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Perihelion Shift

To calculate the perihelion shift, we need to calculate r(ϕ)r(\phi). I will reuse equation for energy from the geodesics chapter:

E2=(drdλ)2+L2r2−2L2r3+2rϵ−ϵ.\mathcal{E}^2 = \left(\frac{dr}{d\lambda}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} + \frac{2}{r} \epsilon - \epsilon.

Considering timelike geodesics, where ϵ=−1\epsilon = -1 and λ=τ\lambda = \tau:

E2=(drdτ)2+L2r2−2L2r3−2r+1.\mathcal{E}^2 = \left(\frac{dr}{d\tau}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} - \frac{2}{r} + 1.

The angular momentum per unit mass is equal to:

L=dϕdτr2  ⟹  dϕdτ=Lr2.\mathcal{L} = \frac{d\phi}{d\tau} r^2 \implies \frac{d\phi}{d\tau} = \frac{\mathcal{L}}{r^2}.

By chain rule, drdτ\frac{dr}{d\tau} is equal to:

drdτ=drdϕdϕdτ=drdϕLr2.\frac{dr}{d\tau} = \frac{dr}{d\phi} \frac{d\phi}{d\tau} = \frac{dr}{d\phi} \frac{\mathcal{L}}{r^2}.

Substituting back into the original equation:

E2=(drdϕLr2)2+L2r2−2L2r3−2r+1=(drdϕ)2L2r4+L2r2−2L2r3−2r+1. \begin{align*} \mathcal{E}^2 &= \left(\frac{dr}{d\phi} \frac{\mathcal{L}}{r^2}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} - \frac{2}{r} + 1 \\ &= \left(\frac{dr}{d\phi}\right)^2 \frac{\mathcal{L}^2}{r^4} + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} - \frac{2}{r} + 1. \end{align*}

Substituting r=1ur = \frac{1}{u}:

r=1u,drdϕ=−1u2dudϕ,E2=(−1u2dudϕ)2u4L2+u2L2−2u3L2−2u+1=(dudϕ)2L2+u2L2−2u3L2−2u+1. \begin{align*} r &= \frac{1}{u}, \\ \frac{dr}{d\phi} &= - \frac{1}{u^2} \frac{du}{d\phi}, \\ \mathcal{E}^2 &= \left(- \frac{1}{u^2} \frac{du}{d\phi}\right)^2 u^4 \mathcal{L}^2 + u^2 \mathcal{L}^2 - 2 u^3 \mathcal{L}^2 - 2u + 1 \\ &= \left(\frac{du}{d\phi}\right)^2 \mathcal{L}^2 + u^2 \mathcal{L}^2 - 2 u^3 \mathcal{L}^2 - 2u + 1. \end{align*}

We then differentiate both sides with respect to ϕ\phi:

0=ddϕ[(dudϕ)2L2+u2L2−2u3L2−2u+1]=2dudϕd2udϕ2L2+2ududϕL2−6u2L2dudϕ−2dudϕ,0=d2udϕ2L2+uL2−3u2L2−1. \begin{align*} 0 &= \frac{d}{d\phi} \left[\left(\frac{du}{d\phi}\right)^2 \mathcal{L}^2 + u^2 \mathcal{L}^2 - 2 u^3 \mathcal{L}^2 - 2u + 1\right] \\ &= 2 \frac{du}{d\phi} \frac{d^2 u}{d\phi^2} \mathcal{L}^2 + 2 u \frac{du}{d\phi} \mathcal{L}^2 - 6 u^2 \mathcal{L}^2 \frac{du}{d\phi} - 2 \frac{du}{d\phi}, \\ 0 &= \frac{d^2 u}{d\phi^2} \mathcal{L}^2 + u \mathcal{L}^2 - 3 u^2 \mathcal{L}^2 - 1. \\ \end{align*}

Another substitution:

w=uL2=L2r,u=wL2,dudϕ=1L2dwdϕ,d2udϕ2=1L2d2wdϕ2,0=d2wdϕ2+w−w23L2−1,d2wdϕ2+w=1+w23L2,=1+αw2, \begin{align*} w &= u \mathcal{L}^2 = \frac{\mathcal{L}^2}{r}, \\ u &= \frac{w}{\mathcal{L}^2}, \\ \frac{du}{d\phi} &= \frac{1}{\mathcal{L}^2} \frac{dw}{d\phi}, \\ \frac{d^2 u}{d\phi^2} &= \frac{1}{\mathcal{L}^2} \frac{d^2 w}{d\phi^2}, \\ 0 &= \frac{d^2 w}{d\phi^2} + w - w^2 \frac{3}{\mathcal{L}^2} - 1, \\ \frac{d^2 w}{d\phi^2} + w &= 1 + w^2 \frac{3}{\mathcal{L}^2}, \\ &= 1 + \alpha w^2, \end{align*}

where α=3L2\alpha = \frac{3}{\mathcal{L}^2}.

If the right side would be equal zero, we will get simple harmonic oscillator:

d2wdϕ2+w=0,w(ϕ)=Asin⁡(ϕ+ϕ0)+Bcos⁡(ϕ+ϕ0). \begin{align*} \frac{d^2 w}{d\phi^2} + w &= 0, \\ w(\phi) &= A \sin (\phi + \phi_0) + B \cos (\phi + \phi_0). \end{align*}

If the right side would be equal to one:

d2wdϕ2+w=1,w(ϕ)=1+Asin⁡(ϕ+ϕ0)+Bcos⁡(ϕ+ϕ0),taking ϕ0=0,w(ϕ)=1+Asin⁡ϕ+Bcos⁡ϕ,w′(ϕ)=Acos⁡ϕ−Bsin⁡ϕ, \begin{align*} \frac{d^2 w}{d\phi^2} + w &= 1, \\ w(\phi) &= 1 + A \sin (\phi + \phi_0) + B \cos (\phi + \phi_0), \qquad \textrm{taking \(\phi_0 = 0\)}, \\ w(\phi) &= 1 + A \sin \phi + B \cos \phi, \\ w'(\phi) &= A \cos \phi - B \sin \phi, \end{align*}

This will become an equation for ellipse with the coordinate system centered at a focus:

Ellipse plot

When ϕ=0\phi = 0, then r=a−c  ⟺  w=L2a−cr = a - c \iff w = \frac{\mathcal{L}^2}{a - c} and w′=0w' = 0 (closest approach). Substituting:

L2a−c=1+Asin⁡0+Bcos⁡0=1+B,B=L2a−c−1,0=Acos⁡0−Bsin⁡0=A,w(ϕ)=1+(L2a−c−1)cos⁡ϕ, \begin{align*} \frac{\mathcal{L}^2}{a - c} &= 1 + A \sin 0 + B \cos 0 \\ &= 1 + B, \\ B &= \frac{\mathcal{L}^2}{a - c} - 1, \\ 0 &= A \cos 0 - B \sin 0 \\ &= A, \\ w(\phi) &= 1 + \left(\frac{\mathcal{L}^2}{a - c} - 1\right) \cos \phi, \\ \end{align*}

In a video by eigenchris, the following equation is provided (where ee is the eccentricity of the ellipse):

w(ϕ)=1+ecos⁡ϕ, \begin{align*} w(\phi) &= 1 + e \cos \phi, \\ \end{align*}

however, I was not able to prove their equivalence. My guess is that with ee is a more general solution for any ellipse and my solution with (L2a−c−1)\left(\frac{\mathcal{L}^2}{a - c} - 1\right) is a solution for a specific ellipse determined by the initial conditions. We will see this term will not matter in the end. I will use A\mathcal{A} for the constant:

w(ϕ)=1+Acos⁡ϕ,A=eorA=(L2a−c−1). \begin{gather*} w(\phi) = 1 + \mathcal{A} \cos \phi, \\ \mathcal{A} = e \quad \textrm{or} \quad \mathcal{A} = \left(\frac{\mathcal{L}^2}{a - c} - 1\right). \end{gather*}

The solution to the original equation is complex due to the extra αw2\alpha w^2 term. We may say this term is a correction for general relativity. Assume there is a power series in terms of α\alpha:

w=∑i=0∞wiαi.w = \sum_{i = 0}^{\infty} w_i \alpha^i.

then, substituting back into the original equation:

d2dϕ2(∑i=0∞wiαi)+(∑i=0∞wiαi)=1+α(∑i=0∞wiαi)2,[d2w0dϕ2+αd2dϕ2(∑i=1∞wiαi−1)]+(w0+α∑i=1∞wiαi−1)=1+α(∑i=0∞wiαi)2,(d2w0dϕ2+w0)+α[d2dϕ2(∑i=1∞wiαi−1)+∑i=1∞wiαi−1]=1+α(∑i=0∞wiαi)2,d2w0dϕ2+w0=1,d2dϕ2(∑i=1∞wiαi−1)+∑i=1∞wiαi−1=(∑i=0∞wiαi)2,w0(ϕ)=1+Acos⁡ϕ. \begin{align*} \frac{d^2}{d\phi^2} \left(\sum_{i = 0}^{\infty} w_i \alpha^i\right) + \left(\sum_{i = 0}^{\infty} w_i \alpha^i\right) &= 1 + \alpha \left(\sum_{i = 0}^{\infty} w_i \alpha^i\right)^2, \\ \left[\frac{d^2 w_0}{d\phi^2} + \alpha \frac{d^2}{d\phi^2} \left(\sum_{i = 1}^{\infty} w_i \alpha^{i - 1}\right)\right] + \left(w_0 + \alpha \sum_{i = 1}^{\infty} w_i \alpha^{i - 1}\right) &= 1 + \alpha \left(\sum_{i = 0}^{\infty} w_i \alpha^i\right)^2, \\[5ex] \left(\frac{d^2 w_0}{d\phi^2} + w_0\right) + \alpha \left[\frac{d^2}{d\phi^2} \left(\sum_{i = 1}^{\infty} w_i \alpha^{i - 1}\right) + \sum_{i = 1}^{\infty} w_i \alpha^{i - 1}\right] &= 1 + \alpha \left(\sum_{i = 0}^{\infty} w_i \alpha^i\right)^2, \\[5ex] \frac{d^2 w_0}{d\phi^2} + w_0 &= 1, \\ \frac{d^2}{d\phi^2} \left(\sum_{i = 1}^{\infty} w_i \alpha^{i - 1}\right) + \sum_{i = 1}^{\infty} w_i \alpha^{i - 1} &= \left(\sum_{i = 0}^{\infty} w_i \alpha^i\right)^2,\\[5ex] w_0(\phi) &= 1 + \mathcal{A} \cos \phi. \end{align*}

For the second equation, we will assume α<<1\alpha << 1, so that the higher order terms will vanish:

d2w1dϕ2+w1=w02=(1+Acos⁡ϕ)2,d2w1dϕ2+w1=1+2Acos⁡ϕ+A2cos⁡2ϕ=1+2Acos⁡ϕ+A22(cos⁡2ϕ+1)=1+2Acos⁡ϕ+A22cos⁡2ϕ+e22=(1+A22)+2Acos⁡ϕ+A22cos⁡2ϕ. \begin{align*} \frac{d^2 w_1}{d\phi^2} + w_1 &= w_0^2 = (1 + \mathcal{A} \cos \phi)^2, \\ \frac{d^2 w_1}{d\phi^2} + w_1 &= 1 + 2\mathcal{A} \cos \phi + \mathcal{A}^2 \cos^2 \phi \\ &= 1 + 2\mathcal{A} \cos \phi + \frac{\mathcal{A}^2}{2} (\cos 2\phi + 1) \\ &= 1 + 2\mathcal{A} \cos \phi + \frac{\mathcal{A}^2}{2} \cos 2\phi + \frac{e^2}{2} \\ &= \left(1 + \frac{\mathcal{A}^2}{2}\right) + 2\mathcal{A} \cos \phi + \frac{\mathcal{A}^2}{2} \cos 2\phi. \end{align*}

The general solution is in the form:

w1(ϕ)=(1+A22)+(aϕ+A)sin⁡ϕ+(bϕ+B)cos⁡2ϕ,w1′(ϕ)=asin⁡ϕ+(aϕ+A)cos⁡ϕ+bcos⁡2ϕ−2(bϕ+B)sin⁡2ϕ,w1′′(ϕ)=acos⁡ϕ+acos⁡ϕ−(aϕ+A)sin⁡ϕ−2bsin⁡2ϕ−2bsin⁡2ϕ−4(bϕ+B)cos⁡2ϕ,=2acos⁡ϕ−(aϕ+A)sin⁡ϕ−4bsin⁡2ϕ−4(bϕ+B)cos⁡2ϕ, \begin{align*} w_1(\phi) &= \left(1 + \frac{\mathcal{A}^2}{2}\right) + (a \phi + A) \sin \phi + (b \phi + B) \cos 2\phi, \\ w_1'(\phi) &= a \sin \phi + (a \phi + A) \cos \phi + b \cos 2\phi - 2 (b \phi + B) \sin 2\phi, \\ w_1''(\phi) &= a \cos \phi + a \cos \phi - (a \phi + A) \sin \phi - 2b \sin 2\phi - 2 b \sin 2\phi - 4 (b \phi + B) \cos 2\phi, \\ &= 2a \cos \phi - (a \phi + A) \sin \phi - 4 b \sin 2\phi - 4 (b \phi + B) \cos 2\phi, \\ \end{align*}

substituting back into the original equation:

2acos⁡ϕ−(aϕ+A)sin⁡ϕ−4bsin⁡2ϕ−4(bϕ+B)cos⁡2ϕ+(1+A22)+(aϕ+A)sin⁡ϕ+(bϕ+B)cos⁡2ϕ=(1+A22)+2Acos⁡ϕ+A22cos⁡2ϕ,2acos⁡ϕ−4bsin⁡2ϕ−3(bϕ+B)cos⁡2ϕ=2Acos⁡ϕ+A22cos⁡2ϕ,2a=2A,a=A,b=0,−3(bϕ+B)=A22,−3B=A22,B=−A26,A=0,doesn’t matter. \begin{align*} 2a \cos \phi - (a \phi + A) \sin \phi - 4 b \sin 2\phi - 4 (b \phi + B) \cos 2\phi \\ + \left(1 + \frac{\mathcal{A}^2}{2}\right) + (a \phi + A) \sin \phi + (b \phi + B) \cos 2\phi &= \left(1 + \frac{\mathcal{A}^2}{2}\right) + 2\mathcal{A} \cos \phi + \frac{\mathcal{A}^2}{2} \cos 2\phi, \\ 2a \cos \phi - 4 b \sin 2\phi - 3 (b \phi + B) \cos 2\phi &= 2\mathcal{A} \cos \phi + \frac{\mathcal{A}^2}{2} \cos 2\phi, \\ 2a &= 2\mathcal{A}, \\ a &= \mathcal{A}, \\ b &= 0, \\ -3(b\phi + B) &= \frac{\mathcal{A}^2}{2}, \\ -3 B &= \frac{\mathcal{A}^2}{2}, \\ B &= -\frac{\mathcal{A}^2}{6}, \\ A &= 0, \qquad \textrm{doesn't matter}. \end{align*}

The solution is then:

w1(ϕ)=(1+A22)+Aϕsin⁡ϕ−A26cos⁡2ϕ.w_1(\phi) = \left(1 + \frac{\mathcal{A}^2}{2}\right) + \mathcal{A} \phi \sin \phi - \frac{\mathcal{A}^2}{6} \cos 2\phi.

Substituting into w=w0+αw1w = w_0 + \alpha w_1:

w(ϕ)=(1+Acos⁡ϕ)+α[(1+A22)+Aϕsin⁡ϕ−A26cos⁡2ϕ]=1+Acos⁡ϕ+α(1+A22)+Aαϕsin⁡ϕ−αA26cos⁡2ϕ=1+α(1+A22)+A(cos⁡ϕ+αϕsin⁡ϕ)−αA26cos⁡2ϕ \begin{align*} w(\phi) &= (1 + \mathcal{A} \cos \phi) + \alpha\left[\left(1 + \frac{\mathcal{A}^2}{2}\right) + \mathcal{A} \phi \sin \phi - \frac{\mathcal{A}^2}{6} \cos 2\phi\right] \\ &= 1 + \mathcal{A} \cos \phi + \alpha \left(1 + \frac{\mathcal{A}^2}{2}\right) + \mathcal{A} \alpha \phi \sin \phi - \alpha \frac{\mathcal{A}^2}{6} \cos 2\phi \\ &= 1 + \alpha \left(1 + \frac{\mathcal{A}^2}{2}\right) + \mathcal{A} (\cos \phi + \alpha \phi \sin \phi) - \alpha \frac{\mathcal{A}^2}{6} \cos 2\phi \\ \end{align*}

the linear term in cos⁡ϕ+αϕsin⁡ϕ\cos \phi + \alpha \phi \sin \phi causes the shift of the perihelion. We can use the small angle approximation (α<<1\alpha << 1) to rewrite it:

sin⁡αϕ=αϕ,cos⁡αϕ=1,cos⁡ϕ+αϕsin⁡ϕ=cos⁡ϕcos⁡αϕ+sin⁡αϕsin⁡ϕ=cos⁡(ϕ+αϕ),w(ϕ)=1+α(1+A22)+Acos⁡ϕ(1+α)−αA26cos⁡2ϕ. \begin{align*} \sin \alpha \phi &= \alpha \phi, \\ \cos \alpha \phi &= 1, \\[2ex] \cos \phi + \alpha \phi \sin \phi &= \cos \phi \cos \alpha \phi + \sin \alpha \phi \sin \phi \\ &= \cos (\phi + \alpha \phi), \\[2ex] w(\phi) &= 1 + \alpha \left(1 + \frac{\mathcal{A}^2}{2}\right) + \mathcal{A} \cos \phi (1 + \alpha) - \alpha \frac{\mathcal{A}^2}{6} \cos 2\phi. \end{align*}

To calculate the (approximate) period we set ϕ(1+α)=2π\phi (1 + \alpha) = 2\pi:

ϕ=2π1+α=2π(1−α)1−α2, \begin{align*} \phi &= \frac{2\pi}{1 + \alpha} \\ &= \frac{2\pi (1 - \alpha)}{1 - \alpha^2}, \end{align*}

and since we are assuming α<<1\alpha << 1, α2→0\alpha^2 \to 0 and we can write:

ϕ=2π(1−α)=2π−2πα,Δϕ=2πα. \begin{align*} \phi &= 2\pi (1 - \alpha) \\ &= 2\pi - 2 \pi \alpha, \\ \Delta \phi &= 2 \pi \alpha. \end{align*}
Δϕ=4.92⋅10−7 rad/period=2.04⋅10−4 rad/century=42.15 arcsec/century \begin{align*} \Delta \phi &= 4.92 \cdot 10^{-7}\ \textrm{rad/period} \\ &= 2.04 \cdot 10^{-4}\ \textrm{rad/century} \\ &= 42.15\ \textrm{arcsec/century} \end{align*}