Another test that Einstein suggested for testing his gravitational theory was the precession of perihelia. This reflects the fact that noncircular orbits in General Relativity are not perfect closed ellipses; to a good approximation they are ellipses that precess.

The strategy is like as in our previous article about light deflection to describe **the evolution of the radial coordinate r as a function of the angular coordinate Φ**; for a perfect ellipse, r(Φ) would be periodic with period 2π, reflecting the fact that perihelion occured at the same angluar position each orbit.

Using then **perturbation theory**, we can show how General Relativity introduces a slight alteration of the period, giving rise to precession.

We recall from our previous article Gravitational deflection of light the relativistic expression of the Binet's equation (in Newtonian physics, the last term in u^{2} is absent) for a particule with mass (note the presence of the term GM/h^{2} on the right-side of the equation)

If we now consider a circular orbit with constant radius r_{c}: r_{c} should be solution of the previous equation so that, with u_{c}=1/r_{c}:

If we now assume that the solution has the form

and subsituting in the above equation, we find

Adding now the hypothesis that w<< 1, we get

This equation looks like the equation of the harmonic oscillator** d ^{2}x/dt^{2} + ω^{2}x = 0** where 1-6GM/r

_{c}= ω

^{2}.

We know that the solution to this equation can be written as **x(t) = A cos(ωt + t _{0})** where A and t

_{0}are constants, thus the solution to our equation looks like

We have then the picture that the planet's oribt oscillates about a circular orbit.

Now, however, the period is not 2π, but if we assume that GM << r_{c}

##### Application to Mercury

Using the above equation, let us calculate the general relativistic portion of Mercury's perihelion advance in seconds of arc per century.

Assuming the following:

Mercury has a period (i.e. the time for one complete orbit) of 87.97 days,

Mercury's orbital radius = 57.9 x 10^{6} km,

the mass of the Sun M = 1.99 x 10^{30} kg,

the gravitational constant G = 6,67 x 10^{-11} Nm^{2}kg^{-2},

one year equals 365.25 days

we can write:

Remark: we dont get exactly 43'' per century as the average orbital radius does not equal exactly r_{c} (see Moore Chapter 11 exercice 11.4)

To get the exact value, it is enough to remind ourselves that in polar coordinates, the ellipse equation is

where a is the semi-major axis and e its eccentricity.

Considering Φ very small, cos Φ approximates to 1 so that

For the motion of Mercury around the Sun, the relevant orbital parameters are

Mercury's orbital radius = a = 57.9 x 10^{6} km

e = 0.2056

The correct value is in fact just the precedent value of 41.11'' which sould be corrected by (1-e^{2})

Einstein triumphantly announced this result in the third of his four November lectures.

"This discover was, i believe, by far the strongest emotional experiment in Einstein's scientific life, perhaps in all his life," Abraham Pais later said. He was so thrilled he had heart palpitations, as if "something had snapped" inside.

Remark: the observed value of Mercury's orbit precession is 5601 arcsecs/100 years. However, much of that (5025 to be precise) is due to the precession of equinoxes in our geodesic coordinate system; the gravitational perturbations of the other planets contribute an additional 532 arcssecs/100 years, leaving **43 arcsecs/100 years to be explained by General Relativity**.

Which it does quite well ;-)