The Proca Equation

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In devising quantum-mechanical equations that describe particles, we employ two entities - a mathematical operator and a state function upon which it operates. We devise our equation by asserting that the state function must have such a form that the statement (operator)(state function)=0 stands true to Reality. We can then use the state function in Born’s theorem to calculate expectation values for the dynamic properties of the particle that we have under study.

Erwin Schrödinger used energy as the operator in the wave/particle equation that he devised in 1924. That was suitable for non-relativistic quantum mechanics. For relativistic quantum mechanics (referring to the special theory, not the general) physicists use the energy-squared relation of relativistic dynamics,

(Eq’n 1)

to provide the operator. In that equation m0 represents the rest mass of the particle being described and pμ=(px,py,pz,E/c) represents the measured momentum-energy four-vector associated with the particle (pν=gνμpμ, gνμ being the metric tensor of the spacetime in which the particle exists). That equation, when applied to the state function of the particle, gives us the Klein-Gordon equation and the Proca equation.

Equation 1 can also be factored, through the use of matrices, into a product of two linear factors. Because the equation zeroes out, one or the other of the factors can be used as the operator in the fundamental quantum equation. The result yields the Dirac equation and the Rarita-Schwinger equation.

The variables in Equation 1 have partial-differential representations,

(Eq’n 2)

Applying that representation to a state function extracts from the state function the value of the variable. Substituting that equivalence into Equation 1 makes the fundamental equation of quantum mechanics a partial differential equation. We know how to solve those, given that we know the basic form of the state function. If we let psi represent the state function, then we can transform Equation 1 into

(Eq’n 3)

which is the Klein-Gordon equation if psi represents a single number, a scalar.

When the Romanian physicist Alexandru Proca (1897 Oct 16 - 1955 Dec 13) produced his own version of the Klein-Gordon equation in 1936, he added an interaction term and asserted that the state function is a four-vector; specifically, the state function is the electromagnetic potentials four-vector, Aμ=(Ax,Ay,Az,ϕ/c), in which Ax, Ay, and Az represent the components of the magnetic vector potential and phi represents the electrostatic potential. One form of Proca’s equation looks like this:

(Eq’n 4)

in which jμ=(jx, jy, jz, cρ) represents the four-current of electricity. But what does that equation mean and how do we obtain it?

As usual, we start with the Lagrangian function. More properly, we use a Lagrangian density, which should consist of a kinetic energy density minus a potential energy density, both associated with a field, if it is to be like the classical Lagrangian and subject to the Euler-Lagrange equations.

If a particle possesses linear momentum p and total energy E, then as it crosses a minuscule distance dx in a minuscule elapse of time dt it will play out a minuscule bit of action

(Eq’n 5)

We have chosen that particular form of the equation because it’s Lorentz invariant, being the dot product of the energy-momentum four-vector and the differential distance-duration four-vector. If we have an electromagnetic field that consists of an electrostatic field ϕ(x,t) and an electrotonic field A(x,t) and if a distribution of electric charge of density ρ(x,t) exists in that field, then we know that the charge possesses a potential linear momentum density of P=ρA and a potential energy density of E=ρϕ. If we multiply those densities by a minuscule element of volume, we can rewrite the action equation as

(Eq’n 6)

In that equation we have the contravariant four-current jν=(ρv,ρc) and the covariant four-potential Aν=(A,-ϕ/c).

That equation gives us the Lagrangian density of the interaction between the field and an intruding current of electric charge. We must add to that the Lagrangian density of the field itself, which means that we must substitute the linear-momentum density and the energy density of the field into the first clause of the equation.

Imagine a sphere that has a quantity of electric charge distributed uniformly over its surface. Gauss’s law tells us that the area density of the charge equals σ=ε0E, in which E represents the electric field intensity (volts per meter) on the outer surface of the sphere (E=0 on the inner surface, of course). Please note that I am using MKS units, in which the fundamental properties of spacetime (the electric permittivity, ε0, and the magnetic permeability, μ0) appear explicitly. The force that the field exerts on a minuscule patch of area da equals dF=σEda=ε0E2da. If the patch comes loose from the sphere, it will move away from the sphere’s center by a distance dr in the time interval dt and it will gain an increment of linear momentum d2p=dFdt. That momentum can only come from the field, so the field must have held that much potential momentum in the element of volume dV=dadr. We thus infer the amount of momentum density in the field through the calculation

(Eq’n 7)

We thus have the action equation as

(Eq’n 8)

Combining that equation with Equation 6 gives us our electromagnetic Lagrangian density,

(Eq’n 9)

(See Appendix for the conversion to the field tensor).

In the quantum theory we associate fields with particles, which are essentially quantized vibrations of the field. The Klein-Gordon, Dirac, and Rarita-Schwinger equations include terms that represent those particles through functions of the particles’ masses. The particle that we associate with the electromagnetic field, the photon, has zero rest mass, so perhaps we should not expect it to appear in the electromagnetic Lagrangian density. But suppose there’s another particle, one that does have a nonzero rest mass, associated wit the electromagnetic field. How would we include that particle in the Lagrangian density?

Modern quantum theory tells us that some particles exist as quantized vibrations of forcefields. We can extract a description of one such particle from the state function describing its field by applying the appropriate operators to that state function. For example, we can obtain the rest energy of the particle by applying the total energy operator to the field,

(Eq’n 10)

The state function also represents the potential of the field, so the time derivative describes the forcefield itself plus the time derivative of the static potential (which latter term corresponds to the F=vdm/dt term in the inertial reaction equation);

(Eq’n 11)

The energy density associated with that field conforms to

(Eq’n 12)

We get the Lagrangian density by subtracting an energy density (as shown in Equation 6), so we get the Proca Lagrangian density as

(Eq’n 13)

That Lagrangian density gives us Proca’s equation by way of the Euler-Lagrange equation,

(Eq’n 14)

In order to apply that equation to our Proca Lagrangian we must so raise and lower the indices in the first term that they match the pattern in the derivative operator; that is, we write

(Eq’ns 15)

so that we can use

(Eq’n 16)

in which the deltas represent Kronecker’s delta. Only the first term on the right side of Equation 13 is subject to that derivative operator, so we can write

(Eq’n 17)

We note that the metric tensor is not a function of the field or of its derivatives, so we treat it as a constant and also make use of the fact that gμνgμν=I, the identity matrix, which simply multiplied all of the other matrices by one.

We also have

(Eq’n 18)

Combining that result and the previous one into Equation 14 and dividing out the minus one gives us Proca’s equation,

(Eq’n 19)

If the mass of the field particle goes to zero, then that equation stands equivalent to the first and fourth of Maxwell’s Equations, the source equations.

To demonstrate the truth of that statement we let Aν=(A,-ϕ/c) and jν=(j,-ρc) and write

(Eq’n 20)

We can split that equation into two independent equations,

(Eq’ns 21)

If we use the Lorenz condition (∂μAμ=0) in the form

(Eq’n 22)

we can transform those equations into

(Eq’ns 23)

A little algebraic reworking transforms those equations into

(Eq’ns 24)

Maxwell’s fourth and first equations respectively.

That’s the standard deduction of Maxwell’s inhomogeneous equations from the Proca equation, but there’s something that feels wrong about it. In assembling the Proca Lagrangian I described the four-current as intruding into a pre-existing four-potential field, implying that the field did not come from that current. But Equations 20 through 24 tacitly assume that the four-current is the source of the field. Certainly the four-current makes a contribution to the field and the field comes, at least in part, from other four-currents. However, Equations 24 refer to certain localized changes in the field and so long as we remember that fact Equations 24 and the derivation that produced them are valid and legitimate.

(We can also get Maxwell’s homogeneous equations, the second and the third, from the Bianchi identity,

(Eq’n 25)

If any two indices in that statement come equal to each other, then the parenthetical part of one term zeroes out and the other two terms cancel each other. So we need only consider the cases in which the indices all differ from one another. Start by letting α=4 (so β,γ=1,2,3); in that case we get

(Eq’n 26)

which, except for the factor of 1/c, expresses Maxwell’s third equation, Faraday’s law. Note that the parenthetical part of the first term is just the curl of the magnetic vector potential and that the terms containing phi (the electrostatic potential) cancel each other out. If we let α=1,2,3, then the parenthetical part of each of all three terms gives us the curl of the magnetic vector potential, the component index of which curl matches that of the outside derivative, so we have a trio of divergences,

(Eq’n 27)

which is Maxwell’s second equation, Gauss’s law of the magnetic field.)

What does Proca’s equation tell us about the particle associated with the field?

The usual discussion around Proca’s equation tells us that it describes a particle carrying one full Dirac unit (aitch-bar) of spin, which spin has three eigenstates associated with it. We want to figure out how to extract that information from Equation 19. Experience with Dirac’s equation tells us to expect the existence of a spinor with six components, three for the positive-energy states (matter) and three for the negative-energy states (antimatter).

As we did with the Dirac equation, we introduce spin matrices into Equation 19 (without the interaction term) in order to factor the second-order differential Proca operator into two linear, first-order operators, one for the matter states and one for the antimatter states. Each of the first-order operators has two terms, just as the second-order operator does, so we need to introduce matrices Γ to eliminate the cross terms from the multiplication; that is, we must have a quartet of Γ matrices such that

(Eq’n 28)

As before, we have the Lorenz condition in play, so those matrices enable us to write

(Eq’n 29)

in which the big phi represents the product of the A-field and the six-element spinor and I represents the 6x6 identity matrix necessitated by the gamma matrices (the matrix equivalent of multiplying by one, so it doesn’t change the value of the mass term). At least one of those two operators must nullify Φ, so we can pick one of them and see whither it leads.

Let’s choose

(Eq’n 30)

That’s one of our two Proca spin equations, so called to distinguish it from Proca’s equation (Equation 19), which does not display the particle’s spin. The gamma matrices must satisfy the relation

(Eq’n 31)

so we have

(Eq’n 32)

in which we have two 3x3 null matrices and the 3x3 Pauli spin matrices (for a spin-1 particle) along with gamma-4,

(Eq’ns 33)

From Equation 2 we know that

(Eq’n 34)

so we can rewrite Equation 30 as

(Eq’n 35)

in which K=(E/c)+m0c. Note that I am using the covariant momenta, so the multiplication by the covariant metric tensor turns the negative sign on the fourth-dimensional element positive.

We describe each of the six spin eigenstates of the particle by setting one element of the spinor equal to one and the other five elements equal to zero. If we make ψ1=1, we get

(Eq’n 36)

If we multiply that equation by its adjoint (the complex conjugate of its transpose), we get

(Eq’n 37)

If we carry out the same calculation with ψ2=1, we get

(Eq’n 38)

Those equations tell us that if the spin vector lies entirely in the x-y plane, the particle has no momentum in the z-direction and that if the spin vector points in the z-direction (with components pointing in the x- and y-directions to satisfy Heisenberg’s rule), the particle can have momenta in all three directions. From those facts we infer the proposition that the particle can only move in the direction in which its spin points. Actually, it’s Equation 38 that tells us that the particle cannot have a component of motion oriented perpendicular to its spin vector. Equation 37 simply confirms that the spin-up state has components of the spin vector oriented in the x- and y-directions.

Equations 37 and 38 plus the other four equations that go with them look strange. But the division by the square of aitch-bar gives the operator part of each equation units of reciprocal distance squared, as we have with the Laplace operator, so these six equations are analogous to Poisson’s equation. We have already determined the imaginary exponential part of the state function, so solving these equations will give us the amplitude of the state function in each of these six cases of spin orientation.

Note that the Proca equation did not tell us anything about the existence of the particle’s spin. We had to assume the existence and magnitude of the spin when we factored the operator in Equation 19 and introduced the spinor. Once we did so, however, the equation told us something unexpected about the relationship between the particle’s spin vector and its linear momentum vector. In this way, little by little, we build up our Rationalist picture of Reality.

Appendix: Electromagnetic Energy Density

In classical electromagnetic theory we describe the energy density of an electromagnetic field as

(Eq’n A-1)

in which E represents the electric field strength (volts per meter) and B represents the magnetic induction field strength (volt-seconds per square meter). In the classical format we have the electric field and the magnetic induction field described through their potentials as

(Eq’ns A-2)

with A=(Ax,Ay,Az) representing the magnetic vector potential (describing the electrotonic field) and phi representing the electrostatic potential. We want to create the tensor equivalents of those equations as a first step in devising a quantum Lagrangian for the electromagnetic field.

Represent the fields themselves as contravariant tensors of the first rank, Ei and Bi, because the fields are measurable vector quantities. The potentials constitute a contravariant four-vector Aμ=(A1, A2, A3, ϕ/c). As usual the Latin indices take the values 1,2,3 and the Greek indices take the values 1,2,3,4. The signature of the spacetime follows the Minkowski metric, gμν=diag (+1,+1,+1,-1). We want to differentiate the potentials with respect to measurable changes in the coordinates, so we use the covariant differentials, ∂μ=∂/∂xμ, with ∂4= /c∂t.

For the electric field we have

(Eq’n A-3)

For the magnetic field we have the standard curl,

(Eq’n A-4)

in which the indices permute in the pattern ikm=123, 231, 312. To combine those two equations into a single electromagnetic field tensor we start by dividing Equation A-3 by the negative of the speed of light and raise the indices on the differentials in both equations.

We know that (∂1,∂2,∂3,∂4)gμν=(∂1,∂2,∂3,-∂4), so Equations A-3 and A-4 become

(Eq’ns A-5)

Those equations give us six elements, three for the magnetic field and three for the electric field. That set lacks only four elements to be a complete permutation set on the Greek indices and those four elements all equal zero because the indices on those elements equal each other. Those facts lead us to write our description of the electromagnetic field as a second-rank contravariant tensor,

(Eq’n A-6)

We can prove and verify the proposition that we have a correct description of the electromagnetic field in that equation by multiplying it on the left with an electric current four-density jμ=(jx, jy, jz, cρ), in which rho represents the static electric charge density. In order to carry out the multiplication properly, we shift the four-current to the right side of the matrix, transposing it from a row vector to a column vector and lowering the indices on it, and get our product as

(Eq’n A-7)

That’s just the force density acting on the four-current plus the rate at which the electric field (divided by the speed of light) does work on the current density.

We also have a covariant version of the field tensor, which we obtain by lowering both indices on the contravariant field tensor. Thus we multiply the contravariant field tensor by the covariant metric tensor twice,

(Eq’n A-8)

All that transformation accomplishes is to interchange the algebraic signs on the electric field components in Equation A-6, row four for column four and vice versa.

Next square the field tensor; that is, calculate the product FμνFμν. That formula and the rules of matrix multiplication tell us to multiply the first row of the contravariant tensor by the first column of the covariant tensor in the manner of a dot product; multiply the second row of the contravariant tensor by the second column of the covariant tensor; multiply the third row by the third column; and the fourth row by the fourth column. When we carry out that procedure, we get

(Eq’n A-9)

(See "The Electromagnetic Field Tensor" under Electromagnetism on this website). Note that the indices are the same on both tensors. This is the tensor analogue of a dot product. If we compare that equation to Equation A-1, we see that it looks like it might be a Lagrangian describing the electromagnetic field, though to give it the proper units of energy density we must multiply it by the electric permittivity of vacuum and the square of lightspeed and divide it by four. We thus have as a Lorentz invariant with units of energy density,

(Eq’n A-10)

eabf

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