There is an interesting derivation of $E=mc^2$ given here which I have updated to use relativistic momentum and relativistic kinetic energy. I find when doing the derivation that it doesn't appear that kinetic energy is conserved.

The setup: Consider an object $A$ moving at a speed $v$ to the right of an object $B$. Let $m_A$ be the mass of $A$, Let $KE_A$ be the kinetic energy of $A$, let $PE_A$ be the potential energy of $A$ and let $P_m$ be the momentum of $A$.

We consider the case of $A$ spontaneously emitting two photons (or generally equivalent pulses of light), one moving vertically up and one vertically down relative to $A$ in a frame where $A$ is at rest.

Frame where $A$ is at rest

$$\begin{matrix} \begin{matrix} \textbf{Before Emission} \\ \text{Potential Energy} = PE_A \\ \text{Horizontal Momentum} = 0 \\ \text{Kinetic Energy} = 0\\ \text{Mass} = m_a \\ \text{Photon Vertical Momentum} = 0 \\ \text{Photon Horizontal Momentum} = 0 \\ \text{Photon Energy} = 0 \end{matrix} & \begin{matrix} \textbf{After Emission} \\ \text{Potential Energy} = PE_A- E \\ \text{Horizontal Momentum} = 0 \\ \text{Kinetic Energy} = 0\\ \text{Mass} = m_a' \\ \text{Photon Vertical Momentum} = \frac{E}{2c} - \frac{E}{2c}=0 \\ \text{Photon Horizontal Momentum} = 0 \\ \text{Photon Energy} = E \end{matrix} \end{matrix} $$

From here we can verify that the total Potential + Kinetic Energy is the same before and after, and the momentum is the same before and after.

Frame where $A$ is moving at speed $v$ to the right

$$\begin{matrix} \begin{matrix} \textbf{Before Emission} \\ \text{Potential Energy} = PE_A \\ \text{Horizontal Momentum} = \frac{m_Av}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \text{Kinetic Energy} = \frac{m_Ac^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_Ac^2 \\ \text{Mass} = m_a \\ \text{Photon Vertical Momentum} = 0 \\ \text{Photon Horizontal Momentum} = 0 \\ \text{Photon Energy} = 0 \end{matrix} & \begin{matrix} \textbf{After Emission} \\ \text{Potential Energy} = PE_A- E \\ \text{Horizontal Momentum} = \frac{m_A'v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \text{Kinetic Energy} = \frac{m_A'c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_A'c^2\\ \text{Mass} = m_a' \\ \text{Photon Vertical Momentum} = \frac{E}{2c} \frac{\sqrt{c^2 - v^2}}{c} - \frac{E}{2c}\frac{\sqrt{c^2 - v^2}}{c} =0 \\ \text{Photon Horizontal Momentum} = \frac{E}{2c} \frac{v}{c} \\ \text{Photon Energy} = E \end{matrix} \end{matrix} $$

The trick then to derive $E=mc^2$ is to simply observe that in the left hand side before emission the total horizontal momentum (sum of horizontal momentum and photon horizontal momentum) was $ \frac{m_a v}{\sqrt{1 - \frac{v^2}{c^2}}} $ and on the right hand side it is $\frac{m_a' v}{\sqrt{1 - \frac{v^2}{c^2}}} + \frac{E}{c} \frac{v}{c} $ and by equating these we find that the change in mass $m_a - m_a'$ depends on the energy $E$ in the famous way.

But I realized with this model there is a problem:

The Question:

If you look at the frame where $A$ is moving at speed $v$. Then one sees that that the sum of Potential Energy + Kinetic Energy + Photon Energy is ONLY conserved if $m_a$ remains the same. Since we have on the left hand side:

$$ PE_A + \frac{m_Ac^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_Ac^2 $$

And on the right hand side:

$$ PE_A - E + \frac{m_A'c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_A'c^2 + E $$

If we assume $m_a$ changes then some energy appears to have disappeared. How do I make this problem go away?

  • $\begingroup$ What is the meaning of -E on the right hand side? $\endgroup$ – stuffu Sep 8 '19 at 7:42
  • $\begingroup$ I tried include the signs of momentum, when they cancel so when you encounter a $a-a$ term it’s clearer it refers to the two photons traveling in opposite directions. There is also a $-E$ in the potential energy on right hand side to account for where the photon was emitted $\endgroup$ – frogeyedpeas Sep 8 '19 at 14:55

To simplify, let me set the $PE_A$ to zero. Which isn't used anyway.

The conservation of energy before is: $m_A c^2 = m_A'c^2 + 2E$

Lorentz transformations don't change the perpendicular part of the the 4-vector to the boost. Also there is no parallel component in your setup. So the new energy conservation equation after the boost is: $\gamma m_A c^2 = \gamma m_A'c^2 + 2 \gamma E$

Canceling the $\gamma$ from both sides, you see that the energy is conserved.

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  • $\begingroup$ Why does the $\gamma$ apply to the photon terms as well? This is not intuitive to me since photons have the same velocity in all inertial frames. So I assumed that the total energy of an individual photon being emitted is still (in your interpretation my question) $E$ and the total momentum is still $\frac{E}{c}$ (up to a choice of direction) even in the moving frame. $\endgroup$ – frogeyedpeas Sep 9 '19 at 15:29
  • $\begingroup$ Photons have the same speed, $c$, in any frame but not the same energy. Roughly speaking, the energy of the photon is the frequency, and because of time-dilation the frequency can change (even if the photon is moving transverse to the motion). It's called the transverse Doppler shift and was first experimentally observed in the Ives-Stilwell experiment. $\endgroup$ – Jase Uknow Sep 9 '19 at 16:13

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