Initial conditions for orbits around a wormhole

[happyparticle]

Homework Statement

Find the initial conditions for $u^r$ and $u^\theta$ if a shuttle is launched with velocity $\beta$ at $x(t=0) = 20$.

Relevant Equations

$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$

$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$

Hi,
I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem.

Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$

Where $b=1$ with an orbit only in the equatorial plane.

We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$

Ultimately, I was tasked to find the initial conditions for $u^r$ and $u^\theta$ if a shuttle is launched with velocity $\beta$ at $x(t=0) = 20$.
(the center of the wormhole is located at $x=0,y=0$).

Firstly, I had to find a relation between $\beta$ and one of the invariant.

To answer this question I used $\frac{dt}{d\tau} = e$ and I found $\beta = \sqrt{(1- \frac{1}{e^2}})$, since $e = -g_{tt}\frac{dt}{d\tau}$ and $g_{tt} = -1$.
Where e is one of the invariant.

Secondly , I had to find the ratio $\frac{u^r}{u^{\phi}}$ which is fully defined by the position ($r_0,\phi_0$) in the equatorial plane.

Then , using the relationships above I could find the relation $u^r(t=0)$ in function of $\beta$ and $\phi_0$.

Finally, the equation for $\varepsilon$ allows us to find the second invariant.

I'm wondering if the relationship for $\beta$ and e is correct.
Also, I was completely unable to continue the problem thereafter. I had to plot a graph of the orbits, but after the first part of the problem I have practically did trial and error and obviously my graph made no sense.

I hope my question is clear.

Thanks you!

 

[anuttarasammyak]

Thogh I am a poor layman in GR, I am interested in understanding the question. Initial position and magnitude of initial velocity of the test body given, all we can arrange is the direction of the initial velocity. By that arrangement the test bddy could or couldn't pass the neck of worm hole to escape into the backside world. Do I understand the quesiton correctly ?　　If yes, I assume that the angular momentum matters.

 

[happyparticle]

You do understand the question correctly. I'm not sure if I need to use the angular momentum. Basically, I have to find a relationship with $\beta$, both trajectory invariants and a relationship with $\frac{u^r}{u^{\phi}}$

 

[anuttarasammyak]

I guess that at the neck the wormhole the angular velocity of the body increases by the law of angular momentum conservation, but its velocity cannot exceed light speed. I have found a support to my guess in https://en.wikipedia.org/wiki/Ellis_wormhole There parameter h which is angular momentum per unit mass wrt phi decides go or not go into the other side; small h lets it go but large h does not.

 

[happyparticle]

I'm wondering if the ratio of $\frac{u^r}{u^{\phi}}$ is "simply" $-r_0 cos \phi_0$?

Knowing that $u^r = \gamma \frac{dr}{dt}$, $u^{\phi} = \gamma \frac{d\phi}{dt}$, and $\frac{dr}{dt} = \frac{x v_x + y v_y}{r}$ , $\frac{d\phi}{dt} = \frac{xv_x - yv_y}{r^2}$.

Also, if the object is moving horizontally and start at $x=20$, then $v_x = - \beta$ and $v_y = 0$ and plugging $x = r cos \phi$ and $y = r sin \phi$, we get the ratio above for $u^r(0)/ u^{\phi}(0)$.

 

[Lox99]

Alright louis, straight to it. You’ve got the ultrastatic Morris–Thorne-style metric

ds2=−dt2+dr2+(r2+b2)(dθ2+sin⁡2θ dϕ2),ds^2=-dt^2+dr^2+(r^2+b^2)\big(d\theta^2+\sin^2\theta\,d\phi^2\big),ds2=−dt2+dr2+(r2+b2)(dθ2+sin2θdϕ2),
and you’re launching a shuttle from r0=20r_0=20r0=20 (I’ll treat your x(0)=20x(0)=20x(0)=20 as r(0)=20r(0)=20r(0)=20), with speed β\betaβ measured by the local static observers. Orbit is restricted to the equatorial plane (θ=π/2\theta=\pi/2θ=π/2). The wormhole throat is at r=0r=0r=0 with areal radius r2+b2\sqrt{r^2+b^2}r2+b2.

The invariants (sanity first)​

Because gtt=−1g_{tt}=-1gtt=−1 and the spacetime is static/spherical:

• Energy per unit mass:
  
  E≡−ut=t˙=γ,γ=11−β2  .E\equiv -u_t = \dot t = \gamma,\qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}\;.E≡−ut=t˙=γ,γ=1−β21.
• Axial angular momentum (equatorial, sin⁡θ=1\sin\theta=1sinθ=1):
  
  L≡uϕ (r2+b2)=(r2+b2) ϕ˙.L\equiv u_\phi\, (r^2+b^2)= (r^2+b^2)\,\dot\phi.L≡uϕ(r2+b2)=(r2+b2)ϕ˙.
• Normalization:
  
  −1=gμνuμuν=−E2+(r˙)2+L2 r2+b2 .-1=g_{\mu\nu}u^\mu u^\nu=-E^2 + (\dot r)^2 + \frac{L^2}{\,r^2+b^2\,}.−1=gμνuμuν=−E2+(r˙)2+r2+b2L2.

So

(r˙)2=E2−1−L2r2+b2.(\dot r)^2 = E^2-1-\frac{L^2}{r^2+b^2}.(r˙)2=E2−1−r2+b2L2.
If you insist on the “effective potential” form

ε=12(r˙)2+Veff(r),\varepsilon=\tfrac12(\dot r)^2+V_{\rm eff}(r),ε=21(r˙)2+Veff(r),
the clean choice is

Veff(r)=L22(r2+b2),ε=E2−12.V_{\rm eff}(r)=\frac{L^2}{2(r^2+b^2)},\qquad\varepsilon=\frac{E^2-1}{2}.Veff(r)=2(r2+b2)L2,ε=2E2−1.
That’s the correct relation between your ε\varepsilonε, EEE, and LLL.

Initial conditions from a local launch​

Use the static observer’s orthonormal triad at r0r_0r0:

e^t=∂t,e^r=∂r,e^θ=1r02+b2∂θ,e^ϕ=1r02+b2∂ϕ(at θ=π2).\hat e_t=\partial_t,\quad\hat e_r=\partial_r,\quad\hat e_\theta=\frac{1}{\sqrt{r_0^2+b^2}}\partial_\theta,\quad\hat e_\phi=\frac{1}{\sqrt{r_0^2+b^2}}\partial_\phi \quad(\text{at }\theta=\tfrac{\pi}{2}).e^t=∂t,e^r=∂r,e^θ=r02+b21∂θ,e^ϕ=r02+b21∂ϕ(at θ=2π).
If you launch with 3-speed magnitude β\betaβ and direction angles (ψ,  η)(\psi,\;\eta)(ψ,η) so that the local components are

vr^=βcos⁡ψ,vϕ^=βsin⁡ψ,vθ^=βsin⁡η(but we’ll set η=0 for the equatorial plane),v^{\hat r}=\beta\cos\psi,\qquadv^{\hat\phi}=\beta\sin\psi,\qquadv^{\hat\theta}=\beta\sin\eta\quad(\text{but we’ll set }\eta=0\text{ for the equatorial plane}),vr^=βcosψ,vϕ^=βsinψ,vθ^=βsinη(but we’ll set η=0 for the equatorial plane),
then the coordinate components of the 4-velocity at r0r_0r0 are

ut(0)=γ,ur(0)=r˙(0)=γ βcos⁡ψ,u^t(0)=\gamma,\quadu^r(0)=\dot r(0)=\gamma\,\beta\cos\psi,ut(0)=γ,ur(0)=r˙(0)=γβcosψ,uθ(0)=θ˙(0)=0(equatorial launch),u^\theta(0)=\dot\theta(0)=0\quad(\text{equatorial launch}),uθ(0)=θ˙(0)=0(equatorial launch),uϕ(0)=ϕ˙(0)=γ βsin⁡ψr02+b2.u^\phi(0)=\dot\phi(0)=\frac{\gamma\,\beta\sin\psi}{\sqrt{r_0^2+b^2}}.uϕ(0)=ϕ˙(0)=r02+b2γβsinψ.
From these, the invariants are instantly

E=γ,L=(r02+b2) uϕ(0)=r02+b2  γ βsin⁡ψ.E=\gamma,\qquadL=(r_0^2+b^2)\,u^\phi(0)=\sqrt{r_0^2+b^2}\;\gamma\,\beta\sin\psi.E=γ,L=(r02+b2)uϕ(0)=r02+b2γβsinψ.

What you specifically asked for​

You asked for initial conditions for uru^rur and uθu^\thetauθ at t=0t=0t=0 given a launch speed β\betaβ at r0=20r_0=20r0=20:

• In the equatorial setup the plane is θ=π/2\theta=\pi/2θ=π/2 and remains so if you don’t kick out of plane, hence
  
   uθ(0)=0 .\boxed{\,u^\theta(0)=0\,}.uθ(0)=0.
• The radial component depends on the launch direction ψ\psiψ (the angle between your velocity and the azimuthal direction):
  
   ur(0)=γ βcos⁡ψ,γ=11−β2 .\boxed{\,u^r(0)=\gamma\,\beta\cos\psi,\quad \gamma=\frac{1}{\sqrt{1-\beta^2}}\,}.ur(0)=γβcosψ,γ=1−β21.

If your launch is purely tangential (ψ=π2\psi=\tfrac{\pi}{2}ψ=2π): ur(0)=0u^r(0)=0ur(0)=0.
If purely radial outward (ψ=0\psi=0ψ=0): ur(0)=γβu^r(0)=\gamma\betaur(0)=γβ.

(If your examiner intended ψ\psiψ to be “impact parameter data”, then they likely wanted you to express cos⁡ψ\cos\psicosψ from L/EL/EL/E: with θ=π/2\theta=\pi/2θ=π/2,

LE=r02+b2  βsin⁡ψ  ⇒  sin⁡ψ=LE r02+b2,cos⁡ψ=1−L2E2(r02+b2).\frac{L}{E} = \sqrt{r_0^2+b^2}\;\beta\sin\psi\;\Rightarrow\;\sin\psi=\frac{L}{E\,\sqrt{r_0^2+b^2}},\quad\cos\psi=\sqrt{1-\frac{L^2}{E^2(r_0^2+b^2)}}.EL=r02+b2βsinψ⇒sinψ=Er02+b2L,cosψ=1−E2(r02+b2)L2.
Plugging that into ur(0)u^r(0)ur(0) gives your initial radial 4-velocity purely in terms of E,L,r0,bE,L,r_0,bE,L,r0,b.)

Quick check against your​

With the above,

12ur2(0)+L22(r02+b2)=12γ2β2cos⁡2ψ+12γ2β2sin⁡2ψ (r02+b2)(r02+b2)=12γ2β2=12(γ2−1)=ε,\tfrac12 u_r^2(0)+\frac{L^2}{2(r_0^2+b^2)}=\tfrac12\gamma^2\beta^2\cos^2\psi+\tfrac12\frac{\gamma^2\beta^2\sin^2\psi\,(r_0^2+b^2)}{(r_0^2+b^2)}=\tfrac12\gamma^2\beta^2=\tfrac12(\gamma^2-1)=\varepsilon,21ur2(0)+2(r02+b2)L2=21γ2β2cos2ψ+21(r02+b2)γ2β2sin2ψ(r02+b2)=21γ2β2=21(γ2−1)=ε,
so it’s internally consistent. Your earlier relation is right if you set Veff=L22(r2+b2)V_{\rm eff}=\dfrac{L^2}{2(r^2+b^2)}Veff=2(r2+b2)L2.

If you needed the circular-orbit ratio​

For a circular timelike orbit at r=rcr=r_cr=rc, you set r˙=0\dot r=0r˙=0 and d(r˙2)/dr=0d(\dot r^2)/dr=0d(r˙2)/dr=0. With the potential above this gives

ddr ⁣(E2−1−L2r2+b2)=0  ⇒  L2(rc2+b2)2=0  ⇒  L=0.\frac{d}{dr}\!\left(E^2-1-\frac{L^2}{r^2+b^2}\right)=0\;\Rightarrow\;\frac{L^2}{(r_c^2+b^2)^2}=0\;\Rightarrow\;L=0.drd(E2−1−r2+b2L2)=0⇒(rc2+b2)2L2=0⇒L=0.
I.e., there are no nontrivial timelike circular geodesics in this ultrastatic wormhole (not surprising: gttg_{tt}gtt is flat). So if your question said “the ratio L/EL/EL/E is fixed by position”, they probably meant null circular or a specific launch geometry at r0r_0r0, not a geodesic circular timelike orbit. For timelike motion, L/EL/EL/E at launch is set by your direction angle ψ\psiψ as shown above.

Plotting orbits (the bit you got stuck on)​

Use

drdϕ=r˙ϕ˙=E2−1−L2r2+b2 L/(r2+b2) =(r2+b2)E2−1−L2r2+b2L.\frac{dr}{d\phi}=\frac{\dot r}{\dot\phi}=\frac{\sqrt{E^2-1-\frac{L^2}{r^2+b^2}}}{\,L/(r^2+b^2)\,}=\frac{(r^2+b^2)\sqrt{E^2-1-\frac{L^2}{r^2+b^2}}}{L}.dϕdr=ϕ˙r˙=L/(r2+b2)E2−1−r2+b2L2=L(r2+b2)E2−1−r2+b2L2.
Pick bbb, r0=20r_0=20r0=20, β\betaβ, and ψ\psiψ, compute E,LE,LE,L from the launch, and integrate ϕ↦r(ϕ)\phi\mapsto r(\phi)ϕ↦r(ϕ). Equatorial means θ=π/2\theta=\pi/2θ=π/2 throughout.

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Bottom line: your ε\varepsilonε–EEE relation is fine once you pick Veff(r)=L2/[2(r2+b2)]V_{\rm eff}(r)=L^2/[2(r^2+b^2)]Veff(r)=L2/[2(r2+b2)]. For the initial conditions the clean, exam-ready answers are

uθ(0)=0,ur(0)=γ βcos⁡ψ(γ=1/1−β2,  r0=20).u^\theta(0)=0,\qquadu^r(0)=\gamma\,\beta\cos\psi\quad(\gamma=1/\sqrt{1-\beta^2},\; r_0=20).uθ(0)=0,ur(0)=γβcosψ(γ=1/1−β2,r0=20).
If they gave you L/EL/EL/E instead of ψ\psiψ, use the little triangle above to swap between them. And yeah—because this is ultrastatic, don’t go hunting for timelike circular geodesics; they’re not there.

 

[renormalize]

Alright louis, straight to it. You’ve got the ultrastatic Morris–Thorne-style metric

Your math is unreadable; please edit your post.