Solutions to General Relativity Problems (Part 5)

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.

Solution:

The difference in frequency must be proportional to the elapsed time so:

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

Or specifically for our purposes:

Dn/ n  = (n2 - n1)/ n  =  Dt/ t

Dt / t    =  [10^-10 s -  10^-24 s]/ 10^-24 s »  10^-10 / 10^-24  »  10 ¹⁴

Dn/ n  =   10 ¹⁴ 

Dn  »  n (10 ¹⁴ )   

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.

Solution: The diagram for reference is:

Rewrite:  q  =    (k M/ r² )  1/ c² ò _{- p/ 2}  ^{p/ 2}   cos q ds

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

(k M/ r²   )  1/ c²  [ sin q] _{- p/ 2}  ^{p/ 2}     =

 (k M/ r²   )  1/ c²    [sin p/ 2  -  sin  (-p/ 2)] = 2 k M/  c² r²  

So: q  = 2k M/ c²  D  =   2GM/ c²  D

Note: The triangle simply shows the relationships of distances r, s, and D with respect to the center of the body of mass M.  Distance s is along the path traveled by the light ray whose bending is given by angle a, and simply shows the ray at a right angle to radial distance D.  It should also be understood that the formula also calls for a negative version of the triangle, defining an angle θ that theoretically varies from – 90º to 0º to + 90º.

The final version or result actually only makes sense by a change of variable in the integral from s to q, thereby obtaining;

q   =    (G M /  D )  1/ c² ò _{- p/ 2}  ^{p/ 2}   cos q dq

= 2GM/ c²  D

Where we recognize: F =  G M /  D

As the gravitational potential. Then we can actually rewrite Einstein’s equation as:

 q =  2 F/  c²