# Energy–momentum relation

The Pythagorean relationship between Energy (E), Mass (m) and Momentum (p) according to special relativity. This clearly shows how E = mc2 is a special case of Einstein's equation.

In physics, particularly the theory of relativity, the energy–momentum relation is a fundamental relation between the energy, momentum and the mass of a particle or system of particles (say, a macroscopic object). In flat Minkowski space, the relation is:

 $E^2 = (pc)^2 + (mc^2)^2$

where c is the speed of light, and E is total energy of the system, m is the invariant mass of the system, and

$\mathbf{p} = \gamma(\mathbf{u})m \mathbf{u}\,, \quad |\mathbf{p}| \equiv p\,,$

is the total relativistic momentum of the system travelling at 3-velocity u. The 3-momentum here includes the Lorentz factor γ(u), not just the classical definition mu.

The invariant mass is an invariant for all frames of reference (hence the name), not just in inertial frames in flat spacetime, but also accelerated frames traveling through curved spacetime (see below). However the total energy of the particle E and its relativistic momentum p are frame-dependent; relative motion between two frames causes the observers in those frames to measure different values of the particle's energy and momentum; one frame measures E and p, while the other frame measures E′ and p′, where E′E and p′p, unless there is no relative motion between observers, in which case each observer measures the same energy and momenta. Although we still have, in flat spacetime;

${E'}^2 = (p'c)^2 + (mc^2)^2 \,.$

The quantities E, p, E′, p′ are all related by a Lorentz transformation. The relation allows one to sidestep Lorentz transformations when determining only the magnitudes of the energy and momenta by equating the relations in the different frames. Again in flat spacetime, this translates to;

$(mc)^2 = \left(\frac{E}{c}\right)^2 - p^2 = \left(\frac{E'}{c}\right)^2 - {p'}^2$

Since m doesn't change from frame to frame, the energy–momentum relation is used in relativistic mechanics and particle physics calculations, as energy and momentum are given in the particle's rest frame (that is, E′ and p′ as an observer moving with the particle would conclude to be) and measured in the lab frame (i.e. E and p as determined by particle physicists in a lab, and not moving with the particles).

It is also the basis of constructing relativistic wave equations, since if the relativistic wave equation describing the particle is consistent with this equation – it is consistent with relativistic mechanics, and is Lorentz invariant from frame to frame.

## Units of energy, mass and momentum

In natural units where c = 1, the energy–momentum equation reduces to

$E^2 = p^2 + m^2 \,,$

In particle physics, energy is typically given in units of electron volts (eV), momentum in units of eV·c−1, and mass in units of eV·c−2. In electromagnetism, and because of relativistic invariance, it is useful to have the electric field E and the magnetic field B in the same unit (Gauss), using the cgs (Gaussian) system of units, where energy is given in units of erg, mass in grams (g), and momentum in g·cm·s−1.

Energy may also in theory be expressed in units of grams, though in practice it requires a large amount of energy to be equivalent to masses in this range. 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. Energies of thermonuclear bombs are usually given in tens of kilotons and megatons referring to the energy liberated by exploding that amount of trinitrotoluene (TNT).

## Norm of the four-momentum

### Special relativity

In Minkowski space, energy and momentum (the latter multiplied by a factor of c) can be seen as two components of a Minkowski four-vector, namely the four-momentum;[1]

$\mathbf{P}=(E/c,\mathbf{p})\,,$

(these are the contravariant components). The Minkowski inner product $\left\langle\,,\,\right\rangle$ of this vector with itself gives the square of the norm of this vector, it is proportional to the square of the rest mass m of the body:

$\left\langle\mathbf{P},\mathbf{P}\right\rangle = |\mathbf{P}|^2 = (mc)^2\,,$

a Lorentz invariant quantity, and therefore independent of the frame of reference. Using the Minkowski metric η with metric signature (+−−−), the inner product in tensor index notation and as matrix multiplication can be calculated as:

$\left\langle\mathbf{P},\mathbf{P}\right\rangle = P^\alpha\eta_{\alpha\beta}P^\beta = \begin{pmatrix} E/c & p_x & p_y & p_z \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix} \begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix} = \left(\frac{E}{c}\right)^2 - p^2\,,$

and so:

$(mc)^2 = \left(\frac{E}{c}\right)^2 - p^2\,,$

which is the energy–momentum relation. If we had the other metric signature (−+++) for η, the inner product would be

$\left\langle\mathbf{P},\mathbf{P}\right\rangle = |\mathbf{P}|^2 = - (mc)^2\,,$

and

$\left\langle\mathbf{P},\mathbf{P}\right\rangle = P^\alpha\eta_{\alpha\beta}P^\beta = \begin{pmatrix} E/c & p_x & p_y & p_z \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix} = -\left(\frac{E}{c}\right)^2 + p^2\,,$

so:

$-(mc)^2 = -\left(\frac{E}{c}\right)^2 + p^2\,,$

which is still the same equation, as it should be because the inner product is an invariant.

### General relativity

In general relativity, the inner product of the 4-momentum with itself is similar,

$\left\langle\mathbf{P},\mathbf{P}\right\rangle = |\mathbf{P}|^2 = (mc)^2\,,$

but the Minkowski metric η is replaced by the metric tensor field g:

$\left\langle\mathbf{P},\mathbf{P}\right\rangle = |\mathbf{P}|^2 = P^\alpha g_{\alpha\beta}P^\beta \,,$

solved from the Einstein field equations, so;[2]

$P^\alpha g_{\alpha\beta}P^\beta = (mc)^2\,.$

This is generally significantly more complicated than the formula quoted at the beginning, as each component of the metric generally has space and time dependence; see metric tensor (general relativity) for more information.

## Special cases

For a body in its rest frame, the momentum is zero, so the equation simplifies to

$E = mc^2 \,,$

If the object is massless then the equation reduces to

$E = pc \,,$

as is the case for a photon.

## Many-particle systems

In the case of many particles with relativistic momenta pn and energy En, where n = 1, 2, ... (up to the total number of particles) simply labels the particles, the four-momenta can be added;

$\sum_n \mathbf{P}_n = \sum_n (E_n /c , \mathbf{p}_n )= \left(\sum_n E_n /c , \sum_n \mathbf{p}_n \right)$

and then take the norm; to obtain the relation for a many particle system:

$\left|\left(\sum_n \mathbf{P}_n \right)\right|^2 = \left(\sum_n E_n/c \right)^2 - \left(\sum_n \mathbf{p}_n \right)^2 = (Mc)^2\,,$

where M is the invariant mass of the whole system, and is not equal to the sum of the rest masses of the particles unless all particles are at rest (see mass in special relativity for more detail). Substituting and rearranging gives the generalization;

$\left(\sum_n E_n \right)^2 = \left(\sum_n \mathbf{p}_n c\right)^2 + (Mc^2)^2 \,.$

## Matter waves

Using the De Broglie relations for energy and momentum

$E=\hbar \omega \,, \quad \mathbf{p}=\hbar\mathbf{k}\,,$

where ω is angular frequency k is wave number, of the matter wave, the energy–momentum relation can be expressed in terms of wave quantities:

$(\hbar\omega)^2 = (c \hbar k)^2 + (mc^2)^2 \,,$

tidying up by dividing by (ħc)2 throughout:

$\left(\frac{\omega}{c}\right)^2 = k^2 + \left(\frac{mc}{\hbar}\right)^2 \,.$

This is also just the magnitude of the four-wavevector

$\mathbf{K} = (\omega/c, \mathbf{k})\,,$

found in a similar way to the four-momentum above.

Since the reduced Planck constant ħ and the speed of light c both occur and clutter this equation, this is where natural units are especially helpful. Normalizing them so that ħ = c = 1, we have:

$\omega^2 = k^2 + m^2 \,.$

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