DEFLECTION OF A RAY OF LIGHT IN A GRAVITATORY FIELD.
Before we study this problem I am going to reproduce some paragraphs of the book: “Electromagnetism & Relativity” by Edmund P. Ney, page 107.
Deflection of Light in the Gravitational Field of a Star.
In 1801 a German mathematician named Soldner, calculated the deflection of light in the gravitational field of the Sun. His result which we reproduce here leads to the prediction of a deflection of 0”.87 seconds of arc. Shortly after developing the general theory, Einstein calculated the deflection and got the same result. however, he later developed the theory and found that general relativity actually predicts a value twice as large, or 1”.75 of arc. The experiment was first performed in 1919, and the result reported was 1.”7 of arc. The experiment is difficult, and more recent experiments have not agreed with this theory as the 1919 experiment, This prediction of the theory can only be considered to be confirmed to an accuracy of perhaps 20 %. The procedure of the experiment, is to photograph a star field around the Sun during a total eclipse (so that the stars are visible). Six months later, when the same stars are visible at night, the star field is photographed again. The displacement of the apparent positions of the stars can be measured by comparing the two photographs.
Soldner’s derivation.
The transverse momentum P^, imparted to an object of mass m can be shown to be given at point P (see figure 68) by:
dP^ = G M m CosQ d Q / c R The total transversal momentum is:
P^ = dP^ = -G M m Cos Q dQ / cR = 2 G M m / c R . . . . (42)
The angle : f = P^ / P = (2 G M m /c R) / m c = 2 G M / c2 R radians . . . . (43’)
Substitution in this formula for the case of the Sun, leads to the “classical” value: f = 0”.87 of arc. Note that although the mass of the photon cancels itself out in the derivation, the mass must be finite, i.e. m ¹ 0, and this was not established in prerelativity physics. It could also be emphasized that Soldner’s value is only half the correct value given by the general theory.
The determination of the angle f by the relativity theory is not going to be effectuate here, because it is not necessary in this study. In the determination of the angle f by Soldner, he gave the gravity force an instantaneous action, but this is not the case, because the gravity has light velocity, and we have to take in account the aberralogy. First we consider the light corpuscles moving from P to Q as in Fig.68. When the corpuscle that we take passes by point P with c velocity, it meets a gravity ray with practically the same velocity that moves in the OP direction, the velocity that meet both rays is equal to the vectorial sum of their velocities. In Fig. 69 we have represented by cx the velocity of the corpuscle of light, and by c the velocity of the gravity that meets the light ray (or the corpuscle of light), and by ca the velocity of reception of the emitted-received fluid by light corpuscle.
Distance OP = R / Cos Q
Distance PP’ = dS = OP d Q / Cos Q = R d Q / Cos2 Q
Gravity force on a corpuscle at point P, according with Newton’s law, is:
F = G M m / (R / Cos Q)2; here: m = 1.47236 x 10—50 Kg. mass of a light corpuscle; M = 1.983 x 1030 Kgs. mass of the Sun; R = 7 x 108 m
The resultant of this gravity force perpendicular to trajectory PQT (but in the same plane PQO) is:
F^ = G M m Cos Q / (R / Cos Q)2 = G M m Cos3 Q / R2 = perpendicular force
The momentum of the corpuscle in the point we are considering or any other point of the trajectory PQT is practically constant (= mcx) and in the same way of the trajectory. The momentum normally the trajectory is obtained multiplying the normal component of the gravity force by dt (time differential). With both magnitudes we make the triangle of Fig. 70 and obtain the deviation angle df for the chosen point. The sum of all these angles df will give us the total angular deviation f , and because this angle is very small, we consider the normal in the trajectory equal to the normal of the straight line PQT, without appreciable error.
d f = G M m Cos3 Q dt / m cx R2
We solve the problem by the principle of impulse and momentum.
dS = PP’ = (PO) d Q / Cos Q = R d Q / Cos2 Q = cx dt
cx = velocity of the corpuscle
dt = differential of time = dS / cx = R d Q / cx Cos2 Q
dF^ = F^ dt = (G M m Cos3 Q / RW2) (R d Q / cx Cos2 Q) = G M m Cos Q d Q / cxR
We are going to consider the corpuscle moving as indicated by the arrow. When the corpuscle is in P, the gravity fluid is moving from O to P with velocity c, equal to light velocity. The corpuscle moves with a velocity equal to cx in the direction PP’Q, so that this corpuscle receives the gravitatory ray with an aberrated velocity:
ca = (cx2 + c2 + 2 c cx Sen Q)0.5 (69)
In accordance with the first law of the double fluid theory of the aberralogy, the increment of energy that produce the radiation (of gravity) is proportional to the square of the increment of velocity with which is received such radiation. So we have that the increment of the action of gravity is equal to: ca / c. Here ca = incremented velocity of gravity (see fugure 69). As velocity of gravity is equal to velocity of light c = cx; In accordance with the principle of impulse and momentum and considering the increment (ca / c), of the gravity force by aberration, and the first law of the double fluid theory, we have:
ca2 / c2 = (cx2 +c2 + 2 cx c Sinq ) / c2 = 2 + 2 Sinq
The angle of deflection is: df = d P^ / P ; dP = dF P = mcx
f = dP^ / P ;
Considering the increment of the action of gravity:
f = P^ dt ca2 / (c2 m c) = (- G M m / c R) Cos q dq (ca2 / c2 m c) =
(- G M / c2 R) (2 + 2 Sin q) Cos q dq = 2 G M / c2 R Sin q
2 G M / c2 R Sin2q / 2 = 4 G M / c2 R + 0 Radians =
4 x 6.673 x 10—11 x 1.983 x 1030 / [ (3 x108)2 x 1.4 x 108 ] = 8.4 x 10—6 Rad. =
206,265 x 8.4 x 10—6 = 1.¨73.
The experiment of the eclipse of the Sun is not very good, because the rays of light that pass near the Sun are affected by the refraction of its atmospnere.
Estanislao de Hoyos; year 1945; C<n<nea Sonora, México.