I Black hole orbital mechanics questions

[alexchamp29]

What are some of the challenges associated with calculating orbital parameters of objects around a black hole (I.e. Orbital velocity, period, semi major axis, kinetic energy). At what point can classical physics no longer provide accurate results?

 

[mathman]

When near a black hole, general relativity must be used.

 

[alexchamp29]

When near a black hole, general relativity must be used.

So is there a distance or mass at which classical mechanics is too inaccurate and GR is used?

 

[PeroK]

So is there a distance or mass at which classical mechanics is too inaccurate and GR is used?

It depends on the accuracy you require. The precession of the perihelion is not predicted by Newtonian gravity, for example. If you want to calculate that you must use GR. Note that it doesn't have to be a black hole. Planetary orbits round the Sun precess slightly.

For a non-rotating black hole the effective potential obtained from GR has an additional term over the Netwonian potential. In general, if that term is significant then you need to use GR.

For rotating black holes, the solution is further removed from the Newtonian approximation.

Also, if you wanted to model the relative motion of the Milky Way and Andromeda galaxies, then you'd have to use GR.

 

[alexchamp29]

So is there an equation in GR for finding orbital velocity?

 

[Nugatory]

So is there an equation in GR for finding orbital velocity?

You may not realize it yet, but at this point you're asking about the geodesics (that is, paths followed by an object in free fall) in the Schwarzschild solution (gravity around a single massive body like the sun) of the equations of general relativity. As @PeroK has said, the equations are the same whether we're orbiting a black hole or not; it's just that they don't differ much from the Newtonian equations except in very strong gravitational fields.

This thread might get some more answers in the relativity subforum, let' try it there.

 

[alexchamp29]

Thanks for the reply and move. I just skimmed the wiki page for geodesics in GR. Looks like this is where my adventure ends lol. I'm a second year chemistry student and was curious about a scenario involving heavy particles and compact objects. I wanted to calculate kinetic energy of particles moving around aBH but I only knew how to do it with classical mechanics. I was hoping to attempt to redo it with relativistic corrections, but after looking at the math involved I think I bit off more than I can chew.

 

[Ibix]

It's not insanely difficult, although the maths to derive the geodesics is complicated. The major problem is nailing down who is measuring what and where they are. For example, kinetic energy is a frame-dependant quantity. GR usually won't let you get away with talking about "the" kinetic energy as if it had a unique value, while Newtonian physics provides a standard interpretation for the term.

If you describe your scenario, we may be able to advise.

 

[alexchamp29]

The frame dependant factor may be what really complicates this then.
Well, I began thinking about trans-plutonic elements which we know as artificial. These elements are too heavy to have been created by super novae, cosmic ray spallation, or s/r processes. As far as we know, they only exist in man-made cyclotrons and some nuclear reactors.
I began wondering what kind of natural, high energy collisions could take place in the cosmos and if any might be capable of synthesis of these 'artificial' elements.
I imagine some extremely compact object (I guess it doesn't have to be a BH) around which a heavy element (particle A) is in a low but stable orbit. Particle B enters the sphere of influence and begins gaining momentum as it falls toward the compact object. The trajectories line up just right and particle A and B collide.
My classical expression of this was
Ek(A) = √GM/r
Ek(B) = Ep(initial) - Ep(final)

The required energy for fusing two elements is
1/4πεo (Za×Zb/r)
εo = permittivity of free space
r = 10^-15

But I know this is very wrong lol. Again my knowledge of physics and math is very limited. I was just curious about this scenario.

 

[stevebd1]

It's worthwhile working through this yourself but you might also find this post (and the thread) useful-

https://www.physicsforums.com/threads/orbit-velocity-in-schwarzschild-metric.963831/#post-6117557

The equation for gravity for an object with tangential velocity around/near a static BH in GR is-

a=\gamma^2\left[\frac{M}{r^2\sqrt{1-2M/r}}-v^2\frac{\sqrt{1-2M/r}}{r}\right]

in geometric units where M=Gm/c^2, \gamma is the Lorentz transformation \gamma=1/\sqrt(1-v^2) and v is tangential velocity as a fraction of the speed of light (i.e. between 0 and 1). For SI units, multiply a by c^2. In the case of a stable orbit where a=0, \gamma becomes redundant and the equation is rearranged to provide the equation for stable tangential velocity (v_s) as shown in the post in the link above. At a great distance from the source of mass, the equation can be reduced to the Newtonian version-

a=\frac{M}{r^2}-\frac{v^2}{r}

At a great distance from the source of mass, Newtonian equations can be used but the closer you get, GR needs to be applied, especially in the case of neutron stars and black holes though a good example of the subtle effects of GR (as already pointed out) is the precession of Mercury's orbit.

 

[Agerhell]

In the case of a stable orbit where a=0, \gamma becomes redundant and the equation is rearranged to provide the equation for stable tangential velocity (v_s) as shown in the post in the link above. At a great distance from the source of mass, the equation can be reduced to the Newtonian version-

a=\frac{M}{r^2}-\frac{v^2}{r}

.

I think that for circular orbits and Schwarzschild metric you always get the same orbital velocity as in classical Newtonian theory if you use "coordinate time"? I mean if you want to calculate the orbits of objects in the solar system you seldom really care about what time the objects themselves percieve? According to GR a clock on Mercury for example will tick slower than a clock on Earth due to the facts that Mercury moves faster as seen from the solar system barycenter and the fact that is more deeply imbedded in the gravitational field of the Sun than the Earth. For a=0 you according to your first formula get:

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

This is true if you have a clock that is static in relation to the central mass but located at the same gravitational potential as the object in circular motion?

 

[Ibix]

I forgot about this thread - apologies.

But I know this is very wrong lol. Again my knowledge of physics and math is very limited. I was just curious about this scenario.

I think your analysis isn't correct even in Newtonian physics. You seem to be imagining an orbiting particle intersecting a radially infalling object. But the energy in that case isn't completely available for a merger, because the particles don't stop (at least, not in the frame you are working in), so a lot of the energy is carried away as kinetic energy.

A better scenario would be two particles moving in circular orbits in opposite directions. Then if they are identical and collide then they stop and all of the energy is available for the merger. The answer is that this is trivially possible - the closer one gets to an "orbital radius" of $3GM/c^2$, where $M$ is the mass of the hole, the closer the kinetic energies of the particles get to infinity. So there must exist a circular orbit where two counter-orbiting particles would have enough energy to fuse.

However, that leads to the question of exactly where these particles got their energy from. They're just spinning round in circular orbits, and then they collide. Why are they in those orbits?

Perhaps a better question is, what happens if we drop two particles from infinity in orbits that graze that $3GM/c^2$ radius, and collide there? If so, I think the energy available (from the loss of gravitational potential) is $2\sqrt{3}mc^2$ (edit: $2(\sqrt 3-1)mc^2$ actually - see post #14), where $m$ is the mass of one particle. If that's enough for fusion, you'll get fusion.

 

[Agerhell]

I forgot about this thread - apologies.

Perhaps a better question is, what happens if we drop two particles from infinity in orbits that graze that $3GM/c^2$ radius, and collide there? If so, I think the energy available (from the loss of gravitational potential) is $2\sqrt{3}mc^2$, where $m$ is the mass of one particle. If that's enough for fusion, you'll get fusion.

I think the energy of a nonmoving object in the Swarzschild approximation is:

E=mc^2(1-\sqrt{1-\frac{2GM}{rc^2}})

In that case you get a total energy loss in the collision of \frac{2mc^2}{\sqrt{3}}

In any case you can not get more energy out from the collision than $2mc^2$ which is the energy at infinity.

 

[Ibix]

I think the energy of a nonmoving object in the Swarzschild approximation is:

I'm not approximating anything. I did, however make a mistake.

An infalling particle has four velocity $u^\mu$. If it collides with another on a reciprocal heading and the resulting system is stationary in whatever coordinate system we are using then the four velocity of the system is zero except for the timelike component, which is $2u^t$. I set up my experiment deliberately so that the resulting system was stationary in Schwarzschild coordinates and so that the collision happened at $r=3GM/c^2$ (strictly, infinitesimally above that). The total energy of the system is, therefore $2m\sqrt{g_{tt}}u^t$.

Carroll's equation 7.43 tells us that $u^t=E/(1-2GM/c^2r)$ (Carroll uses $c=1$), where $E$ is the energy per unit mass of one particle at infinity, so just $c^2$ for a particle dropped from rest at infinity. And $g_{tt}=1-2GM/c^2r$. That gives an expression for the total energy of the final system of $2mc^2/\sqrt{1-2GM/c^2r}$, which comes out to $2\sqrt 3mc^2$ as I stated above.

The mistake I made was failing to subtract the rest mass energy of the particles before collision - only the kinetic energy is available to overcomb the Coulomb repulsion. So the available energy for this is only $2(\sqrt 3-1)mc^2$.

E=mc^2(1-\sqrt{1-\frac{2GM}{rc^2}})

In that case you get a total energy loss in the collision of \frac{2mc^2}{\sqrt{3}}

I'm not sure where these formulae come from. The first one looks like mass energy less gravitational potential, but I don't immediately see how that helps here. Part of the problem with this whole topic is being certain that the quantities you are using are relevant to the physics at hand. I don't think I've made a mistake, but we clearly don't agree. Perhaps you could post your reasoning in more detail.

In any case you can not get more energy out from the collision than $2mc^2$ which is the energy at infinity.

Why is the energy at infinity relevant except as an initial condition? We aren't working at infinity and our fused nucleus drops into the black hole.

 

[Agerhell]

Why is the energy at infinity relevant except as an initial condition? We aren't working at infinity and our fused nucleus drops into the black hole.

I thought that at infinity you have the energy $E=mc^2$ for each of the two particles. After the collision you have $E=mc^2\sqrt{1-\frac{2GM}{rc^2}}$ for each of the two particles. Inserting $r=3GM/c^2$ into the latter expression you get $E=\frac{mc^2}{\sqrt{3}}$ for each of the two particles. The difference in energy must be lost as heat in the collision and we have two particles, thus an energy lost in the collision of $2mc^2-2mc^2/\sqrt{3}$. According to that discussion I should have written that the total energy loss in the collision should be $2mc^2(1-\frac{1}{\sqrt{3}})$. I read "we drop two particles from infinity" but of course as you wrote in order for the dropped particles to have an orbit that goes to $r=3GM/c^2$ and back up again you need to have some initial velocity/angular momenta that will contribute to the total energy available in the collision.

 

[Agerhell]

Carroll's equation 7.43 tells us that $u^t=E/(1-2GM/c^2r)$ (Carroll uses $c=1$), where $E$ is the energy per unit mass of one particle at infinity, so just $c^2$ for a particle dropped from rest at infinity. And $g_{tt}=1-2GM/c^2r$. That gives an expression for the total energy of the final system of $2mc^2/\sqrt{1-2GM/c^2r}$, which comes out to $2\sqrt 3mc^2$ as I stated above.

I realized we are both right. At a lower gravitational potential the perception of energy is different. People seldom talk of "proper energy" and "coordinate energy" but that is basically what it is. My expression is correct for a distant observer and your expression is correct for an observer at the same gravitational potential as the place of the collision. However as energy levels for an atom or a molecule are lower deeper in a gravitational potential your expression makes more sense if we are dealing with the amount of energy reguired to initate nuclear reactions at the place of collision.

 

[PeterDonis]

nergy levels for an atom or a molecule are lower deeper in a gravitational potential

Not for local observers. And the energy measured by local observers is what's relevant for initiating fusion. The fact that that energy appears redshifted to observers far away doesn't change that.