Specific Orbital Energy at Black holes

In summary: E. it is sometimes called the "energy at infinity". Similarly ~L is the conserved angular momentum. So any orbit will have constant values for ~E and ~L (if we ignore loses due to gravitational radiation).
  • #1
kmarinas86
979
1
If I want determine the specific orbital energy of something orbiting a black hole, will the vis-viva equation still apply? Or will it apply if I replace the euclidean coordinate [itex]r[/itex]with a radial coordinate? None of the above?

If the answer to the second question is yes, then what do I make of <math>a</math>, the semi-major axis? Is it converted to a schwarzschild-type coordinate as well?
 
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  • #2
Orbital velocity in nearly Keplerian orbits?

Hi, kmarinas86,

I bet you just came from the nexus of evil http://en.wikipedia.org/wiki/Vis-viva_equation, or so I guess because you used Wikipedia style markup instead of PF style markup (which is why your symbol [tex]a[/tex] didn't get formatted properly--- left click on the symbol I just wrote to see how I obtained it).

kmarinas86 said:
If I want determine the specific orbital energy of something orbiting a black hole, will the vis-viva equation still apply?

The potential energy term won't be happy because Newtonian potential energy doesn't play nice with curved spacetimes (the kinetic energy of a test particle fares better).

I'll go out on a limb here and guess that you want to compute the (square of the) orbital velocity of a test particle in a quasi-Keplerian bound orbit around a spherically symmetric nonrotating massive object, as modeled using the Schwarzschild vacuum solution of the EFE. If so, I think you immediately run smack into the problem we have been discussing in some other recent threads: the multiplicity of "distances in the large", and thus, "velocities in the large". If you formulate the question very carefully, you should be able to find an answer: the problem lies in deciding what you want to compute and how to interpret the result of said computation. Or rather: the difficulty lies, I guess, in explaining why this is not at all straightforward!

About the "quasi" in "quasi-Keplerian": don't forget that test particle orbits in the Schwarzschild spacetime are not perfectly elliptical; people often say that "the long axis of the ellipse very slowly rotates" which gives the right idea of the so-called "geodetic precession effect" (aka "de Sitter precession"). But don't forget that while the projection of the world lines of orbiting test particles in the Schwarzschild vacuum may arise from bound orbits (i.e. orbits confined to a region near the source of the field), they are certainly not closed curves. That fact that in Newtonian gravitation, at least in the two body idealization, bound orbits traverse closed curves and in fact ellipses, is a very special propery of Newton's theory. As Newton himself pointed out, if you change the exponent in his force law, this property is destroyed.

Chris Hillman
 
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  • #3
Chris Hillman said:
But don't forget that while the projection of the world lines of orbiting test particles in the Schwarzschild vacuum may arise from bound orbits (i.e. orbits confined to a region near the source of the field), they are certainly not closed curves.

In very, very special circumstances, there are closed "spirograph" orbits.

A condition for a closed orbit is that the precession angle divides evenly into an integral multiple of 360 degrees, i.e., n*360/(precession angle) = m, where n and m are integers. If this is true, then the total precession after m aphelia is n complete circles, hence the repetition.

I wrote a Java applet for Scwharzschild orbits that illustrates this and other things.
 
  • #4
Yup

George Jones said:
In very, very special circumstances, there are closed "spirograph" orbits.

Good point, I agree!

(Since my Ph.D. thesis concerned a connection between rational approximation and dynamics, I really should have seen that one coming!)

Chris Hillman
 
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  • #5
kmarinas86 said:
If I want determine the specific orbital energy of something orbiting a black hole, will the vis-viva equation still apply? Or will it apply if I replace the euclidean coordinate [itex]r[/itex]with a radial coordinate? None of the above?

If the answer to the second question is yes, then what do I make of <math>a</math>, the semi-major axis? Is it converted to a schwarzschild-type coordinate as well?

Take a look at http://www.fourmilab.ch/gravitation/orbits/ for the orbital equations of a test particle orbiting a black hole. (Or consult a GR textbook, the equations on this webpage are pretty much taken directly from MTW's "Gravitation").

A conserved energy parameter for the orbit is written in this webpage as
~E. it is sometimes called the "energy at infinity". Similarly ~L is the conserved angular momentum. So any orbit will have constant values for ~E and ~L (if we ignore loses due to gravitational radiation).

Irrelevant latex question: why doesn't
[tex]\stackrel{~}{E}[/tex] work, and how do I fix it?
 
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  • #6
pervect said:
Irrelevant latex question: why doesn't
[tex]\stackrel{~}{E}[/tex] work, and how do I fix it?

Try \tilde{E} which shows up as:

[itex]\tilde{E}[/itex]
 
  • #7
pervect said:
Take a look at http://www.fourmilab.ch/gravitation/orbits/ for the orbital equations of a test particle orbiting a black hole. (Or consult a GR textbook, the equations on this webpage are pretty much taken directly from MTW's "Gravitation").

A conserved energy parameter for the orbit is written in this webpage as
~E. it is sometimes called the "energy at infinity". Similarly ~L is the conserved angular momentum. So any orbit will have constant values for ~E and ~L (if we ignore loses due to gravitational radiation).

How is this?

Givens:

[itex]\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}[/itex]

[itex]dt=\frac{d\tau}{\sqrt{1-2M/r}}[/itex] (is this still applicable here?)

Results:

[itex]\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}[/itex]

[itex]\left(\sqrt{1-2M/r}\right)\frac{dt}{d\tau}=1[/itex]

[itex]\sqrt{1-2M/r}=\tilde{E}[/itex]

Is it really that simple, or did I do something wrong?

Also, after converting natural units to "regular" units, we get:

[itex]\sqrt{1-2GM/rc^2}=\frac{PE_{r=\infty}}{m_0 c^2}[/itex]

Therefore:

[itex]1-r_s/r=\left(\frac{PE_{r=\infty}}{m_0 c^2}\right)^2[/itex]

[itex]1-\left(\frac{PE_{r=\infty}}{m_0 c^2}\right)^2=r_s/r[/itex]

and

[itex]\frac{dt}{d\tau}=\frac{m_0 c^2}{PE_{r=\infty}}[/itex]

?? what did I do wrong ?
 
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  • #8
I think you have made some mistake:

How do you get from [itex]E^2=\left(mc^2\right)^2+\left(pc\right)^2[/itex] to [itex]\frac{E^2}{\left(mc^2\right)^2}=1+\left(pc\right)[/itex] ? That would imply that [itex]{\left(mc^2\right)^2} = pc[/itex], where did that come from?

I could be being ignorant of some relation.
 
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  • #9
Jheriko said:
I think you have made some mistake:

How do you get from [itex]E^2=\left(mc^2\right)^2+\left(pc\right)^2[/itex] to [itex]\frac{E^2}{\left(mc^2\right)^2}=1+\left(pc\right)[/itex] ? That would imply that [itex]{\left(mc^2\right)^2} = pc[/itex], where did that come from?

I could be being ignorant of some relation.

I think you need to refresh your browser cache ;).
 
  • #10
So I do...
 
  • #11
I am not quite sure I follow your derivation, but does knowing that ~E = 1 for a stationary object at r=infinity help?

(1-2GM/rc^2) * (dt/dtau)

is the correct expression for the "energy at infinity" which is conserved. Actually I probably should have called it the "specific energy at infinty", i.e. the energy at infinity per unit mass. That's what the tilde is for.

When I say mass above in "energy per unit mass", I mean invariant mass. The invariant mass of an object does not depend on its state of motion. If you think of a 1 kg test mass like they use at the bureau of standards, a platinium-iridium bar, you can imagine the 1kg mass I mean above as the mass of a physical copy of that platinium-iridium bar.

Note that at r=infinity, for a stationary object dt/dtau = 1, hence the energy at infinity per unit mass is 1 as I said before. So if we have a 1kg platinum iridium standard bar motionless at r=infinity, this equation says that its energy is equal to its mass in geometric units, or mass*c^2 in non-geometric units.

For a rapidly moving object, dt/dtau > 1 and hence the energy at infinity per unit mass is greater than 1. In fact, it is equal to the factor gamma, so that if our bar were moving with a time dilation factor gamma , it's energy at infinity would be gamma times its rest mass (i.e. gamma times its invariant mass).

The correct relationship between dt, dtau, dr, etc. is given by the Schwarzschild metric - I think this may have been what you were trying to express in your second equation, but I'm not sure, i.e:

dtau^2 = (1-2M/r) dtau^2 - dr^2 / (1-2M/r) - r^2 dtheta^2 - r^2 sin^2(theta) dphi^2
 
  • #12
pervect said:
I am not quite sure I follow your derivation, but does knowing that ~E = 1 for a stationary object at r=infinity help?

(1-2GM/rc^2) * (dt/dtau)

is the correct expression for the "energy at infinity" which is conserved. Actually I probably should have called it the "specific energy at infinty", i.e. the energy at infinity per unit mass. That's what the tilde is for.

Yep that's what I got. From the previous post:

kmarinas86 said:
pervect said:
Take a look at http://www.fourmilab.ch/gravitation/orbits/ for the orbital equations of a test particle orbiting a black hole. (Or consult a GR textbook, the equations on this webpage are pretty much taken directly from MTW's "Gravitation").

A conserved energy parameter for the orbit is written in this webpage as
~E. it is sometimes called the "energy at infinity". Similarly ~L is the conserved angular momentum. So any orbit will have constant values for ~E and ~L (if we ignore loses due to gravitational radiation).

How is this?

Givens:

[itex]\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}[/itex]

[itex]dt=\frac{d\tau}{\sqrt{1-2M/r}}[/itex] (is this still applicable here?)

Results:

[itex]\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}[/itex]

[itex]\left(\sqrt{1-2M/r}\right)\frac{dt}{d\tau}=1[/itex]

[itex]\sqrt{1-2M/r}=\tilde{E}[/itex]

Is it really that simple...

The part that agrees is:

[itex]\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}[/itex]

Converting from natural units in regular units, we know that [itex]M[/itex] becomes [itex]M*G/c^2[/itex].

Notice the lack of a square root.

pervect said:
When I say mass above in "energy per unit mass", I mean invariant mass. The invariant mass of an object does not depend on its state of motion. If you think of a 1 kg test mass like they use at the bureau of standards, a platinium-iridium bar, you can imagine the 1kg mass I mean above as the mass of a physical copy of that platinium-iridium bar.

Note that at r=infinity, for a stationary object dt/dtau = 1, hence the energy at infinity per unit mass is 1 as I said before. So if we have a 1kg platinum iridium standard bar motionless at r=infinity, this equation says that its energy is equal to its mass in geometric units, or mass*c^2 in non-geometric units.

For a rapidly moving object, dt/dtau > 1 and hence the energy at infinity per unit mass is greater than 1. In fact, it is equal to the factor gamma, so that if our bar were moving with a time dilation factor gamma , it's energy at infinity would be gamma times its rest mass (i.e. gamma times its invariant mass).

This where my "results" i had ealier disagreed with the standard results. This is why:

kmarinas86 said:
Givens:

[itex]\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}[/itex]

This agrees what with you said, but notice how the RHS (right hand side) is not written gamma. If [itex]\tilde{E}[/itex] was gamma, this would make:

[itex]\frac{dt}{d\tau}=\frac{\gamma}{1-2M/r}[/itex] ?

kmarinas86 said:
[itex]dt=\frac{d\tau}{\sqrt{1-2M/r}}[/itex] (is this still applicable here?)

Results:

[itex]\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}[/itex]

[itex]\left(\sqrt{1-2M/r}\right)\frac{dt}{d\tau}=1[/itex]

[itex]\sqrt{1-2M/r}=\tilde{E}[/itex]

Is it really that simple...

Notice that the third-to-last equation (variant of the first given) is divided by the second to last equation (variant of the second given) to get the first-to-last equation. So either the second given is incorrect for some reason or the results are right.

I'm thinking that I've been mixing "stationary" with "moving", so I'm not sure if what I did was correct.

pervect said:
The correct relationship between dt, dtau, dr, etc. is given by the Schwarzschild metric - I think this may have been what you were trying to express in your second equation, but I'm not sure, i.e:

dtau^2 = (1-2M/r) dtau^2 - dr^2 / (1-2M/r) - r^2 dtheta^2 - r^2 sin^2(theta) dphi^2

Actually :devil: its:

http://upload.wikimedia.org/math/d/0/4/d04cde2566a8e949e0e88b2bd98815e7.png

I know. I know =P.

Still, I think we agree that:

[itex]\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}[/itex] is correct when the energy is at infinity.

and

[itex]\left(1-2M/r\right)\frac{dt}{d\tau}=\tilde{E}[/itex] is correct when the energy is at infinity.

What I just now realized is that (when the object is at [itex]r=\infty[/itex]):

[itex]\left(1-2M/r\right)=1[/itex]

Making:

[itex]\tilde{E}=\frac{dt}{d\tau}[/itex]

But is there a case where both time:

[itex]\frac{dt}{d\tau}=\frac{\tilde{E}}{1-2M/r}[/itex]

from (http://www.fourmilab.ch/gravitation/orbits/) .

and, Gravitational time Dilation

[itex]dt=\frac{d\tau}{\sqrt{1-2M/r}}[/itex]

...can be satisfied (exactly)... or is this impossible?

Are their constraints contradictory?

But apparently(!) this can only hold at infinity if [itex]\tilde{E}=1[/itex].

As for the first equation, why have r, if the equation only applies for infinity?

The weird thing is this, looking at the time part of the first equation, it is the time as perceived from infinity. With the [itex]\tilde{E}[/itex] part, it is energy at infinity.

And if [itex]\tilde{E}[/itex] is equal to [itex]\gamma[/itex], then I'm afraid that the two equations cannot co-exist with each other unless if [itex]\gamma=1[/itex] and [itex]r=\infty[/itex].

Once more, if both equations are correct then [itex]\tilde{E}[/itex] must equal [itex]\sqrt{1-2M/r}[/itex]. If scientists are right about these equations, and IF they can apply for the same test particle, what are we to make of [itex]\tilde{E}=\gamma[/itex]?

Mr. Hillman, are you there? I believe you will know plenty about this... help us resolve this...
 
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  • #13
The two constraints can both be met when r=infinity for ~E=1. For ~E<1, there is some R which satisfies the equations.

The first equation is the equation which defines the energy-at-infinty, a number that is constant for any object following a natural orbit (i.e. geodesic motion).

The second equation appears to me to be the equation for a stationary object (dr, dtheta, dphi=0). At least that's where I think it came from - do you concur?

Sovling both equations is asking where (at what radius) the object is stationary.

This only has a solution when the angular momentum of the object is zero. Think of the Newtonian analogy.,

An object free-falling from some radius R is stationary at r=R in this Newtonian example. An object in an elliptical or hyperbolic orbit is never stationary, but is always moving (you can make the radial component of its velocity zero, but it always has some non-zero velocity if it's orbiting in an elliptical orbit).
 
  • #14
I can't find a convenient web reference, but I think Kmarnis could use a good introduction to the "effective potential" methods in classical mechanics (i.e. not GR). I know there is one in Goldstein, "Classical mechanics", but that's a textbook which might not be convenient (also it's somewhat advanced). All the web references I could find were about the GR case. I could look further, but I have to run and do some shopping.

I think I've made some posts about this in the past in some other forum (Astorphysics?).
 
  • #15
pervect said:
The two constraints can both be met when r=infinity for ~E=1. For ~E<1, there is some R which satisfies the equations.

From the two equations, we determine that:

[itex]r=2M/(1-\tilde{E}^2)[/itex]

pervect said:
The first equation is the equation which defines the energy-at-infinty, a number that is constant for any object following a natural orbit (i.e. geodesic motion).

The second equation appears to me to be the equation for a stationary object (dr, dtheta, dphi=0). At least that's where I think it came from - do you concur?

So for very small velocities (considering roles of [itex]\theta,\phi,r,dr[/itex]), [itex]\frac{dt}{d\tau}=\frac{1}{\sqrt{1-2M/r}}[/itex] is "approximately" correct. This also means that [itex]r=2M/(1-\tilde{E}^2)[/itex] is approximately correct. I'm wondering about circular orbits, which I think would correspond to [itex]r=3M/(1-\tilde{E}^2)[/itex]...

pervect said:
Sovling both equations is asking where (at what radius) the object is stationary.

This only has a solution when the angular momentum of the object is zero. Think of the Newtonian analogy.,

An object free-falling from some radius R is stationary at r=R in this Newtonian example. An object in an elliptical or hyperbolic orbit is never stationary, but is always moving (you can make the radial component of its velocity zero, but it always has some non-zero velocity if it's orbiting in an elliptical orbit).
 
  • #16
To get a sensible approximation, you need to include at least the first velocity term in dt/dtau.

For instance, in a Minkowski space-time in geometric units

dt/dtau = (1-v^2)^(-1/2) [itex]\approx[/itex] 1 + v^2/2

This is just a taylor series expansion.

Here v is the "ordinary" three-velocity. Thus you can say that E is very close to one when you are moving slowly, but you need to add in at least the Newtonian kinetic energy term to get a sensible result - a term proportional to 1/2 m v^2 for a mass m. (Remember that ~E is really specific energy - or energy at infinity - so it's energy per unit mass).

At least, this is what you'd do if you were in flat space-time. If you are far enough away from the black hole (r is large) space-time is almost flat, so you can probalby compute v by familiar seeming methods.Unfortunately we are not in general in a Minkowski space-time, so the notion of velocity isn't quite as simple as my example makes it out to be if you are close to the black hole.

You can take a look at
https://www.physicsforums.com/showthread.php?t=146755

to see how velocities are computed in GR in a local frame-field which is approximately Minkowskian, but I'm not sure how much it will help you.

I'll sketch an outline of another approach below.

In general, you'll have to specify your trajectory somehow. If your trajectory is already parameterized via proper time, you'll have four functions that determine your trajectory.

t(tau), r(tau), theta(tau), phi(tau)

You can then compute dt/dtau directly.

If you happen to have r(t), theta(t), and phi(t), things are more complicated, but you can then use the metric to compute dtau given dr, d theta, and dt. You can then compute dt/dtau. While it will be close to 1, you have to keep at least the fisrt term which makes it different from one to get a reasonable approximation.
 
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  • #17
What is the question?

Hi, kmarinas86,

What are you trying to compute here? Are you trying to compute "invariants of motion" associated with the orbit of a particular test particle in a (bound?) orbit around a static spherically symmetric isolated massive object, as modeled by the Schwarzschild vacuum solution in gtr?

I already pointed out that you should not confuse "effective potential" with "Newtonian gravitational potential". You should probably think of "effective potential" a useful mathematical trick, which can be used in a manner mathematically analogous to how various "potentials" are often used elsewhere in physics, but which is not generally applicable and which does not possesses a general physical significance in gtr (unlike the role of potential energy in Newtonian physics).

Chris Hillman
 
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1. What is specific orbital energy at black holes?

The specific orbital energy at black holes is a measure of the energy per unit mass required for an object to stay in a stable orbit around a black hole. It is determined by the mass and radius of the black hole and the distance of the object from the black hole.

2. How is specific orbital energy related to escape velocity at black holes?

The specific orbital energy at black holes is directly proportional to the escape velocity at the event horizon of the black hole. This means that a higher specific orbital energy is required for an object to escape the gravitational pull of a more massive black hole.

3. Can specific orbital energy be negative at black holes?

Yes, specific orbital energy can be negative at black holes. This occurs when an object is in a bound orbit, meaning it is gravitationally bound to the black hole and will not escape. A positive specific orbital energy indicates an unbound orbit, where the object has enough energy to escape the black hole's gravitational pull.

4. How does the spin of a black hole affect specific orbital energy?

The spin of a black hole can affect the specific orbital energy of objects in orbit around it. A spinning black hole has a larger event horizon, and therefore a greater escape velocity, which requires a higher specific orbital energy for objects to stay in orbit. Additionally, the spin of a black hole can alter the shape of its event horizon, making it more difficult for objects to maintain a stable orbit.

5. How is specific orbital energy calculated at black holes?

The specific orbital energy at black holes is calculated using the equation E = -GM/(2r), where G is the gravitational constant, M is the mass of the black hole, and r is the distance of the object from the black hole's center. This equation takes into account the gravitational potential energy and kinetic energy of the object in relation to the black hole.

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