[Still off-top] Physics [was Requests author discusses MentalHealthError exception]

[Gene Heskett]

On Saturday 05 March 2016 08:11:46 Oscar Benjamin wrote:

> On 5 March 2016 at 02:51, Gregory Ewing <greg.ewing at canterbury.ac.nz> 
wrote:
> >  The masslessness of photons comes from an extrapolation
> >
> >> that leads to a divide by infinity: strictly speaking it's just
> >> undefined.
> >
> > No, it's not. The total energy of a particle is given by
> >
> >    E**2 == c**2 * p**2 + m**2 * c**4
> >
> > where p is the particle's momentum and m is its mass.
> > For a photon, m == 0. No division by zero involved.
> >
> > For a massive particle at rest, p == 0 and the above
> > reduces to the well-known
> >
> >    E == m * c**2
>
> The distinction I'm drawing is between physical fact and mathematical
> convenience. For other particles we can say that the 1st formula above
> holds with m taken to be the mass of the particle at rest. We can
> extend that formula to the case of photons which are never at rest by
> saying that in the case of photons m=0. That's nice and it's
> mathematically convenient in the calculations. It's analogous to
> extending the natural definition of the factorial function by saying
> that 0!=1. We can't prove that 0!=1 but it's useful to define it that
> way. It wouldn't be a disaster to simply leave 0! undefined: it would
> just make some equations a little more complicated.
>
> Since the generally accepted physical fact is that photons are never
> at rest we are free to define their "rest mass" (use any term you
> like) to be anything that is mathematically convenient so we define it
> as zero because that fits with your equation above. Turning full
> circle we can then use the equation above to say that they are
> massless since they would hypothetically be massless in some other
> situation even though genuinely massless photons are not thought to
> exist in physical reality (unless I'm really out of date on this!).
>
> >> Something I don't know is if there's some theoretical reason why
> >> the binding energy could never exceed the sum of the energies of
> >> the constituent particles (resulting in an overall negative mass).
> >
> > Conservation of energy would be one reason. If you
> > put two particles together and got more energy out than
> > went in, where did the extra energy come from?
>
> That's the point: the energy balance would be satisfied by the
> negative energy of the bound particles. The binding energy can be
> defined as the energy required to unbind the particles (other
> definitions such as André's are also possible). From this definition
> we see that the binding energy depends on the binding interaction
> (electromagnetic or whatever) that binds the particles together.
>
> The only examples I know of where the binding energy is computed
> approximately for e.g. a hydrogen atom predict that the binding energy
> is proportional to the (rest) mass of the bound particle(s). If it's
> guaranteed that the binding energy always somehow comes out
> proportional to the mass of the particles with a coefficient
> necessarily smaller than 1/c**2 then you could say that the bound
> product could never have negative energy. I just can't see off the top
> of my head an argument to suggest that this is impossible.
>
> --
> Oscar

I've never heard of a massless photon, and they do exert a push on the 
surface they are reflected from, its even been proposed to use it as a 
space drive.  The push is miniscule indeed at normal illumination levels 
but some have calculated how much laser power it would take to move 
something like a solar sail. Practically, the cost of the energy and the 
size of the laser needed are impractical.

Cheers, Gene Heskett
-- 
"There are four boxes to be used in defense of liberty:
 soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)
Genes Web page <http://geneslinuxbox.net:6309/gene>