Question on Effective Potential

[mysearch]

I am trying to draw one of those nice plots of effective potential [Vr] against radius [r], which suggest the position of stable and quasi-stable orbits. Have no trouble getting the classical curve showing a minimum coinciding with a radius corresponding to a given angular momentum [L], but having trouble getting the equivalent min/max curve associated with general relativity. By way of reference, I was initially looking at the following Wikipedia page:

http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

This page has two sub-titles that are relevant:
`Relation to classical mechanics and precession of elliptical orbits`
`Circular orbits and their stability`

Under the first sub-title, the following equation for the effective potential is listed as:

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

The first term relates to gravitational potential energy, the second relates to kinetic energy of rotation, while the third is described as "an attractive energy unique to general relativity". It is stated that the 3 terms, containing different powers of radius [r], combine to give a curve with a minimum corresponding to a stable orbit and a maximum that relates to a quasi-knife-edge orbit. However, when I put the figures into a spreadsheet I don’t get this min/max curve.

Key values used:
Gravitational Constant [G] = 6.67E-11
Black Hole Mass [M] = 7.92E+30 (4 solar masses)
Light [c] = 2.99E8
Schwarzschild Radius [Rs] = 1.18E+04 2GM/c^2
Angular Momentum [L] = 4.08E+12 (r=2Rs??)

Assumptions about [L] for verification:
In classical physics, the angular momentum of a circular orbit L = mvr and, as such, [L] must change for each fixed circular orbit [r]. Only in a closed system is angular momentum [L] constant and therefore any decrease in radius [r] means an increase in velocity [v]. Normally, the 'natural` orbit of a satellite can be determined by balancing the outward `centrifugal` force [mv^2/r] with the inward pull of gravity [GMm/r]. The effective potential (V) curve can show this balance in terms of energy plotted for fixed values of angular momentum [L]. In the classical case, the value of [L] inserted is only correct at one value of [r] corresponding to an energy rate of change [F=dE/dr=0]. With the addition of the relativistic component, the curve is said to produce a max/min curve, which I am not getting, but I can’t see my mistake.

Some other points for verification:
If the assumption L=mvr for a circular orbit is valid, then the equation for the effective potential above could be simplify to the form:

Vr = -\frac{GMm}{r} + 1/2 mv^2} - \frac{GMm}{r}(\frac{v^2}{c^2})

Where [v] is the orbital velocity, while the radial velocity [dr/dt] is set to zero for a circular orbit. If so, can this equation also be written in the form?

Vr = 1/2mv^2} - \frac{GMm}{r}(1 + \frac{v^2}{c^2})

This equation appears analogous to the classical form, but with an additional relativistic component. However, can this equation also be converted into an equivalent expression showing the balance between the `centrifugal` force and gravitational pull associated with a circular orbit?

\frac{mv^2}{r} = \frac{GMm}{r^2}(1 + \frac{v^2}{c^2})

If the assumptions made are correct, it would suggest that a larger gravitational force/curvature is required to counter the `centrifugal` force. This factor would appear to only range between [1:2]. Equally, this expression could be solved without the use of the quadratic approach normally employed, e.g.

v^2 = \frac{Rs*c^2}{(2r-Rs)}

Therefore, I not sure any of the assumptions made are correct as I have not seen this approach presented in any standard text. Therefore, I would appreciate any help that could be given to clarify the situation.

 

[yuiop]

This link might help: http://www.fourmilab.ch/gravitation/orbits/

It has a nice applet illustrating the effective potential and precession of an eleiptical orbit.

The equations might look a little different because they are given in geometric units where Newton's gravitational constant G, the speed of light c, and Boltzmann's constant k all equal to 1.

 

[mysearch]

Question on Effective Potential: reply

Kev, thanks for the link to http://www.fourmilab.ch/gravitation/orbits/
I have seen the equation below in a number of texts, so I accept it must be right, but still having problems drawing a graph that looks like the max/min effective potential curve used in so many texts.

[1]...V^2 = (1-\frac{2M}{r}) (1+\frac{L^2}{r^2})

I also still need to equate this form to the equation given in the wikipedia reference
http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

[2]...Vr = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3}

While noting that [1] is [V^2], the units of each term in equation [2] correspond to energy [m^2 kg/s^2] and there are only 3 terms. The expansion of [1] seems to lead to the form:

V^2 = (1 + \frac{L^2}{r^2} - \frac{2M}{r} - \frac{2ML^2}{r^3})

Now there seems to be 4 terms, none of which represent energy. I presume the geometric units are intended to normalise everything to metres, e.g.

Geometric [M] = \frac{ GM }{c^2}

Personally, I find the following equivalent to be easier:

(1-\frac{2M}{r}) = (1-\frac{Rs}{r})

Given that this term has no units, i.e. it is just a ratio; I assume the geometric units must be resolved in the second term:

(1+\frac{L^2}{r^2})

Again, find this form non-intuitive when trying to understand what is really going on, especially when trying to insert `real-world`values into a spreadsheet. However, I haven’t worked through this yet, but would still like to know whether [2] is a valid statement of the effective potential [V]. If so, I would also like to know what you have to do to make this expression produce the effective potential curve so often presented in relativity texts. Again, thanks for the reference, I guess I need to do a bit more investigating.

 

[mysearch]

I have attempted to upload a .jpg file showing the graph obtained when plugging some example black hole values into the wikipedia equation. Hoping somebody might be able to tell me why the the effective potential, i.e. the sum of the other 3 curves does not produce the min/max curve expected.

 

[yuiop]

Starting with the same equation that Wikipedia use for the deriviation:

\left( \frac{dr}{d\tau} \right)^{2} = <br /> \frac{E^{2}}{m^{2}c^{2}} - c^{2} + \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

and noting that dr/dt is a radial velocity

\left( \frac{dr}{d\tau} \right)^{2} = V^2

So
<br /> V^{2} = <br /> c^{2} + \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}


Replacing r_{s} with 2GM/c^2

<br /> V^{2} = <br /> c^{2} + \frac{2GM}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}

Using L_{o} = L/m for angular momentum per unit mass and units of G=c=1

V^2 = (1 - \frac{L_{o}^{2}}{r^2} +\frac{2M}{r} + \frac{2M L_{o}^{2}}{r^3})

which is similar to the expansion of the fourmilab equation you got:

V^2 = (1 + \frac{L^2}{r^2} - \frac{2M}{r} - \frac{2ML^2}{r^3})

except for a sign problem.

Now the fourmilab equation if velocity squared per unit mass. What we want is the effective potential. The potential energy (PE) of an object at infinity is converted to kinetic energy (KE) as the object falls. The potential energy at a given radius (PE_r)[/itex] is the PE at infinity ( PE_{inf}) minus the KE. <br /> <br /> So PE_r = PE_{inf} - KE = PE_{inf} - \frac{mV^2}{2}<br /> <br /> If we assume the PE at infinity is mc^2 then<br /> <br /> PE_r = mc^2 - \frac{mV^2}{2} <br /> <br /> --&gt; PE_r = m(c^2 - (V^2)/2)<br /> <br /> Replacing v^2 with the equation derived earlier<br /> <br /> PE_r = m\left(c^2 - \left( c^{2} + \frac{2GM}{r} - &lt;br /&gt; \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\frac{1}{2}\right)<br /> <br /> If you ignore the 1/2 then the c^2 cancel out, giving the 3 terms of the wikipedia solution but that is not really satisfactory.<br /> <br /> Now divide both sides by m to get the PE per unit mass that fourmilab uses (and multiply by 2 to make things clearer)<br /> <br /> \frac{2 PE_r}{m}= \left(2c^2 - \left( c^{2} + \frac{2GM}{r} - &lt;br /&gt; \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\right)<br /> <br /> --&gt; \frac{2 PE_r}{m}= \left(c^2 - \frac{2GM}{r} + &lt;br /&gt; \frac{L^{2}}{m^{2} r^{2}} - \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\right)<br /> <br /> which is the fourmilab solution multiplied by a factor of two. (Remember they are using angular momentum per unit mass so the m&#039;s disappear) <br /> <br /> Neither solution seems to be completely satifactory but I hope that I have been of some help.<br /> <br /> It should also be noted that that wikpedia is using the Newtonian formula KE = mv^2/2 rather than the relativistic equation for kinetic energy and that makes the solution an aproximation.

 

[yuiop]

My last post does not seem to solve your problem of thereis no min max curve.
Looking at the fourmilab page again, the effective potential is given as V^2(L,r) suggesting it is a function of the variables L and r. In other words the the angular momentum is not constant. Remember momentum is not always conserved in GR while momentum-energy is. Further down the page he gives L=r^2d\theta / d\tau. That should sort the problem out (I think). If all else fails, download the code for the applet and analyse that.

If you find the answer, let me know

 

[mysearch]

Kev: Appreciate the detailed feedback; it is indeed helpful to have somebody check whether I had made an obvious mistake, although I am sure it is there somewhere. I was not sure from your last post, whether you had tried plotting the effective potential curve? Just by way of reference, I have attached 3 .jpg files to this first post, which might be useful as a reference with respect to the discussion in the next post, which tries to follow on from some of the points you have raised.

The first two diagrams relate to effective potential of an example black hole based on Newtonian assumptions, i.e. without the additional gravity terms. This plot produces a minimum that coincides with the value of angular momentum [L] being held constant. As such, the minimum [Veff] corresponds to a classical radius of a stable orbit with this angular momentum at r/Rs=2.5. Another point worth noting is that the kinetic energy associated with the orbital velocity is always half the potential energy of gravitation. This point is raised because I was wondering whether this ratio holds true in the relativistic equivalent?

The last diagram is the data for the relativistic plot submitted in an earlier post. It is essentially the same example black hole, but includes the additional relativistic terms, as per the equation sourced by Wikipedia. However, rather than the relativistic term, driven by 1/r^3, causing a min/max peak corresponding to stable and quasi-stable orbits, the effective potential appears simply to roll off, implying no stable orbit, which would appear to be obviously wrong.

 

[yuiop]

Kev: Appreciate the detailed feedback; it is indeed helpful to have somebody check whether I had made an obvious mistake, although I am sure it is there somewhere. I was not sure from your last post, whether you had tried plotting the effective potential curve? Just by way of reference, I have attached 3 .jpg files to this first post, which might be useful as a reference with respect to the discussion in the next post, which tries to follow on from some of the points you have raised.

The first wto diagram relate to effective potential of an example black hole based on Newtonian assumptions, i.e. without the additional gravity terms. This plot produces a minimum that coincides with the value of angular momentum [L] being held constant. As such, the minimum [Veff] corresponds to a classical radius of a stable orbit with this angular momentum at r/Rs=2.5. Another point worth noting is that the kinetic energy associated with the orbital velocity is always half the potential energy of gravitation. This point is raised because I was wondering whether this ratio holds true in the relativistic equivalent?

The last diagram is the data for the relativistic plot submitted in an earlier post. It is essentially the same example black hole, but includes the additional relativistic terms, as per the equation sourced by Wikipedia. However, rather than the relativistic term, driven by 1/r^3, causing a min/max peak corresponding to stable and quasi-stable orbits, the effective potential appears simply to roll off, implying no stable orbit, which would appear to be obviously wrong.

I notice you have 2 columns in your spreadsheet both headed as L = mvr, one of which is constant and the other varies. Which L are using in your plots? Constant or variable L?

P.S. I am inclined to prefer the foumilab equations because they need to get the equations right to get the applet to "work". Also bear in mind that there is potential confusion due to Wikipedia using V to represent velocity while fourmilab use V to represent effective potential energy. With a bit of luck a real expert will jump in and help us both out here :P When I get time I will I will have a look at the code for the applet. The answer is in there!

 

[mysearch]

The plots are based on the constant value. In the classical case, the minimum value coincides with the corresponding value of radius [r/Rs=2.5]. I was trying to post another reply, but the preview keeps crashing on me!

The problem I have with the Fourmilab equation is that doesn't seem to be consistent with the classical derivation of effective potential I found. This is what I was trying to post. The reason for posting this information was simply that an 'expert` might be able to resolve the issue by looking at the data.

 

[mysearch]

Need to split this reply into 2 parts as the preview keeps crashing. Is there a size limit?

Part1:
In this post, I wanted to pick up on some of your comments concerning classical versus relativistic assumptions, as I thought it might be useful to compare the derivation from Newtonian mechanics against the Schwarzschild metric to see if it might highlight where the issue lies. For comparison and easier crosschecking of units, I won’t use geometric units. There is a Newton derivation of effective potential in Alonso-Finn, which has been a standard physics text for years. In my version, which is a bit old, the effective potential curve is only shown for the classical case. This has been reproduced without any problem, see attachment in previous post.

Effective Potential: Classical

[1] Total Energy = Kinetic + Potential

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

[2] Et = 1/2m\left( vr^2 + vo^2\right)-GMm/r

Where [vr =dr/d\tau] is the radial velocity. Note, this velocity can be set to zero for a circular orbit, while [vo] is associated with the angular velocity [w]:

vo = wr = r.d\phi/d\tau; vo = mvr/mr = L/mr

On the basis on these assumptions, we can write [2] as:

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

Alonso-Finn defines effective potential V[r] as:

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

Which based on the earlier substitution becomes:

[5] Veff = \frac{L^2}{2mr^2}-\frac{GMm}{r}

The form of the equation above, i.e. [5], is compatible with the Wikipedia equation for relativistic effective potential, except for the inclusion of the additional relativistic terms.

[6] Veff = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3}

My interpretation of this additional term is not that there is `an attractive energy unique to general relativity`, but simply that a stable orbit requires a stronger gravitational pull to counter an increase in kinetic energy associated with a relativistic orbital velocity. This would make sense in terms of a relativistic increase in the orbital mass [m]. This statement is being based on the assumption that the relative perspectives of a stationary observer, at the same radius as the orbital observer, would only be subject to special relativity and therefore velocity would be invariant as space and time are both subject to the same `contraction/dilation`. However, I am digressing into another topic. In your previous post, you cite that:

\left( \frac{dr}{d\tau} \right)^{2} = \frac{E^{2}}{m^{2}c^{2}} - c^{2} + \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

followed by

\left( \frac{dr}{d\tau} \right)^{2} = V^2 = c^{2} + \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

I assume [V] is velocity in this case, not effective potential, but the standard definition seems to suggest that:

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

We seem to have lost the total energy in one of the steps. I believe this form can be derived directly from the following re-arrangement of the Schwarzschild metric:

\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)

Substituting for d\phi/d\tau = L/mr^2:

See Part2:

 

[mysearch]

part2

Part2:

\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{L}{mr^2}\right)^2 + c^2 \right)

\left( \frac{dr}{d\tau} \right)^{2} = c^2\left(1-\frac{Rs}{r}\right)^2 \left(\frac{dt}{d\tau}\right)^2 - \left( \frac{L^2}{m^2r^2} - \frac{RsL^2}{m^2r^3} - \frac{Rs c^2}{r} + c^2 \right)

Now in the Wikipedia derivation, the c^2 term on the right is lumped in with the total energy terms in the form of (dt/d\tau), which is why it reduces immediately to 3 terms, not 4. This assumption does seem to be supported by the classical derivation and appears to be supported by the data Ek=-Ep/2, but I need to check this more closely. I was also a bit uncertain about the next bit of your derivation, for a circular orbit in which dr/d\tau could be set to zero and therefore we would only be left with the orbital velocity. This velocity is not proportional to radius [r] in the same way as a free-falling velocity, i.e.

Free-fall v^2 = \frac{2GM}{r}
Orbital v^2 = \frac{GM}{r}

The former corresponding to the conversion of potential energy at infinity into kinetic energy at any given radius, while the latter is the balance of a centrifugal force with gravity. However, if we step back to the equation of the form


\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(\frac{L^2}{m^2r^2} + c^2 \right)

We might re-arrange around dr/d\tau = 0:

c^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}{m^2r^2} + c^2 \right)

Multiplying through by mass [m] would convert to energy

mc^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}{mr^2} + mc^2 \right)

Certainly, if we follow your step of converting into geometric unit and setting mass [m] to unity, we get:

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

Where the right-hand side appears to be of the form of the Fourmilab expression, but I would still question whether this represents the total energy (left) equalling the effective potential (right) for the case of an orbit, where dr/d\tau=0].

Maybe I should simply ask anybody in the forum for a definition of effective potential, which then produces the max/min curve detailed in so many sources. For unless I am making an obvious mistake in the spreadsheet data, see previous posting, I can’t get the Fourmilab or Wikipedia variants to produce this curve.

 

[mysearch]

Summary:

Without further input, this thread has almost run its course. Therefore, I wanted to summarise the issues discussed and leave them as open questions. Initially, the equation for effective potential was taken at face value from a Wikipedia source: http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

Veff = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3}

However, when inserting this equation into a spreadsheet, the expected max/min curve corresponding to the stable and quasi-stable orbits could not be reproduced. Based on feedback, it was highlighted that there was a discrepancy with the form of this expression in comparison with other sources, e.g. Fourmilab, which forwarded the expression for the effective potential as:

Veff = (1-\frac{2M}{r}) (1+\frac{L^2}{r^2})

It is believed that the source of this expression can be derived from the Schwarzschild metric starting with the form:

\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)

\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(\frac{L^2}{m^2r^2} + c^2 \right)

Where dr/d\tau corresponds to the radial velocity [vr]. To convert this equation into an energy expression, we can multiply through by [½ m]

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)

However, for a circular orbit, it is assumed that dr/d\tau=0 and therefore we can reduce the form to:

\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)

So the first question is whether the expression on the right aligns to the effective potential, as the form appears to correspond to the Fourmilab format. However, if we expand this expression, we get:

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

Substituting for Rs=2GM/c^2:

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

Now the similarity to the Wikipedia equation re-emerges, albeit with the mc^2/2 addition, which in the Wikipedia derivation is moved to the other side of the equation. However, the classical derivation of effective potential suggests:

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

Of course, in the case under consider, i.e. a circular orbit, the term dr/d\tau=0 and therefore we get the situation where the total energy [Et] must equal the effective potential, which in turn equals the sum of the kinetic energy associated with the orbital velocity plus the potential energy of gravitation. If this is the case, it seems to support the fact that mc^2/2 is not a component of the effective potential, but rather a component of the total energy, as implied by the Wikipedia derivation. However, the problem that I am raising is that neither seems to produce the expected max/min curve.

One of the advantages of using a constant value of angular momentum [L] for all values of radius [r] is that you do not have to know the explicit value of the orbital velocity. The implication is that the values within a spreadsheet will only tie-up when the angular momentum is matched against the appropriate radius [r], i.e. this corresponds to the minimum dip in the curve, at least, in the case of the classical solution shown in an earlier posting. Another point worth raising is that a classical orbit has a specific ratio between the kinetic and potential energy, i.e. [Ek=-Ep/2]. In my example, the black hole mass [M] was set to 4 solar masses and a value of angular momentum [L=4.08E12] was selected to be in the right range for a radius spread between 1-10Rs. For the values indicated, the kinetic energy [Ek=-Ep/2] at a radius of 2Rs.

It is accepted that there must be some flaw in this logic, but it is still unclear why the relativistic effective potential does not produce the max/min curve. Should any member of the forum know the answer, it would be much appreciated.

 

[1effect]

So
<br /> V^{2} = <br /> c^{2} + \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

This can't be correct since \frac{E^2}{m^2c^2}-c^2 is not equal to c^2

 

[mysearch]

I agree, but the use of geometric units can often lead to apparent inconsistencies unless you are careful, which is why I reverted back to a conventional derivation from the Schwarzschild metric in the summary. Can you see any problems in the summary or what is being overlooked as far as effective potential is concerned?

 

[Jorrie]

Relativistic Effective Potential

It is accepted that there must be some flaw in this logic, but it is still unclear why the relativistic effective potential does not produce the max/min curve. Should any member of the forum know the answer, it would be much appreciated.

I haven't gone through all your derivations, but I use a variant of what is found in MTW, with c and G reinstated:

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

where V and L are per unit mass.

I get the curves as attached, normalized and with g_tt = 1-2GM/r. The Newtonian curve has c^2 added to put it at comparable level to the relativistic one. Note that if you choose L < 3.4642GM/c, you won't get a lower turning point in the relativistic case.

Hope it helps.

 

[mysearch]

Thanks very much. I will take a closer look and let you know how I get on.

 

[yuiop]

Jorrie is right that you will not get the min max curve if L is less than about 3.46 as can be seen by using the fourmilab applet. I think that may be the main problem you are having. Did you get the private message I sent?

 

[mysearch]

Yes, did you receive the email & spreadsheet I sent?

 

[yuiop]

Yes, did you receive the email & spreadsheet I sent?

OK, found it! My junk mail sorter can be over zealous :P

 

[yuiop]

This is the derivation I made in post #5 edited to correct the error pointed out by 1effect.

( I am using v to represent velocity to avoid confusion with V for gravitational potential)

=======================================================

Starting with the same equation that Wikipedia use for the deriviation:

\left( \frac{dr}{d\tau} \right)^{2} = <br /> \frac{E^{2}}{m^{2}c^{2}} - c^{2} + \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

and noting that dr/dt is a radial velocity

\left( \frac{dr}{d\tau} \right)^{2} = v^2

So
<br /> v^{2} = \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}


Replacing r_{s} with 2GM/c^2

<br /> v^{2} = \frac{2GM}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}

THis is the radial (vertical) velocity squared. What we want is the effective potential. The potential energy (PE) of an object at infinity is converted to kinetic energy (KE) as the object falls. The potential energy at a given radius (PE_r)[/itex] is the PE at infinity ( PE_{inf}) minus the KE. <br /> <br /> So PE_r = PE_{inf} - KE = PE_{inf} - \frac{mv^2}{2}<br /> <br /> (We can use the Newtonian aprximations because the infalling radial velocity is relativly low)<br /> <br /> If we assume the PE at infinity is mc^2 then<br /> <br /> PE_r = mc^2 - \frac{mv^2}{2} <br /> <br /> --&gt; PE_r = m(c^2 - (v^2)/2)<br /> <br /> Replacing v^2 with the equation derived earlier<br /> <br /> PE_r = m\left(c^2 - \left( \frac{2GM}{r} - &lt;br /&gt; \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\frac{1}{2}\right)<br /> <br /> PE_r = m\left(c^2 - \frac{GM}{r} +&lt;br /&gt; \frac{L^{2}}{2m^{2} r^{2}} -\frac{GM L^{2}}{m^{2} r^{3} c^2}\right)<br /> <br /> <br /> In Newtonian physics the gravitational potential energy is -\frac{GMm}{R} and the gravitational potential is -\frac {GM}{R} so we can obtain the potential (V) from PE/m. (This is probably not rigorously accurate)<br /> <br /> V = \frac{PE_r}{m} = c^2 - \frac{GM}{r} +&lt;br /&gt; \frac{L^{2}}{2m^{2} r^{2}} -\frac{GM L^{2}}{m^{2} r^{3} c^2}<br /> <br /> When using geometrical units and angular momentum per unit mass the factored equation becomes:<br /> <br /> V = \left( 1 -\frac{M}{r} \right) \left(1+\frac{L^{2}}{2 r^{2}} \right)<br /> <br /> Wikipedia seems to have simply assumed the effective potential is simply the negative of the radial kinetic energy.<br /> <br /> My result does not agree exactly with the fourmilab or wikipedia result (but its close to both) but I have at least shown the steps I used to get there. Hopefully someone will post a more rigorous derivation :)

 

[yuiop]

Deleted duplicate post

 

[1effect]

This is the derivation I made in post #5 edited to correct the error pointed out by 1effect.

( I am using v to represent velocity to avoid confusion with V for gravitational potential)

=======================================================

Starting with the same equation that Wikipedia use for the deriviation:

\left( \frac{dr}{d\tau} \right)^{2} = <br /> \frac{E^{2}}{m^{2}c^{2}} - c^{2} + \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

and noting that dr/dt is a radial velocity

\left( \frac{dr}{d\tau} \right)^{2} = v^2

So
<br /> v^{2} = \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

I think is still wrong. E is not mc^2, it is \gamma mc^2

 

[yuiop]

I think is still wrong. E is not mc^2, it is \gamma mc^2

You have a point, but it seems to be the method used by fourmilab and wikipedia. I am assuming they are deviding both sides of the equation by \gamma m when they talk about potential per unit mass. Neither of the derivations seem to take into account relativistic mass increase due to velocity or gravity. I guess the equations get too messy if you try to do that, or they have assumed it has canceled out. I welcome any improvements on my attempt ;)

Note that Wikipedia has obtained:

\left( \frac{dr}{d\tau} \right)^{2} = 0 + \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

from:

\left( \frac{dr}{d\tau} \right)^{2} = \frac{E^{2}}{m^{2}c^{2}} - c^{2} + \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}

Clearly, wikipedia IS assuming E=mc^2 if they obtain \frac{E^{2}}{m^{2}c^{2}} - c^{2} =0

Presumably they are using \frac{E^2}{(\gamma m)^2 c^2} = \frac{(\gamma m)^2 c^4}{(\gamma m)^2 c^2}

In other words all references to m are not references to m_0 but to \gamma m_0 (It would seem). As I said before, I would like to see a thorough and clean derivation but it seems hard to find one.

 

[yuiop]

I just noticed in another forum that someone mentions that the virial theorem says that on average 2.KE + PE = 0 (not PE + KE = const). Presumably it is this relationship (PE = -2.KE) that is being used in the derivations? I would question that assumption, because I think the virial theorem refers to whole systems (eg a gas cloud collapsing gravitaionally) rather than an individual falling point particle. Any experts care to comment?

 

[1effect]

Clearly, wikipedia IS assuming E=mc^2 if they obtain \frac{E^{2}}{m^{2}c^{2}} - c^{2} =0

Presumably they are using \frac{E^2}{(\gamma m)^2 c^2} = \frac{(\gamma m)^2 c^4}{(\gamma m)^2 c^2}

In other words all references to m are not references to m_0 but to \gamma m_0 (It would seem). As I said before, I would like to see a thorough and clean derivation but it seems hard to find one.

bad physics leads to bad results :-)

 

[yuiop]

... I was also a bit uncertain about the next bit of your derivation, for a circular orbit in which dr/d\tau could be set to zero and therefore we would only be left with the orbital velocity. This velocity is not proportional to radius [r] in the same way as a free-falling velocity,,,

The relativistic equations seem to show that exactly stable orbits are unlikely, so we can not simply assume dr/d\tau can be set to zero.

 

[yuiop]

bad physics leads to bad results :-)

Pervect writes a lot of these articles in Wikipedia, so maybe this is one his. Maybe he would care to comment on the type of physics used to obtain the results in Wikipedia?

To be honest, I do not know the final answer. Do you?

 

[1effect]

Pervect writes a lot of these articles in Wikipedia, so maybe this is one his. Maybe he would care to comment on the type of physics used to obtain the results in Wikipedia?

To be honest, I do not know the final answer. Do you?

The wiki page contains the correct solution, with no newly introduced errors :-)

 

[yuiop]

The wiki page contains the correct solution, with no newly introduced errors :-)

In post #22 you declared


\frac{E^{2}}{m^{2}c^{2}} - c^{2} is not equal to zero,

because E is not equal to mc^2 so you are disagreeing with the Wikipedia solution that you say is correct.

 

[yuiop]

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

=======================================================

Starting with the Wikipedia equation for radial velocity:

v^{2} = \left( \frac{dr}{d\tau} \right)^{2} = \left[<br /> \frac{E^{2}}{m^{2}c^{2}} - c^{2} \right] + \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}} = 0+<br /> \frac{r_{s}c^{2}}{r} - <br /> \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}<br />

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

PE = -m v^{2} = -\frac{r_{s}c^{2}m}{r} +<br /> \frac{L^{2}}{m r^{2}} -\frac{r_{s} L^{2}}{m r^{3}}


Replacing r_{s} with 2GM/c^2

<br /> PE = -\frac{2GMm}{r} +<br /> \frac{L^{2}}{m r^{2}} - \frac{2GM L^{2}}{m r^{3} c^2}

It is obvious from the equation that the potential energy at infinity is zero, but this is arbitary because the constant of integration has been assumed to be zero. We can if we like (if we are consistent in its use) set the adjusted potential energy at infinty PE_0 to m c^2[/tex] and get:<br /> <br /> &lt;br /&gt; PE_0 = m c^2 -\frac{2GMm}{r} +&lt;br /&gt; \frac{L^{2}}{m r^{2}} - \frac{2GM L^{2}}{m r^{3} c^2}<br /> <br /> Assuming the gravitational potential (V) is simply PE/m then:<br /> <br /> V = \frac{PE_o}{m} = c^2 -\frac{2GM}{r} +&lt;br /&gt; \frac{L^{2}}{m^2 r^{2}} - \frac{2GM L^{2}}{m^2 r^{3} c^2}<br /> <br /> The factored equation becomes:<br /> <br /> V = \left( 1 -\frac{2GM}{r c^2} \right) \left(c^2+\frac{L^{2}}{m^2 r^{2}} \right)<br /> <br /> Using L_o = L/m and geometrical units of G=c=1 we obtain the fourmilab result:<br /> <br /> V = \left( 1 -\frac{2M}{r } \right) \left(1+\frac{L_o^{2}}{r^{2}} \right)<br /> <br /> By the way, did you get my email, mysearch?

 

[1effect]

In post #22 you declared \frac{E^{2}}{m^{2}c^{2}} - c^{2} is not equal to zero,

because E is not equal to mc^2 so you are disagreeing with the Wikipedia solution that you say is correct.

Why do you always have such a hard time admitting you made an error.
You already know that E is not equal to mc^2 but to \gamma mc^2
I already pointed out the solution in wiki that is correct, here it is again: http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity#Geodesic_equation

I never agreed with the one that includes the error, i.e. the one that you claim supports your derivation. Here is the wrong one: http://en.wikipedia.org/wiki/Kepler...mechanics_and_precession_of_elliptical_orbits

Clear now?

 

[yuiop]

Why do you always have such a hard time admitting you made an error.
You already know that E is not equal to mc^2 but to \gamma mc^2
I already pointed out the solution in wiki that is correct, here it is again: http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity#Geodesic_equation

I never agreed with the one that includes the error, i.e. the one that you claim supports your derivation. Here is the wrong one: http://en.wikipedia.org/wiki/Kepler...mechanics_and_precession_of_elliptical_orbits

Clear now?

Your correct link points to an equation for geodesics lines and a little further down in the same section I see the equation for the motion of a particle but nowhere within that section do I see a definition of effective potential.

The effective potential is only derived in the section that you say is wrong, and is derived from the equation of motion given in the section you say is correct.

So what is the equation for effective potential that you consider correct. Can you use latex to spell it out here?

 

[1effect]

Your correct link points to an equation for geodesics lines and a little further down in the same section I see the equation for the motion of a particle but nowhere within that section do I see a definition of effective potential.

It is not the SAME section, they are two DIFFERENT captions. The first one is correct, the second one is not. I even spelled out the captions for you.

The effective potential is only derived in the section that you say is wrong, and is derived from the equation of motion given in the section you say is correct.

So what is the equation for effective potential that you consider correct. Can you use latex to spell it out here?

Easy, calculate: (\frac{E}{mc})^2-c^2=c^2(\gamma^2-1).
Right? :-)

 

[yuiop]

It is not the SAME section, they are two DIFFERENT captions. The first one is correct, the second one is not. I even spelled out the captions for you.



Easy, calculate: (\frac{E}{mc^2})^2-c^2=c^2(\gamma^2-1).
Right? :-)

I don't see your interpretation used anywhere in the section titled "geodesic equation" nor do I see the equation for effective potential derived anywhere in the Wikpedia article using your interpretation of E and m.

 

[1effect]

I don't see your interpretation used anywhere in the section titled "geodesic equation"

Of course not, it is not necessary.

nor do I see the equation for effective potential derived anywhere in the Wikpedia article using your interpretation of E and m.

..because the wiki page is most likely wrong in the section calculating the potential. This might explain the discrepancy that you were trying to find. Either way, Ichpublished a more correct formula.
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. :-)

 

[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&lt;&lt;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&lt;&lt;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:

<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.