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/ r^{2 }cos q ds
Show that this would actually be equal to: a = 2 F/ c

^{2}
Where
F is the gravitational
potential.

__Solution__: The diagram for reference is:

Rewrite: a = (k
M/ r

^{2 })^{ }1/^{ }c^{2}^{ }**ò**_{- }_{p/ 2}^{p/ 2}^{ }cos q ds
a = (k M/ r

^{2 }) 1/^{ }c^{2}^{ }**ò**_{- }_{p}_{/ 2}^{p}^{/ 2}^{ }cos q ds =
(k
M/ r

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

^{2 }) 1/^{ }c^{2}^{ }[sin p/ 2 - sin (-p/ 2)] = 2 k M/ c^{2}^{ }r^{2 }
So:
a = 2k M/ c

^{2 }D = 2GM/ c^{2 }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**to radial distance D. It should also be understood that the formula also calls for a***at a right angle***of the triangle, defining an angle***negative version**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;

a = (G
M

**/**D^{ })^{ }1/^{ }c^{2}^{ }**ò**_{- }_{p/ 2}^{p/ 2}^{ }cos q dq
= 2GM/ c

^{2 }D
Where we recognize: F = G M

**/**D
As the

*gravitational potential*. Then we can actually rewrite Einstein’s equation as:
a
= 2 F/ c

^{2}
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