Bertrand s Theorem. October 8, µr 2 + V (r) 0 = dv eff dr. 3 + dv. f (r 0 )

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1 Bertrand s Theorem October 8, Circlar orbits The eective potential, V e = has a minimm or maximm at r i and only i so we mst have = dv e L µr + V r = L µ 3 + dv = L µ 3 r r = L µ 3 At this radis, there is no net radial orce, so that circlar orbits are possible. Sch orbits may be stable or nstable, depending on the sign o the second derivative o the eective potential. Stable orbits occr only i < d V e = 3L µr 4 = 3L µr 4 + d V d r and since r = L, we have µ 3 d r < 3 r r General case Eliminating = L µr, the remaining, radial eqation o motion is µ r L µr 3 r =

2 Write this as an orbit eqation, sing ṙ = dt = L µr d r = dt = L µr L µr L µr 3 + L d r µr Sbstitting, = L µ r 5 + L µ r 4 d r Now let r = The eqation o motion now becomes = L µr 5 or = L 5 µ = L µ = L µ = µ r L µr 3 r = L µr 5 + L d r µr 4 L µr 3 r = d d r = d + L µr 4 d r d d + L µ = 3 d d d L µr 3 r + L 4 d µ 3 d L µ d L 3 µ r Finally, write the orce in terms o the potential, d + = µ L = dv = d = dv d d L µr 3 r d L 3 µ r dv d

3 Now we have simply d + = µ dv L d This is the eqation o a iven harmonic oscillator. This striking set o transormations is what happens when people spend 3 years working on a problem. Now consider circlar orbits. This means that does not change with at all, so we have d = and thereore, i we set h = µ dv L d then = h Next, expand and the orce or small pertrbations abot, = + η µ dv L d = h Sbstitting these into the eqation o motion, = h + h η + h η + d η + + η = h + h η + h η + = + h η + h η + d η + [ h ] η = h η + I h <, then the eqation has exponential instead o oscillatory soltions, and the circlar orbits are not stable. For stable orbits, we thereore set λ = h > I we neglect the qaatic and higher terms on the right side o the eqation, with initial condition η = when =, we have soltions η = A sin λ In order to reprodce the initial conditions ater some integer nmber, q, o complete orbits, we reqire so that η = η πq = A sin πqλ qλ = p with p another integer. We see that λ mst be rational, λ = p q 3

4 Assming the orce is a continos nction o position, h is continos and λ is also continos. Thereore, λ = p q or all nearly circlar orbits. Retrning to the deinition o λ, and sing the constancy o λ, Since the circlar orbit satisies λ = h = d d λ = µ L 3 µ L + µ L = h = µ L = L µ 3 so that λ = µ L 3 L λ = + µ L λ 3 = µ 3 + µ L L µ 3 This mst hold regardless o the vale o, so we may op the sbscript and integrate d d = λ 3 d = λ 3 r r d = λ 3 r r = Ar λ 3 and in order to have stable, pertrbatively closed orbits, the orce law mst be a rational power law. This give h as Now retrn to the ll eqation o motion h = µ A L λ 3 = µa L λ d η + + η = + h η + h η + 6 h η 3 + 4

5 and expand in a Forier series, We also need powers o η. Keeping p to third order, η = η + η cos λ + η cos λ + η 3 cos 3λ + η + η cos λ + η cos λ + η 3 cos 3λ + = η + η η cos λ + η η cos λ + η η 3 cos 3λ +η η cos λ + η cos λ + η η cos λ cos λ +η η cos λ + η η cos λ cos λ + η η 3 cos 3λ η + η cos λ + η cos λ + η 3 cos 3λ + 3 = η 3 η + η η cos λ + η η cos λ + η η 3 cos 3λ + and sing addition ormlas Thereore, +η η cos λ + η η cos λ + η η η cos λ cos λ +η η η cos λ + η η η cos λ cos λ +η η 3 cos 3λ +η η cos λ + η η cos λ + η η η cos λ cos λ +η η cos λ + η 3 cos 3 λ +η η η cos λ cos λ +η η cos λ + η η η cos λ cos λ +η η η cos λ cos λ +η η 3 cos 3λ cos λ = + cos λ cos λ cos λ = [cos λ + λ + cos λ λ] = [cos 3λ + cos λ] cos 3 λ = cos λ + cos λ = cos λ + 4 cos 3λ + cos λ 4 = 3 4 cos λ + cos 3λ 4 η + η cos λ + η cos λ + η 3 cos 3λ + = η + η + η η + η η cos λ + η η + η cos λ + η η 3 + η η cos 3λ η + η cos λ + η cos λ + η 3 cos 3λ + 3 = η 3 + 3η η cos λ + 3η η cos λ + 3η η 3 cos 3λ +3η η cos λ + 6η η η cos λ cos λ + η 3 cos 3 λ = η 3 + 3ηη cos λ + 3η η cos λ + 3η η 3 cos 3λ +3η η + cos λ + 6η η η [cos 3λ + cos λ] +η cos λ + 4 η3 cos 3λ = η η η + 3η η + 3η η η η3 cos λ 5

6 Sbstitting and expanding, + 3ηη + 3 η η cos λ + 3η η η3 + 3η η η cos 3λ d η + + η = + h η + h η + 6 h η 3 + η λ cos λ 4λ η cos λ 9λ η 3 cos 3λ + + η = h η + η cos λ + η cos λ + η 3 cos 3λ + Eqating like terms, + h η + η cos λ + η cos λ + η 3 cos 3λ h η + η cos λ + η cos λ + η 3 cos 3λ + 3 η = h η + h η + h η + 6 h η h 3 η η [ η λ + η cos λ = h η + h η η + η η + 6 h 3ηη + 3η η η + 3 ] 4 η3 cos λ [ 4λ η cos λ + η cos λ = h η + h η η + η + 6 h 3ηη + 3 ] η η cos λ [ 9λ η 3 cos 3λ + η 3 cos 3λ = h η 3 + h η η 3 + η η + 6 h 3η η ] η3 + 3η η η cos 3λ Now we keep only the lowest order terms rom each eqation, i.e., we have η η η and so on. We set h = λ and solve = λ η + h η + 6 h η h η + 4 h η or, smmarizing, = λ η + 4 h η η = 4λ h η = h η η + η η + 8 η3 h = h λ h η 3 λ h η η3 h = η 3 4λ 5 h + 3λ h η = λ h η η 3 = [ 8λ h η η + ] 4 η3 h η = 4λ h η = 4λ η3 5 h + 3λ h η = η 3 = 8λ λ h η [ h η η + 4 η3 h 6 ]

7 where η is constant. Now, sing we may ind the derivatives, h = µa L λ so we may solve the eqality λ = h = λ µa L λ µa L λ = h = λ λ µa L λ = λ λ h = + λ λ λ = η 3 4λ 5 h + 3λ h = λ 5λ 4 λ + 3λ + λ λ λ = 5λ λ + 3λ + λ λ = 8λ λ 4 + λ 6 = λ λ 4 λ and we see that the only candidate power laws, r = Ar λ 3, or stable, closed orbits are: r = Ar 3 r = Ar r = Ar The irst o these yields only perectly circlar orbits, so the only nontrivial cases are the inverse sqare law and Hookes law. This condition only shows that these power laws are necessary. That they are sicient to prodce closed orbits reqires solving or their orbits exactly. 7