Reading Steven Zheng's note on manipulating some of the equations of special relativity, I wondered how enlightening the calculations would be for someone who' d never studied relativity before. Therefore, I want to collect some simple, if not completely rigorous, arguments for mass-energy equivalence that make it clear why physics should have a relation between mass and energy. I'll make my own contributions in a few days and will select the clearest demonstrations from the comments. The connection between mass and energy is by far the most well known result in physics and is deeply curious, its fame notwithstanding.

**Photon ejection**

By the mid-1800s, it was known that light has energy, and a momentum, related by \(p = \frac{E}{c}\). This was established by two independent lines of argument, one being the Poynting vector that can be derived from Maxwell's equations, and the other being experiments that shone light on a vane of metal. With this connection, we can motivate the equivalence of mass and energy if we analyze a block that ejects a single photon.

Suppose that a block of mass \(M\), sits at rest. Let us call the total energy of this state. Then, the block ejects a photon of energy \(E_\gamma\), and momentum \(p=E_\gamma/c\) to the right. For the time being, we write the mass of the block after ejecting the photon as \(m\), not necessarily different from \(M\). Due to momentum conservation, the block moves with velocity \(p/m = E_\gamma/cm\) to the left.

Now, we look at this situation from the perspective of an observer who moves exactly downward at the speed \(v\). Momentum in any direction is conserved before and after the photon ejection (regardless of frame), so we can write down \(Mv = mv + E_\gamma^\prime/c\times \sin\theta\) (conservation of momentum in the \(y\)-direction). The photon's energy according to the moving observer is written as \(E_\gamma^\prime\) to reflect the blueshift from the relative motion (see below).

From the photoelectric effect (known to be frequency dependent in 1902), we have \(p=E_\gamma/c = hf/c\) in the original frame. To the observer, \(f\) is shifted a la \(\displaystyle f\rightarrow f\frac{c}{c-v_s}\approx f\left(1+v/c\right)\), so that \(p^\prime_\gamma = p_\gamma \left(1+v/c\right)\). The momentum in the \(y\)-direction is then \(p_\gamma^\prime \sin\theta\) which is \(p_\gamma^\prime v/c = p_\gamma \left(1+v/c\right)v/c \approx p_\gamma v/c\) (since \(v\ll c\)).

Thus, conserving momentum in the \(y\)-direction, we have \(Mv = mv + E_\gamma v/c^2\) or \(\Delta M = \Delta E / c^2\). Incredibly, this is the condition for momentum conservation. The block's loss of mass upon photon ejection is inescapable as long as momentum is conserved. In other words, if we hadn't allowed for differing \(M\), and \(m\), we would have had to sacrifice the conservation of momentum.

Einstein's original derivation of \(E=Mc^2\) involved the Lorentz transformations, and was rather needlessly complicated. This derivation is more in line with a version he published in 1945, long after his original works. It is interesting to realize that mass energy equivalence can be obtained, if a bit loosely, without the complex framework of special relativity.

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## Comments

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TopNewestHow is the picture related to this?

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It's the cover of this relativity book: Einstein Gravity in a Nutshell

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I wrote a new note on that includes the derivation of relativistic energy from first principles (make special relativity work under basic Newtonian definitions). Check it out here Discussion: Why E = mc2

However, I would say it is rather reductionist. So if you want to explore more of the

why? I'd be interested to see what you can find.Log in to reply

Yes, it is fine given \(p=\gamma m v\) with the form for \(\gamma\) and all that. But that formula isn't known by someone who hasn't studied relativity. I'm thinking of arguments to motivate the mass energy equivalence without assuming special relativity.

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