In special relativity, fourmomentum is the generalization of the
classical threedimensional momentum to fourdimensional spacetime.
Momentum
p = ( p 0 , p 1 , p 2 , p 3 ) = ( E c , p x , p y , p z ) . displaystyle p=(p^ 0 ,p^ 1 ,p^ 2 ,p^ 3 )=left( E over c ,p_ x ,p_ y ,p_ z right). The quantity mv of above is ordinary nonrelativistic momentum of the particle and m its rest mass. The fourmomentum is useful in relativistic calculations because it is a Lorentz vector. This means that it is easy to keep track of how it transforms under Lorentz transformations. The above definition applies under the coordinate convention that x0 = ct. Some authors use the convention x0 = t, which yields a modified definition with p0 = E/c2. It is also possible to define covariant fourmomentum pμ where the sign of the energy is reversed. Contents 1 Minkowski norm 2 Relation to fourvelocity 3 Derivation 4 Conservation of fourmomentum 5 Canonical momentum in the presence of an electromagnetic potential 6 See also 7 References Minkowski norm[edit]
Calculating the Minkowski norm squared of the fourmomentum gives a
Lorentz invariant
p ⋅ p = η μ ν p μ p ν = p ν p ν = − E 2 c 2 +
p
2 = − m 2 c 2 displaystyle pcdot p=eta _ mu nu p^ mu p^ nu =p_ nu p^ nu = E^ 2 over c^ 2 +mathbf p ^ 2 =m^ 2 c^ 2 where η μ ν = ( − 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) displaystyle eta _ mu nu =left( begin matrix 1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1end matrix right) is the metric tensor of special relativity with metric signature for definiteness chosen to be (–1, 1, 1, 1). The negativity of the norm reflects that the momentum is a timelike fourvector for massive particles. The other choice of signature would flip signs in certain formulas (like for the norm here). This choice is not important, but once made it must for consistency be kept throughout. The Minkowski norm is Lorentz invariant, meaning its value is not changed by Lorentz transformations/boosting into different frames of reference. More generally, for any two fourmomenta p and q, the quantity p ⋅ q is invariant. Relation to fourvelocity[edit] For a massive particle, the fourmomentum is given by the particle's invariant mass m multiplied by the particle's fourvelocity, p μ = m u μ , displaystyle p^ mu =mu^ mu , where the fourvelocity u is u = ( u 0 , u 1 , u 2 , u 3 ) = γ v ( c , v x , v y , v z ) , displaystyle u=left(u^ 0 ,u^ 1 ,u^ 2 ,u^ 3 right)=gamma _ v left(c,v_ x ,v_ y ,v_ z right), and γ v = 1 1 − v 2 c 2 displaystyle gamma _ v = frac 1 sqrt 1 frac v^ 2 c^ 2 is the
Lorentz factor
p i = ∂ S ∂ q i , E = − ∂ S ∂ t , displaystyle p_ i = frac partial S partial q_ i ,quad E= frac partial S partial t , it is immediate (recalling x0 = ct, x1 = x, x2 = y, x3 = z and x0 = −x0, x1 = x1, x2 = x2, x3 = x3 in the present metric convention) that p μ = − ∂ S ∂ x μ = ( E c , − p ) displaystyle p_ mu = frac partial S partial x^ mu =left( E over c ,mathbf p right) is a covariant fourvector with the threevector part being the (negative of) canonical momentum. Observations Consider initially a system of one degree of freedom q. In the derivation of the equations of motion from the action using Hamilton's principle, one finds (generally) in an intermediate stage for the variation of the action, δ S = [ ∂ L ∂ q ˙ δ q ]
t 1 t 2 + ∫ t 1 t 2 ( ∂ L ∂ q − d d t ∂ L ∂ q ˙ ) δ q d t . displaystyle delta S=left.left[ frac partial L partial dot q delta qright]right_ t_ 1 ^ t_ 2 +int _ t_ 1 ^ t_ 2 left( frac partial L partial q  frac d dt frac partial L partial dot q right)delta qdt. The assumption is then that the varied paths satisfy δq(t1) = δq(t2)
= 0, from which
Lagrange's equations
δ S = ∑ i ∂ L ∂ q ˙ i δ q i = ∑ i p i δ q i . displaystyle delta S=sum _ i frac partial L partial dot q _ i delta q_ i =sum _ i p_ i delta q_ i . Observing that δ S = ∑ i ∂ S ∂ q i δ q i , displaystyle delta S=sum _ i frac partial S partial q _ i delta q_ i , one concludes p i = ∂ S ∂ q i . displaystyle p_ i = frac partial S partial q_ i . In a similar fashion, keep endpoints fixed, but let t2 = t vary. This time, the system is allowed to move through configuration space at "arbitrary speed" or with "more or less energy", the field equations still assumed to hold and variation can be carried out on the integral, but instead observe d S d t = L displaystyle frac dS dt =L by the fundamental theorem of calculus. Compute using the above expression for canonical momenta, d S d t = ∂ S ∂ t + ∑ i ∂ S ∂ q i q ˙ i = ∂ S ∂ t + ∑ i p i q ˙ i = L . displaystyle frac dS dt = frac partial S partial t +sum _ i frac partial S partial q_ i dot q _ i = frac partial S partial t +sum _ i p_ i dot q _ i =L. Now using H = ∑ i p i q ˙ i − L , displaystyle H=sum _ i p_ i dot q _ i L, where H is the Hamiltonian, leads to, since E = H in the present case, E = H = − ∂ S ∂ t . displaystyle E=H= frac partial S partial t . Incidentally, using H = H(q, p, t) with p = ∂S/∂q in the above equation yields the Hamilton–Jacobi equations. In this context, S is called Hamilton's principal function. The action S is given by S = − m c ∫ d s = ∫ L d t , L = − m c 2 1 − v 2 c 2 , displaystyle S=mcint ds=int Ldt,quad L=mc^ 2 sqrt 1 frac v^ 2 c^ 2 , where L is the relativistic Lagrangian for a free particle. From this, glossing over these details, The variation of the action is δ S = − m c ∫ δ d s . displaystyle delta S=mcint delta ds. To calculate δds, observe first that δds2 = 2dsδds and that δ d s 2 = δ η μ ν d x μ d x ν = η μ ν ( δ ( d x μ ) d x ν + d x μ δ ( d x ν ) ) = 2 η μ ν δ ( d x μ ) d x ν . displaystyle delta ds^ 2 =delta eta _ mu nu dx^ mu dx^ nu =eta _ mu nu left(delta left(dx^ mu right)dx^ nu +dx^ mu delta left(dx^ nu right)right)=2eta _ mu nu delta left(dx^ mu right)dx^ nu . So δ d s = η μ ν δ d x μ d x ν d s = η μ ν d δ x μ d x ν d s , displaystyle delta ds=eta _ mu nu delta dx^ mu frac dx^ nu ds =eta _ mu nu ddelta x^ mu frac dx^ nu ds , or δ d s = η μ ν d δ x μ d τ d x ν c d τ d τ , displaystyle delta ds=eta _ mu nu frac ddelta x^ mu dtau frac dx^ nu cdtau dtau , and thus δ S = − m ∫ η μ ν d δ x μ d τ d x ν d τ d τ = − m ∫ η μ ν d δ x μ d τ u ν d τ = − m ∫ η μ ν [ d d τ ( δ x μ u ν ) − δ x μ d d τ u ν ] d τ displaystyle delta S=mint eta _ mu nu frac ddelta x^ mu dtau frac dx^ nu dtau dtau =mint eta _ mu nu frac ddelta x^ mu dtau u^ nu dtau =mint eta _ mu nu leftlbrack frac d dtau left(delta x^ mu u^ nu right)delta x^ mu frac d dtau u^ nu rightrbrack dtau which is just δ S = [ − m u μ δ x μ ] t 1 t 2 + m ∫ t 1 t 2 δ x μ d u μ d s d s displaystyle delta S=left[mu_ mu delta x^ mu right]_ t_ 1 ^ t_ 2 +mint _ t_ 1 ^ t_ 2 delta x^ mu frac du_ mu ds ds δ S = [ − m u μ δ x μ ] t 1 t 2 + m ∫ t 1 t 2 δ x μ d u μ d s d s = − m u μ δ x μ = ∂ S ∂ x μ δ x μ = − p μ δ x μ , displaystyle delta S=left[mu_ mu delta x^ mu right]_ t_ 1 ^ t_ 2 +mint _ t_ 1 ^ t_ 2 delta x^ mu frac du_ mu ds ds=mu_ mu delta x^ mu = frac partial S partial x^ mu delta x^ mu =p_ mu delta x^ mu , where the second step employs the field equations duμ/ds = 0, (δxμ)t1 = 0, and (δxμ)t2 ≡ δxμ as in the observations above. Now compare the last three expressions to find p μ = − ∂ μ [ S ] = − ∂ S ∂ x μ = m u μ = m ( c 1 − v 2 c 2 , v x 1 − v 2 c 2 , v y 1 − v 2 c 2 , v z 1 − v 2 c 2 ) , displaystyle p^ mu =partial ^ mu [S]= frac partial S partial x_ mu =mu^ mu =mleft( frac c sqrt 1 frac v^ 2 c^ 2 , frac v_ x sqrt 1 frac v^ 2 c^ 2 , frac v_ y sqrt 1 frac v^ 2 c^ 2 , frac v_ z sqrt 1 frac v^ 2 c^ 2 right), with norm −m2c2, and the famed result for the relativistic energy, E = m c 2 1 − v 2 c 2 = m r c 2 , displaystyle E= frac mc^ 2 sqrt 1 frac v^ 2 c^ 2 =m_ r c^ 2 , where mr is the now unfashionable relativistic mass, follows. By comparing the expressions for momentum and energy directly, one has p = E v c 2 , displaystyle mathbf p =E frac mathbf v c^ 2 , that holds for massless particles as well. Squaring the expressions for energy and threemomentum and relating them gives the energy–momentum relation, E 2 c 2 = p ⋅ p + m 2 c 2 . displaystyle frac E^ 2 c^ 2 =mathbf p cdot mathbf p +m^ 2 c^ 2 . Substituting p μ ↔ − ∂ S ∂ x μ displaystyle p_ mu leftrightarrow  frac partial S partial x^ mu in the equation for the norm gives the relativistic Hamilton–Jacobi equation,[3] η μ ν ∂ S ∂ x μ ∂ S ∂ x ν = − m 2 c 2 . displaystyle eta ^ mu nu frac partial S partial x^ mu frac partial S partial x^ nu =m^ 2 c^ 2 . It is also possible to derive the results from the Lagrangian directly. By definition,[4] p = ∂ L ∂ v = ( ∂ L ∂ x ˙ , ∂ L ∂ y ˙ , ∂ L ∂ z ˙ ) = m ( γ v x , γ v y , γ v z ) = m γ v = m u , E = p ⋅ v − L = m c 2 1 − v 2 c 2 , displaystyle begin aligned mathbf p &= frac partial L partial mathbf v =left( partial L over partial dot x , partial L over partial dot y , partial L over partial dot z right)=m(gamma v_ x ,gamma v_ y ,gamma v_ z )=mgamma mathbf v =mmathbf u ,\E&=mathbf p cdot mathbf v L= frac mc^ 2 sqrt 1 frac v^ 2 c^ 2 ,end aligned which constitute the standard formulae for canonical momentum and
energy of a closed (timeindependent Lagrangian) system. With this
approach it is less clear that the energy and momentum are parts of a
fourvector.
The energy and the threemomentum are separately conserved quantities
for isolated systems in the Lagrangian framework. Hence fourmomentum
is conserved as well. More on this below.
More pedestrian approaches include expected behavior in
electrodynamics.[5] In this approach, the starting point is
application of
Lorentz force law
The fourmomentum p (either covariant or contravariant) is conserved.
The total energy E = p0c is conserved.
The
3space
Note that the invariant mass of a system of particles may be more than the sum of the particles' rest masses, since kinetic energy in the system centerofmass frame and potential energy from forces between the particles contribute to the invariant mass. As an example, two particles with fourmomenta (5 GeV/c, 4 GeV/c, 0, 0) and (5 GeV/c, −4 GeV/c, 0, 0) each have (rest) mass 3 GeV/c2 separately, but their total mass (the system mass) is 10 GeV/c2. If these particles were to collide and stick, the mass of the composite object would be 10 GeV/c2. One practical application from particle physics of the conservation of the invariant mass involves combining the fourmomenta pA and pB of two daughter particles produced in the decay of a heavier particle with fourmomentum pC to find the mass of the heavier particle. Conservation of fourmomentum gives pCμ = pAμ + pBμ, while the mass M of the heavier particle is given by −PC ⋅ PC = M2c2. By measuring the energies and threemomenta of the daughter particles, one can reconstruct the invariant mass of the twoparticle system, which must be equal to M. This technique is used, e.g., in experimental searches for Z′ bosons at highenergy particle colliders, where the Z′ boson would show up as a bump in the invariant mass spectrum of electron–positron or muon–antimuon pairs. If the mass of an object does not change, the Minkowski inner product of its fourmomentum and corresponding fouracceleration Aμ is simply zero. The fouracceleration is proportional to the proper time derivative of the fourmomentum divided by the particle's mass, so p μ A μ = η μ ν p μ A ν = η μ ν p μ d d τ p ν m = 1 2 m d d τ p ⋅ p = 1 2 m d d τ ( − m 2 c 2 ) = 0. displaystyle p^ mu A_ mu =eta _ mu nu p^ mu A^ nu =eta _ mu nu p^ mu frac d dtau frac p^ nu m = frac 1 2m frac d dtau pcdot p= frac 1 2m frac d dtau left(m^ 2 c^ 2 right)=0. Canonical momentum in the presence of an electromagnetic potential[edit] For a charged particle of charge q, moving in an electromagnetic field given by the electromagnetic fourpotential: A = ( A 0 , A 1 , A 2 , A 3 ) = ( ϕ c , A x , A y , A z ) displaystyle A=left(A^ 0 ,A^ 1 ,A^ 2 ,A^ 3 right)=left( phi over c ,A_ x ,A_ y ,A_ z right) where Φ is the scalar potential and A = (Ax, Ay, Az) the vector potential, the components of the canonical momentum fourvector P is P μ = p μ + q A μ . displaystyle P^ mu =p^ mu +qA^ mu .! This, in turn, allows the potential energy from the charged particle
in an electrostatic potential and the
Lorentz force
Physics portal Fourforce Fourgradient Pauli–Lubanski pseudovector References[edit] ^ Landau & Lifshitz 2002, pp. 25–29 ^ Landau & Lifshitz 1975, pp. 139 ^ Landau & Lifshitz 1975, p. 30 ^ Landau & Lifshitz 1975, pp. 15–16 ^ Sard 1970, Section 3.1 ^ Sard 1970, Section 3.2 ^ Lewis & Tolman 1909 Wikisource version Goldstein, Herbert (1980). Classical mechanics (2nd ed.). Reading,
Mass.: Addison–Wesley Pub. Co. ISBN 0201029189.
Landau, L. D.; Lifshitz, E. M. (1975) [1939]. Mechanics. Translated
from Russian by J. B. Sykes and J. S. Bell. (3rd ed.). Amsterdam:
Elsevier. ISBN 9780750628969.
Landau, L.D.; Lifshitz, E.M. (2000). The classical theory of fields.
4th rev. English edition, reprinted with corrections; translated from
the Russian by Morton Hamermesh. Oxford: Butterworth Heinemann.
ISBN 9780750627689.
Rindler, Wolfgang (1991). Introduction to
Special
