Question on Effective Potential

[yuiop]

Of course not, it is not necessary.



..because the wiki page is most likely wrong.
I thought that the fix was very obvious: in your incorrect derivation, replace the early mistake $$\frac{E^2}{m^2c^2}-c^2=c^2$$ or the later,subtler mistake $$\frac{E^2}{m^2c^2}-c^2=0$$ with $$\frac{E^2}{m^2c^2}-c^2=c^2(\gamma^2-1)$$ in the equation for the potential. :-)

If we do as you suggest we end up with an equation for effective gravitational potential that does not agree with the Wiki solution or the fourmilab solution or the MTW solution. I guess they all must be wrong.

 

[1effect]

If we do as you suggest we end up with an equation for effective gravitational potential that does not agree with the Wiki solution or the fourmilab solution or the MTW solution. I guess they all must be wrong.

I guess :-) You only need to know what is E , last I checked it was still $$\gamma mc^2$$ :-)
Don't feel bad, all the formulas you cited are still good for $$v<<c$$ , not so good at relativistic speeds :-)

 

[yuiop]

Jorrie posted this version of the equation for effective potential from MTW:


$$V_{eff} = c^2\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

If we assume he made a typo and meant


$$V_{eff} = c\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

then the MTW solution is in agreement with the fourmilab solution

 

[yuiop]

I guess :-) You only need to know what is E , last I checked it was still $$\gamma mc^2$$ :-)
Don't feel bad, all the formulas you cited are still good for $$v<<c$$ , not so good at relativistic speeds :-)

I look forward to the 1effect solution ;)

 

[1effect]

Jorrie posted this version of the equation for effective potential from MTW:$$V_{eff} = c^2\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

If we assume he made a typo and meant $$V_{eff} = c\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

then the MTW solution is in agreement with the fourmilab solution

...but in disagreement with your solution as well as with the wiki solution. Try calculating $$V_{eff}^2$$.
Doesn't change anything about your mistakes, your derivation is still wrong :-)

 

[yuiop]

...but in disagreement with your solution as well as with the wiki solution. Try calculating $$V_{eff}^2$$.
Doesn't change anything about your mistakes, your derivation is still wrong :-)

My derivation might well be wrong. It is based on other people's equations for motion and I have assumed Potential Energy and Kinetic Energy is related as per the virial theorum but I can not be sure that theorem applies here. I am trying to find out what that relationship is in GR for a falling/orbiting particle. As I said, I would welcome any alternative derivations that are clearly stated. You are also right that the different sources do not agree with each other, so hopefully some of the more experienced members of PF can shed some light on the situation.

 

[Jorrie]

This is how I figure fourmilab got their solution for effective gravitational potential:

...
Assuming the virial theorem relationship for potential energy and kinetic energy PE = -2.KE then
...

I think this is a weak field approximation only and does not hold for relativistic energies.

Assuming the gravitational potential (V) is simply PE/m then:
...

This is not the definition of effective potential, so I think your derivation may be flawed on two counts. According to MTW, effective potential is the total orbital energy (potential and kinetic) at the points of the orbit where dr/dt = 0 (peri- and apoapsis for closed orbits). MTW gives it in standard energy form and Fourmilab in energy squared form, hence the difference in appearance.

From MTW, using geometric units and working per unit orbiting mass, we have the normalized total orbital energy for dr/dt=0:

$$E = \frac{1-2M/r}{\sqrt{1-2M/r-r^2d\phi^2/dt^2}}$$

and

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

(Note: equation edited to correct erroneous r to r^2 above the line)

These two can be solved by eliminating $r d\phi/dt$ to give the (MTW) effective potential (energy) as:

$$V_{eff} = \sqrt{(1-2M/r)(1+L^2/r^2)}$$

equivalent to the Fourmilab result, which is just the value squared.

 

[Jorrie]

Jorrie posted this version of the equation for effective potential from MTW:


$$V_{eff} = c^2\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

If we assume he made a typo and meant


$$V_{eff} = c\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

then the MTW solution is in agreement with the fourmilab solution

I understand MTW's V_eff as energy per unit orbiting mass, so the c^2 is required if you want it in SI units of joules/kg.

The fourmilab solution in SI units must then be in (joules/kg)^2, not so?

 

[mysearch]

Part 1 of 2

There has been a lot of activity since I last looked at this thread last night (my time). I don’t have much time today, so apologises upfront if I have misinterpreted any statement of position, but wanted to try to comment on the exchanges that have taken place.

1. kev: pointed out that the value I was using (4.08E12) in my spreadsheet seem to produce a max/min curve if doubled. I also cross-referenced this with Jorrie statement that [L] should be greater that 3.4642GM/c. You're both right, I needed to use a minimum value of 6.08E12. However, this only seems to produce the quasi-stable maximum, which then tails off without a definitive minimum. However, I was rushing and may have made a mistake.

2. Also tried to quickly use Jorrie’s equation:

$$V_{eff} = c^2\sqrt{(1-2GM/r)(1+L^2/(r^2c^2))}$$

Again, didn’t get the result expected, but I need to spend some time tidying up and checking my spreadsheet. So I will assume I have made a mistake.

However, I would also like to see if we could standardise towards some set of terms that then leads to a consistent and agreed derivation of effective potential, e.g.

Effective Potential (Veff)
Total Energy (Et); Kinetic Energy (Ek); Potential Energy (Ep)
Radial velocity ( $$v_r = dr/d\tau$$)
Orbital velocity ( $$v_o = \omega*r$$)
Angular velocity $$[\omega] = d\phi/d\tau = v_o/r = mv_or/mr = L/mr$$)
Angular Momentum [L] = $$mv_or$$ Only valid for circular orbits


I believe that there is some value in stepping back and looking at the derivation of effective potential in a purely classical form. I have several sources, which appear unambiguous and lead directly to the classical minimum Veff curve associated with the appropriate value of angular momentum [L] required to balance the orbit at a given radius [r]:

Et = Ek – Ep

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

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

Based on the assumptions above:

$$Et = 1/2m\left( \frac{L^2}{(mr)^2}\right)-\frac{GMm}{r} = \frac{L^2}{2mr^2}-\frac{GMm}{r}$$

We might wish to normalise to unit mass, just to cross-reference standard text:

$$\frac{Et}{m} = \frac{(L/m)^2}{2r^2}-\frac{GM}{r}$$

However, from a classical perspective:

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

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

$$\frac {Veff}{m} = \frac{(L/m)^2}{2r^2}-\frac{GM}{r}$$

End of Part-1: See Part-2

 

[mysearch]

Part 2 of 2:

Sorry about splitting the reply, but the preview kept blowing up. Presumably the size was too big. Anyway, continuing...


From the outset, it was noted that the previous equations corresponds with the form in the Wikipedia reference, albeit with an additional relativistic term:

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

In an earlier posting I did a derivation from the Schwarzschild metric, which led to the following form. Note, the equation has been multiplied by m/2 to be consistent with the definition of kinetic energy $$Ek = 1/2mv^2$$

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

I argued that because we were only considering circular orbits, $$dr/d\tau$$ could be set to zero. However, kev has raised the issue as to whether this is valid for the orbital radii under consideration. Any thoughts on this point? However, if we proceed with this assumption, we get:

$$\frac{mc^2}{2}\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 = \left(1-\frac{Rs}{r}\right) \left(\frac{L^2}{2mr^2} + \frac{mc^2}{2} \right)$$

As far as I can see, this is the root of the form:

$$\left(1-\frac{Rs}{r}\right) \left(1 + \frac{L^2}{r^2}\right)$$

Which then gets translated into several different forms based on the substitution of the Schwarzschild radius [Rs] and the introduction of geometric units. However, the expansion of this term leads to the form:

$$\frac{L^2}{2mr^2}+\frac{mc^2}{2}-\frac{Rs}{r}*\frac{L^2}{2mr^2}-\frac{Rs}{r}*\frac{mc^2}{2}$$

$$\frac{L^2}{2mr^2}+\frac{mc^2}{2}-\frac{GML^2}{c^2mr^3}-\frac{GMm}{r}$$

The difference between the Wikipedia + classical derivation and the Fourmilab + relativistic derivations appears to be linked to which side of the equation you put the $$mc^2/2$$ term. On the basis of the classical definition, Veff is defined as the sum of the orbital kinetic energy plus the potential energy and therefore this term must belong on the left with total energy [Et], i.e. I think Wikipedia is right. Again, on the basis that $$dr/d\tau$$ can be set to zero, we have a direct correlation between total energy [Et] and Veff

$$Et = Ek + Ep$$

$$Veff = Ek(v_o) + Ep = Et$$ if $$v_r = 0$$

Of course, it is possible that the re-arrangement used in most relativistic text doesn’t actual change the basic tenets of the argument for the effective potential curves being suggested. However, it would be nice to use an equation, which is consistent and can show its derivation to be compatible with the classical form.

Really appreciate all the input. Any further thoughts?

 

[yuiop]

Hi mysearch.

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. :P

 

[yuiop]

Hi Jorrie,

thanks for your valued and informed input ;)

I understand MTW's V_eff as energy per unit orbiting mass, so the c^2 is required if you want it in SI units of joules/kg.

The fourmilab solution in SI units must then be in (joules/kg)^2, not so?

You are right about the units. Gravitational potential should be in units of joules/kg or the equivalent (meter/second)^2 which is satisfied by c^2.

What was confusing me was that the values inside the brackets of

$$V_{eff} = \sqrt{(1-2M/r)(1+L^2/r^2)}$$

are not pure ratios and I was trying to fix that.

I think this is a weak field approximation only and does not hold for relativistic energies.

This is not the definition of effective potential, so I think your derivation may be flawed on two counts.

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.

As for using gravitational potential = gravitational PE/m let's replace the expression (gravitational potential) with (gravitational PE per unit mass) in line with the various texts on the subject. It less emotive then ;)

According to MTW, effective potential is the total orbital energy (potential and kinetic) at the points of the orbit where dr/dt = 0 (peri- and apoapsis for closed orbits). MTW gives it in standard energy form and Fourmilab in energy squared form, hence the difference in appearance.

From MTW, using geometric units and working per unit orbiting mass, we have the normalized total orbital energy for dr/dt=0:

$$E = \frac{1-2M/r}{\sqrt{1-2M/r-r^2d\phi^2/dt^2}}$$

and

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

These two can be solved by eliminating $r d\phi/dt$ to give the (MTW) effective potential (energy) as:

$$V_{eff} = \sqrt{(1-2M/r)(1+L^2/r^2)}$$

equivalent to the Fourmilab result, which is just the value squared.

I hate to say it, but there seems to be an error in the MTW derivation (although it more likey the error is mine and 1effect will soon find it :P )

Eliminating $r d\phi/dt$ in

$$E = \frac{1-2M/r}{\sqrt{1-2M/r-r^2d\phi^2/dt^2}}$$

yields

$$\sqrt{(1-2M/r)(1+L^2)}$$

which is not the MTW result you quoted.



It seems that $r^2$ should be removed from the expression inside the square root. Analysis of the units seems to support this.

We now have:

$$E = \frac{1-2M/r}{\sqrt{1-2M/r-d\phi^2/dt^2}}$$

and

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

which would then reduce to the result you got:

$$\sqrt{(1-2M/r)(1+L^2/r^2)}$$





On that basis, adding G, m and c back into the equations, so that the units balance correctly, the derivation would read:

$$E = mc^2 \frac{1-2GM/(rc^2)}{\sqrt{1-2GM/rc^2-d\phi^2/(dt^2c^2)}}$$

and

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

which would then reduce to the restored fourmilab equation

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

where V is potential energy per unit mass.

 

[1effect]

Hi Jorrie,

thanks for your valued and informed input ;)



You are right about the units. Gravitational potential should be in units of joules/kg or the equivalent (meter/second)^2 which is satisfied by c^2.

What was confusing me was that the values inside the brackets of

$$V_{eff} = \sqrt{(1-2M/r)(1+L^2/r^2)}$$

are not pure ratios and I was trying to fix that.




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.

As for using gravitational potential = gravitational PE/m let's replace the expression (gravitational potential) with (gravitational PE per unit mass) in line with the various texts on the subject. It less emotive then ;)




I hate to say it, but there seems to be an error in the MTW derivation (although it more likey the error is mine and 1effect will soon find it :P )

Eliminating $r d\phi/dt$ in

$$E = \frac{1-2M/r}{\sqrt{1-2M/r-r^2d\phi^2/dt^2}}$$

yields

$$\sqrt{(1-2M/r)(1+L^2)}$$

which is not the MTW result you quoted.

Yes, more precisely, you get
$$E=\sqrt{(1-2M/r)(1+L^2)}$$

 

[yuiop]

Yes, more precisely, you get
$$E=\sqrt{(1-2M/r)(1+L^2)}$$

Do you agree that is not a balanced equation even when we allow for the normalised units?
r is not a normalised variable and has not been set to one like G and x.
The only way to balance that equation in normal units to get pure ratios is like this:

$$\frac{E}{mc^2} = \sqrt{\left(1-\frac{2GM}{rc^2}\right)\left(1+\frac{L^2c^2}{G^2m^2}\right)}$$

which expands to

$$\frac{E^2}{m^2} = c^4\left(1+\frac{L^2c^2}{G^2m^2}-\frac{2GM}{rc^2}-\frac{2ML^2c^2}{G m^2 c^4}\right)$$

which I don't think gives the required min max potential curve because there are no rpowers of r greater than one.

 

[Jorrie]

I hate to say it, but there seems to be an error in the MTW derivation (although it more likey the error is mine and 1effect will soon find it :P )

Eliminating $r d\phi/dt$ in

$$E = \frac{1-2M/r}{\sqrt{1-2M/r-r^2d\phi^2/dt^2}}$$

yields

$$\sqrt{(1-2M/r)(1+L^2)}$$

which is not the MTW result you quoted.

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:

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

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.

 

[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:

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

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:


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

 

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

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:

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

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):

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

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

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

Divide both sides by c^2 and rearrange:

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

For constant radius this becomes the familiar

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

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 $$r->infinity$$ . Obviously, this makes $$v->infinity$$. In other words, you forgot that $$\frac{L}{mcr}->infinity$$. 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 $$r->infinity$$ . Obviously, this makes $$v->infinity$$. In other words, you forgot that $$\frac{L}{mcr}->infinity$$. 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.