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

[Oscar Benjamin]

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exist in physical reality (unless I'm really out of date on this!).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