Energy–momentum relation

Special cases

Centre-of-momentum frame (one particle)

For a body in its rest frame, the momentum is zero, so the energy-momentum relation (1) reduces to $${\displaystyle E=E_0=m{c}^2,}$$ where E₀ denotes the rest energy of the body.

Massless particles

If the object has mass m = 0, as is the case for a photon, then the energy-momentum relation (1) reduces to $${\displaystyle E=pc.}$$

Massless particles also have constant speed v = c. This has consequences for the Lorentz factor γ, $${\displaystyle \gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{0},}$$ which in turn gives indeterminate forms to some formulas involving this factor:

$${\displaystyle \begin{array}{cc}E=\gamma m{c}^2& =\frac{0}{0},\text{and}\\ p=\gamma mv& =\frac{0}{0}.\end{array}}$$

The workaround for these indeterminate forms for energy and momentum lies in the de Broglie relations, discussed in § Photons.

Classical correspondence

Rewriting the energy-momentum relation (1) for massive particles $${\displaystyle E=m{c}^2\sqrt{1+{\left(\frac{p}{mc}\right)}^2},}$$ and expanding into power series by the binomial theorem (or a Taylor series) around ${\displaystyle {\left(\frac{p}{mc}\right)}^2=0}$, $${\displaystyle E=m{c}^2\left[1+\frac{1}{2}{\left(\frac{p}{mc}\right)}^2-\frac{1}{8}{\left(\frac{p}{mc}\right)}^4+\cdots \right].}$$ In the limit that v ≪ c, we have γ ≈ 1 so the momentum has the classical form p ≈ mv. Then to first order in ${\displaystyle {\left(\frac{p}{mc}\right)}^2}$ we have $${\displaystyle \begin{array}{cc}E& \approx m{c}^2\left[1+\frac{1}{2}{\left(\frac{mv}{mc}\right)}^2\right],\text{or}\\ E& \approx m{c}^2+\frac{1}{2}m{v}^2,\end{array}}$$ where the first term is the rest energy E₀ and the second term is the classical kinetic energy.

This approximation is not valid for massless particles, since the expansion required the division of momentum by mass. Incidentally, there are no massless particles in classical mechanics.

Tachyons and exotic matter

Particles of ordinary matter must have positive mass m > 0 and subluminal speed v < c, yielding the usual relativistic energy–momentum relation (1). Massless particles are a marginal case where m = 0 and v = c, requiring special treatment discussed in § Massless particles. However, we can hypothesize about forms of matter that exceed these mass and speed restrictions in other ways.

Hyphothetical tachyons are particles that would move exclusively at superluminal speeds v > c, and whose energy–momentum relation would be $${\displaystyle E^2=(pc{)}^2-{\left(m{c}^2\right)}^2.}$$

Hypothetical exotic matter particles would have negative mass m < 0, resulting in yet another form of the energy–momentum relation: $${\displaystyle E^2=-(pc{)}^2+{\left(m{c}^2\right)}^2.}$$

Many-particle systems

Addition of four-momenta

Consider a system of many particles with relativistic momenta p_n and energy E_n measured in a particular frame, where n simply labels the particles n = 1, 2, ... up to the total number of particles. The four-momenta in this frame can be added: $${\displaystyle \sum _n{\mathbf{P}}_n=\sum _n\left(\frac{E_n}{c},{\mathbf{p}}_n\right)=\left(\sum _n\frac{E_n}{c},\sum _n{\mathbf{p}}_n\right).}$$ Taking the norm obtains the energy-momentum relation for the whole system: $${\displaystyle {\left|\left(\sum _n{\mathbf{P}}_n\right)\right|}^2={\left(\sum _n\frac{E_n}{c}\right)}^2-{\left(\sum _n{\mathbf{p}}_n\right)}^2={\left(Mc\right)}^2.}$$ Here M is the total mass of the system, and is not equal to the sum of the individual masses of the particles unless all particles are at rest (see Mass in special relativity § The mass of composite systems for more detail). Substituting and rearranging gives the generalization of (1);

The energies and momenta in the equation are all frame-dependent, while M is frame-independent.

Centre-of-momentum frame

In the centre-of-momentum frame, by definition we have $${\displaystyle \sum _n{\mathbf{p}}_n=\mathbf{0},}$$ with the implication from (2) that the total energy of the center-of-momentum frame is the total mass M multiplied by c²: $${\displaystyle {\left(\sum _n{E}_n\right)}^2={\left(M{c}^2\right)}^2\Rightarrow \sum _n{E}_{\mathrm{C}\mathrm{O}\mathrm{M}n}=E_{\mathrm{C}\mathrm{O}\mathrm{M}}=M{c}^2.}$$ This is true in all reference frames since M is invariant.

The energies E_{COM n} are those in the centre-of-momentum frame, not the lab frame. However, for many familiar bound systems the centre-of-momentum frame is the lab frame: if the system itself is not in motion, the individual momenta must add to zero. For example, when placing a container of gas on a scale, there is internal energy in the vibration of atoms in the container walls, and in the kinetic motion of the gas molecules. Because the container is at rest relative to the scale, all of these energies read as mass.

Relating total mass to individual masses

Either the energies or momenta of the particles, as measured in some frame, can be eliminated using the energy-momentum relation for each particle, $${\displaystyle E_n^2-{\left({\mathbf{p}}_{n}c\right)}^2={\left(m_n{c}^2\right)}^2,}$$ allowing the total mass M to be expressed either in terms of individual energies and masses, or individual momenta and masses. In a particular frame, the squares of sums can be rewritten as sums of squares (and products): $${\displaystyle \begin{array}{cc}{\left(\sum _n{E}_n\right)}^2& =\left(\sum _n{E}_n\right)\left(\sum _k{E}_k\right)\\ & =\sum _{n,k}{E}_n{E}_k=2\sum _{n<k}{E}_n{E}_k+\sum _n{E}_n^2,\end{array}}$$ $${\displaystyle \begin{array}{cc}{\left(\sum _n{\mathbf{p}}_n\right)}^2& =\left(\sum _n{\mathbf{p}}_n\right)\cdot \left(\sum _k{\mathbf{p}}_k\right)\\ & =\sum _{n,k}{\mathbf{p}}_n\cdot {\mathbf{p}}_k=2\sum _{n<k}{\mathbf{p}}_n\cdot {\mathbf{p}}_k+\sum _n{\mathbf{p}}_n^2,\end{array}}$$ so substituting the sums, we can introduce the individual masses m_n in (2): $${\displaystyle \sum _n{\left(m_n{c}^2\right)}^2+2\sum _{n<k}\left(E_n{E}_k-c^2{\mathbf{p}}_n\cdot {\mathbf{p}}_k\right)={\left(M{c}^2\right)}^2.}$$

The energies can be eliminated by substituting for E_n and E_k using $${\displaystyle \begin{array}{cc}{E}_n& =\sqrt{{\left({\mathbf{p}}_{n}c\right)}^2+{\left(m_n{c}^2\right)}^2}\text{and}\\ E_k& =\sqrt{{\left({\mathbf{p}}_{k}c\right)}^2+{\left(m_k{c}^2\right)}^2}.\end{array}}$$ Alternatively, the momenta can be eliminated by substituting for p_n · p_k using $${\displaystyle \begin{array}{cc}{\mathbf{p}}_n\cdot {\mathbf{p}}_k& =\left|{\mathbf{p}}_n\right|\left|{\mathbf{p}}_k\right|\cos {\theta }_{nk},\\ \left|{\mathbf{p}}_n\right|& =\frac{1}{c}\sqrt{E_n^2-{\left(m_n{c}^2\right)}^2}\text{and}\\ \left|{\mathbf{p}}_k\right|& =\frac{1}{c}\sqrt{E_k^2-{\left(m_k{c}^2\right)}^2},\end{array}}$$ where θ_nk is the angle between the momentum vectors p_n and p_k.

Rearranging: $${\displaystyle {\left(M{c}^2\right)}^2-\sum _n{\left(m_n{c}^2\right)}^2=2\sum _{n<k}\left(E_n{E}_k-c^2{\mathbf{p}}_n\cdot {\mathbf{p}}_k\right).}$$

Since the total mass M of the system and the individual masses m_n of the particles are all invariants, the right hand side is also an invariant, even though the energies and momenta are all measured in a particular frame.

Relation to quantum theory

Central to the quantum treatment of both massive and massless particles are the de Broglie relations, $${\displaystyle \begin{array}{cc}E& =h\nu =\hslash \omega ,\\ \mathbf{p}& =\hslash \mathbf{k}\text{and}\\ p& =\frac{h}{\lambda },\end{array}}$$ where h is the Planck constant; ν is the frequency; ħ is the reduced Planck constant h/2π; ω is the angular frequency; k is the wavevector, whose magnitude is the wavenumber k; and λ is the wavelength.

Matter waves

The energy–momentum relation (1) can be expressed in terms of wave quantities: $${\displaystyle {\left(\hslash \omega \right)}^2={\left(\hslash ck\right)}^2+{\left(m{c}^2\right)}^2.}$$ Dividing by (ħc)² throughout,

This can also be derived from the magnitude of the four-wavevector $${\displaystyle \mathbf{K}=\left(\frac{\omega }{c},\mathbf{k}\right),}$$ in a similar way to the four-momentum in § Norm of the four-momentum.

Photons

As mentioned in § Massless particles, photons have mass m = 0 and a simplified energy-momentum relation E = pc. Substituting the de Broglie momentum $${\displaystyle p=\frac{h}{\lambda }=\hslash k,}$$ we get $${\displaystyle E=\frac{hc}{\lambda }=\hslash ck.}$$ The de Broglie energy also applies directly: $${\displaystyle E=h\nu =\hslash \omega .}$$ The same relations would apply to any other massless particle.

Relativistic wave equations

In the history of relativistic quantum mechanics, the energy-momentum relation (1) was the basis for constructing relativistic wave equations, ultimately leading to the development of the Dirac equation, which incorporates the concepts of antimatter and spin. In quantum field theory, the relativistic energy-momentum relation applies to all particles and fields.

Units of energy, mass and momentum

Natural units

In natural units where c = 1, the energy–momentum relation (1) reduces to $${\displaystyle E^2=p^2+m^2.}$$ One way to conceptualize this system is through units of distance such as the light-year or light-second.

Choosing natural units where ħ = c = 1 simplifies the wavenumber relation (3) to $${\displaystyle {\omega }^2=k^2+m^2.}$$

The resulting formulas, although simpler in appearance, only work for this particular choice of units. After much manipulation, it can be challenging to re-insert the appropriate constants to revert to a universal form.

Particle physics units

In particle physics, energy is typically given in units of electronvolts (eV), momenta are expressed in eV/c, and masses in eV/c².

In informal contexts, mass is sometimes conflated with rest energy and given in eV (possibly implying some version of natural units). For example, if a proton is referred to as having a "mass" of 938 MeV, in context this means MeV/c².

Mass-equivalent

Energy may also be given in mass-equivalent units such as grams (strictly speaking grams·c²), thereby illustrating the mass defect in especially powerful releases of energy. For example, the first atomic bomb liberated about 1 gram of heat, and the largest thermonuclear bombs have generated a kilogram or more of heat.

This is not to be confused with TNT equivalent units such as kilotons or megatons: those units refer to the energy released by detonating a specified mass of the explosive TNT, rather than the energy itself being scaled into mass units by a factor of 1/c².

Alternate conventions for mass

The familiar mass–energy relation E = mc² can be seen as a special case of the energy-momentum relation (1) where the momentum p = 0, and the total energy E equals the rest energy E₀: $${\displaystyle E_0=m{c}^2.}$$ Alternatively, a relativistic mass m_rel may be defined that varies with the frame of reference and increases in proportion to the total energy, and must be contrasted with the invariant mass m₀: $${\displaystyle \begin{array}{cc}E& =m_{\mathrm{r}\mathrm{e}\mathrm{l}}{c}^2\text{and}\\ E_0& =m_0{c}^2,\end{array}}$$ where the two are related as follows: $${\displaystyle m_{\mathrm{r}\mathrm{e}\mathrm{l}}=\gamma m_0.}$$ In different sources, either of these quantities might be denoted as m depending on context and the author's preference.

Elsewhere in this article, the term 'mass' and the symbol m refer exclusively to invariant mass, and relativistic mass is not used. The total energy E can be used instead to illustrate the relevant concepts.