Hi Jorrie,
thanks for your valued and informed input ;)
Jorrie said:
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.
Jorrie said:
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 ;)
Jorrie said:
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.
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:
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.