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Since photons have an energy given by $E=h\nu$, we could define a particle whose rest mass is such that it has the same energy than the photon: $E=m_0c^2 \Longrightarrow m_0=\frac{h\nu}{c^2}$. We now have a way of computing the linear momentum of the photon with an analogous massive particle, which is given by: $p=mv=\frac{h\nu}{c^2}\cdot c \Longrightarrow p_\gamma=\frac{E_\gamma}{c}$, which yields the correct expression for the linear momentum of a photon.

My question is, is this reasoning correct? Is it valid to deduce the properties of a boson by making an analogous reasoning with its equivalent baryon in terms of energy? What do you think are the limitations of this reasoning? (Of course, one of the strongest drawbacks is a massive particle cannot travel at the speed of light as I have assumed when calculating the linear momentum. However, I do think it might be useful in some cases to define a "photonic mass" as to simplify calculations)

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  • $\begingroup$ which is given by: đť‘ť=đť‘šđť‘Ł ...... is the variable v velocity or mu frequency? $\endgroup$
    – PhysicsDave
    Commented Jun 17, 2023 at 11:43
  • $\begingroup$ Which calculations does this simplify ? for a massless particle you get $p=E/c$ directly from $m^2=0$. This analogy just seems to obfuscate the actual physics $\endgroup$
    – user341440
    Commented Jun 17, 2023 at 12:14
  • $\begingroup$ When I learned Physics many years ago your equation was used to define a "mass" for the photon, and similarly relativistic mass and rest mass were distinguished. However the word "mass" is no longer used in this way; it now refers only to the rest mass. $\endgroup$
    – Peter
    Commented Jun 17, 2023 at 12:19
  • $\begingroup$ A photon always travels at the speed of light. A massive particle never does. So you would be comparing apples to oranges. $\endgroup$
    – mmesser314
    Commented Jun 17, 2023 at 16:08

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The full Einstein mass-energy-momentum relation is $$\tag1E=\sqrt{(m_0c^2)^2+(pc)^2}$$ and so for photons $$\tag2\text{(massless, e.g. photons)}\qquad\therefore m_0=0\qquad\implies\qquad\frac Ec=p$$ There is thus no need to invoke whatever it is you wanted to do.

My question is, is this reasoning correct?

It is not. You have a double inconsistency: First of all, photons are massless, as in, its rest mass should be zero, but you are pretending for it to have mass.

And then you are wanting its mass to be inconsistent. If $m_0$ is defined as you did, then the mass in momentum $p=mv$ is actually supposed to have $m=\gamma m_0\geqslant m_0$ and diverges to infinity as $v\to c,$ yet you have violently decided that $m=m_0$ when it cannot be.

However, I do think it might be useful in some cases to define a "photonic mass" as to simplify calculations

The calculations were already simple and your thing makes it even more complicated.

However, when photons go into materials and interact with materials, it can slow down from the universal speed limit. When that happens, it is useful to define a photonic mass. It says nothing of the most important fundamentals, though. Just a mathematical trick to cheaply cover some stuff. Nobody assigns this any actual real-world weight.

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