We know light is made of photons and so it should not have mass, but light is a form of energy (light has energy) and has velocity ($c$), so according to $E=mc^2$, light should have mass... So what is correct?
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$\begingroup$ This might help. physics.stackexchange.com/q/540485 $\endgroup$– mmesser314Commented Dec 17, 2021 at 13:18
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/2229/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Dec 17, 2021 at 14:50
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4 Answers
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This the complete Einstein equation: \begin{equation} E^2=m^2c^4+p^2c^2 \end{equation} Photons don't have mass but they still carry energy due to their momentum.
answered Dec 17, 2021 at 12:38
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2$\begingroup$ It would be worth explaining, for OP's sake as well as mine, how the concept of "momentum" can exist without mass $\endgroup$– FShrikeCommented Dec 17, 2021 at 12:39
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1$\begingroup$ +1 In some exotic cases photons can be attributed a non-zero mass: see Photon effective mass in plasma. $\endgroup$– Roger V.Commented Dec 17, 2021 at 12:49
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$\begingroup$ It's because the definition of momentum in Newtonian mechanics $p=mu$ is not correct in special relativity. The correct expression for momentum is given by Energy-Momentum in my post. There is a way to make momentum look like the Newtonian momentum. In special relativity $p=\gamma m u$ where $\gamma$ is the Lorentz factor. At first glance, it may still look like the photon should have zero momentum but because it travels with the speed of light the Lorentz factor blows up and you have something of the form $p=\frac{0}{0)$. $\endgroup$ Commented Dec 17, 2021 at 13:13
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1$\begingroup$ Even in the absence of special relativity and E =mc^2 momentum must still not only be present in matter, but also in the EM fields themselves. Since there is a propagation delay in the EM field. Conservation of momentum in the form just "mv" is actually violated. Net Force is the rate of change of momentum of a body. If the EM field can produce a net momentum increase/a force. Then the field itself MUST have momentum! $\endgroup$ Commented Dec 17, 2021 at 13:37
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The value $E/c^2$ has sometimes been called relativistic mass, but we make an effort nowadays to discourage that term because it refers to a frame-dependent quantity, and it's more helpful to focus on the invariant mass. A free particle has energy-momentum relation $E^2=m_0^2c^4+p^2c^2$ (see here for a generalization to a non-free particle, but we'll overlook that for now), with $m_0$ a symbol we have to use for invariant mass, instead of the preferred $m$, if someone's already claimed $m$ for $E/c^2$ in the discussion. I'll stick with $m_0$ for now, since we're already "in this deep".
With that preamble out of the way, a photon in the vacuum has $m_0=0,\,E=|p|c>0$. The photon has zero invariant mass, and no rest frame. This makes "rest mass", another term that's been used for invariant mass, unnecessarily confusing! (It's less of a problem for particles with $m_0>0$, which do have a rest frame; indeed, light in a medium with refractive index greater than $1$ has been modelled in terms of $m_0>0$ photons.) But any photon, regardless of the context, has a frame-dependent $E/c^2>0$, which was discussed above.
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$\begingroup$ I will make the same comment as I did on the post below; both OP and myself share the confusion over how a massless quantity, or perhaps a quantity with zero $m_0$, may have a momentum (considering $\rho=mv$) $\endgroup$– FShrikeCommented Dec 17, 2021 at 12:42
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1$\begingroup$ @FShrike for formula $\vec{p}=m\vec{v}$ for momentum is true in Newtonian mechanics. When considering relativity, the momentum is part of a $4$-vector, together with the energy $(E/c, p_x, p_y, p_z)$. The magnitude of this vector is invariant, and exactly gives the "invariant" mass $E^2 - |\vec{p}|^2 c^2 = m^2_0 c^4$. For massive particles with velocities much smaller than the speed of light, you can convince yourself that $E \simeq m_0 c^2 + p^2/2m$ which exactly gives the kinetic energy term. If we identify it with $E = m_0c^2+mv^2/2$ we get $p=mv$ in the much-slower-than-light approximation $\endgroup$– user275556Commented Dec 17, 2021 at 12:57
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$\begingroup$ @yyy Thank you. I am currently writing a question about the very subject of definitions; how is momentum fundamentally defined here, since $p=mv$ is not an adequate definition? $\endgroup$– FShrikeCommented Dec 17, 2021 at 13:00
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$\begingroup$ @FShrike as a definition, usually the momentum will be defined via the action (see here en.wikipedia.org/wiki/Four-momentum under "derivation"), or via the components of the Stress-Energy tensor (en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor). Another definition is through symmetry and Noether's theorem, as the quantity that corresponds to invariance under continuous translations in space $\endgroup$– user275556Commented Dec 17, 2021 at 13:09
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$\begingroup$ @FShrike If I may use relativistic mass $m$ without upsetting anyone,$$\vec{p}=m\vec{v},\,E=mc^2\implies\vec{p}=c^{-2}E\vec{v}.$$The last equation can be used even when $m_0=0$. Indeed, it implies $E^2-c^2p^2=E^2(1-v^2/c^2)$, which agrees with $m_0^2c^4$ at $v=c$ (then both are $0$) as well as $v<c$ (then both are $m_0^2c^4>0$). $\endgroup$– J.G.Commented Dec 17, 2021 at 15:18
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Does light have mass or not
Light is the word we use for classical electromagnetic radiation at optical frequencies.
Electromagnetic radiation is emergent from a large number of photons. The figure in this experiment is a clear proof that the addition of photons, elementary particles of zero mass and energy equal to $hν$ where $ν$ is the frequency of the classical electromagnetic wave arising from the confluence of very many photons .
Photons are described by four vectors in the special theory of relativity . The addition of the four vectors of two non collinear photons has an invariant mass even if each individual photon has zero mass,
thus the built up light will have an invariant mass, within the formalism of special relativity.
Experimental proof is the decay of the pi0 to two gamma. The added four vectors of the two gamma have the invariant mass of the pi0. (Related )
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When the value of mass is given as mc^2=hf, or m=hf/c^2, this is the equivalent newtonian mass, which appears in momentum for example.
Under Special relativity, it has no mass.
