# Assumption that the rest mass energy is mc^2

• I
• snate
In summary, Einstein's original derivation of the mass-energy equivalence showed that the existence of rest energy is necessary for the conservation of momentum and total energy. This leads to the equation E = E_0 + E_k, where E_0 is the invariant rest energy and E_k is the frame-dependent kinetic energy. This equation has experimental support and is used in many derivations of E2=p2c2+m2c4. However, there are some derivations that do not make this assumption, such as in general relativity where the energy-momentum-stress tensor becomes observable through the gravitational field.
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?

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:
Battlemage!, Physics Footnotes and snate
That kinda makes it sound like the choice for rest mass energy is arbitrary though.

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.

Isn't the relation something that has experimental support?

snate said:
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?

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.

Battlemage!, Simon Bridge and snate
Simon Bridge and snate
snate said:
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?

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.

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.

## 1. What is the meaning of the equation E=mc^2?

The equation E=mc^2 represents the relationship between energy (E), mass (m), and the speed of light (c). It states that the energy of an object is equal to its mass multiplied by the speed of light squared.

## 2. How did Einstein come up with the idea of mass-energy equivalence?

Einstein's theory of special relativity, which led to the famous equation E=mc^2, was based on his belief that the laws of physics should be the same for all observers, regardless of their relative motion. Through mathematical calculations and thought experiments, he concluded that mass and energy are two forms of the same thing.

## 3. Does this equation only apply to particles at rest?

No, the equation E=mc^2 applies to all particles, whether they are at rest or in motion. However, it is most commonly used to describe the rest energy of particles, which is the energy they possess even when they are not moving.

## 4. How does the equation E=mc^2 relate to nuclear energy?

Nuclear energy is released when the nuclei of atoms split or combine to form new atoms. The mass of these new atoms is slightly less than the mass of the original atoms, and this difference in mass is converted into energy according to the equation E=mc^2. This is the principle behind nuclear power and nuclear weapons.

## 5. Is the assumption that the rest mass energy is mc^2 universally accepted by scientists?

The concept of mass-energy equivalence and the equation E=mc^2 are well-established theories in physics and have been verified through numerous experiments. However, there are still ongoing debates and research surrounding the exact mechanisms of mass-energy conversion and its implications in areas such as quantum mechanics and cosmology.

• Special and General Relativity
Replies
55
Views
4K
• Special and General Relativity
Replies
2
Views
322
• Special and General Relativity
Replies
8
Views
1K
• Special and General Relativity
Replies
2
Views
1K
• Special and General Relativity
Replies
17
Views
1K
• Special and General Relativity
Replies
14
Views
2K
• Special and General Relativity
Replies
3
Views
3K
• Special and General Relativity
Replies
62
Views
5K
• Special and General Relativity
Replies
4
Views
1K
• Special and General Relativity
Replies
7
Views
1K