Orbits of particles with mass. Effective potential

Recall:
(1)The (length^2) of the 4-velocity is -1.
(2) (Subscript[u, 0])is conserved because the metric is independent of time.
(3) (Subscript[u, 3])(in the ϕ direction) is conserved because the metric is independent of ϕ.
From (1–3), we derived
	(((e^2)-1)/2)=(1/2)(((((d r)/(d τ))))^2)+(Subscript[V, eff])(r),
where (Subscript[V, eff])(r)=-(M/r)+((l^2)/(2(r^2)))-((M (l^2))/(r^3))
Scale r and angular momentum l by dividing by M.

vEff[r_,l_]:=-1/r+((l^2)/(2(r^2)))-(( (l^2))/(r^3));vEffNewton[r_,l_]:=-1/r+((l^2)/(2(r^2)))

plotVEffective[l_,e_,OptionsPattern[PlotRange->Automatic]]:=Module[{r,tp,rl,pr},r=.;tp=r/.Solve[e==vEff[r,l],r][[2;;3]];
pr=OptionValue[PlotRange];
rl=If[Length@Dimensions@pr>1,pr[[1]],{2,40}];Plot[{vEff[r,l],vEffNewton[r,l]},{r,rl[[1]],rl[[2]]},BaseStyle->{FontFamily->"Helvetica",FontSize->Medium},AxesLabel->{"r/M","(Subscript[V, eff])"},PlotRange->OptionValue[PlotRange],Epilog->{Dashed,If[Im[tp[[1]]]≠0,Line[{}],Line[{#,e}&/@tp]]}]
]

plotVEffective[4.3,-.025,PlotRange->{-.05,.05}]

Caption: Effective potential for Einstein's theory (blue) and for Newton's theory (purple)

Orbit exists if (((e^2)-1)/2)>(Subscript[V, eff])(r) for some r, and there are two turning points. The turning points are where (((e^2)-1)/2)=(Subscript[V, eff])(r).

Questions

For an orbit of a planet that just grazes the sun, what is the order of magnitude of r/M?

For the sun,

rSun=6.96*^5;mSun=1.48;
rSun/mSun

When orbits are not possible

Find extrema of (Subscript[V, eff])(r)

Solve[D[vEff[r,l],r]==0,r]

If l>(12^(1/2))M, there are two extrema. If l<(12^(1/2))M, the extrema are imaginary.

plotVEffective[3.5,-.053,PlotRange->{-.06,.02}]

How to find the orbit

Use a new variable u=1/r, and substitute ((d r)/(d τ))=((d u)/(d ϕ))((d ϕ)/(d τ))((d r)/(d u))=((d u)/(d ϕ))(l/(r^2))(-(r^2)) to get
	((((du/(d ϕ))))^2)=(((e^2)-1)/(l^2))+((2M u)/(l^2))-(u^2)+2M (u^3).
Integrate to find u(ϕ). This needs to be done numerically.

To find ϕ(τ), numerically integrate
	d τ=(1/l)((r(ϕ))^2)d ϕ

Particles without mass. Light

For planets, we used
	(u^μ)=(((d t)/(d τ)),((d r)/(d τ)),((d θ)/(d τ)),((d ϕ)/(d τ))).
We could just as well have used 4-momentum (p^μ)=m (u^μ).
For photons, that is not valid because proper time is 0.

Instead of proper time, use a parameter λ. Then
	(p^μ)=(((d t)/(d λ)),((d r)/(d λ)),((d θ)/(d λ)),((d ϕ)/(d λ))).
We will solve a particular orbit and find out what λ is in that case.

Recall:
(1) The length of the 4-momentum p is 0.
(2) (Subscript[p, 0])is conserved because the metric is independent of time. Define 
	e=(Subscript[p, 0])=(1-((2M)/r))((d t)/(d λ)).
(3) (Subscript[p, 3])(in the ϕ direction) is conserved because the metric is independent of ϕ.
Define
	l=(r^2)((d ϕ)/(d λ))
From (1–3), we get
	-(((1-((2M)/r)))^(-1))(e^2)+(((1-((2M)/r)))^(-1))(((((d r)/(d λ))))^2)+((l^2)/(r^2))=0.
	((((e/l)))^2)=(1/(l^2))(((((d r)/(d λ))))^2)+(Subscript[W, eff])(r),
where (Subscript[W, eff])(r)=(1/(r^2))(1-((2M)/r)).

A photon is headed for r=b from the star. The parameter b is called the impact parameter.  sin ϕ=b/r, and cos ϕ ((d ϕ)/(d λ))=-(b/(r^2))((d r)/(d λ)). Rewrite as
	cos ϕ l=-b((d r)/(d λ))
As r->∞, LHS->l and RHS->b e. Therefore b=l/e.

(1/(b^2))=(1/(l^2))(((((d r)/(d λ))))^2)+(Subscript[W, eff])(r),

New Slide

wEff[r_]:=-2(r^(-3))+(r^(-2));

plotWEffective[b_,OptionsPattern[PlotRange->Automatic]]:=Module[{r,tp,rl,pr},r=.;tp=r/.Solve[(b^(-2))==wEff[r],r][[2;;3]];
pr=OptionValue[PlotRange];
rl=If[Length@Dimensions@pr>1,pr[[1]],{2,40}];Plot[wEff[r],{r,rl[[1]],rl[[2]]},BaseStyle->{FontFamily->"Helvetica",FontSize->Medium},AxesLabel->{"r/M","(Subscript[W, eff])"},PlotRange->OptionValue[PlotRange],Epilog->{Dashed,If[Im[tp[[2]]]≠0,Line[{}],Line[{{tp[[2]],(b^(-2))},{rl[[2]],(b^(-2))}}]]}]
]

plotWEffective[10.,PlotRange->{{2,40},{0,.04}}]

When b=10M, photons go to in to r=8.6M and then back out again.

Find the peak of (Subscript[W, eff]).

r/.Solve[D[wEff[r],r]==0,r][[1]]

Solve for b

b=Sqrt[1/wEff[%]]

plotWEffective[3Sqrt[3],PlotRange->{{2,40},{0,.04}}]

If b<(27^(1/2))M, photons go toward the mass and never go back out.

If the sun were a point mass, then the critical impact parameter for capture is

mSun=1.48"km";Sqrt[27]mSun

What is the parameter λ? Path of a radial light ray

e=(1-((2M)/r))((d t)/(d λ)) and l=(r^2)((d ϕ)/(d λ)). What is λ?

We have differential equations for ((d r)/(d λ)) and ((d ϕ)/(d λ)), which we can solve. Do the simple case of an almost radial light ray, for which l≪1. In that case ((((e/l)))^2)=(1/(l^2))(((((d r)/(d λ))))^2)+(Subscript[W, eff])(r) becomes
	(e^2)=(((((d r)/(d λ))))^2).
The solution is
	r=e λ.

Surprise: the parameter λ is not time. For a radial light ray, the parameter λ is the radial coordinate divided by the energy at ∞. (Recall e is the energy at r->∞.)

Calculate the coordinate time.
	e=(1-((2M)/r))((d t)/(d λ))
=(1-((2M)/r))((d t)/(d r))((d r)/(d λ))
For radial paths, (e^2)=(((((d r)/(d λ))))^2). Substitute to get
	d t=±d r (((1-((2M)/r)))^(-1))
=±d r (1+((2M)/(r-2M)))
Use + for outgoing paths and – for incoming paths. Substitute r=e λ to get
	Δ t=e ((Subscript[λ, 2])-(Subscript[λ, 1])+2M log (((Subscript[λ, 2])-2M/e)/((Subscript[λ, 1])-2M/e)))

If the energy of the photon is bigger, the parameter λ changes more slowly as r and t change. However the path r(t) is independent of energy.

New Slide

Calculate the maximum orbital speed of a planet in circular orbit around a point mass.