Circular Orbits of a Black Hole

[mysearch]

In part, this thread is a tangential extension of an earlier thread entitled `Questions on Effective Potential`. However, this thread now moves the discussion in the direction of interpreting the results of the effective potential in terms of the circular orbital velocity, as derived from the Schwarzschild metric.

The difference between the 2 effective potential (Veff) plots attached to this post is not really that important to this new thread, the main point to highlight is that the shape of the max/min curve is essentially identical, although the vertical scales differ. Both Veff curves correspond to a given value of angular momentum [L=7.42E12]. The value was selected as suitable based on the assumption the L>3.4642GMm/c. The max and min points of the curve correspond to a quasi-stable inner orbit and a stable outer orbit, while the graphs allude to the approximate value, the actual values are said to be calculated from the equations:

$$r_{outer} = \frac {a^2}{Rs}\left(1+\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

$$r_{inner} = \frac {a^2}{Rs}\left(1-\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

where $$a = \frac{L}{mc}$$

See 3rd attachment shows a graph against different values of normalised [L]
Details of the derivation can be found on the Wikipedia site:
http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

For the example value of angular momentum [L=4.2GMm/c] the outer and inner radius, as a ratio to the Schwarzschild radius [Rs], are 6.9 and 1.92 respectively. However, one of the questions being raised to the PF forum concerns the velocity that corresponds to these radii. Clearly, at these radii, the orbit is subject to relativistic effects due to gravity and orbital velocity. How these effects apply is assumed to depend on the observer:

1. Distant [dt]
2. Stationary/shell at [r]
3. Orbiting at [r] $$d\tau$$

As a generalisation, the assumption is that distant observer is essentially unaffected by gravity or velocity, while the stationary (shell) observer is affected by gravity. However, the orbiting observer is affected by both gravity and velocity. So the first question is:

Who measures the radii [r] calculated from the equations?

The answer to this question is required in order to determine the frame of reference of the angular momentum. For a circular orbit, the normal assumption is that [L=mvr]. Therefore, by rearranging, we can determine a velocity [v=L/mr]. However, if we apply this equation to both radii produced above, the outer orbital velocity [v=9.10E7], while the implication is that the inner velocity would exceed the speed of light [c] at [v=3.28E8]. Taking another approach, we might try solve the Schwarzschild metric for $$[d\phi/dt]$$ and $$[d\phi/d\tau]$$, which is assumed to correspond to the distant and orbiting observers. The following solutions are assumed, although only the $$[d\phi/dt]$$ case has been checked:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

Again, using the value in the specific example cited leads to a value of v=8.35E7 for the outer orbit, as perceived by the orbiting observer and the shell observer. This assumption is based on the fact that these two observers sit at the same radius, i.e. same gravity effects, and therefore are only separated by velocity. Velocity under special relativity is assumed to be invariant to both observers. The velocity with respect to the distant observer appears to come out at [v=8.05E7]. If these equations are applied with [r=inner radius] then the respective velocities are [1.78E8] and [1.53E8] and so, at least, fall below the speed of light [c]. However, focusing on just the values calculated for the outer orbit by $$[d\tau]$$ and [dt] using the original assumption [L=mvr], we then appear to get an inconsistency to the value of [L=7.42E12] derived from the relationship [L=N*GMm/c] where [N=4.2] in this example, i.e.

V=8.35E7; L=mvr=6.82E12
V=8.05E7; L=mvr=6.56E12

So how do you correlate angular moment [L] to velocity and who measures it?

Initially, I assumed that although the distant observer is in flat spacetime, other results from the Schwarzschild metric suggest that measurements taken by this observer are distorted by the gravity and velocity effects acting on spacetime at [r]. This leaves the other 2 observers sitting at the same radius [r], although they are separated by the orbital velocity. Now relativity seem to suggest that time with respect to the distant observer will run slower, while distant in the radial direction will be expanded due to gravity. The addition of a relativistic velocity would cause time for the orbiting observer to appear to run even slower with respect to both the distant and shell observer, while the velocity causes the circumference distant to be contracted maintaining the invariance of velocity. However:

Would this change the perception of [r] based on $$r=C/2\pi$$?

Sorry if this might appear as a very convoluted set of questions, but would appreciate any help in trying to understand exactly how these issues are resolved by the standard text. Thanks

 

[mysearch]

Missing attachments

Not sure what happen to the attachments. They were there on preview. Here goes again:

 

[Jorrie]

For the example value of angular momentum [L=4.2GMm/c] the outer and inner radius, as a ratio to the Schwarzschild radius [Rs], are 6.9 and 1.92 respectively. However, one of the questions being raised to the PF forum concerns the velocity that corresponds to these radii. Clearly, at these radii, the orbit is subject to relativistic effects due to gravity and orbital velocity. How these effects apply is assumed to depend on the observer: ...

Your equations and interpretations are correct as far as I can establish.

You asked:

Who measures the radii [r] calculated from the equations?

It does not matter in Schwarzschild coordinates, as long as one understand it as a circumferential radius $r=C/2\pi$ and the observer is either static in the field, or in orbit around the body. The length contractions and time dilations cancel out in the measurement of the circumference C.

The following solutions are assumed, although only the $$[d\phi/dt]$$ case has been checked:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

I think both are correct. Any Schwarzschild transverse velocity $$(r d\phi/dt)$$ transforms by a factor $$1/\sqrt{1-Rs/r}$$ to the get the local transverse velocity.

-J

 

[mysearch]

Response to Jorrie

Apologises for not responding sooner, but I didn’t appear to get any email notification. I agree that the velocity equations do appear to be supported by derivation from the Schwarzschild metric. However, in my mind, who measures a given radius [r], which is then input into the equation [L=mvr] would seem to be important to determining the frame of reference of the angular momentum. One of the problems I am having is that I could not get the value of [L] calculated from [mvr] to correlate to radius [r] or velocity [v] of any of the observers cited.

To summarise, at one level, the effective potential (Veff) defines an outer radius [r], which you can then calculate a velocity [v]. However, if you input these values into [L=mvr] you don’t get the same value of [L] implied by the effective potential.

 

[Jorrie]

To summarise, at one level, the effective potential (Veff) defines an outer radius [r], which you can then calculate a velocity [v]. However, if you input these values into [L=mvr] you don’t get the same value of [L] implied by the effective potential.

You must remember that effective potential is an energy and orbital energy is observer dependent. The fact that $L/m = r^2 d\phi/d\tau$ tells us that as well, because $d\tau/dt$ varies with observer in Schwarzschild coordinates.

The Schwarzschild radial parameter r is however observer independent, by its definition as a circumferential radius.

-J

 

[yuiop]

Apologises for not responding sooner, but I didn’t appear to get any email notification. I agree that the velocity equations do appear to be supported by derivation from the Schwarzschild metric. However, in my mind, who measures a given radius [r], which is then input into the equation [L=mvr] would seem to be important to determining the frame of reference of the angular momentum. One of the problems I am having is that I could not get the value of [L] calculated from [mvr] to correlate to radius [r] or velocity [v] of any of the observers cited.

To summarise, at one level, the effective potential (Veff) defines an outer radius [r], which you can then calculate a velocity [v]. However, if you input these values into [L=mvr] you don’t get the same value of [L] implied by the effective potential.

Hi mysearch,

I tend to agree with Jorrie's comments. R should always be calculated as circumference/(2Pi). If a circular orbit is available the circumference can be measured by a local observer who measures the velocity of an orbiting sattelite and then multiplies the orbital velocity by the orbital period. This figure will agree with the circumference (and the implied radius) by an observer at any other altitude calculated from his measurement of the orbital velocity and period of the same sattelite. Where a circular orbit is not available (below the photon orbit radius) circumference can be measured (in principle) by a local ruler. Trying to measure radius by timing vertiacal signals or direct vertical measurement with a local ruler will give very inconsistent results.

Using L=mvr for the angular momentum seemed to cause problems with the energy associated with angular motion at infinity and may well be causing the problem you are having here. Probably best to stick with $d\tau/dt$ as Jorrie suggested.

 

[mysearch]

Who measures the radii [r] calculated from these equations?

I agree that effective potential is observer dependent, our previous discussions have shown that Veff is a function of angular momentum [L], mass [m] and radius [r] and if we assumed that [L=mvr] for a circular orbit, then Veff is also a function of [v,m, r]. However, I am not sure about the statement:

“The Schwarzschild radial parameter r is however observer independent, by its definition as a circumferential radius”.

I would have thought the circumference would be observer dependent as it depends on the velocity of observer, i.e. relativistic velocity will cause length contraction in the direction of motion. Picking up on Kev’s statement:

“If a circular orbit is available the circumference can be measured by a local observer who measures the velocity of an orbiting satellite and then multiplies the orbital velocity by the orbital period”

I agree, but want to highlight a caveat for verification. A stationary `shell’ observer sits at the same radius [r] as the `orbiting` observer. While both will agree on the orbital velocity [v] due to the invariance of velocity under special relativity, they disagree on the orbit time and circumference. As such, both have a different perception of the radius [r]. I believe we can predict the effects on each of my observers by applying one or both of the following assumptions as appropriate to their position:

o As velocity [v] approaches the speed of light [c], time slows and length contracts in the direction of motion, at least, with respect to a `stationary` observer.

o On approaching a gravitational mass [M], time slows and length expands in the direction of gravitational pull, as a function of radius [r], at least, with respect to a `distant` observer.

If so, then all observers measure a different radius [r]. As indicated, velocity will be invariant between the shell and orbiting observer, but this is not true for the distant observer who will determine a different velocity. Presumably, mass [m] is also affected by the relativistic effects of velocity [v]? I am not sure if there is any relativistic effect on mass [m] due to gravity? Now while the following equation for Veff is not one normally used, I believe it is equivalent in form:

$$Veff = \left[1/2mv_o^2-\frac{GMm}{r} \left(1+\frac{vo^2}{c^2}\right)\right]$$

So can any of our observers plug in their values of [m,v & r] and determine a Veff curve that has max/min that correlates to the values predicted by the equations:

$$r_{outer} = \frac {a^2}{Rs}\left(1+\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

$$r_{inner} = \frac {a^2}{Rs}\left(1-\sqrt{1-\frac{3Rs^2}{a^2}\right)$$

where $$a = \frac{L}{mc}$$

As such, we may have we arrived back to the question in my first posting relating to the outer and inner radii forwarded by the two equations produced by the quadratic solution in the Wikipedia reference, i.e.

Who measures the radii [r] calculated from these equations?

 

[Jorrie]

I agree, but want to highlight a caveat for verification. A stationary `shell’ observer sits at the same radius [r] as the `orbiting` observer. While both will agree on the orbital velocity [v] due to the invariance of velocity under special relativity, they disagree on the orbit time and circumference.

They disagree on orbit time but agree on the circumference, because surprisingly they don't agree on velocity.

The distant observer measures circular orbital velocity at r as $v_d = \sqrt{m/r}$

The shell observer measures circular orbital velocity as $v_s = \sqrt{m/(r-2m)}$

The orbiting observer measures circular orbital velocity as $v_o = \sqrt{m/(r-3m)}$

simply due to their respective proper times $d\tau/dt$ that differ. Remember that the observers are in curved spacetime where they cannot simply measure a Doppler shift and so determine their relative velocity. They are forced to lay a set of static rods, say of proper length L each, along the orbit and measure velocity relative to that using their own clocks. That fixes the circumference for everybody, I would think.

-J

 

[mysearch]

Thanks Jorrie, let me check out your equations

 

[mysearch]

Puzzling?

I have found a derivation of sorts in Taylor & Wheeler, which supports your statement regarding the shell observed velocity. I have not had time to work through the details; but will assume it is correct. However, I haven't yet resolved the apparent anomaly of why the separation between the orbiting observer and the shell observer is not explained by special relativity.

Just to restate this issue, I assumed that because these observers were both positioned at the same radius [r] the gravitational effects on time and space must be equal, leaving only the effects of special relativity to be taken into account. While the velocity of the orbiting observer causes an additional time dilation, it would also cause a comparable length contract of the orbit circumference. So while time is running slower, the circumference distant is shorter, leading to the velocity being invariant between these two observers. While, I have yet to resolve this anomaly in my mind, it would be very useful to clarify a few point on your statement of the equations:

The distant observer measures circular orbital velocity at r as
$v_d = \sqrt{m/r}$
The shell observer measures circular orbital velocity as
$v_s = \sqrt{m/(r-2m)}$
The orbiting observer measures circular orbital velocity as
$v_o = \sqrt{m/(r-3m)}$

I am assuming these equations are in geometric units, but are the distant observer velocity $$r d\phi/dt$$ and orbiting velocity $$r d\phi/d\tau$$ still meant to correspond to:

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

If so, are the following equations equivalent in geometric units (?):

The first equation seems compatible in form to the distant observer:

$$c\sqrt{\frac{Rs}{2r}} = c\sqrt{\frac{2GM/c^2}{2r}} = c\sqrt{\frac{GM}{rc^2}} -> \sqrt{\frac{M}{r}}$$

I haven’t actually derived this equation from the Schwarzschild metric yet, but noted its equivalence to the classical equation:

$$\frac{mv_o^2}{r} = \frac{GMm}{r^2}$$

$$v_o = c\sqrt{\frac{GM}{r}} = c\sqrt{\frac{Rs}{2r}}$$


However, had trouble in matching the orbit velocity equations, i.e. not sure where the 3M comes from:

$$c\sqrt{\frac{Rs}{2r-Rs}} = c\sqrt{\frac{2GM/c^2}{2r-2GM/c^2}} = c\sqrt{\frac{GM}{rc^2-GM}} -> \sqrt{\frac{M}{r-M}} (?)$$

Finally, want to confirm the equivalence of the form of the shell velocity equation:

$$v_s = \sqrt{m/(r-2m)} = c\sqrt{\frac{GM/c^2}{r-2GM/c^2}} = c\sqrt{\frac{Rs}{2(r-Rs)}} (?)$$

I recognise that I now need to sit down and work through the issues, but would appreciate any help/references, which explain the apparent anomalies.

 

[Jorrie]

I am assuming these equations are in geometric units, but are the distant observer velocity $$r d\phi/dt$$ and orbiting velocity $$r d\phi/d\tau$$ still meant to correspond to:

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

Yes. If you stick to geometric units, so that Rs=2m, it's easy to spot the correspondence.

However, had trouble in matching the orbit velocity equations, i.e. not sure where the 3M comes from:

$$c\sqrt{\frac{Rs}{2r-Rs}} = c\sqrt{\frac{2GM/c^2}{2r-2GM/c^2}} = c\sqrt{\frac{GM}{rc^2-GM}} -> \sqrt{\frac{M}{r-M}} (?)$$

Start with the distant observer's orbital velocity (the Schwarzschild coordinate circular orbital speed) $v_d = \sqrt{m/r}$ and multiply it by $dt/d\tau$, where $d\tau$ represents the propertime of the orbiting observer, giving (in geometric units):

$$v_o^2 = \frac{m/r}{1-2M/r-v_d^2} = \frac{m}{r-3m}$$

Remember this v_o is based on a a fixed number of meter sticks, laid out along the orbit and is not length contracted, because we are working with the time dilated clock of the orbiting observer.

-J

 

[Jorrie]

$$v_o^2 = \frac{m/r}{1-2M/r-v_d^2} = \frac{m}{r-3m}$$

Remember this v_o is based on a a fixed number of meter sticks, laid out along the orbit and is not length contracted, because we are working with the time dilated clock of the orbiting observer. (Note: where v_o is the circular orbiting speed measured by an observer orbiting at r)

On second thoughts, this may not be the proper way of expressing the orbiting observer's own observed velocity, because it means $v_o \rightarrow \infty$ as $r \rightarrow 3m$.

I suppose as the orbiting and static shell observers flash past each other, they can determine their relative speed and both should obtain

$$v_o^2 = \frac{m}{r-2m}$$

If this is correct, then (as 'mysearch' has hinted) the orbiting observer's time is dilated relative to the shell observer's time by a factor

$$\frac{d\tau_o}{d\tau_s}=\sqrt{1-v_o^2} =\sqrt{1-\frac{m}{r-2m}} =\sqrt{\frac{r-3m}{r-2m}}$$

One must then conclude that the orbiting observer's own measurement of the circumference of the orbit is Lorentz contracted and no longer 'r', but rather

$$r_o = r \sqrt{\frac{r-3m}{r-2m}}$$

approaching zero as r approaches 3m (the 'light radius'), which sounds right.(?)

However, this does not detract from the fact that the Schwarzschild radial parameter remains 'r' for all the equations of the metric.

 

[mysearch]

Radius [r] in the metric?

Jorrie, thanks for both inputs. I am hoping to get some time today to work through all the issues you’ve raised in my mind, but will try to respond as soon as possible. However, your last sentence raised one immediate train of thought:

“However, this does not detract from the fact that the Schwarzschild radial parameter remains 'r' for all the equations of the metric.”

Just showing a simple form of the metric for reference, I wanted to clarify some basic assumptions:

$$c^2 d\tau = c^2\left(1-\frac{Rs}{r}\right)dt^2-\left(1-\frac{Rs}{r}\right)dt^{-1}dr^2-r^2d\phi^2$$

It would seem that we have 4 key variables that define the context of the observer in question, albeit described in very simplified terms in the following summary:

[dt]: far-away time in essentially flat spacetime
$$[d\tau]$$: wristwatch time of the observer
[dr]: associated with radial velocity via [dt] or $$[d\tau]$$
$$[d\phi]$$: associated with angular velocity via [dt] or $$[d\tau]$$

Therefore, depending on how you manipulate the basic metric into any given solution, e.g.

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

I would have though the “Schwarzschild radial parameter [r]” would be the radial distance as measured by the observer in question? If so, in the example above, the radius [r] would correspond to that measured by the orbiting observer?

 

[Jorrie]

Just showing a simple form of the metric for reference, I wanted to clarify some basic assumptions:

$$c^2 d\tau = c^2\left(1-\frac{Rs}{r}\right)dt^2-\left(1-\frac{Rs}{r}\right)dt^{-1}dr^2-r^2d\phi^2$$

Haven't you got a spurious $dt^{-1}$ in the second coefficient? I think it should read:

$$c^2 d\tau = c^2\left(1-\frac{Rs}{r}\right)dt^2-\left(1-\frac{Rs}{r}\right)^{-1}dr^2-r^2d\phi^2$$

Therefore, depending on how you manipulate the basic metric into any given solution, e.g.

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

Correct.

I would have though the “Schwarzschild radial parameter [r]” would be the radial distance as measured by the observer in question? If so, in the example above, the radius [r] would correspond to that measured by the orbiting observer?

No, AFAIK, the “Schwarzschild radial parameter [r]” is defined by the metric as the same for all static observers, at least.

 

[mysearch]

Sorry about the typo, don't seem to have edit access to correct it.

Therefore, depending on how you manipulate the basic metric into any given solution, e.g.

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

Correct.

I would have though the “Schwarzschild radial parameter [r]” would be the radial distance as measured by the observer in question? If so, in the example above, the radius [r] would correspond to that measured by the orbiting observer?

No, AFAIK, the “Schwarzschild radial parameter [r]” is defined by the metric as the same for all static observers, at least.

So if I say that r=3*Rs, the orbit velocity $$[v_o]$$ would be 0.44c?
Who perceives this velocity?
Who perceives the orbiting observer to be at a radius r=3Rs?

 

[yuiop]

Sorry about the typo, don't seem to have edit access to correct it.



So if I say that r=3*Rs, the orbit velocity $$[v_o]$$ would be 0.44c?
Who perceives this velocity?
Who perceives the orbiting observer to be at a radius r=3Rs?

As Jorrie mentioned, R is determined by a static shell observer who measures the circumference and assumes R= C/(2Pi). Any other static shell observer at any other altitude will agree with that measurement. The same is not true for orbiting observers.

As for your equation:

for r = (3/2)*Rs (the photon orbit) I get

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{3Rs-Rs}} = 0.707c$$

using the other equation for coordinate time rather than proper time

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{2r}}$$

for r = (3/2)*Rs (the photon orbit) I get:

$$v_o = r\left(\frac {d\phi}{dt}\right)=c\sqrt{\frac{Rs}{3Rs}} = 0.577c$$

neither of which seem quite right for a photon.

[EDIT} The observer at infinity perceives the coordinate speed of light at the photon orbit radius to be:

$$c_0 \sqrt{1-(2Rs)/(3Rs)} = 0.577 c_0$$ so the second equation seems OK.

The first equation where dtau would presumably be zero for a photon does not seem to make sense.

 

[Jorrie]

The equation that you gave actually also contains an error that I didn't spot initially:

Should read:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}}$$

So if I say that r=3*Rs, the orbit velocity $$[v_o]$$ would be 0.44c?
Who perceives this velocity?
Who perceives the orbiting observer to be at a radius r=3Rs?

The corrected value is then 0.5c and it is measured by the shell static observer, who also knows that the the orbit is at r=3Rs, because he measures the circumference by his meter rods.

 

[yuiop]

The equation that you gave actually also contains an error that I didn't spot initially:

Should read:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}}$$

Using your corrected equation and r= (3/2)Rs for the photon orbit radius


$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}} = c$$

indicating dtau is the time measured by a local static shell observer rather than the time measured by the clock of an orbiting observer.

 

[Jorrie]

Using your corrected equation and r= (3/2)Rs for the photon orbit radius

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}} = c$$

indicating dtau is the time measured by a local static shell observer rather than the time measured by the clock of an orbiting observer.

Hi kev, yep - if dtau was for the orbiting observer, r should also have been as determined by the orbiting observer, I think.

I have to be off - laters...

 

[mysearch]

Kev, just wanted to clarify a couple of points on one of your recent posts. I was using the following equation for a mass object:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2r-Rs}}$$

Therefore, if I insert r=3Rs, the equation becomes

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{6Rs-Rs}}$$

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{1}{5}} = 0.447c$$

One of Jorrie’s recent posts appears to question this equation by quoting:

$$v_o = r\left(\frac {d\phi}{d\tau}\right)=c\sqrt{\frac{Rs}{2(r-Rs)}}$$

As far as I can see, the equation I have used is derived from the Schwarzschild metric when dividing through by $$d\tau$$, i.e. it is the orbiting velocity. While I thought the equation quoted by Jorrie related to the shell’s observer perception of the orbiting velocity? While I conceded this equation is supported by standard text, I am in the process of trying to check why.

$$v_s = \sqrt{m/(r-2m)} = c\sqrt{\frac{GM/c^2}{r-2GM/c^2}} = c\sqrt{\frac{Rs}{2(r-Rs)}} (?)$$

 

[Jorrie]

As far as I can see, the equation I have used is derived from the Schwarzschild metric when dividing through by $$d\tau$$, i.e. it is the orbiting velocity. While I thought the equation quoted by Jorrie related to the shell’s observer perception of the orbiting velocity? While I conceded this equation is supported by standard text, I am in the process of trying to check why.

$$v_s = \sqrt{m/(r-2m)} = c\sqrt{\frac{GM/c^2}{r-2GM/c^2}} = c\sqrt{\frac{Rs}{2(r-Rs)}} (?)$$

This derivation is correct for both shell and orbiting observers. You have divided through by the dtau of the shell observer, but the orbiting observer's velocity is in any case the same as the shell observer's, as we have determined earlier.

 

[mysearch]

Velocity & Effective Potential?

Over the previous posts, a number of equations have been forwarded regarding the circular orbital velocity. I believe there now appears to be agreement that the shell and orbiting velocities are the same due to the invariance of velocity under special relativity. However, I would also like to correlate the results to the max/min effective potential curve attached in post #2 and the radii calculated from the first set of equations in post #1.

I have attached two files; the first (Veff1.jpg) shows the data corresponding to a specific effective potential curve with a value of angular momentum [L=7.42E12]. I have used the Wikipedia variant because I want to cross-reference this derivation in a subsequent post, although the max/min results correspond to the Fourmilab variant. I have also highlighted the max/min radii that are confirmed by the first set of equations in post #1.

The second attachment (Veff2.jpg) shows 3 different sets of velocity results based on different equations, which are then used to calculate a value of angular momentum [L=mvr]. Again, the max/min values have been highlighted, but only the results associated with v3 seem to align to the value of angular momentum [L=7.42E12] at both the max/min radii. The equations for [v1, v2 & v3] were:

$$v1 = c\sqrt{\frac{Rs}{2(r-Rs)}}$$

$$v2 = c\sqrt{\frac{Rs}{(2r-Rs)}}$$

$$v3 = c\sqrt{\frac{Rs}{(2r-3Rs)}}$$

However, before outlining the derivation of [v3], I wanted to verify my understanding of linear and angular velocity in connection with relativistic orbit velocities. This is done in my next post followed by the derivation in a 3rd post.

 

[mysearch]

Linear & Angular Velocity

It is the assertion of special relativity that time dilates and length contracts as a function of velocity, at least, as perceived by a stationary observer. If we only consider circular orbits, we may define the classical relations between linear velocity [v] and angular velocity [w] via the simple equation: w = v/r, where [r] corresponds to the radius of the orbit. For example, we might consider an object orbiting a central point at radius [r], where [r = 1] and the linear velocity to be 6.284m/s. Given that we normally assume that the circumference of a circle equals 2pr, then the circumference in our example is 6.284 and, at the stated velocity, the orbit time would be 1s.

But what happens when we introduce a relativistic velocity?

If we consider the effects when [v=0.866c], which conveniently gives a relativistic factor [g] of 2, the onboard time [dt] would dilate or tick at half the rate of a stationary clock and, likewise, the distance would contract by half. Now, in terms of linear velocity, the two effects cancel out and the velocity remains invariant to both orbiting and stationary frames of reference.

But what of angular velocity?

Angular velocity [w] is measured in degrees or radians of rotation per sec. It has been assumed that [w = v/r] and linear velocity [v] is invariant under special relativity. If we transpose the equation for the circumference of a circle, i.e. C=2pr, we get that r=C/2p. However, the orbiting measure of circumference is affect by length contraction and, in our example, the circumference would be half that of the stationary observer. As such, an orbiting observer might calculate the radius as r=0.5 in comparison to the stationary observer measure of r=1. Therefore, given [w = v/r], the angular velocity would double for the orbiting observer in our example.

How does this affect the conservation of angular momentum [L]?

On the assumption that the angular momentum for a circular orbit is defined by [L=mvr], we have shown that radius [r] is dependent on the observer. It might also be assumed that the mass also increases as a function of linear velocity (?). If so , we might modify the expression of angular momentum to:

$$L = (\gamma m)* v * (r/\gamma) = mvr$$

$$L = (\gamma m) * (r/\gamma )^2 * (\gamma \omega) = mr^2 \omega$$

For a stationary observer [g=1], but increases to [g=2] for our orbiting observer, thus the value of [L] remains constant. As such, angular momentum appears to be conserved between the two frames of reference and supports the use of L=mvr.

 

[mysearch]

Derivation of Velocity [v3] Equation

The premise of the derivation is that the effective potential curve should align to the angular momentum [L=mvr]. The derivation starts with the form of the effective potential equation in Wikipedia:

http://en.wikipedia.org/wiki/Kepler_...ral_relativity

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

This form is modified by substituting for:

$$Rs = \frac{2GM}{c^2}$$

$$Vr = \frac{L^2}{2mr^2} - \frac{Rs mc^2}{2r} - \frac{Rs L^2}{2mr^3}$$

However, F=dVeff/dr and the force is zero for a stable circuit orbit:

$$F = -\frac{2L^2}{2mr^3} + \frac{Rs mc^2}{2r^2} + \frac{3Rs L^2}{2mr^3} = 0$$

$$\frac{L^2}{mr^3} = \frac{Rs mc^2}{2r^2} + \frac{3Rs L^2}{2mr^3}$$

If [L=mvr] is now substituted, the equation is rationalise to:

$$\frac{mv^2}{r} = \frac{Rs mc^2}{2r^2} + \frac{3Rs mv^2}{r^2}$$

Finally, with a little more rearranging, I believe we arrive at:

$$v = c \sqrt {\frac{Rs}{2r-3Rs}}$$

The attraction of this equation, at least at first glance, is that the results of the effective potential curve appear to align to angular momentum [L=mvr]. However, there is an anomaly with the velocity calculated at the minimum radii [r/Rs=1.92] in that the light speed threshold appears to be crossed at [r/Rs=2]

I don’t know if all the assumptions are correct, but would be interested in any thoughts on the issues raised. Thanks

 

[mysearch]

Typo Corrections

Sorry: noticed some typo’s on the previous posts:

Post #23: Forgot to latex symbol for $$\pi$$ & $$g=\gamma$$
$$C=2pr -> C=2\pi r$$

Post #24: Is a bit more serious as I dropped a power when differentiating from Veff to Force in the last term, although it corrected itself when L=mvr was substituted

$$F = -\frac{2L^2}{2mr^3} + \frac{Rs mc^2}{2r^2} + \frac{3Rs L^2}{2mr^4} = 0$$

$$\frac{L^2}{mr^3} = \frac{Rs mc^2}{2r^2} + \frac{3Rs L^2}{2mr^4}$$