# Conversion of mass and energy

First of all I am not a scientist and all these doubts are coming from my curiosity.

When Googling about Einstein's $E = mc^2$. I understand that mass and energy are convertible.

What it exactly means?

Also can we have greater amount of energy from smaller mass objects?

And if this conversion is possible then why we cant convert 1kg of iron to large energy?

(I know that the question is really worst. But I am a little bit weak in English and almost all Google searches disappoint me.)

Any simple explanation?

• – Qmechanic Jun 22 '13 at 18:37
• @Qmechanic But not exactly :( – Red Jun 22 '13 at 18:39
• I'd say you're wrong. If a block of anti-uranium meets uranium and a block of wood meets anti-wood, the energy released would be the same. In real life, we dont get 100% mass transformation, because we use desintegration of Uranium. Wood of course is much more difficult to transform into energy (burning energy is due to bond formation). – jinawee Jun 22 '13 at 19:21
• Actually chemical bonds do contribute to rest mass. So in a chemical reaction the mass will slightly change. – SpiderPig Jun 22 '13 at 22:20
• @Red: yes. The key point is that, even with Uranium fission, only a small amount of mass is actually converted to energy. There is no practical everyday case where you have a block of some type of matter, and you completely get rid of it to make energy. – Jerry Schirmer Jun 23 '13 at 3:57

I know you're mostly looking for an intuitive explanation, but I think it's important to at least get a feel for the mathematical truth of this equation first. So I'll give a (much) shortened derivation of it first and get to the intuition part later.

## The mathematics

Einstein's theory of special relativity tells us the energy $E$ of a particle is given by

$$E^2 = |\vec{p}|^2c^2+m^2c^4,$$

which can be derived from the free-particle Lagrangian in Minkowski space (a flat version of spacetime) and the definition of canonical (four-)momentum $p=(p^0,\vec{p})$. Working out this definition for the free-particle case gives you the components of the four-momentum:

$$p^0 = \frac{mc}{\sqrt{1-\frac{v^2}{c^2}}} \hspace{2cm} (*)$$ $$\vec{p} = \frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}} \hspace{2cm} (**)$$

From this you can find an expression for the relativistic energy by considering the time component $p^0$ in the limit $v\ll c$:

$$p^0 = \frac{mc}{\sqrt{1-\frac{v^2}{c^2}}} \approx mc\left(1+\frac{1}{2}\frac{v^2}{c^2}\right)$$

so

$$cp^0 \approx mc^2+\frac{mv^2}{2}\hspace{2cm}(1)$$

The second term looks familiar. Indeed, it's our old friend the kinetic energy $T$. Since we're talking about a free particle, there's no potential energy. So what's this $mc^2$ term doing there? And how should we interpret $cp^0$? Well, rewriting equation $(1)$ gives us

$$T = cp^0 - mc^2.$$

From this, it should seem obvious that $cp^0$ is the total energy $E$ of the free particle and $mc^2$, which depends only on an intrinsic property of the particle (its mass), is some sort of rest energy the particle cannot get rid of. The only way for the particle to change that part of its energy is to somehow changes its mass, which is hard to do, even with the strictest of diets.

Of course particles don't go on diets (excuse that humorous outburst of mine just a moment ago) but they can change their mass if they're prepared to give up on their entire identity. But more on that in the intuition part below. Let's first get our focus back on the mathematics.

Remember that equation $(1)$ only holds for small $v$. Going back to general $v$, we can now use $p^0 = E/c$ to calculate the length of the four-momentum. Recall that the length of a vector $\vec{a}$ is the square root of the inner product of $\vec{a}$ with itself, i.e. in 3D Euclidian space

$$|\vec{a}| = \sqrt{\vec{a}\cdot\vec{a}}$$

or

$$|\vec{a}|^2 = \vec{a}\cdot\vec{a} = \sum_{i=1}^3{a_i^2}$$

In Minkowski space, things are slightly different. The technical details aren't that important, the relevant consequence for us is this: the basic form of the previous formula is the same, but it is slightly adjusted$^1$:

$$|a|^2 = \sum_{\alpha=0}^3{g_{\alpha\alpha}a_{\alpha}^2}.$$

Note that I use a simple letter to denote a four-vector and a letter with an arrow over it for a 3D vector. Furthermore, I've used the convention that Greek letters ($\alpha$) serve as indices for components of four-vectors, in contrast to the Latin letters ($i$) for three-vector components.

Note also that there is some factor $g_{\alpha\alpha}$ appearing in the new expression for vector lengths. This factor has two indices and thus constitutes a rank 2 tensor (or matrix) called the metric tensor. This object sort of draws out what you could call the geometry of the space in which we're working. It tells us how to measure lengths. In particular, this metric tensor consists only of diagonal elements $g_{\alpha\alpha}$ which are either $+1$ or $-1$. It's a matter of convention which elements are positive and which negative, we'll choose $g_{00} = +1$ and the rest $-1$.

All this to show you that the length of a four-vector in Minkowski space is calculated as

$$|a|^2 = a_{0}^2 - \sum_{i=1}^3{a_i^2}.$$

For our momentum vector $p = (p^0,\vec{p}) = (E/c,\vec{p})$ this becomes

\begin{align} |p|^2 &= E^2/c^2 - \sum_{i=1}^3{p_i^2} \\ &= E^2/c^2 - |\vec{p}|^2 \end{align}

Keeping in mind that $E = cp^0$ we now use $(*)$ and $(**)$ to find \begin{align} |p|^2 &= E^2/c^2 - |\vec{p}|^2 \\ &= \frac{m^2c^2}{1-\frac{v^2}{c^2}} - \frac{m^2v^2}{1-\frac{v^2}{c^2}} \\ &= m^2c^2\left(\frac{1}{1-\frac{v^2}{c^2}} - \frac{v^2/c^2}{1-\frac{v^2}{c^2}}\right) \\ &= m^2c^2 \end{align}

Combining the first and last lines we get

$$E^2/c^2 - |\vec{p}|^2 = m^2c^2$$

which can be rewritten into the equation we were searching for

$$E^2 = |\vec{p}|^2c^2 + m^2c^4$$

or if the particle is at rest ($\vec{p} = \vec{0}$):

$$E = mc^2.$$

## The intuition

Now, that's all grand, but what does it mean? Well, it means mass and energy are not two separate quantities. In fact, they are the same up to a constant factor of $c^2$ (let's consider particles at rest). So what happens in fission reactors? Simply put, an atom's core collides with a neutron, briefly exciting the core into a higher isotope. This excited state is unstable and the core splits into two smaller, (more) stable cores. These two cores have a lower combined mass than the total mass of the original excited core. The difference in mass is released as energy in accordance with the formula we have just derived.

Then why isn't this possible with iron for example? That's a very good question. To answer it, let's think about what the core of an atom actually is. It consists of protons and neutrons, held together by the strong nuclear force. Aha! Wait a minute! Protons and neutrons are not point particles. They have a finite volume. So the more there are in a core, the bigger the core will be. But protons are also charged particles and they repel each other through the electrostatic interaction!

Now, the strong nuclear force is a short-distance attractive force, the electrostatic interaction is long-distance and repelling for equal charges. So the larger the core, the closer the protons at the edge of the core will get to feeling equally strong repelling and attracting forces. That should give you an intuition as to why some cores are unstable (it's also why there is an end to the periodic table, higher cores are too unstable to exist in nature).

This puts a lower bound on the cores that are 'fissionable'. There are models for this, but the important thing is that iron is definitely below the lower bound. In fact, it is the most stable core in existence. To get it to split, you would need to put a whole lot more energy into it than you would get out.

$^1$ I'm neglecting some notational details, which are in fact important but not in this context. Here I stayed as close to the form of the better-known equation for the length of a 3D vector to avoid having to discuss covariance and contravariance. So some of my notation is strictly speaking incorrect and I'm slightly inconsistent by switching between denoting the components of the four-vector $p$ with an upper index and those of the four-vector $a$ with a lower index. But the notational inaccuracies have no further consequences and the discussion itself is not affected.

As a child (and enthusiastic reader of Sci-Fi) I recall being puzzled by what the matter-energy transformation actually meant. It's all very well to say $E = mc^2$, but what actually happens when matter is converted into energy or vice versa?

I think the best way to understand what is going on is by considering quantum field theory. This is a hugely complex area (that I only dimly understand) so bear in mind that what follows is vastly oversimplified and possibly misleading.

Anyhow, in QFT the fundamental object is the field. The particles are excitations in the field, so if we excite the field by adding energy to it we are creating a particle. Likewise a field can lose energy by destroying a particle. This is how energy creates particles and particles can disappear and turn into energy.

There is a different field for every particle, and these fields can interact. So for example electrons (and positrons) are described by the electron field and photons are described by the photon field. Take the classic example of a matter to energy transformation of an electron and positron annihilating to create a photon. What happens is that the electron field interacts with the photon field and transfers two electron masses worth of energy to it. We observers see the two particles converting to energy and disappearing to produce a photon, but what is actually going on is just a transfer of energy between two quantum fields.

Or take another example of colliding two quarks in the LHC. The quarks are excitations in the quark field. When they collide the quark field suddenly has the original two quarks worth of energy plus lots of extra energy from the kinetic energy of the quarks. Because the quark field can interact with other fields energy can be exchanged and we get excitations of photon fields, electron fields and the quark fields of other flavours, and we see the excitations of these fields as a shower of particles coming out of the collision. From our perspective it looks as if the kinetic energy has changed into particles, but again it's just quantum fields exchanging energy with each other.

The quantum fields can't just interact in any fashion. They can only interact in specific ways described by the Standard Model. This is why your kg or iron can't suddenly turn into energy. The rules of the Standard Model don't allow it. However, and this is where things get speculative, there might be ways in which your kg of iron could turn into energy. The Standard Model is (probably) a low energy approximation to some more complex theory. For example, at about the time I was doing my degree there was much excitement about grand unified theories that allowed protons to decay and release energy. So far the experiments to measure proton decay haven't found anything, but if it existed your kg of iron would decay and release energy if you waited long enough. Sadly this is of no use for power generation as the rate of decay (if it happens at all) is fantastically slow.

This is about mass energy equivelance. Which means a body of mass m can have energy E by this relation. The reason why can't we get this from iron is just because it would produce so much heat that is unbearable.

• i think that heat is another form of energy ?? – Red Jun 22 '13 at 19:14
• Will that is the problem . Producing energy is uncontrolable in the current time – narn Jun 22 '13 at 19:16
• What you mean by uncontrollable energy ? – Red Jun 22 '13 at 19:18
• FYI : i am not the one who down voted ur answer without comments – Red Jun 22 '13 at 19:22
• This answer is nonsense. If you could transform iron into energy so easily, iron would be the deadliest weapon in the world. – jinawee Jun 22 '13 at 23:22

The most general relation in relativity (for free paricles in a flat spacetime) is:

$E = m^*c^2$, which is the Einstein relation.

Where $m^*$ is the relativistic mass, defined as $m^* = \gamma m$, with $\gamma = \frac{1}{\sqrt{1-\left(v/c\right)^2}}$.

If we take the classical limit, this means that the velocity becomes much smaller than the speed of light (which is the case in classical physics), then the energy relation becomes:

$E = mc^2 + \frac{1}{2}mv^2$.

We see that the energy gets an extra contribution from the mass, which is called the relativistic energy. This is the relation that you have written down in your question. Denote that this relation is only valid in the limit of small velocities and is interpreted as the energy that follows from the mass of the particle. The most general relation (without approximations) is given by:

$E^2=m^2c^4+\vec{p}^2c^2$.

Considering the conversion of mass to energy. Because of the above relation energy and mass are set on an equal footing, we can change mass into energy and the other way around. A simple experiment which shows this is the annihilation of electrons and positrons: $e^-e^+\rightarrow\gamma$. Due to the conservation of energy and momentum the mass of the electron and positron is completely converted into energy (the photon doesn't have any mass and is pure energy).

• I know that its not easy to explain all these things easily, However i am expecting something more simply structured answers. – Red Jun 23 '13 at 7:17
• Well it's basically applying conservation of energy. Only in this case mass plays a role in the energy and can be converted. The full derivation of E = mc² demands the understanding of the Lagrangian (as in the comment of Wouter) or simply by postulating the 4-momentum tensor and starting from there. I have skipped out on that in my comment because I tought that might lead us to far, also these tools demand a bit of physical background (espacially the Lagrangian) and I'm not quite aware of what yours might be ? – Nick Jun 23 '13 at 12:18