Question

1. Let us employ eccentric geodesics in Schwarzschild spacetime to obtain the perihelion precession of Mercury. Defining the perihelion precession per orbit as \(\Delta \phi = \Phi(2\pi) - 2\pi,\) where \(\Phi(X) = \int_0^X \frac{d\phi}{dx} dx\) (1.23) for an eccentric orbit parametrized by its dimensionless semilatus rectum \(p\) and eccen- tricity \(e\) via \(r(x) = \frac{pM}{1 + e \cos x},\) with \(x \in [0, 2\pi]\) being referred to as the orbital anomaly. \(\phi\) is the azimuthal coordinate of the timelike geodesics and \(d\phi/dx\) can be obtained from the Euler-Lagrange equa- tions for an eccentric geodesic with energy \(E\) and angular momentum \(L\). To obtain these, use \(u^a u_a = -1\) and the fact that \(\dot{r} = dr/dr = 0\) at two specific points along the elliptical orbit. Then solve for \(E\) and \(L\) as functions of \(p\) and \(e\). As a check, evaluate these at \(p = R, e = 0\) to recover the circular-orbit expressions from ACM40750. You can either numerically evaluate \(\Phi(2\pi)\) or use the fact that Eq. (1.23) is given by a constant times the incomplete elliptic integral of the first kind, i.e., EllipticK. You will need to extract the values of \(p, e\) etc. for Mercury. How does the precession rate compare with that of the double pulsar (PSR_J0737-3039)? You may at first find a discontinuity at \(x = \pi\) in your expressions. So carefully think about the sign of each term as the orbit goes from \(x \in [0, \pi]\) to \(x \in [\pi, 2\pi].\)

          1. Let us employ eccentric geodesics in Schwarzschild spacetime to obtain the perihelion
precession of Mercury. Defining the perihelion precession per orbit as
\(\Delta \phi = \Phi(2\pi) - 2\pi,\)
where
\(\Phi(X) = \int_0^X \frac{d\phi}{dx} dx\)
(1.23)
for an eccentric orbit parametrized by its dimensionless semilatus rectum \(p\) and eccen-
tricity \(e\) via
\(r(x) = \frac{pM}{1 + e \cos x},\)
with \(x \in [0, 2\pi]\) being referred to as the orbital anomaly. \(\phi\) is the azimuthal coordinate
of the timelike geodesics and \(d\phi/dx\) can be obtained from the Euler-Lagrange equa-
tions for an eccentric geodesic with energy \(E\) and angular momentum \(L\). To obtain
these, use \(u^a u_a = -1\) and the fact that \(\dot{r} = dr/dr = 0\) at two specific points along the
elliptical orbit. Then solve for \(E\) and \(L\) as functions of \(p\) and \(e\). As a check, evaluate
these at \(p = R, e = 0\) to recover the circular-orbit expressions from ACM40750.
You can either numerically evaluate \(\Phi(2\pi)\) or use the fact that Eq. (1.23) is given by a
constant times the incomplete elliptic integral of the first kind, i.e., EllipticK.
You will need to extract the values of \(p, e\) etc. for Mercury. How does the precession
rate compare with that of the double pulsar (PSR_J0737-3039)?
You may at first find a discontinuity at \(x = \pi\) in your expressions. So carefully think
about the sign of each term as the orbit goes from \(x \in [0, \pi]\) to \(x \in [\pi, 2\pi].\)

1. Let us employ eccentric geodesics in Schwarzschild spacetime to obtain the perihelion
precession of Mercury. Defining the perihelion precession per orbit as
Δϕ = Φ(2π) - 2π,
where
Φ(X) = ∫0^X (dϕ)/(dx) dx
(1.23)
for an eccentric orbit parametrized by its dimensionless semilatus rectum p and eccen-
tricity e via
r(x) = (pM)/(1 + e cos x),
with x ∈ [0, 2π] being referred to as the orbital anomaly. ϕ is the azimuthal coordinate
of the timelike geodesics and dϕ/dx can be obtained from the Euler-Lagrange equa-
tions for an eccentric geodesic with energy E and angular momentum L. To obtain
these, use u^a ua = -1 and the fact that ṙ = dr/dr = 0 at two specific points along the
elliptical orbit. Then solve for E and L as functions of p and e. As a check, evaluate
these at p = R, e = 0 to recover the circular-orbit expressions from ACM40750.
You can either numerically evaluate Φ(2π) or use the fact that Eq. (1.23) is given by a
constant times the incomplete elliptic integral of the first kind, i.e., EllipticK.
You will need to extract the values of p, e etc. for Mercury. How does the precession
rate compare with that of the double pulsar (PSRJ0737-3039)?
You may at first find a discontinuity at x = π in your expressions. So carefully think
about the sign of each term as the orbit goes from x ∈ [0, π] to x ∈ [π, 2π].

Added by Alicia Y.

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