# $E = mc^2$ derivation thought experiment seems to not have conserved energy

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?

• What is the meaning of -E on the right hand side? – stuffu Sep 8 '19 at 7:42
• 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 – 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.
• 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. – frogeyedpeas Sep 9 '19 at 15:29
• 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. – Jase Uknow Sep 9 '19 at 16:13