From: darice@yoyo.cc.monash.edu.au (Fred Rice)
Subject: Re: What are the conditions for Mass <-> Energy conversion
Organization: Monash University, Melb., Australia.
Date: Mon, 28 Dec 1992 02:42:10 GMT

In <1992Dec26.134900.7041@hubcap.clemson.edu> jtbell@hubcap.clemson.edu (Jon Bell) writes:
>In article <8091@tekig7.PEN.TEK.COM> bhides@tekig1.PEN.TEK.COM (Sandhiprakas J Bhide) writes:
>>
>>Now, if it is indeed true, why is that sunrays or for that matter
>>any other energy such as heat does not by itself convert into mass?
>>That is while I am shaving in front of the mirror, why is that the
>>light emnating from the lamp does not get converted into mass?
>>
>Various "conservation laws" have to be satisfied.  For example, a single
>photon (quantum of light energy) traveling through "empty" space 
>cannot spontaneously convert into a material particle (or convert part
>of its energy into one) because it would violate conservation of 
>momentum.  However, if a high-energy (e.g. gamma-ray) photon passes near
>an atomic nucleus, it can produce an electron-positron pair, because 
>the nucleus can recoil slightly to conserve momentum.

In more detail, according to special relativity, the following relation
holds:

E^2 = (p^2)(c^2) + (m^2)(c^4)   [1]

where E=energy, p=momentum, m=mass,
and c=speed of light, which is a constant.

We require conservation of energy and momentum, i.e. of E and p.

Since a photon has no mass, it's energy squared according to [1] is

E(photon)^2 = (p^2)(c^2)   [2]

whereas the energy of a single particle is the full equation given in [1].
Since we require energy to be conserved (and thus energy squared to be
conserved), equating [2] and [1], for the initial photon before
conversion to mass and the particle after the photon is converted to
mass, we get

(p^2)(c^2) = (p^2)(c^2) + (m^2)(c^4)   [3].

The momentum (p) on the left hand side (for the photon) and on the right
hand side (for the particle) are identical because we require
conservation of momentum. 

Equation [3] simplifies to

0 = (m^2)(c^4)   [5].

Since c is a nonzero constant, in equation [5] we require m=0, i.e. a 
single photon cannot spontaneously convert into a single particle with mass.

How about a photon converting into two particles?

Let's say our two particles, 1 and 2, have masses m1 and m2 and momenta
p1 and p2 respectively, and let's leave p to be the momentum of the
photon.

Then equation [3] becomes:

(p^2)(c^2) = (p1^2)(c^2) + (p2^2)(c^2) + (m1^2)(c^4) + (m2^2)(c^4)
           = (p1^2 + p2^2)(c^2) + (m1^2 + m2^2)(c^4)               [6]

However, due to conservation of momentum, we require _p_ = _p1_ + _p2_, 
where _a_ means that _a_ is a vector,
and therefore _p_^2 = (_p1_ + _p2_)^2 = p1^2 + p2^2 + 2p1p2cos(A), where A
is the angle between _p1_ and _p2_.  Thus [6] becomes

(p1^2 + p2^2 + 2p1p2cos(A))(c^2) = (p1^2 + p2^2)(c^2) + (m1^2 + m2^2)(c^4)

=> 2p1p2cos(A) = (m1^2 + m2^2)(c^2)  [7]

Equation [7] is the equation that needs to be fulfilled in
order to have a photon spontaneously convert into two particles (such as
an electron-positron pair).  

This seems to me to be possible.

For example, let's say an electron-positron pair is created, each having
a mass of m0.  (By the way, in this case m0 = 0.511 MeV/c^2.)
For simplification, let's also assume that the angle A -> 0, and thus 
cos(A) -> 1.  And, for further simplification, let's assume that the
magnitude of the momenta of particles 1 and 2 are equal, i.e., that
p1 = p2.  Then [7] becomes

2(p1^2) = 2(m0^2)(c^2)  [8]

According to Relativity, the relation of the momentum, p, to the
velocity, v, of a particle of mass m0, is

p =  m0 * v                [9]
    -----------------
    [1-(v/c)^2]^(1/2) 

Substituting this into [8], we get

2(m0^2)(v^2)
------------ = 2(m0^2)(c^2)
[1-(v/c)^2]

=>   (v^2)/(c^2) = 1 - (v^2)/(c^2)    [assuming m0 does not equal 0]

=>   2(v^2) = c^2

=>   v^2 = 1/2(c^2)

which is a possible result (i.e. v < c).
     
So, it seems to me that a photon *can* spontaneously convert into two
particles with mass, as long as it has enough energy.

So, could Jon Bell or someone else remark on this?
My math, assuming it is mistake-free, contradicts what Jon Bell was claiming.

>In this example, an electron-positron pair has to be produced (not just an
>electron or just a positron) so that electric charge is conserved.

>Actually, in the case of light rays coming from your shaving lamp, the
>photons have so little energy (a few eV apiece) that they cannot possibly
>produce even the least massive particles (electrons/positrons, 511,000 eV
>apiece).

>Jon Bell / Dept. of Physics & C.S. / Presbyterian College / Clinton SC USA
>would violate 

 Fred Rice
 darice@yoyo.cc.monash.edu.au   
Newsgroups: sci.physics