always a good practice for catching out errors.

-J

Edit: you can reinstate the G's and c's and then the terms inside the square root must still be dimensionless in conventional units, because it is subtracted from 1, a dimensionless scalar. Remember that velocity is dimensionless in geometric units...

 

[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

 

[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.