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When I learned the theory of relativity, I was taught that the equation of energy conservation for a moving particle is: $$E^2= (mc^2)^2 + (pc)^2.$$ When $p = 0$, this equation becomes $E=mc^2$, but I fail to see how this explains that $m$ (the mass at rest) can be converted to pure energy, and how it can be done (by atom nuclei fission or fusion for example). The only thing we can say is that if you have a certain distribution of energy, or its equivalent distribution of mass, the space-time would curve in the same way.

The equation above assumes a fixed value of $m$ originally, there is nothing that indicates that $m$ can vary.

Edit: In the other "duplicate" linked question it is assumed that Einstein's theory explains that mass and energy flow back and forth. I want to know how it explains that.

Further comments: In the course, after we derived that formula, the professor said that that's the explanation of why matter is energy. However, I was never satisfied with this because in the whole theory of special relativity there is no description of any flow between matter and energy. Like I said, only that they would be equivalent in certain situations, like when calculating the curvature of space-time. What I understand is that we need additional theories like the ones mentioned in the answers to fully explain it. Special relativity alone is not enough.

Agustín
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3 Answers3

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In contrast to your claim, the formula $$ E= c \sqrt{m^2 c^2+ \vec{p}^2} \tag{1} \label{1} $$ is simply the relativistic energy-momentum relation of a particle with mass $m$ and not "the equation of energy conservation for a moving particle". Energy-momentum conservation in a scattering (decay) process with $n_i$ particles in the initial state and $n_f$ particles in the final state is expressed by $$ \sum\limits_{k=1}^{n_i} p_k= \sum\limits_{\ell=1}^{n_f} q_\ell \tag{2} \label{2} $$ in terms of the four-momenta $p_k=(E_k/c, \vec{p}_k)$ and $q_\ell=(E_\ell/c, \vec{q}_\ell)$, respectively. The formula \eqref{2} alone cannot tell you if a specific reaction can occur in nature. But it tells you that if a certain scattering (or decay) process occurs, the associated energies and momenta are restricted by \eqref{2}.

A simple example of a decay process observed in nature is the (electromagnetic) decay of a neutral pion into a photon pair: $\pi^0 \to \gamma \gamma$. In the rest frame of the pion, eq. \eqref{2} corresponds to $$ (M_\pi c, \vec{0})= (| \vec{q}_1|, \vec{q}_1) +(|\vec{q}_2|, \vec{q}_2), \tag{3} \label{3} $$ telling you that the two photons in the final state have opposite momenta ($\vec{q}_1=-\vec{q}_2$) and equal energies $|\vec{q}_1| c = |\vec{q}_2| c= M_\pi c^2/2$, i.e. the rest energy $M_\pi c^2$ of the pion (in its rest frame) in the initial state is converted to the total energy of the two-photon system in the final state.

An example of a ficticious process not occuring in nature could be the decay of a proton into two photons. Although not forbidden by energy-momentum conservation (simply replace $M_\pi$ by $m_p$ in the previous example), it is forbidden by charge conservation, conservation of baryon number and several other conservation laws.

Addendum concerning the questions in your edit:

  1. "...it is assumed that Einstein's theory explains that mass and energy flow back and forth. I want to know how it explains that." - See my example with the decay $\pi^0 \to \gamma \gamma$.

  2. The phrase "matter is energy" is confusing and therefore strongly discouraged. The precise terminology is that matter particles have energy described by the relativistic energy-momentum relation \eqref{1}. This has nothing to do with curvature of space time.

Hyperon
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$E=mc^2 $does not say mass can be converted to energy or energy to mass. So it does not claim antimatter, it just says "If you annihilate mass m you get the energy $E=mc^2" That conversion is possible has to be proved by experiment.

trula
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Einstein's $E=mc^2$ law does not predict anything in a general sense, it just says that rest mass and energy are sort of equivalent forms of entities (to the degree of $c^2$).

Now, if you want some verification of this law,- for the matter conversion to pure energy see for example electron-positron annihilation reaction $e^- + e^+ \to 2 \gamma.$

If you want reverse proof,- check cases of particle pair production reactions,- for example $ \gamma \to e^- + e^+$ from pure energy (usually happens in a strong external electromagnetic fields). Hawking radiation from black hole horizon is also some sort of pair production process, because BH gravitational energy is just as successfully materialized into particles.

So it goes BOTH ways,i.e. $E \leftrightarrow m.$

Agnius Vasiliauskas
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