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Unformatted text preview: 66 CHAPTER 6. GRAVITATION AS GEOMETRY b) The fourvelocity in the (t, :13) frame can be calculated from the expression
for m(t’) and y(t’) given in Problem 6, together with the line element derived
in part (a) of that problem which gives the connection between t’ and proper
time 7' along a curve of constant 113’, y’, and z’, viz., drz<1+£> dt'. CZ The resulting four—velocity of a curve at constant x’ has components dt gt’ dx __ _ gt'
a; — cosh , E —c smh (Components in the y— and z—direction vanish). The components of the
four—acceleration are Thus, which decreases with x’ 68. [S] It’s not legitimate to mix relativistic with nonrelativistic concepts,
but imagine that that a photon with frequency an is like a particle with
gravitational mass huh/c2 and kinetic energy K : ha). Using Newtonian
ideas, calculate the “kinetic” energy loss to a photon that is emitted from the
surface of a spherical star of radius R and mass M and escapes to inﬁnity
From this calculate the frequency of the photon at inﬁnity How does this
compare with the gravitational redshift in (6.14) to ﬁrst order in 1/02? Solution: In Newtonian physics, the loss in kinetic energy of a particle that
moves from the surface of the star to inﬁnity is equal to the difference in 66 PROBLEM 6.9 67 potential energy between those locations. That diﬁerence is GmM / R for a
particle of mass m. Thus, if KR is the kinetic energy at R and Koo the kinetic energy at inﬁnity,
KR ~ KDO = GmM/R. Equating K 2 ha; and m : had/c2 for a photon, we ﬁnd This is exactly the gravitational redshift to ﬁrst order in 1/02. 69. A GPS satellite emits signals at a constant rate as measured by an on
board clock. Calculate the fractional difference in the rate at which these are
received by an identical clock on the surface of the Earth. Take both the effects
of special relativity and gravitation into account to leading order in 1/02. For
simplicity assume the satellite is in a circular equatorial orbit, the ground
based clock is on the equator, and that the angle between the propagation
of the signal and the velocity of the satellite is 90° in the instantaneous rest frame of the receiver. Solution: The parameters of the orbit of a satellite are given in (6.17 The velocity of the Earth’s surface, which is also relevant, is V63 2 (27rR$/24 hr) =
0.46 km / s. With these parameters the relativistic effects due to to time dilation and the gravitational potential can be estimated as follows: Time dilation. We are considering the special case when angle between
the direction of propagation and the velocity of the satellite is a’ = 7r / 2 in
the rest frame of the receiver. Then the Doppler shift is transverse, or, put
differently, the difference in rates is due to the time dilation arising from the
relative velocity of the emitting and receiving clocks. In the rest frame of the
receiver the velocity of the clock is V, — V69 plus small corrections of order
(ve/C)2 (cf. (4.27)]. From (5.73) with 0/ : 7r/2 or from (4.15), the difference
in rates is proportional to 1 — (1 /2)(V5  V6;)2/c2 plus corrections that are
negligible for the small velocities relevant here. The fractional difference in 67 PROBLEM 6.10 69 of a radioactive element with a decay time of 4 billion years were present to
start, how much more of that element would be present at the center than the
surface? Assume the density of the Earth is constant. Solution: We give two different approaches to a solution:
First version. Suppose for simplicity that the density of the earth p is
constant over its radius. The gravitational force on a particle of mas m at radius r is G M 4
F, = ———nl2—@ = —m (—qur) .
r 3 The gravitational potential difference between the center and the surface R69
is therefore R F, 2
A<D=<I><Rt>—<I><o>=—/O $——dr:— m 3 1 GM
ﬂGpRéT—i ( Ref) . Thus, (see useful constants) ACE/c2 = (.443 cm)/(2 x 6.38 x 108 cm)
2 3.47 x 10—10 From (623) we can ﬁnd the difference in elapsed times,
A70 2 (1 + 3.47 X 10—10) TREE.
The abundance of a radioactive species will be
N QC 6”” Where T is the decay time. Thus the ratio of abundances is Ncenter — _———e_(T°/T) ~ exp (T3292)
Nsurface 5
Ncenter/Nsurf % 1 ‘l‘ 1 X X 10—10) —~ a very small difference! 69 70 CHAPTER 6. GRAVITATION AS GEOMETRY Second version:
We ﬁrst calculate the gravitational potential difference between the center of the earth and its surface. Let the radius of the surface be Re; and the mass
of the earth be M63. The potential difference is R _.
A<I>E<D(Re)—<I>(O)=—/ ®FdF. (1)
0
Here 13 is the gravitational force per unit mass, i.e.
F" = —GMQ(T) a
7" where M (7") is the mass inside a radius 7‘. Assuming a constant density p63 4 7" 3
Mzgnper3:M@( ). 56;
so that GM
4 r
F: — (B — “ .
Re (Re) 6’"
Inserting in the integral (1) gives
no 21 GMEB .
2 R33 The surface is at a higher in gravitational potential than the center. A clock
at the surface therefore runs faster by a factor of A<I> 1 GM69
1 —— = 1 — m1 . 1—10
(+62) (+2 CgRe) +35>< 0 The rocks at the center are therefore younger by
(5 x 109 yr) (3.5 x 10—10) :17 yr. The abundance of a radioactive element with a halflife 71/2 : 4 x 109yrs
decays as
N = N0 €_t/Tl/2 .
There will be more of the element at the center than at the surface by +1.73”: 1.7 <4wa rs) z 1 z 1 4. 10"10 .
e y + 4 X109 + 3 X 70 ...
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This note was uploaded on 12/01/2011 for the course PHYS 5523 taught by Professor Kennefick during the Fall '11 term at Arkansas.
 Fall '11
 Kennefick
 Theory Of Relativity

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