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, 2019 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 $\endgroup$ Sep 8, 2019 at 14:55

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Sep 9, 2019 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, 2019 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.