Looking At General Relativity (Part 5): Observational Aspects- Tests

Observational Aspects of General Relativity

1. Slowing Clocks in Gravitational Fields.

Earlier we saw that  Einstein's general theory of relativity predicts that clocks  run slower in gravitational fields (a phenomenon called 'gravitational time dilation') In this case, for the Earth, one would have the fractional difference in proper time, as a fraction of time passage t[1]:

Dt/ t »    dt2 - dt1/ dt   »  GM(1/r1 - 1/r2)

where G is the Newtonian gravitational constant, M is the Earth's mass, and g is the acceleration of gravity (g = 980 cm/ sec² in cgs) and c = 3 x 10¹⁰ cm/sec.

In terms of frequency (or clock tick rate) this amounts to a shift in frequency. A large elapsed time (dt2 - dt1) implies a fast tick rate so high frequency. This shows the difference in frequency must be proportional to the elapsed time so:

Dn/ n  = (n2 - n1)/ n  =  Dt/ t »  GM [1/r1 - 1/r2]

Generalizing, any atom placed inside a significant gravitational potential will display a redshift compared with a similar atom outside the potential. Thus, inside the potential (at smaller radius r1) the tick rate slows compared to that measured at radius r2. This means the frequency is slower or redshifted.

    Near to Earth’s surface it is common to make the approximation:

GM [1/r1 - 1/r2] » GM Dr/ R²  =  g Dr

Let us say that in one experiment (r2 - r1) = 0.001 mm = 0.0001cm, then:

Dt/ t ~ (980 cm/s²)(10^-4 cm)/ (3 x 10¹⁰ cm/sec )²

Dt/ t »  10^-22

and for an interval say t = 0.01 sec, Dt =

(10^-22 )(0.01 sec) = 10^-24 sec

2. Deflection of Light in Gravitational Field

The deflection of star light in a gravitational field was tested during solar eclipse in 1919, and was actually described in Eddington’s book in detail[2].  A rough illustration of the effect is shown below:

Fig. 1. Deflection of star light as predicted by General Relativity

In the diagram the light from the star at actual position S2 is seen to deflect by some angle  a, thereby altering the image position to that seen at S1. This is a direct result of the effect of the gravitational field of the Sun on the light rays. The true direction is thus alone the ray ES2 while the deflected position is along the ray ES1.

Theoretically and quantitatively, one get obtain an estimate of the magnitude of deflection by incorporating another parameter – call it b – as shown in the diagram below:

Fig. 2: Determining the linear deflection parameter, b.

Then we obtain for the deflection angle, a:

a  = - 4 GM/ b   or (in cgs units): 

a       = - 4 GM/ b c² 

Einstein, in his own paper: ‘On the Influence of Gravitation on the Propagation of Light’, gives the result as:

aq  = 2k M/ c²  D

Fig. 3 Diagram for Einstein’s derivation of a (q).

Here k = G, the gravitational constant and D = the distance of the ray from the center of the body. It is obtained from the diagram of Fig. 2.

Einstein used the cosine of the angle and integrated from the negative angle 

q = - p/ 2 to q =  p/ 2:

a  =    1/ c²  ò _{- p/ 2}  ^{p/ 2}   k M/ r²   cos q ds

3. Advance of Mercury’s Perihelion

Another prediction from General Relativity is the advance of Mercury’s perihelion. This is a phenomenon in no way predictable from Newtonian gravitation and is illustrated below, with the orbits separated in exaggerated fashion:

Fig. 4. Advance of Mercury’s Perihelion

The amount of rotation of 'the planetary ellipse' due to the effects of gravitation in general relativity was set out in the following equation of Einstein’s:

e = [24 (p)³ a²]/ T² c² [1 - e²]


where e is the advance (or rotation) in seconds of arc, T is the period of revolution in seconds, c the velocity of light and e the eccentricity of the orbit.

Einstein, on page 164 of the same paper, asserts that for Mercury e= 43 seconds per century

Problem: Validate the magnitude of the perihelion advance of Mercury from Einstein’s given equation.

The key to the solution is to note that Einstein framed his equation in cgs units, so we must have:

c = 3 x 10¹⁰ cm/s

a = (1.5 x 10¹³  cm) (0.387) = 5.8 x 10¹² cm

Based on using 1 astronomical unit (AU) = 1.5 x 10¹³ cm. But from a Table of distances, Mercury's semi-major axis a = 0.387 AU.

The period, T (in seconds) is just the length of Mercury's year (in days = 87.96, again from a Table ) multiplied by the seconds-length of an Earth day, or 86,400 s:


T = 7.6 x 10⁶ s

The eccentricity, e from a similar Table is e = 0.205.


Substituting all these values into the given equation yields:


e = 5.036 x 10^-7  radian


To get the equivalent seconds of arc (or arcsec) we use 1 rad (radian) = 57.3 degree where  one degree has 3600 seconds. Thus, 1 radian will have:

2.063 x 10⁵ arcsec


So, the associated arcsec for e will be:

(5.036 x 10^{- 7}   rad) x (2.063 x 10⁵ arcsec/ rad) =

0.104 arcsec

We are still not finished because the quantity is defined per century.


At this point, you need to recall the PERIOD of Mercury is 0.2405 YRS.

So, the number of arcsec of perihelion advance per Earth years is:


0.104 arcsec/ 0.2405 years =  0.432

Or 43 seconds of arc per century.

Other Problems:

1) The tick rate of a clock is found to slow from 10^-24 sec   to 10^-10 sec when placed inside a strong gravity well. Find the amount by which the frequency is redshifted.

2) Using Einstein’s relation for the deflection of starlight in a strong gravitational field:

a  =   ò _{- p/ 2}  ^{p/ 2}   k M/ r²   cos q ds

Show that this would actually be equal to: a  =  2 F/  c²

(Where F is the gravitational potential.)

 

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[1] Cf. Ohanian, H.C. and Ruffini, R. (1995), :Gravitation and Spacetime, p. 182.

[2] Eddington, A.S., op.. cit., p. 102.