Known that $E=hf$, $p=hf/c=h/\lambda$, then if $p=mc$, where $m$ is the (relativistic) mass, then $E=mc^2$ follows directly as an algebraic fact. Is this the case?
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As you may know, photons do not have mass. Relating relativistic momentum and relativistic energy, we get: $E^2 = p^2c^2+(mc^2)^2$. where $E$ is energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light. As mass is zero, $E=pc$. Now, we know that $E=hf$. Then we get the momentum for photon. Note that there is a term called effective inertial mass. Photon does have it. |
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Here's another way to think about it (personally, I think this addresses the question most directly): $E = hf$ and $p = \frac{hf}{c}$ both apply to photons. What those get you is simply that $E = pc$, so you can conclude that $E = pc$ should be valid for photons. And it is. Now, your question is worded to ask whether you can start with $p = mc$ and plug in $E = pc$ to get $E = mc^2$. But I think what you really want to know is, can you start with $E = mc^2$ and use it with $E = pc$ to derive $p = mc$? The answer is, of course, no. $E = mc^2$ doesn't apply to photons. In fact, there is no case in which $E = mc^2$ and $E = pc$ both apply to the same object. So you can never validly combine them. The former is for objects at rest, for which $p = 0$, and the latter is for massless objects, for which $m = 0$, and which always move at the speed of light. As others have shown, they're both special cases of $E^2 = p^2 c^2 + m^2 c^4$. Incidentally, I can't think of a single physical system for which $p = mc$ is satisfied. |
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As a further elaboration, let's look at this from the angle of relativistic momentum. Recall that momentum, in relativistic mechanics, is not a linear function of velocity as it is in Newtonian mechanics where $p = mv$. In relativistic mechanics: $p = \gamma m v$ $m$ is the invariant mass $\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$ Clearly, for a non-zero $m$, $p \rightarrow \infty$ as $v \rightarrow c$ Now, please keep in mind that $p$ in the relativistic energy relationship is not just $mv$ but is the relativistic momentum $\gamma m v$: $E^2 = (\gamma m v c)^2 + (mc^2)^2$ From this, it's clear that the relativistic energy is: $E = \gamma m c^2$ So, if we fix $E$ and let $m \rightarrow 0$, we find that $v \rightarrow c$ in the limit. |
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According to Special Relativity the relativistic energy for a particle is: $E^2= m^2c^4+p^2c^2$ The invariant quantity under relativistic transformations is the rest mass $m$ of the particle. For a photon $m=0$ Using some simple algebra it is found $E=pc$ for a photon. You will see this preserves the frequency and energy relationship. The error in the question is that momentum $p$ is always related to mass&velocity ($p=mv$ where $c$ is placed in as $v$ for the photon)- which for a massless particle this does not apply. |
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