advanced30 min studyLesson 815 in Physics

Timelike orbits and innermost stable circular orbits

Derive circular-orbit energy and stability in Schwarzschild geometry and distinguish an ISCO from the last possible circular orbit.

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What you’ll understand

  • Normalize the conserved energy and angular momentum of a test body.
  • Test circular-orbit existence and radial stability separately.
  • State what ideal ISCO binding energy does and does not predict.

A circular orbit can solve the geodesic equations and still be unstable. This distinction explains the innermost stable circular orbit, usually abbreviated ISCO, without confusing it with the horizon or the circular photon orbit. We consider an uncharged, nonspinning test body in Schwarzschild geometry, neglecting its own gravitational field, radiation reaction, and material forces. Spherical symmetry lets its orbit lie in the equatorial plane. The mass length m is positive, and the exterior time coordinate retains its normalization at infinity.

Conserved quantities give a one-dimensional radial equation

Use proper length parameter s=cτs=c\tau along the massive body’s worldline. Normalize its energy as total Killing energy divided by rest energy, and its angular momentum as angular momentum divided by rest mass times c. The first quantity is dimensionless and the second has length units. The unit timelike normalization then combines those constants into a radial equation with an effective potential. Keeping this normalization explicit avoids importing a Newtonian energy convention whose additive constant or factor of two differs.

A circular orbit requires zero radial velocity and zero radial force, so its energy squared equals the potential and the radial derivative of the potential vanishes. That derivative must be taken while holding angular momentum fixed. Solving it gives an angular momentum that depends on the chosen circular radius. Substituting this radius-dependent result before differentiating would instead compare different orbits and produce an incorrect stability test. The positivity and finiteness of the resulting constants already restrict the circular family.

(drds)2=E2−V(r),V(r)=(1−2mr)(1+ℓ2r2),E=Em0c2,ℓ=Lm0c\left(\frac{dr}{ds}\right)^2=\mathcal E^2-V(r),\qquad V(r)=\left(1-\frac{2m}{r}\right)\left(1+\frac{\ell^2}{r^2}\right),\qquad \mathcal E=\frac E{m_0c^2},\quad\ell=\frac L{m_0c}
m₀ is the body’s rest mass, E its conserved total energy, and L its conserved angular momentum. The radial equation and potential are dimensionless; r, s, m, and ell have length units.↔ Scroll wide equations horizontally.

Existence and radial stability give different boundaries

The circular-family formulas below require radius greater than three m for finite timelike energy and angular momentum. Their divergence as that radius is approached signals the photon-orbit boundary, not the stable-orbit boundary. Linearizing radial motion about a circular solution shows that positive second derivative of the potential gives a restoring displacement. The sign changes at six m: outer circular orbits are radially stable, inner timelike circular orbits are unstable, and the boundary orbit is marginal at this order.

Thus there are timelike circular geodesics between three and six m, even though there are no stable ones there. The ISCO marks the loss of a local radial minimum, not the disappearance of every circular solution. Stability here means small radial geodesic perturbations of the ideal test-body orbit; an accreting plasma, a finite-mass companion, or a spinning body introduces additional dynamics. The coordinate angular frequency and the locally measured speed also use different clocks and cannot be equated by simply multiplying by radius.

ℓ2=mr2r−3m,E2=(r−2m)2r(r−3m),V′′∣circ=2m(r−6m)r3(r−3m)\ell^2=\frac{mr^2}{r-3m},\qquad \mathcal E^2=\frac{(r-2m)^2}{r(r-3m)},\qquad V''\big|_{\mathrm{circ}}=\frac{2m(r-6m)}{r^3(r-3m)}
The derivative is taken at fixed ell before imposing the circular relation. It has inverse square length units; its sign determines radial stability for this geodesic problem.↔ Scroll wide equations horizontally.

An orbital observable needs an observer and a model

The angular frequency relative to asymptotic coordinate time has the familiar square-root dependence on mass and radius, but its local interpretation changes with clock normalization. A static observer at the orbit measures a tangential speed larger than coordinate radius times coordinate frequency by the inverse square root of f. At the ISCO that speed is half the speed of light, despite the orbit remaining well outside the horizon. Such a local measurement is defined only for static observers in the exterior.

The difference between rest energy and conserved circular-orbit energy measures ideal binding energy relative to infinity. Evaluated at the Schwarzschild ISCO, it is about 5.72 percent of rest energy. Calling this a universal luminosity efficiency would add unjustified assumptions. A thin-disk estimate needs a stress prescription, radiative transport, and conditions near its inner edge; radiation reaction also changes the test-body evolution. The oscillator lab captures restoring motion with positive stiffness only, so it can illustrate the stable side but cannot locate an ISCO or display the unstable geodesic branch.

Ω=dφdt=cmr3,vstaticc=mr−2m,1−EISCO=1−89\Omega=\frac{d\varphi}{dt}=c\sqrt{\frac m{r^3}},\qquad \frac{v_{\mathrm{static}}}{c}=\sqrt{\frac m{r-2m}},\qquad 1-\mathcal E_{\mathrm{ISCO}}=1-\sqrt{\frac89}
Omega has s⁻¹ units and v_static is measured by a local static observer. The binding fraction describes the ideal Schwarzschild circular geodesic family, not every accretion flow.↔ Scroll wide equations horizontally.

Evaluate the ISCO without confusing its scales

For a Schwarzschild test-body orbit at r=6mr=6m, calculate the normalized angular momentum, normalized energy, local static-observer speed, and binding fraction.

  1. Insert six m into the circular angular-momentum expression: ℓ2=12m2\ell^2=12m^2, so ∣ℓ∣=12 m|\ell|=\sqrt{12}\,m. The sign chooses the orbital direction and has no dynamical distinction in Schwarzschild geometry.

  2. The energy is E=8/9≃0.94281\mathcal E=\sqrt{8/9}\simeq0.94281, taking the future-directed positive-energy branch. The local speed is c1/(6−2)=c/2c\sqrt{1/(6-2)}=c/2.

  3. The binding fraction is 1−8/9≃0.057191-\sqrt{8/9}\simeq0.05719. The second derivative vanishes here, identifying the marginal stability boundary rather than a finite-frequency restoring minimum.

The result

The ISCO has |ℓ| = √12 m, normalized energy about 0.94281, local speed c/2, and ideal binding fraction about 5.719 percent.

Explore the model

Interactive lab

Interactive lab · oscillator

A stable restoring displacement only

Reset and set damping ζ=0\zeta=0, natural frequency ω0=1\omega_0=1; Step once to lab time 0.5 and predict x=cos⁡(0.5)x=\cos(0.5), v=−sin⁡(0.5)v=-\sin(0.5). Repeat from Reset with both controls ζ=0\zeta=0, ω0=2\omega_0=2 and one Step; predict x=cos⁡1x=\cos1, v=−2sin⁡1v=-2\sin1. Initial conditions are fixed at x = 1, v = 0. Positive stiffness illustrates stable restoring motion only; the lab cannot reach zero or negative stiffness or solve Schwarzschild radial motion.

A stable restoring displacement onlyUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forIncreasing a positive restoring frequency changes oscillation; the existence of an unstable branch requires a different physical equation.

Check your understanding

Which statement correctly describes Schwarzschild timelike circular geodesics?

Try it yourself

Evaluate ℓ\ell and E\mathcal E for a future-directed circular timelike orbit at r=4mr=4m, and determine radial stability. Does energy equal to rest energy guarantee a stable orbit?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply both circular formulas and then test the sign of the fixed-angular-momentum second derivative.

Compare with the solution

The formulas give ℓ2=16m2\ell^2=16m^2, hence ∣ℓ∣=4m|\ell|=4m, and E2=1\mathcal E^2=1, hence E=1\mathcal E=1. However V′′=−1/(16m2)V''=-1/(16m^2) at this radius, so the orbit is radially unstable. Its conserved total energy equals the body’s rest energy, but the effective potential has a maximum. Energy normalization alone cannot establish stability; a radial perturbation can drive departure from the circular solution.

Take it with you

  • Circular existence follows from the first derivative; stability follows from the second at fixed angular momentum.
  • The ISCO at 6m leaves an unstable timelike circular family down to 3m.
  • Coordinate frequency, local speed, and ideal binding energy require distinct measurement and modeling conventions.

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