# Question on Effective Potential

#### mysearch

Gold Member
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

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??)

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

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.

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#### mysearch

Gold Member

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-worldvalues 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

Gold Member
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.

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#### yuiop

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

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

and noting that dr/dt is a radial velocity

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

So
$$V^{2} = c^{2} + \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}$$

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

$$V^{2} = c^{2} + \frac{2GM}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}$$

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 $$(PE_r)[/itex] is the PE at infinity $( PE_{inf})$ minus the KE. So [tex] PE_r = PE_{inf} - KE = PE_{inf} - \frac{mV^2}{2}$$

If we assume the PE at infinity is $$mc^2$$ then

$$PE_r = mc^2 - \frac{mV^2}{2}$$

--> $$PE_r = m(c^2 - (V^2)/2)$$

Replacing v^2 with the equation derived earlier

$$PE_r = m\left(c^2 - \left( c^{2} + \frac{2GM}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\frac{1}{2}\right)$$

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.

Now divide both sides by m to get the PE per unit mass that fourmilab uses (and multiply by 2 to make things clearer)

$$\frac{2 PE_r}{m}= \left(2c^2 - \left( c^{2} + \frac{2GM}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\right)$$

--> $$\frac{2 PE_r}{m}= \left(c^2 - \frac{2GM}{r} + \frac{L^{2}}{m^{2} r^{2}} - \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\right)$$

which is the fourmilab solution multiplied by a factor of two. (Remember they are using angular momentum per unit mass so the m's disappear)

Neither solution seems to be completely satifactory but I hope that I have been of some help.

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

Gold Member
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.

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#### 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!

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#### mysearch

Gold Member
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.

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#### mysearch

Gold Member
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:

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#### mysearch

Gold Member
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.

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#### mysearch

Gold Member
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 $$Rs=2GM/c^2$$:

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

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
$$V^{2} = c^{2} + \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}$$
This can't be correct since $$\frac{E^2}{m^2c^2}-c^2$$ is not equal to $$c^2$$

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#### mysearch

Gold Member
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

Gold Member
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:

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

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.

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#### mysearch

Gold Member
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?

Gold Member

#### yuiop

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:

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

and noting that dr/dt is a radial velocity

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

So
$$v^{2} = \frac{r_{s}c^{2}}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{r_{s} L^{2}}{m^{2} r^{3}}$$

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

$$v^{2} = \frac{2GM}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}$$

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 $$(PE_r)[/itex] is the PE at infinity $( PE_{inf})$ minus the KE. So [tex] PE_r = PE_{inf} - KE = PE_{inf} - \frac{mv^2}{2}$$

(We can use the Newtonian aprximations because the infalling radial velocity is relativly low)

If we assume the PE at infinity is $$mc^2$$ then

$$PE_r = mc^2 - \frac{mv^2}{2}$$

--> $$PE_r = m(c^2 - (v^2)/2)$$

Replacing v^2 with the equation derived earlier

$$PE_r = m\left(c^2 - \left( \frac{2GM}{r} - \frac{L^{2}}{m^{2} r^{2}} + \frac{2GM L^{2}}{m^{2} r^{3} c^2}\right)\frac{1}{2}\right)$$

$$PE_r = m\left(c^2 - \frac{GM}{r} + \frac{L^{2}}{2m^{2} r^{2}} -\frac{GM L^{2}}{m^{2} r^{3} c^2}\right)$$

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)

$$V = \frac{PE_r}{m} = c^2 - \frac{GM}{r} + \frac{L^{2}}{2m^{2} r^{2}} -\frac{GM L^{2}}{m^{2} r^{3} c^2}$$

When using geometrical units and angular momentum per unit mass the factored equation becomes:

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

Wikipedia seems to have simply assumed the effective potential is simply the negative of the radial kinetic energy.

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 :)

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