Some geometry questions re Schwarzschild metric

[Buzz Bloom]

I would like to ask what I hope are two simple questions about what I recognize to be a complicated subject. I did make an effort to search the Internet for the answers, but the two most promising looking sources I found did not help.

https://www.physicsforums.com/threads/null-geodesics-in-schwarzschild-spacetime.895174/
https://preposterousuniverse.com/wp-content/uploads/grnotes-seven.pdf​

The math in these sources is way over my head. There are also issues with not being able to find (by searching the material) definitions of the symbols for some variables. Examples are: italicized λ and L.

Q1: I understand that a mass-less particle moving at velocity c can exist in a circular orbit around a black hole (BH) at radius

R = 2 R_S,​

where R_S is the Schwartzchild radius.
What would the shape of its trajectory be if it passed by the BH with its closest position to the BH being

R = R_S + ε,​

where ε is a very small distance? My guess is that it would be a hyperbola, and I would appreciate knowing if this is correct or not.

Q2. Is a parabolic trajectory possible? If so, would its closest distance to the BH be

R < 2 R_S?​

 

[Ibix]

Q1: I understand that a mass-less particle moving at velocity c can exist in a circular orbit around a black hole (BH) at radius
R = 2 RS,where RS is the Schwartzchild radius.

The circular orbit is at $1.5R_S=3GM/c^2$.

What would the shape of its trajectory be if it passed by the BH with its closest position to the BH being
R = RS + ε,where ε is a very small distance? My guess is that it would be a hyperbola, and I would appreciate knowing if this is correct or not.

In another recent thread, PeterDonis made the point that if there is a circular orbit at lightspeed at some radius then any unpowered orbit dipping below that must fall in, since it would have to be doing more than lightspeed just to maintain orbit. And $R_S+\epsilon<1.5R_S$. If you actually meant to write $R=1.5R_S+\epsilon$ then the orbit would be open, yes. I don't believe that orbits in GR are actually conic sections, unlike Newtonian gravity, but it's certainly the analogous form.

Q2. Is a parabolic trajectory possible? If so, would its closest distance to the BH be

R < 2 R_S?​

Again, I don't believe GR orbits are conic sections. There is, of course, a limiting case between open and closed orbits - these graze $3GM/c^2$, if I've followed the maths correctly.

 

[pervect]

I would like to ask what I hope are two simple questions about what I recognize to be a complicated subject

...

Q1: I understand that a mass-less particle moving at velocity c can exist in a circular orbit around a black hole (BH) at radius

R = 2 R_S,​

where R_S is the Schwartzchild radius.
What would the shape of its trajectory be if it passed by the BH with its closest position to the BH being

R = R_S + ε,​

where ε is a very small distance? My guess is that it would be a hyperbola, and I would appreciate knowing if this is correct or not.

Q2. Is a parabolic trajectory possible? If so, would its closest distance to the BH be

R < 2 R_S?​

If we start a light pulse near r=3M (equivalanetly, near r=1.5 times the Schwarzschild radius $r_s$, as $r_s=2M$) , , I believe it will either spiral outwards or spiral inwards.

The case of interest would probably be a time reversal of the outward spiral for a particle starting at $r=3M+\epsilon$ and spiraling out to infinity. The time-reversal would be a light pulse starting at infinity, and spiraling in.

The trajectories won't be conic sections as they are in Newtonian orbits.

As far as writing the differential equations for the orbit, I'm more familiar with writing them for massive particles. I could probably eventually write the differential equations down, but I'm not sure that'd be helpful. There won't really be that much difference between the two - the orbit of the massless particle and the orbit of a particle of very small mass moving close to the speed of light will be very similar. However, the differential equations for the massive particle are usually written in terms of proper time. The light pulse doesn't have proper time, though it does have an affine parameter that serves the same mathematical function. However, it's less easy to interpret physically.

 

[Buzz Bloom]

If we start a light pulse near r=3M (equivalanetly, near r=1.5 times the Schwarzschild radius rsr_s, as rs=2Mr_s=2M) , , I believe it will either spiral outwards or spiral inwards.

Hi pervect:

Thank you for your post.

To keep this relatively simple, assume the BH has no angular momentum. Then, imagine a photon at a position at

radius r = r_s + ε, and time t = 0,​

and also at the t = 0 moment, about to move along a vector perpendicular to a radial line. Wouldn't its trajectory, moving both forward and backward in time, be symmetrical? It so, then it could not be a spiral. If quadratic forms are not possible with GR, then perhaps it could be asymptotic to a "V" shape, hyperbolic-like, as the trajectory is followed both forward and backward in time. Another conceptual possibility is that the the trajectory might cross itself, and perhaps a finite multiple times. Eventually though, such a trajectory would stop crossing itself again, and it would move toward infinity becoming asymptotically a "V" shape, in both forward and backward time.

Now, it the starting vector at time t = 0 is not perpendicular to a radial direction, then perhaps the trajectory could be a spiral. Would such a spiral necessarily cross the horizon? Or might its spiraling remains outside of the horizon? In the outward direction, would it necessarily spiral indefinitely, it might it at some point continue asymptotically to a line to infinity?

Does any of this make sense?

I apologize for mis-remembering the photon's orbital radius. I find that as I age I more and more have these senior moments when I think I remember something accurately, and I don't bother to recheck it.

Regards,
Buzz

 

[pervect]

MTW's "Gravitation" has the equations for motion of a massless particle on page 674, box 25.7

Geometric units (with G=c=1) are used throughout.

We have for the equations of motion

$$\left( \frac{dr}{d\lambda} \right)^2 + B^{-2}(r) = b^{-2}$$

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

b is a constant of motion called the "impact parameter". For the case of a massive particle, we have conserved quantities E and L, "energy per unit rest mass" and "angular momentum per unit rest mass" that stay constant in the orbit. For a massless particle, we don't have a rest mass, but we can regard b as being the ratio of L/E in the limit as the rest mass of the particle approaches zero. Thus a particle with a high impact parameter has more angular momentum, with enough angular momentum (a high enough value of b), it escapes the black hole. With too little angular momentum, it falls in.

M is the mass of the black hole in geometric units
r is the schwarzschild r coordinate
$\lambda$ is the affine parameter, which replaces proper time for a photon.

Conceptually, the equations of motion for the massless particle are a solution for functions $r(\lambda), t(\lambda), \phi(\lambda)$. Here r, $\phi$, t are the schwarzschid coordinates of the massless particle. The curve through space-time that the particle takes is traced out as $\lambda$ evolves through time. The coordinate time t increases monotonoically as the affine parameter $\lambda$ increases (at least with th sensible sign choices), so every value of time t corresponds to some value of the affine parameter $\lambda$ and vica-versa. However, the differential equations of motion are written so that we consider the evolution of the system as a function of $\lambda$, not as a function of t. $\lambda$ can be regarded as representing time, because it's a monotonic function of t (and vica versa), but it's not the same numerically as the coordinate time t. The proper time doesn't exist for the case of a massless particle, as I mentioned, so all we have are the coordinate time and the affine parameter $\lambda$.Supplementary equations for the orbit are also needed.

$$\frac{d\phi}{d\lambda} = r^{-2} \quad \frac{dt}{d\lambda} = b^{-1}(1-\frac{2M}{r})^{-1}$$

Here $\phi$ is the angle, and t is the schwarzschild coordinate time

For the description of the motion in section B, MTW writes:

"For $0<b-3\sqrt{3}M << M$, the particle circles the star many times ("unstable circular orit) at r $\approx$ 3M before flying back o r=infinity"

And they mention that for $b<3\sqrt{3}M$ the particle falls into the black hole. They don't mention it "spiraling in", but figure 25.6 on pg 677 shows such a spiral.

MTW also has some other equations for the orbit in which they eliminate $\lambda$ and consider r as a function onf $\phi$, but I don't know if they are of interest.

 

[Buzz Bloom]

We have for the equations of motion

Hi pervect:

Thanks for the reference to MTW and also for the content giving the equations of motion.

The only thing that seems to be missing is a definition for "b". It appears that "b" is an independent parameter, or it is a constant. There are three equations for the three derivatives of r, t, and φ with respect to λ. Two of the three include the parameter b. I have been looking at several references (not too reliable) which call "b" the "impact parameter", but I have not found consistent definitions of this parameter, either in English or by a formula.
One formula is

https://physics.stackexchange.com/q...nd-distance-of-closest-approach-of-a-light-ra
b = r₀ / √(1-r_s/r₀)​

Another is

https://arxiv.org/pdf/gr-qc/9907034.pdf
If space-time is asymptotically flat, b is simply the impact parameter which can be expressed
in terms of the radial coordinate of the point (r0, φ0) of closest approach
b = (D(r₀) / A(r₀)) × r₀.​

However, D and A are defined only as functions of "r" in the equation for the line element definition equation for ds². However, it is not clear that the paper is using the Schwartzchild metric.

After some more searching on the Internet, I have confirmed that using the Schwartzchild metric gives:

D/A = 1/√(1-r_s/r₀).​

So, the two references agree.

Regards,
Buzz

 

[pervect]

Hi pervect:

Thanks for the reference to MTW and also for the content giving the equations of motion.

The only thing that seems to be missing is a definition for "b". It appears that "b" is an independent parameter, or it is a constant. There are three equations for the three derivatives of r, t, and φ with respect to λ. Two of the three include the parameter b. I have been looking at several references (not too reliable) which call "b" the "impact parameter", but I have not found consistent definitions of this parameter, either in English or by a formula.

A massive particle in flat space-time will have some linear momentum p, and some energy E.

A masless particle in flat space-time will have a fixed relationship between p and E, E=pc. Or in geometric units, p=E, since c=1. So it has one less degree of freedom than the massive particle. One can think of this extra degree of freedom of the massive particle as being related to its velocity, which is variable for a massive particle, but fixed at c for a massless particle.

To specify the orbit of a massive particle falling into a black hole in a Schwarzschild space-time, we need two parameter, the energy E and the angular momentum L. These have a different definition in General relativity than in Newtonian physics, but conserved quantities with roughly the same physical significance exist in both theories.

Because the massless particle falling into a black hole has one less degree of freedom, there is only one parameter of the particle that needs to be specified for a black hole of given mass. This parameter is b.

The easiest general way to compute b is to pick some point on the trajectory, and use the equation
$$\left( \frac{dr}{d\lambda} \right)^2 + B^{-2}(r) = b^{-2}$$ to compute it. So if you know $dr/d\lambda$ at some point r on the trajectory, and you can compute b. If there is a point of closest approach to the black hole r*, we can conveniently set dr/d\lambda=0 at that point, and relate the impact parameter b to the r-coordinate of closest approach by $b = r* / \sqrt{1-2M/r*}$.

Note that I've only talked about Schwarzschild black holes. The analysis of a rotating black hole would be considerably more complex. I'm not familiar with it, though you might find some PF posts about it.

 

[Buzz Bloom]

Hi @pervect:

I have been working with the equations you posted, and I have put them into a form that I expect will be easier for me to use. I plan to develop on a spreadsheet some graphs showing the relationship (based on initial conditions) between (a) radius and (b) the angle between trajectory and a radial line.

The equation forms are below. If you can spot any errors, I would much appreciate your posting me about them.

Regards,
Buzz

 

[PeterDonis]

@Buzz Bloom please learn to use the PF LaTeX feature. We can't respond usefully to equations you post if we can't quote them, and we can't quote them if they're in an image. The PF rules are clear about this.

Help for the PF LaTeX feature is here:

https://www.physicsforums.com/help/latexhelp/

 

[Buzz Bloom]

We can't respond usefully to equations you post if we can't quote them, and we can't quote them if they're in an image.

Hi Peter:

I apologize for not reading about the LaTeX before this. After a quick perusal, It does not look any more confusing than the feature I use in LibreOffice. I will work on it, and I will re post my last post after I become somewhat comfortable with it.

Regards,
Buzz

 

[Buzz Bloom]

Hi @pervect:

I have been working with the equations you posted, and I have put them into a form that I expect will be easier for me to use. I plan to develop on a spreadsheet some graphs showing the relationship (based on initial conditions) between (a) radius and (b) the angle between trajectory and a radial line.

If you can spot any errors, I would much appreciate your posting me about them.

$(1) { } { } \left( \frac{dr}{d\lambda} \right)^2 = b^{-2} -r^{-2} \left(1-\frac{2M}{r} \right)$
$(2) \frac{d\phi}{d\lambda} = r^{-2}$
$(3) r^2\left( \frac{d\phi}{dr} \right)^2 = {b^{-2} -r^{-2} \left(1-\frac{2M}{r} \right)}$
(1) and (2) are from post #5. (3) is r² times the square of (2) divided by (1).

The following are some new variables I will substitute into (3) to simplify the expressions for my convenience.
$(4) \tau^2 = \tan^2 \theta = r^2\left( \frac{d\phi}{dr} \right)^2$
At radius r, θ is the angle between the trajectory of a light beam and the radial line.
$(5) r_s = 2 M \rho = r/r_s, \beta = b/r_s$

$(6) 1/\tau^2 = {\frac {\rho_2} {\beta^2} - \left(1-\frac{1}{\rho} \right)}$
(6) is from (3) and (4) with substitutions from (5).
This establishes an algebraic relationship between τ and ρ which depends on the variable β. The value of β is determined by the initial condition values for τ and ρ as shown in (7).
$(7) \beta^2 = \frac {\rho_{0}^{2}} {(1/\tau_{0}^{2}) + (1 - 1/\rho_0)}$

Regards,
Buzz

 

[pervect]

In 25.55, MTW gives the equation when one eliminates $\lambda$

$$\left( \frac{1}{r^2} \frac{dr}{d\phi} \right)^2 + \frac{1-\frac{2M}{r}}{r^2} = \frac{1}{b^2}$$

This basically follows from the previous equations if one notes that
$$\frac{dr}{d\phi} = \frac{ \frac{dr}{d\lambda} } { \frac{d\phi}{d\lambda} }$$

They introduce a change of variable, replacing r with a new variable, u(r).

Unfortunately it's not clear what u is from the text. I believe they define u(r) = M/r

Letting $u=Mr^{-1}$ we get
$$ du = -Mr^{-2}dr$$

so
$$\frac{du}{d\phi} = \frac{du}{dr} \frac{dr}{d\phi} = -Mr^{-2} \frac{dr}{d\phi}$$

After rewriting
$$\left( \frac{M}{r^2} \frac{dr}{d\phi} \right)^2 + \left( 1-\frac{2M}{r} \right) \left( \frac{M}{r} \right)^2 = \frac{M^2}{b^2}$$

we can find the differential equation which writes u as a function of $\phi$, namely

$$ \left( \frac{du }{d\phi } \right) ^2 + u^2 \left( 1 - 2u \right) = \left( \frac{M }{b } \right)^2 $$

which they give in the text.

It's a bit simpler looking, but it's written in terms of u, which is inversely proportional to r, rather than in terms of r.

 

[Buzz Bloom]

Hi @pervect:

Thank you for your post.
Your next to last equation on your post #12 directly converts to my (6) with the substitutions from my (4) and (5). My (7) derives from (6) by solving for β².

When I complete my spreadsheet calculations, I will create some graphs for [ρ,θ] pairs for a variety of β values. After that I plan to do some numerical integration to develop some additional graphs for corresponding pairs [ρ,φ] and [x,y] which will be displayed as light trajectories. I expect there will be three kinds.
(a) From θ = 90⁰ closest to ρ=1 (radius r_s) for some φ=0, asymptotically to a radius ρ for some φ < 180^o.
(b) Similar to (a) except the asymptotic radius with be for some φ > 180^o. This means that the outward moving trajectory would cross the inward moving part of the same trajectory at least once.
(c) The inward part of the trajectory would cross ρ=1.

Please let me know if you would like me to post the graphs.

Regards,
Buzz

 

[pervect]

One other thing that might help. Look up the method of "effective potentials" and apply it to the problem at hand.

The basic idea is that if you have a ball rolling along a hill, one can write the (Newtonian) differential equation

$$\frac{1}{2}m \dot{x}^2 + m \, gh(x) = E$$

where x is a position, $\dot{x}$ is a velocity, and gh(x) is a potential function (potential per unit mass).

The point is that one can use this physical problem to gain insight into the solution of the differential equation

$$\dot{u} ^2 + u^2(1-2u) = \frac{M^2}{b^2}$$

by the formal similarities between the two equations, i.e a change of variable x=u and t=$\phi$, setting m=2, letting $gh(u) = \frac{1}{2} u^2(1-2u)$ and $E=M^2/b^2$

One can then leverage one's intuition of how balls roll up hills to understand the solution of the differential equation. Basically, the potential function will have a peak at R=3M, so a ball placed right on the peak will stay there. A ball coming in from r=infinity (u=0) with just the right amount of 'energy' ($M^2/b^2$ in the analogy) can roll right up to the peak of the hill. With a perfect choice of energy the ball could stop there, realistically it's only a metastable state and the ball would roll off the peak in one direction or the other.

This gives insight into the orbits when one plots u vs $\phi$ or r vs $\phi$. During the time when the "ball" is in the metastable state at r=3M, the correspoinding orbit can circle the black hole several times if the metastable state lasts long enough.