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According to Einstein's mass energy equivalence equation,

$$ E = mc^2$$

where $m$ is the rest mass and $c$ is speed of light.

According to Planck's equation,

$$ E=hv$$

If Planck's equation suggests that energy can only come in discrete packets, and if mass is equivalent to energy, then wouldn't this mean mass only comes in discrete packets?

I saw this post and I found one answer discussing my question, however it was downvoted with no explanation on why it's wrong. The accepted answer's give evidence for why it should be the case that mass is continuous, however, there are a few 'downvoted' answers which suggest the idea I said before and have been downvoted. So, I'm trying to understand why exactly those answers are wrong.

Edit: The answer where I saw it in , see answer by user "howyoudoin" and by user "hhh",

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    $\begingroup$ Does this answer your question? Is (rest) mass quantized? $\endgroup$
    – Dale
    Commented Aug 2, 2020 at 14:41
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    $\begingroup$ Did you ever read my question? that was exactly the question I had linked my text $\endgroup$
    – Cathartic Encephalopathy
    Commented Aug 2, 2020 at 14:45
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/330050/2451 Related: physics.stackexchange.com/q/169209/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Aug 2, 2020 at 14:53
  • $\begingroup$ FWIW energy of photons is quantised in units of $h\nu$ for a given frequency of $\nu$. It doesn’t necessarily imply that energy in itself is quantised because $\nu$ can be varied continuously (for a free charge). $\endgroup$
    – Superfast Jellyfish
    Commented Aug 2, 2020 at 17:01
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    $\begingroup$ Basically, I am asking the reason for the rejected answers being wrong $\endgroup$
    – Cathartic Encephalopathy
    Commented Aug 2, 2020 at 18:54

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There is a slight mistake in the way you are defining the energy. In general, the energy of a relativistic particle is $E^2 = p^2c^2 + m_0^2c^4$. Therefore, for a massless particle--like the photon--the energy becomes $E= pc$, not $E = mc^2$!

So in reality, what Plank's relation is telling us is that momentum is quantized for light, not the mass.

Edit: By momentum is quantized, I mean for a given frequency, there is a specified momentum the light can have. So for example, if you have a beam of light all at the same frequency and you could somehow measure the momentum (which is not absurd because since light has momentum it can exert a force, a phenomenon that is utilized in solar sails), the momentum would be a multiple of the frequency times some constant.

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    $\begingroup$ This seems to miss the point. The EM spectrum is continuous - all momenta and energies are possible. What $E=h\nu$ says is that for a given frequency $\nu$, the available sharp momenta are multiples of the corresponding momentum, but since frequency is continuous, all momenta are possible. $\endgroup$
    – ACuriousMind ♦
    Commented Aug 2, 2020 at 18:20
  • $\begingroup$ This is a good point; I will add this to the post. Thanks! $\endgroup$
    – Tabin
    Commented Aug 2, 2020 at 18:29
  • $\begingroup$ what exactly is a 'sharp momenta'? $\endgroup$
    – Cathartic Encephalopathy
    Commented Aug 2, 2020 at 18:58
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    $\begingroup$ @DDD4C4U sharpness generally refers to the “spread” (standard deviation) from the mean of an observable being relatively very small. $\endgroup$
    – Superfast Jellyfish
    Commented Aug 2, 2020 at 21:41

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