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Is there any Matter identified as being without mass? I have heard that Photon's meet this description.

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Is there any Matter identified as being without mass? I have heard that Photon's meet this description.

The answer depends somewhat on what you mean by "matter." Historically, people used to classify things as light or as matter, but that classification isn't really seen as the right one anymore. Relativists these days talk about "matter fields," by which they mean pretty much anything that has energy or momentum. By this usage, both electromagnetic waves and static electromagnetic fields are referred to as matter.

Just to confuse things further, there are two different conventions for talking about mass. Some older books use mass to mean what is now referred to as "relativistic mass," which is a quantity that varies with motion. The universal convention today, except in popularizations, is to define "mass" using the equation $m^2=E^2-p^2$ (in units with $c=1$), which does not vary with motion.

So by modern conventions, electromagnetic fields, including electromagnetic waves are "matter" (matter fields) that have no "mass" (invariant mass).

BTW, there is nothing quantum mechanical about this issue, so if you want to talk about photons, you can, but you can also just talk about electromagnetic fields. Another example of a massless matter field would be the gluon. Gravitational waves are massless, but relativists don't usually refer to them as a matter field, because they aren't explicitly written in the Einstein field equations as part of the stress-energy tensor -- but they do contribute to the gravitational mass of an object as measured by an observer at a distance.

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Yes, photons have energy and are massless. The photon energy depends on the frequency and hence the wavelength of the photon. The equation is $E = h\nu$ where $h$ is Planck's constant and $\nu$ is the frequency. Since $c = \lambda\nu$ where $c$ is speed of light and $\lambda$ is wavelength we can write $\nu = c/\lambda$ hence $E = hc/\lambda$. This is just the energy in terms of the wavelength $\lambda$.

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  • $\begingroup$ Are photons the only phenomena without mass? $\endgroup$ – Marshall O'Donovan Dec 18 '17 at 19:36
  • $\begingroup$ That i'm not sure about. It's the only one I know of. As far as we know, gravitational waves are massless too. $\endgroup$ – jmh Dec 18 '17 at 19:38
  • $\begingroup$ We once thought neutrinos were massless but we have since learned they do have a very small mass. $\endgroup$ – jmh Dec 18 '17 at 19:40
  • $\begingroup$ @MarshallO'Donovan Gluons are also massless. $\endgroup$ – probably_someone Dec 18 '17 at 19:46
  • $\begingroup$ @MarshallO'Donovan Also, spin waves in ferromagnets are quantized in terms of massless Goldstone bosons. $\endgroup$ – probably_someone Dec 18 '17 at 19:53
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The photon, gluon, and graviton are massless particles. https://en.wikipedia.org/wiki/Massless_particle#References.

From special relativity, the energy, $E$ and momentum $\vec{p}$ of a massless particle are related by the equation: $$E = \left| {\vec p} \right|c$$

Here's an interesting passage from "An Introduction to Elementary Particle Physics" by David Griffiths, pgs 90-91:

You may well ask...what does determine the momentum and energy of a massless particle? Not the mass (that's zero by assumption); not the speed (that's always c). How, then, does a photon with an energy of $2 \mathrm{eV}$ differ from a photon with an energy of $3 \mathrm{eV}$? Relativity offers no answer to this question, but curiously enough quantum mechanics does, in the form of Planck's formula (v in the formula below stands for frequency) : $$E = hv$$ It is the frequency of the photon that determines its energy and momentum: The $2 \mathrm{eV}$ photon is red, and the $3 \mathrm{eV}$ photon is purple!

Sticking with Griffiths use of $v$ for frequency... if $\lambda$ stands for the wavelength, $$c = v \times \lambda$$ $$v = \frac{c}{\lambda}$$ $$E = hv = \frac{hc}{\lambda}$$

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  • $\begingroup$ The material about wavelength, frequency, and quantum mechanics doesn't have anything to do with the question. The first two sentences are the answer. $\endgroup$ – Ben Crowell Dec 18 '17 at 22:11

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