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From: AI.Mayank@MCC.ARPA
Newsgroups: net.physics
Subject: Re: Faster than light
Message-ID: <368@sri-arpa.ARPA>
Date: Tue, 9-Jul-85 16:03:51 EDT
Article-I.D.: sri-arpa.368
Posted: Tue Jul  9 16:03:51 1985
Date-Received: Sun, 14-Jul-85 09:06:03 EDT
Lines: 55

From:  Mayank Prakash 

>         Mayank, I think that some people might ascribe too much meaning to the
>wave function from your discription.  After all, a wave function is
>fundamentally different from an electromagnetic wave in that the EM wave
>has physical reality (i.e. physical energy density = mass) at every
>point of the wave all the time independently of any observer.  The QM
>wave function only has "reality" when it is observed.

Not true. Let us not confuse between the classical and quantum electromagnetic
fields. The quantum EM fields are as much unobservable as the wave function.
The classical EM field, on the other hand, is just the average of a large
number of photons. The quantum EM field does not have any physical energy
density. The energy density of the classical EM field at any point is the
average energy density of the photons at that point = energy of a photon *
mod-squared of the wave function of the photon at that point. It is hard for me
to see how one can ascribe *more reality* (what does it mean anyway) to the EM
field than the wave function of the photons.

>        Apropo of your discussion on simultaneity.  Consider two observers
>with synchronized perfect clocks starting out, in a relativistic way, from
>the same spot on the equator, in opposite directions, parallel to the
>equator, at the same instant that your photon leaves the north pole.  From
>symmetry, the photon reaches each at the "same time".  Each makes a
>measurement at that "instant".  However, from each observer's point of
>view, the other observer's clock is running slow.  So although each agrees
>that each made her measurement when her own clock read time T, each also
>insists that she made her measurement before the other did, since she
>observed that the other's clock was running slow.  Hence the wave function
>"instantaneously" collapsed with the "first" measurement and the other
>observer couldn't have made the measurement because the wave funtion had
>collapsed prior to time T on the other observer's clock.  But since the
>situation is completely symmetric each observer makes the same argument
>that it is impossible for the other to make the measurement!
>        How can this be???

Please be advised that making a measurement is not the same as seeing the
photon. The measurement in this case can have two outcomes - either you find
the photon, or you don't. Secondly, the photon does not reach the two observers
at the same time, only the wave function does. Only one of them can *see* the
photon, not both, despite the symmetry. The only catch here is that no matter
who makes the measurement first in whatever frame, the collapse of the wave
function in each frame is consistent, i.e., that it collapses to the same
region (which may be different from the positions of both observers, or  may
coincide with the position of one of them). This is the sticking point - how
does the wave function know how to collapse. As I have stated before, as long
as it does know however, there will not be any observable contradictions.

- mayank.

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II  Mayank Prakash  AI.Mayank@MCC.ARPA      (512) 834-3441		II
II  9430 Research Blvd., Echelon 1, Austin, TX 78759.			II
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