GENERAL RELATIVITY:
This is a difficult subject, but at its core is a simple idea that can be used to solve certain types of problems. The idea is that the way things behave in a gravitational field is the same as in an accelerated system with no gravity. Imagine being in an elevator at rest in a gravitational field g. You know how things are- the same as in everyday life. Now suppose you could turn off gravity and give the elevator an upward acceleration equal to g. Things would behave exactly the way they did with gravity. The normal force on something resting on the elevator floor would be mg, a dropped object would go distance (1/2)gt2 in time t, etc. (Really the elevator would accelerate upward to meet it.)
Similarly, being in free fall in a gravitational field is the same as being at rest with no gravity.
We can now understand how gravity bends light. A light beam enters the upward accelerating elevator. Suppose it is traveling horizontally with respect to the elevator. It travels a horizontal distance x in time t = x/c, where c is the speed of light. But during this time, the elevator's upward velocity increased by gt, hence relative to the elevator, the light now has a downward component of velocity equal to gt. The distance the light drops below horizontal is simply (1/2)gt2. We get the same result if we imagine the elevator in free fall. The light must undergo free fall also, so that the result is the same as the no-gravity case: the light does not bend relative to the elevator.
Radius of curvature
Radius of curvature R for any horizontally moving projectile is easy to calculate. Since the acceleration g is perpendicular to the velocity, equate it to v2/R and solve for R and get R = v2/g. For horizontal light, there is another issue due to time warp in a gravitational field which introduces a factor of 2 in the equation. Velocity of light is c so R = c2/2g. See circular motion if you need to. It is easy to show that g is GM/r2 , where r is the distance to the center of a body of mass M possessing spherical symmetry, and G is the universal gravitation constant. See gravity if you need to. Do not confuse r, the distance to the center, with R, the radius of curvature of the light. Show that
R = c2r2/2GM
On Earth the bending of light due to g is insignificant: R is about a quarter of the way to the nearest star (other than the sun).
Now let us consider a black hole. You can easily find the r for which the light is in circular orbit around the black hole. (Let r = R and do the algebra.) You will find that
R = 2GM/c2.
It also can be shown that light closer to the black hole than this cannot escape (which is why they are called black holes).
The time warp causes what is known as the gravitational red shift. Time is slower in the high gravitational field of a star, so light frequencies are lower. Calculating the shift is easy if you cheat in the following way: Think of the photon as a bullet of energy hf, where h is Planck's constant and f is the frequency, you can see that if it loses energy it decreases in frequency.
If a bullet is fired with enough energy to escape a planet
of mass M and radius r (with no atmosphere), it will lose energy GMm/r. So the minimum v it needs to start with is
such that
1/2mv2 = GMm/r, or v
= (2GM/r)1/2. The same is true
of the photon, although this is not the conventional way of looking at it.
Gravity does not slow down the photon, but if you replace v with c in the
equation for the minimum speed of the bullet, you get the relationship between
the mass and maximum radius of a black hole. Square it and get
r = 2GM/c2.
E = mc2 expresses the equivalence of mass and energy. Thus the photon acts like it has mass = energy/c2
(although gravity does not slow it down). But the energy of a photon is hf, where h is Planck's constant and f is the frequency,
so hf - GMhf/(rc2)
= hf ' where f ' is the frequency after making its
escape. Do a little algebra on this and show that the fractional loss of
frequency,
D f/f , is
GM/(rc2). Red is the lowest visible frequency, so this is called the
gravitational red shift.
The maximum D f is f, then the
frequency is zero and the photons have lost all their energy and would cease to
exist. Thus the black hole:
D f/f = 1 = GM/(rc2) or r =
GM/c2 would appear to be the maximum radius of a black hole. But the
change of time zones or something make it
r = 2GM/c2.
There is another red shift caused by the Doppler effect, not an application
of general relativity. The galaxies are moving away from each other (the
aftermath of the big bang, not some kind of cosmic halitosis thing), and if the
relative speed of recession is v, we observe
f ' = f[(c-v)/(c+v)]1/2.
This is a much bigger effect than the gravitational red shift for distant
galaxies, and it is used to figure the v, which is proportional to the distance
away (Hubble's law). I am nervous about the practice of calculating the
distance to quasars by this method, since we don't know what they are. In both
red shifts the new frequency is proportional to the original one, so you cannot
tell the difference between the shifts. How do we know that quasars do not have
a much larger gravitational red shift than other things?
F. Gram,
My main pages:
mechanics
fluids, heat, electricity and magnetism
vibrations and waves
quantum
index
Comments, questions: fredrick.gram @tri-c.edu
Don't click on back, or you will find yourself in a wormhole to another universe.