Energy–momentum relation

For a massive object moving at three-velocity u = (u_x, u_y, u_z) with magnitude |u| = u in the lab frame:[1]

${\displaystyle E=\gamma _{(\mathbf {u} )}m_{0}c^{2}}$

is the total energy of the moving object in the lab frame,

${\displaystyle \mathbf {p} =\gamma _{(\mathbf {u} )}m_{0}\mathbf {u} }$

is the three dimensional relativistic momentum of the object in the lab frame with magnitude |p| = p. The relativistic energy E and momentum p include the Lorentz factor defined by:

${\displaystyle \gamma _{(\mathbf {u} )}={\frac {1}{\sqrt {1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}}={\frac {1}{\sqrt {1-\left({\frac {u}{c}}\right)^{2}}}}}$

Some authors use relativistic mass defined by:

${\displaystyle m=\gamma _{(\mathbf {u} )}m_{0}}$

although rest mass m₀ has a more fundamental significance, and will be used primarily over relativistic mass m in this article.

Squaring the 3-momentum gives:

${\displaystyle p^{2}=\mathbf {p} \cdot \mathbf {p} ={\frac {m_{0}^{2}\mathbf {u} \cdot \mathbf {u} }{1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}={\frac {m_{0}^{2}u^{2}}{1-\left({\frac {u}{c}}\right)^{2}}}}$

then solving for u² and substituting into the Lorentz factor obtains its alternative form in terms of 3-momentum and mass, rather than 3-velocity:

${\displaystyle \gamma ={\sqrt {1+\left({\frac {p}{m_{0}c}}\right)^{2}}}}$

Inserting this form of the Lorentz factor into the energy equation:

${\displaystyle E=m_{0}c^{2}{\sqrt {1+\left({\frac {p}{m_{0}c}}\right)^{2}}}}$

followed by more rearrangement yields (1). The elimination of the Lorentz factor also eliminates implicit velocity dependence of the particle in (1), as well as any inferences to the "relativistic mass" of a massive particle. This approach is not general as massless particles are not considered. Naively setting m₀ = 0 would mean that E = 0 and p = 0 and no energy–momentum relation could be derived, which is not correct.

In Minkowski space, energy (divided by c) and momentum are two components of a Minkowski four-vector, namely the four-momentum;[6]

${\displaystyle \mathbf {P} =\left({\frac {E}{c}},\mathbf {p} \right)\,,}$

(these are the contravariant components).

The Minkowski inner product ⟨ , ⟩ 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:

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=\left(m_{0}c\right)^{2}\,,}$

a Lorentz invariant quantity, and therefore independent of the frame of reference. Using the Minkowski metric η with metric signature (− + + +), the inner product is

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=-\left(m_{0}c\right)^{2}\,,}$

and

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =P^{\alpha }\eta _{\alpha \beta }P^{\beta }={\begin{pmatrix}{\frac {E}{c}}&p_{x}&p_{y}&p_{z}\end{pmatrix}}{\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}{\begin{pmatrix}{\frac {E}{c}}\\p_{x}\\p_{y}\\p_{z}\end{pmatrix}}=-\left({\frac {E}{c}}\right)^{2}+p^{2}\,,}$

so

${\displaystyle -\left(m_{0}c\right)^{2}=-\left({\frac {E}{c}}\right)^{2}+p^{2}.}$

In general relativity, the 4-momentum is a four-vector defined in a local coordinate frame, although by definition the inner product is similar to that of special relativity,

${\displaystyle \left\langle \mathbf {P} ,\mathbf {P} \right\rangle =|\mathbf {P} |^{2}=\left(m_{0}c\right)^{2}\,,}$

in which the Minkowski metric η is replaced by the metric tensor field g:

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

solved from the Einstein field equations. Then:[7]

${\displaystyle P^{\alpha }g_{\alpha \beta }P^{\beta }=\left(m_{0}c\right)^{2}\,.}$

Performing the summations over indices followed by collecting "time-like", "spacetime-like", and "space-like" terms gives:

${\displaystyle \underbrace {g_{00}{\left(P^{0}\right)}^{2}} _{\text{time-like}}+2\underbrace {g_{0i}P^{0}P^{i}} _{\text{spacetime-like}}+\underbrace {g_{ij}P^{i}P^{j}} _{\text{space-like}}=\left(m_{0}c\right)^{2}\,.}$

where the factor of 2 arises because the metric is a symmetric tensor, and the convention of Latin indices i, j taking space-like values 1, 2, 3 is used. As each component of the metric has space and time dependence in general; this is significantly more complicated than the formula quoted at the beginning, see metric tensor (general relativity) for more information.

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

${\displaystyle E_{0}=m_{0}c^{2}\,,}$

where m₀ is the rest mass of the body.

If the object is massless, as is the case for a photon, then the equation reduces to

${\displaystyle E=pc\,.}$

This is a useful simplification. It can be rewritten in other ways using the de Broglie relations:

${\displaystyle E={\frac {hc}{\lambda }}=\hbar ck\,.}$

if the wavelength λ or wavenumber k are given.

Rewriting the relation for massive particles as:

${\displaystyle E=m_{0}c^{2}{\sqrt {1+\left({\frac {p}{m_{0}c}}\right)^{2}}}\,,}$

and expanding into power series by the binomial theorem (or a Taylor series):

${\displaystyle E=m_{0}c^{2}\left[1+{\frac {1}{2}}\left({\frac {p}{m_{0}c}}\right)^{2}-{\frac {1}{8}}\left({\frac {p}{m_{0}c}}\right)^{4}+\cdots \right]\,,}$

in the limit that u ≪ c, we have γ(u) ≈ 1 so the momentum has the classical form p ≈ m₀u, then to first order in (p/m₀c)²
(i.e. retain the term (p/m₀c)²ⁿ
for n = 1 and neglect all terms for n ≥ 2) we have

${\displaystyle E\approx m_{0}c^{2}\left[1+{\frac {1}{2}}\left({\frac {m_{0}u}{m_{0}c}}\right)^{2}\right]\,,}$

or

${\displaystyle E\approx m_{0}c^{2}+{\frac {1}{2}}m_{0}u^{2}\,,}$

where the second term is the classical kinetic energy, and the first is the rest mass of the particle. 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.

In the case of many particles with relativistic momenta p_n and energy E_n, where n = 1, 2, ... (up to the total number of particles) simply labels the particles, as measured in a particular frame, the four-momenta in this frame can be added;

${\displaystyle \sum _{n}\mathbf {P} _{n}=\sum _{n}\left({\frac {E_{n}}{c}},\mathbf {p} _{n}\right)=\left(\sum _{n}{\frac {E_{n}}{c}},\sum _{n}\mathbf {p} _{n}\right)\,,}$

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

${\displaystyle \left|\left(\sum _{n}\mathbf {P} _{n}\right)\right|^{2}=\left(\sum _{n}{\frac {E_{n}}{c}}\right)^{2}-\left(\sum _{n}\mathbf {p} _{n}\right)^{2}=\left(M_{0}c\right)^{2}\,,}$

where M₀ 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 of (1);

${\displaystyle \left(\sum _{n}E_{n}\right)^{2}=\left(\sum _{n}\mathbf {p} _{n}c\right)^{2}+\left(M_{0}c^{2}\right)^{2}}$

(2)

The energies and momenta in the equation are all frame-dependent, while M₀ is frame-independent.

In the center-of-momentum frame (COM frame), by definition we have:

${\displaystyle \sum _{n}\mathbf {p} _{n}={\boldsymbol {0}}\,,}$

with the implication from (2) that the invariant mass is also the centre of momentum (COM) mass–energy, aside from the c² factor:

${\displaystyle \left(\sum _{n}E_{n}\right)^{2}=\left(M_{0}c^{2}\right)^{2}\Rightarrow \sum _{n}E_{\mathrm {COM} \,n}=E_{\mathrm {COM} }=M_{0}c^{2}\,,}$

and this is true for all frames since M₀ is frame-independent. The energies E_{COM n} are those in the COM frame, not the lab frame.

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 M₀ to be expressed in terms of the energies and rest masses, or momenta and rest masses. In a particular frame, the squares of sums can be rewritten as sums of squares (and products):

${\displaystyle \left(\sum _{n}E_{n}\right)^{2}=\left(\sum _{n}E_{n}\right)\left(\sum _{k}E_{k}\right)=\sum _{n,k}E_{n}E_{k}=2\sum _{n<k}E_{n}E_{k}+\sum _{n}E_{n}^{2}\,,}$

${\displaystyle \left(\sum _{n}\mathbf {p} _{n}\right)^{2}=\left(\sum _{n}\mathbf {p} _{n}\right)\cdot \left(\sum _{k}\mathbf {p} _{k}\right)=\sum _{n,k}\mathbf {p} _{n}\cdot \mathbf {p} _{k}=2\sum _{n<k}\mathbf {p} _{n}\cdot \mathbf {p} _{k}+\sum _{n}\mathbf {p} _{n}^{2}\,,}$

so substituting the sums, we can introduce their rest masses m_n in (2):

${\displaystyle \sum _{n}\left(m_{n}c^{2}\right)^{2}+2\sum _{n<k}\left(E_{n}E_{k}-c^{2}\mathbf {p} _{n}\cdot \mathbf {p} _{k}\right)=\left(M_{0}c^{2}\right)^{2}\,.}$

The energies can be eliminated by:

${\displaystyle E_{n}={\sqrt {\left(\mathbf {p} _{n}c\right)^{2}+\left(m_{n}c^{2}\right)^{2}}}\,,\quad E_{k}={\sqrt {\left(\mathbf {p} _{k}c\right)^{2}+\left(m_{k}c^{2}\right)^{2}}}\,,}$

similarly the momenta can be eliminated by:

${\displaystyle \mathbf {p} _{n}\cdot \mathbf {p} _{k}=\left|\mathbf {p} _{n}\right|\left|\mathbf {p} _{k}\right|\cos \theta _{nk}\,,\quad |\mathbf {p} _{n}|={\frac {1}{c}}{\sqrt {E_{n}^{2}-\left(m_{n}c^{2}\right)^{2}}}\,,\quad |\mathbf {p} _{k}|={\frac {1}{c}}{\sqrt {E_{k}^{2}-\left(m_{k}c^{2}\right)^{2}}}\,,}$

where θ_nk is the angle between the momentum vectors p_n and p_k.

Rearranging:

${\displaystyle \left(M_{0}c^{2}\right)^{2}-\sum _{n}\left(m_{n}c^{2}\right)^{2}=2\sum _{n<k}\left(E_{n}E_{k}-c^{2}\mathbf {p} _{n}\cdot \mathbf {p} _{k}\right)\,.}$

Since the invariant mass of the system and the rest masses of each particle are frame-independent, the right hand side is also an invariant (even though the energies and momenta are all measured in a particular frame).