Here is an email exchange with Professor Clifford Will, of the department of Physics at Washington University in St. Louis. He is the author of a very fine book on the general theory of relativity that he wrote for lay people. Ram Ramarao loaned me the book. It was very nice and easy reading, absorbing, and I went through it quickly from cover to cover, but for one hiccup, which was the subject of the following exchange.

  In his mail Professor Will uses the same symbol h for both height and the Planck's constant. Don't let that confuse you. Also read the Roman n in equations as the Greek "nu" for frequency.
 

Wed, 20 Aug 1997 

Dear Professor Will,

    I was a student at Wash U some years ago. Recently, one of my friends lent me the second edition of your book "Was Einstein Right?" that I am reading through now. I have enjoyed it very much so far, but I would like to point out one stumbling block. On page 71, line 7, you write "Thus, the angle of entry was deflected downward, by the phenomenon of aberration." I find this very difficult to understand.

    I do understand the phenomenon of aberration, why a moving person would see the rain slant more towards her, but I can't for the life of me, understand from your explanation why Observer 2 would measure a steeper angle from the vertical. This may need more explanation in a later edition.

    I would have thought that a more straightforward explanation, both of this fact, and of the gravitational red-shift, could follow from the equations E = mc² = hn, that you would find legitimate to use, having assumed some knowledge of the Special Theory. Am I correct in this line of thinking? If you have the time, I would very much like some more explanation of that sentence you wrote on p. 71, line 7.

    Also, could this effect not be explained using a single elevator? When Observer A measures an identical angle of entry and exit, that should suffice to tell her that the photon was deflected down, since she has been in free fall. But how would she know whether she was falling up or falling down? I am confused.

Regards,
Arun

Dear Arun: 

    To see the change in apparent angle from one elevator to another think of a rain analogy. Observer 1 sits on a bridge over a freeway in a rainfall. For simplicity let the rain fall vertically. A passenger in a car with an open sunroof passes underneath. The rain enters the car at an angle. This is easy to see. But here there is no "real'' deflection, because the whole problem can be analyzed globally from the viewpoint of the observer on the bridge. In the case of light deflection, we can only analyse things locally (where the equivalence principle works). Thus, the apparent deflection from one freely frame to the next is a real deflection. You can't get by with only one elevator for exactly the reason you mention: inside a single elevator, the observer can't tell whether she is going up or down -- she is in total free fall, the windows are closed, there is no gravity -- but inside her elevator, the light ray moves on a straight line. It's only by connecting one elevator with the next and the next and so on, that you can add up the relative deflections to get the net effect. 

    The idea of using E=mc² can also work: in some sense it's the basis of the Newtonian-von Soldner calculation of 1/2 the bending (pp. 66-67), although von Soldner (1803) predated special relativity. It can also be used to derive the gravitational redshift using a simple conservation of energy argument: at a height h in a gravitational field, take two particles of mass m₁ and m₂, and combine them into a bound particle of mass M = m₁+ m₂ - E/c², plus a photon of energy E. Let the atom fall to the floor, and let the photon propagate to the floor. The atom picks up kinetic energy KE = Mgh, while the photon's energy changes to E'. At the floor, use the photon's energy and the kinetic energy to tear apart the atom, and send the two particles back up. The energy left over must be exactly enough, (m₁+m₂)gh to get the two particles up with no kinetic energy left over (the system returns to its original state, so there must be no energy left, by conservation of energy). Thus equating the total rest mass energy at the floor with that of the two particles plus the energy require to boost them back up, we get 
Mc² + Mgh + E' = (m₁+m₂) c² + (m₁+m₂) gh, or 
(m₁+m₂-E/c²)c² + (m₁+m₂-E/c²)gh + E' = (m₁+ m₂)c² + (m₁+ m₂)gh, or
E' = E(1+gh/c²).

    Since E=hn, we have n' = n(1+gh/c²). This is the gravitational redshift formula! Thus there are several different routes to obtaining the effects that come from the equivalence principle. The only thing that you can't derive this way is anything that specifically depends on space curvature --- for that you need the full equations of the gravitational theory, such as GR [Genreal Relativity]. 

Cliff Will 
 
 

Dear Professor Will, 

    Since I wrote to you last I have read your book all the way through, and enjoyed it immensely, and found in it much to think about. I thank you for the book, and I thank you more for taking the time to reply to my questions. 

    I still have a nagging doubt about those elevators. Let Observer OA in Elevator EA measure angle AA, while Observer OB measures angle AB. Now I would like to say that the velocity vB of the photon seen by OB is a vector sum of the velocity vA with which the photon enters EA and the downward velocity vDA of EA. Then the angle AB will be different from the angle AA and the photon would appear drawn to the Sun. This is reasoning exactly as I would in the rainfall example. 

    Two questions crop up. First. Even if EA were not there, observer OB would still measure angle AB. Thus AB is independent of the downward velocity vDA of EA. In other words, vB is independent of vDA. It follows that the addition of vector vDA needs to be explained by something other than the fall of elevator EA. This appears to leave the phenomenon unexplained again. 

    Second question. The speed of the photon measured by OB will be higher than the speed measured by OA, if time were absolute. I expect this speedup gets eaten up by SR [Special Relativity] time dilation. Is that right? 

    I would love to have you respond, but I will understand if you do not have the time. 

Regards, Arun 

Dear Arun: 

    1. Remember, we are trying to understand the deflection (at least, 1/2 of it) using purely local arguments, where the EP is valid. Thus you absolutely have to have the two local elevators. Of course, if you wanted to determine the angle of entry into just one elevator without the other, you could do so, but you would have to have the full mathematical machinery of GR to do so. If you did, you would still get the same answer (ignoring the space curvature part, which is an added effect). The multiple elevator argument is simply a way to evade the detailed math of GR. 

    2. The speed (as measured in either elevator) is still c, even though the individual components, and hence the angle changes. If you add the vector velocity of the light and the downward velocity of the first elevator using the proper SR addition formulas (which have time dilation built in), you get the correct change in angle, with the magnitude still c. 

Cliff