Energy–momentum relation

In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating total energy (which is also called relativistic energy) to invariant mass (which is also called rest mass) and momentum.

It is the extension of mass–energy equivalence for bodies or systems with non-zero momentum.

This equation holds for a body or system, such as one or more particles, with total energy E, invariant mass m0, and momentum of magnitude p; the constant c is the speed of light.

It assumes the special relativity case of flat spacetime[1][2][3] and that the particles are free.

For bodies or systems with zero momentum, it simplifies to the mass–energy equation

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: The quantities E, p, E′, p′ are all related by a Lorentz transformation.

[4] The energy–momentum relation goes back to Max Planck's article[5] published in 1906.

It was used by Walter Gordon in 1926 and then by Paul Dirac in 1928 under the form

[6][7] The equation can be derived in a number of ways, two of the simplest include: For a massive object moving at three-velocity u = (ux, uy, uz) with magnitude |u| = u in the lab frame:[1] is the total energy of the moving object in the lab frame, 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: Some authors use relativistic mass defined by: 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: then solving for u2 and substituting into the Lorentz factor one obtains its alternative form in terms of 3-momentum and mass, rather than 3-velocity: Inserting this form of the Lorentz factor into the energy equation gives: followed by more rearrangement it 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.

Naively setting m0 = 0 would mean that E = 0 and p = 0 and no energy–momentum relation could be derived, which is not correct.

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: a Lorentz invariant quantity, and therefore independent of the frame of reference.

Using the Minkowski metric η with metric signature (− + + +), the inner product is and so or, in natural units where c = 1, 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, in which the Minkowski metric η is replaced by the metric tensor field g: solved from the Einstein field equations.

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.

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).

If the object is massless, as is the case for a photon, then the equation reduces to This is a useful simplification.

It can be rewritten in other ways using the de Broglie relations: if the wavelength λ or wavenumber k are given.

Rewriting the relation for massive particles as: and expanding into power series by the binomial theorem (or a Taylor series): 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 or where the second term is the classical kinetic energy, and the first is the rest energy 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.

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

Either the energies or momenta of the particles, as measured in some frame, can be eliminated using the energy momentum relation for each particle: 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): so substituting the sums, we can introduce their rest masses mn in (2): The energies can be eliminated by: similarly the momenta can be eliminated by: where θnk is the angle between the momentum vectors pn and pk.

Rearranging: 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).

Using the de Broglie relations for energy and momentum for matter waves, where ω is the angular frequency and k is the wavevector with magnitude |k| = k, equal to the wave number, the energy–momentum relation can be expressed in terms of wave quantities: and tidying up by dividing by (ħc)2 throughout:

This can also be derived from the magnitude of the four-wavevector in a similar way to the four-momentum above.

Since the reduced Planck constant ħ and the speed of light c both appear and clutter this equation, this is where natural units are especially helpful.

Normalizing them so that ħ = c = 1, we have: The velocity of a bradyon with the relativistic energy–momentum relation can never exceed c. On the contrary, it is always greater than c for a tachyon whose energy–momentum equation is[10] By contrast, the hypothetical exotic matter has a negative mass[11] and the energy–momentum equation is