NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/physics/21858

Text File | 1992-12-29 | 4.4 KB | 111 lines

Newsgroups: sci.physics
Path: sparky!uunet!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!sunova!sscvx1.ssc.gov!draper
From: [e-mail]
Subject: Re: What are the conditions for Mass <-> Energy conversion
Message-ID: <[e-mail]>
Lines: 98
Sender: [e-mail] (News Admin)
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Organization: Superconducting Super Collider Laboratory
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Tue, 29 Dec 1992 21:41:12 GMT

Sorry for the bandwidth, but I need to include alot of what I'm replying to
for clarity...
In article <[e-mail]>, [e-mail] (Fred Rice) writes:
> In <[e-mail]> [e-mail] (Jon Bell) writes:
>>In article <[e-mail]> [e-mail] (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.
>
That's because equation 6 is wrong.
What you probably meant to say is that conservation of energy demands
E^2 = (E1 + E2)^2
so that
E^2 = p^2 = p1^2 + m1^2 + p2^2 + m2^2 + 2E1E2
and since by conservation of momentum
p = p1 + p2
we then get
p1^2 + p2^2 + 2p1p2cos(A) = p1^2 + m1^2 + p2^2 + m2^2 + 2E1E2.
Solving for cos(A) gives
cos(A) = [m1^2 + m2^2]/2p1p2 + E1E2/p1p2
Now as long as m1 and m2 are nonzero, the first term on the right is
positive-definite and the second term is greater than 1. Thus there is no real
solution for A.

But I generally only take off a couple of points for algebra errors.

Paul Draper
University of Texas at Arlington
Mythink, not UTAthink, not SSCthink, not GOVthink