# I Assumption that the rest mass energy is mc^2

Tags:
1. Jun 18, 2017

### snate

In all the derivations of E2=p2c2+m2c4 ,that I've stumbled on, it's assumed that the rest mass energy is m*c2 just because kinetic energy is mc2λ-mc2. Was it originally assumed? If so, can someone explain me why is it a logical assumption? Are there any derivations without such assumptions?

2. Jun 18, 2017

### vanhees71

Of course, except in general relativity, where the absolute value of the energy-momentum-stress tensor becomes observable through the gravitational field (described as the curvature of spacetime in GR), in general you can only measure energy differences. So what you call $E=0$ of a system is more or less arbitrary. In classical relativistic point-particle mechanics, it's however very convenient to choose the energy and momentum such as to make up a four-vector in Minkowski space.

You come to this idea when you ask how to generalize the Newtonian equation of motion
$$m \ddot{\vec{x}}=\vec{F}$$
to special relativity. From the point of view of Minkowski space and Lorentz transformations, the Newtonian expression has a very complicated transformation law. It's much simpler to look for a formulation of a relativistic equation of motion that is manifestly covariant under Lorentz transformations, i.e., formulated in terms of four-vectors and tensors.

The first idea is to use a simpler parameter to parametrize the spacetime trajectory than the coordinate time $t$. For a massive particle, which always moves along a time-like worldline, you can use its proper time. It's the time some ideal clock measures which comoves with the particle. It's related to the space-time trajectory by
$$c^2 \mathrm{d} \tau^2=\eta_{\mu \nu} \mathrm{d}x^{\mu} \mathrm{d} x^{\nu}=\left [c^2-\left (\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \right)^2 \right ] \mathrm{d} t^2.$$
Here $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ is the Minkowski pseudometric. Obviously $\mathrm{d} \tau$ is a Lorentz scalar, i.e., it doesn't change under Lorentz transformations. Indeed, it's the time measured by an observer at rest relative to the particle, i.e., it's a time measured in a preferred frame of reference (which however in general is not an intertial system).

Now you can define the four-velocity which is a Minkowski vector:
$$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
Note that
$$u_{\mu} u^{\mu} = \eta_{ \mu \nu} u^{\mu} u^{\nu}=1.$$
From this, using the invariant mass $m>0$ of the particle, we define the energy-momentum vector of the particle
$$p^{\mu}=m c u^{\mu}.$$
To see, how it's related to the kinetic energy and momentum of a particle in the non-relativistic limit, we express this in terms of the usual coordinate time, $t$
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau}=m \gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix},$$
where
$$\gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}, \quad \vec{v} = \frac{\mathrm{d} \vec{x}}{\mathrm{d} t}.$$
Now assume that $|\vec{v}/c| \ll 1$ (non-relativistic limit). Then you get
$$p^0=m c \gamma=m c (1-\vec{v}^2/c^2)^{-1/2}=m c \left (1+\frac{1}{2} \frac{\vec{v}^2}{c^2}\right) = \frac{1}{c} \left (m c^2 + \frac{m}{2} \vec{v}^2 \right ).$$
Now set
$$E=c p^0=m c^2 + \frac{m}{2} \vec{v}^2.$$
This is, up to the additive constant $m c^2$ just the non-relativistic kinetic energy. As we've seen, this limit occurs, because we wanted a four-vector description of energy and momentum, and that's why it's convenient to include the "rest energy" $E_0=m c^2$ in the energy of the particle.

We also immediately
$$p_{\mu} p^{\mu}=m^2 c^2 \; \Rightarrow \; \frac{E^2}{c^2}-\vec{p}^2=m^2 c^2$$
or
$$E=c \sqrt{m^2 c^2+\vec{p}^2}.$$

A good guess for an equation of motion, generalizing Newton's 2nd Law to a relativistic law now is
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu},$$
where $K^{\mu}$ is the Minkowski four-force. However, not all four components are independent of each other, because from
$$p_{\mu} p^{\mu}=m^2 c^2=\text{const} \; \Rightarrow \; p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=p_{\mu} K^{\mu}=0.$$
You find more on relativistic mechanics of classical point particles in my special-relativity FAQ article:

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Last edited: Jun 18, 2017
3. Jun 18, 2017

### Simon Bridge

That kinda makes it sound like the choice for rest mass energy is arbitrary though.

4. Jun 18, 2017

### vanhees71

In a way it's not that arbitrary, because it combines energy and momentum to a four-vector, which is a very important concept to formulate the fundamental dynamical laws of relativistic physics.

5. Jun 18, 2017

### Simon Bridge

Isn't the relation something that has experimental support?

6. Jun 18, 2017

### SiennaTheGr8

See Einstein's original derivation of the mass–energy equivalence:

https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

The gist of it is this (but note that he doesn't make all of this explicit, and he doesn't use the same symbols I'll use):

It was already known that light carries momentum and energy. Einstein uses this fact (and the assumption that momentum and total energy are indeed conserved) to posit the existence of rest energy: if a body at rest emits identical light waves in opposite directions, their momenta must cancel, leaving the body's motion unaffected. But this means that the body must have had some energy to lose in the first place, even though it had no kinetic energy (in its rest frame). Hence rest energy. And the following equation arises naturally:

$E = E_0 + E_k$

(total energy is the sum of the invariant rest energy and the frame-dependent kinetic energy).

Using the relativistic Doppler equation that he'd already derived in his first paper on special relativity, Einstein considers that light-emitting body from the perspective of a different inertial frame (one in which the body has some non-zero constant velocity). This leads to the following relation between $E$ and $E_0$:

$E = \gamma E_0$,

where of course $\gamma = ( 1 - \beta^2 )^{-1/2}$, and $\beta = v/c$.

Nice. Now back to kinetic energy:

$E_k = E - E_0 = E_0 ( \gamma - 1 )$.

When $\beta \ll 1$, we have (via binomial expansion):

$\gamma \approx 1 + \dfrac{1}{2} \beta^2$.

Plugging that in above, we get the classical approximation for $E_k$:

$E_k \approx E_0 \left[ \left( 1 + \dfrac{1}{2} \beta^2 \right) - 1 \right] = \dfrac{1}{2} E_0 \beta^2$.

That had better damn well equal the Newtonian $\frac{1}{2} m v^2$, and so:

$E_0 = mc^2$.

Hope that helps.

7. Jun 18, 2017

### vanhees71

8. Jun 18, 2017

### pervect

Staff Emeritus
As other posters have mentioned, in classical mechanics energy is only defined up to an additive constant. The best argument for $E_0 = m_0 c^2$ is probably the energy released by particle anti-particle annihiliation, which is basically a quantum phenomenon.

9. Jun 18, 2017

### Mister T

To derive something means to show that it follows from some assertions, so there is nothing in the result that wasn't already present in those assertions. In that sense it's nothing new. Assertions like the mass-energy equivalence can be derived from other assertions but only if those other assertions already contain the mass-energy equivalence. This is precisely what Einstein did in his derivation. He showed that for the special case of light emission those emissions must reduce the mass of the emitter.

But the mass-energy equivalence is more general than that. Like the First Law of Thermodynamics it's a grand generalization from observation, and its value is in its utility and vast limits of validity.