Photon mass (was: [Still off-top] Physics)

[Thomas 'PointedEars' Lahn]

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.

The physical fact is that as far as we know the photon mass is zero.  That 
is, the upper limit of the photon mass is so close to zero (at the time of 
writing, Adelberger et al. place it at < 10⁻²⁶ eV∕c² ≈ 1.783 × 10⁻⁵⁹ g 
≈ 1.957 × 10⁻³² m_e; see below) that we can assume it to be zero without 
introducing significant error in our *physical* calculations.

<http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html>;
but also <http://pdg.lbl.gov/2015/listings/rpp2015-list-photon.pdf> pp.,
and in general <http://pdg.lbl.gov> → “Summary Tables” → “Gauge and Higgs 
Bosons”.

> […]
> Since the generally accepted physical fact is that photons are never
> at rest

ISTM that this is rather a mathematical fact following from that energy–
momentum relation and the assumption that the photon (γ) mass m_γ is zero.  
It follows rather obviously then that E(m = m_γ = 0, p = 0) = 0.  But 
everything that exists has to have some (non-zero) energy.  So photons, and 
in general any objects, that have no mass *and* no momentum relative to a 
frame of reference cannot exist for an observer at rest relative to that 
frame.

> 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.

No.  We know that if our present understanding of physics is a sufficiently 
complete/correct description of reality, m_γ = 0 is a requirement.

Several reasons for that can be given; here are some:

1) If m_γ ≠ 0, the speed of light in vacuum could not be c as, according to 
special relativity, nothing that has mass m ≠ 0 can move at c relative to a 
frame of reference ­­– infinite energy in terms of that frame would be 
required for accelerating it to that speed: The kinetic energy of such a 
body is

  Eₖ = E_rel − E₀

where

  E₀ = E(m ≠ 0, p = 0) = mc²

by the energy–momentum relation (see above), and

  lim E_rel(v) = lim mc²∕√(1 − (v∕c)²) = +∞
  v→c            v→c

–, and photons are the particles of *light*.  So m = m_γ ≠ 0 would have 
profound and measurable implications on the propagation of light, that would 
*at the very least* be inconsistent with our knowledge of optics that 
precedes special relativity.

<https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies>

2) When one looks into

  E² = (p c)² + (m c²)²

further – which is not just an arbitrary equation that happens to fits 
observations very well, but is derived from the norm of the four-momentum –, 
one realizes that if, and only if,

  m_γ = 0,

then also

  E_γ = E(m = m_γ, p) = p c = ℏk c = ℎc/λ = ℎf

(ℏ = ℎ∕2π, k = ||k⃗|| = 2π∕λ — wavenumber, k⃗ — wave vector), which has been 
experimentally confirmed even *before* the postulation of special relativity 
in observations of the photoelectric effect (the kinetic energy imparted by 
light to electrons in a metal’s surface is proportional to its wavelength 
λ/frequency f).  So if m_γ ≠ 0, this would also have profound and measurable 
implications on the energy transported by light of different 
wavelengths/frequencies/colors that would be inconsistent with our knowledge 
of electromagnetic radiation that precedes special relativity.

<https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation#Norm_of_the_four-momentum>
<https://en.wikipedia.org/wiki/Planck_constant>

Finally,

3) if m_γ ≠ 0, the range of the electromagnetic interaction would not be 
infinite, as because of the uncertainty principle the range r of an 
interaction depends on the mass m of the particle that carries it:

  r(m) ≈ c∆t ≈ ℏ∕2mc.

This would have profound and measurable implications on fundamental 
interactions that would be inconsistent with our experimentally very well 
confirmed (and daily applied) knowledge of quantum mechanics.

<http://hyperphysics.phy-astr.gsu.edu/hbase/forces/exchg.html>


X-Post & F'up2 sci.physics.relativity

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.