# Why does E=mc^2

1. Sep 9, 2013

### ZeroPivot

explain it like im 12 years old.

2. Sep 9, 2013

### Staff: Mentor

So you don't know any calculus?

3. Sep 9, 2013

### Turion

Good luck with that.

4. Sep 9, 2013

Staff Emeritus
5. Sep 10, 2013

### ZeroPivot

i do know upto multivariable.

6. Sep 10, 2013

### Staff: Mentor

OK, then you are prepared for something much better than a 12-year-old level explanation. There are a ton of good derivations online - find one, study it, come back with a more specific question.

One hint: You're really looking for the derivation of the equation $E^2=(m_0{c}^2)^2+(pc)^2$, as $E=mc^2$ is just the $p=0$ special case of that more general relationship.

7. Sep 10, 2013

### Staff: Mentor

OK, try this. It's a draft of an FAQ that I've been working on.

In both non-relativistic and relativistic physics, we can define an object's (kinetic) energy in terms of the work done on the object by a force, using the work-energy theorem. In the non-relativistic case, this leads to the familiar $E = \frac{1}{2}mv^2$; in the relativistic case, this leads to the familiar $E = mc^2$ (using an old-fashioned interpretation of $m$).

Warning: This derivation uses calculus. It also assumes that we already know the non-relativistic and relativistic formulas for an object's momentum.

Beginning physics students learn early on that when a constant force $F$ acts on an object, causing the object to move a distance $\Delta x$, it does work $W = F \Delta x$. If the force is not constant, we have to consider it as a function of position, and integrate to find the work:
$$W = \int^{x_2}_{x_1} {Fdx}$$
Inserting Newton's Second Law of Motion in the form $F = dp/dt$ (where $p$ is momentum) and playing some games with the differentials:
$$W = \int^{x_2}_{x_1} {\frac{dp}{dt} dx} = \int^{x_2}_{x_1} {\frac{dp}{dv} \frac{dv}{dt} dx} = \int^{v_2}_{v_1} {\frac{dp}{dv} \frac{dx}{dt} dv} = \int^{v_2}_{v_1} {\frac{dp}{dv} v dv}$$
So far, this applies in both non-relativistic and relativistic mechanics. To proceed further, we need an equation for $p$ in terms of $v$. Here the relativistic and non-relativistic cases diverge.

Non-relativistically, $p = mv$, so $\frac{dp}{dv} = m$ and
$$W = \int^{v_2}_{v_1} {m v dv} = \frac{1}{2} mv^2_2 - \frac{1}{2} mv^2_1$$
We define the energy of the object as
$$E = \frac{1}{2} mv^2$$
so that $W = E_2 - E_1$ (the work-energy theorem). When $v = 0$, $E = 0$, so the energy that we've defined here is due only to the object's motion, and we therefore call it kinetic energy.
$$E = K = \frac{1}{2} mv^2$$

Relativistically,
$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$
where $m$ is what is often called the "rest mass." I leave it as a calculus exercise for the reader to evaluate the derivative:
$$\frac{dp}{dv} = \frac{m}{(1-v^2/c^2)^{3/2}}$$
so that
$$W = \int^{v_2}_{v_1} {\frac {mvdv}{(1-v^2/c^2)^{3/2}}}$$
I leave it as another calculus exercise for the reader to evaluate the integral above to get:
$$W = \frac{mc^2}{\sqrt{1-v_2^2/c^2}} - \frac{mc^2}{\sqrt{1-v_1^2/c^2}}$$
We define the energy of the object as
$$E = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$
so that $W = E_2 - E_1$ (the work-energy theorem), just like in the non-relativistic case. When $v = 0$, $E = mc^2$, so the energy that we've defined here is not due only to the object's motion. We can separate it into two pieces: the rest energy
$$E_0 = mc^2$$
which doesn't depend on the motion, and the kinetic energy which does:
$$K = \frac{mc^2}{\sqrt{1-v^2/c^2}} - mc^2$$
so that
$$E = E_0 + K = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$
In the early days of relativity, physicists called the $m$ that I used above, the "rest mass" $m_0$; and defined the "relativistic mass" as
$$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$
so that $E = mc^2$ and $E_0 = m_0 c^2$. Nowadays, most physicists don't do this, but popular-level books and even some introductory textbooks still do.

Last edited: Sep 11, 2013
8. Oct 19, 2014

### robphy

It seems to me that the work-energy theorem
is a relation between the net-work done by the forces on the object
and the change in Kinetic-Energy (not the change in [total-]Energy,
in spite of this common name for the theorem.).

In intro physics, we obtain the total-Mechanical-Energy when we identify the
"portion of the net-work that is done by conservative forces"
as "minus the change in Potential Energy". When we move this portion to the other side,
we then have
the work-done-by-nonconservative-forces is equal to the change-in-total-Mechanical-Energy.

Let's not make that split now.

If you consider the case when it starts from rest to final velocity V, then the net-work is $(1/2)mV^2$, which we can identify as the classical kinetic energy.

Following the same scheme for the relativistic case,
you get (using your integration) $\frac{mc^2}{\sqrt{1-(V/c)^2}}-mc^2=(\frac{1}{\sqrt{1-(V/c)^2}}-1)mc^2$, which is precisely the relativistic-kinetic-energy, which is zero when V=0.
(From that you can build an expression for the relativistic-energy by grouping K and mc^2.).

Then, more generally,
you can express the work done in changing the velocity from $v_1$ to $v_2$
as the difference of their relativistic-kinetic-energies.

It's true that each of those relativistic-kinetic-energies has a constant $mc^2$, which
cancel when the difference is calculated, ending up with something that resembles the difference of relativistic-energies. However, that result is secondary, in my opinion.

In addition, the work-energy theorem suggests that we can apply forces to change the kinetic energy of a particle of constant-mass m, possibly bringing the kinetic energy to zero.
However, we can't apply forces to bring the relativistic-energy of a particle of constant-mass m (here, \$\gamma mc^2) to zero.

So, it seems to me that
the work-energy-theorem is really about changes in kinetic-energy.

9. Oct 19, 2014

### Staff: Mentor

So is it just a lucky coincidence that the constant term turns out to be the correct rest-energy?

10. Oct 19, 2014

### Hertz

How does one conclude this "zero velocity kinetic energy" is the totality of energy in the system? If you have some sort of particle at rest, which is unable to exchange energy with its environment, and which takes up some volume; is $mc^2$ necessarily the entire amount of energy contained within the volume that the particle resides? Is the concept of "total energy in some volume in some frame of reference" subjective by nature? Or is there literally no such thing as negative energy (indicating that there is some definite "zero point" of energy)?

Perhaps this derivation indicates that there is an intrinsic amount of "kinetic energy" in an object with some mass. I.e. $K=mc^2$, and this "kinetic energy" increases as velocity is added to the object. Maybe the "total rest energy" of an object is it's "rest kinetic energy" plus any internal potential energy or other possible forms of energy present.

(furthermore...)
Furthermore, I would intuitively think that the "total energy in some volume" would be dependent of what is immediately outside that volume. Obviously, in a more pressurized environment, it takes more energy to open up some volume of space. I find it puzzling that the entirety of energy in some object at rest could be dependent only on the mass of the object, and not any other properties of the object and its environment.