Degrees Of Freedom Of Classical Fields

Degrees of freedom of classical fields Sergei Winitzki (Dated: DRAFT December 3, 2006) This is a pedagogical account of the concept of “independent degrees of freedom” for classical fields. I propose two definitions and show how to compute the number of degrees of freedom for the electromagnetic and the gravitational fields.

I.

DEGREES OF FREEDOM IN MECHANICS AND IN FIELD THEORY

In classical mechanics, one says that a system has n degrees of freedom if it can be unambiguously described by specifying n real-valued functions of time q1 (t), ..., qn (t), called generalized coordinates, and if it cannot be described by a fewer number of coordinates. (Of course, a system can always be described by a larger number of coordinates than strictly necessary.) For example, two point masses m1 , m2 moving in three-dimensional space and connected by a rigid, infinitely thin massless rod comprise a mechanical system with five degrees of freedom. The state of this system can be specified by the values of three Euclidean coordinates x1 , y1 , z1 of the first point mass and two spherical angles φ, θ describing the orientation of the rod. To compute the number of degrees of freedom (DOF) in a mechanical system, one starts with a description of the system in some set of generalized coordinates, looks at the equations of motion, and determines the number of independent integration constants needed to describe the general solution. This number is equal to twice the number of the DOF. (The factor 2 comes from having coordinates qi and velocities q˙i for each DOF.) For example, the equations of motion x¨ = y,

y¨ = 1

(1)

allow four independent constants of integration, and therefore the system has two degrees of freedom. The computation of the DOF in mechanics is relatively simple because every system is made of a finite number of point masses. In field theory, the basic object of description is not a point mass but a field which is a function of space and time. For example, the state of a scalar field (e.g. charge density) is described by a function ρ(t, x), while an electromagnetic field can be described by the 4-vector potential Aµ (t, x). According to the definition used in mechanics, any field has infinitely many degrees of freedom since one needs infinitely many functions of time (such as Aµ (t, x1 ), Aµ (t, x2 ), etc., for every possible vector x) to describe the states of the system. One would like to have a more discriminating definition of the DOF, so that different fields can be distinguished by their number of DOF. For instance, we would like to say that the scalar field ρ(t, x) has “fewer” DOF than the vector field Aµ . Indeed, the definition of DOF in field theory is different from that in ordinary mechanics. We shall first consider

a straightforward definition of DOF and later consider a different meaning of “DOF” and give a second definition. Definition 1: A field has n DOF if its physical states can be unambiguously described by n real-valued functions of space and time, and no fewer number of functions of space and time can suffice. According to this definition, a scalar field has one DOF, while a vector field, such as Aµ , has four DOF because the four components A0 , A1 , A2 , A3 are realvalued functions. However, this conclusion is not quite correct for the electromagnetic field because of the gauge invariance of electrodynamics. A gauge transformation changes Aµ (t, x) to another function, Aµ (t, x) → A˜µ (t, x) = Aµ (t, x) + ∂µ α(t, x),

(2)

where α(t, x) is an arbitrary function of space and time. However, the physically observable effects of the fields Aµ and A˜µ are the same, i.e. we cannot distinguish between the states of the electromagnetic field described by Aµ and A˜µ . In other words, the two different functions Aµ (t, x) and A˜µ (t, x) describe the same physical state of the electromagnetic field. Since the gauge freedom contains an arbitrary function of space and time, we expect that the electromagnetic field actually has fewer than four DOF. In the particular case of the gauge transformation (2), it is easy to see that the function α can be chosen to set A0 (t, x) = 0. The remaining three functions form a 3-vector potential A ≡ (A1 , A2 , A3 ) and are still subject to the residual gauge freedom, ˜ x) = A(t, x) + ∇α(x). A(t, x) → A(t,

(3)

However, the function α now may depend only on x (but not on t). We only count functions of all four coordinates, and thus the residual gauge transformations (3) do not reduce the effective number of the DOF. We conclude that the electromagnetic field has three DOF. In Definition 1, the phrase “physical states” must be understood as “field configurations up to gauge freedom.” II.

DEGREES OF FREEDOM IN VACUUM

In some situations, it is necessary to consider a more restrictive definition of the number of DOF. In particular, care should be taken when dealing with fields in backgrounds or those having higher-order equations of motion.

2 An important issue is to describe vacuum solutions, i.e. nontrivial solutions in the absence of sources. For example, electromagnetic waves are vacuum solutions of electrodynamics because these solutions can propagate in vacuum, in the absence of any charges or currents. We expect that in a given theory there are infinitely many vacuum solutions and that all these solutions are described by a certain number of arbitrary functions. This number is what we call the number of vacuum DOF. We shall first count the number of vacuum DOF for electrodynamics and then formulate a more general definition which is applicable also to nonlinear field theories such as general relativity. A.

Vacuum DOF in electrodynamics

The 4-vector potential Aµ (t, x) is a solution of the Maxwell equations which may be written as ∂ν ∂ ν Aµ − ∂µ (∂ ν Aν ) = jµ ,

(4)

where jµ (t, x) is the 4-vector describing the sources (charges and currents). Note that the current jµ must satisfy the conservation law ∂ µ jµ = 0,

(5)

which follows from Eq. (4) by applying ∂ µ to both sides; if the current jµ is not conserved, Eq. (4) cannot be solved. We shall now consider the vacuum situation, jµ = 0. An example of a vacuum solution is the plane wave Aµ (t, x) = aµ cos (kν xν ) ,

(6)

where aν is a 4-vector describing the amplitude of the wave and kν is the wave vector. Let us find the number of independent solutions of the form (6). By substituting into Eq. (4) with jµ = 0 one can verify that the ansatz (6) is a vacuum solution only if kν k ν = 0 and aν k ν = 0, i.e. kµ must be a null 4-vector and aµ must be orthogonal to kµ . All 4-vectors aµ orthogonal to kµ form a three-dimensional subspace spanned by kµ itself and two spacelike vectors that we shall denote lµ and mµ . To find the components of the vectors lµ and mµ explicitly, consider the components of the vector kµ ≡ (k0 , k) and choose any pair of 3-vectors l, m which are orthogonal to each other and to the 3-vector k; then the vectors lµ ≡ (0, l) and mµ ≡ (0, m) are orthogonal to each other and to kµ . Therefore the admissible vacuum solutions of the form (6) must have the amplitude vector aµ of the form aµ = αkµ + βlµ + γmµ ,

(7)

changes the amplitude by aµ → aµ + λkµ , where λ is an arbitrary constant. Now the term αkµ in Eq. (7) can be removed by a gauge transformation (8) with λ = −α, and the remaining freedom in choosing the amplitude aµ consists of choosing the two coefficients β, γ. Thus the admissible plane waves of the form (6) are described by just two free parameters. We are getting near the (correct) conclusion that the vacuum solutions of the Maxwell equations have only two DOF. However, this conclusion is not yet fully justified because the plane waves of the form (6) are not the only possible vacuum solutions. Let us therefore imagine the set of all vacuum solutions and try to analyze the structure of this set. Since Eq. (4) with jµ = 0 is linear in Aµ and homogeneous, all vacuum solutions form a linear space (linear combinations of vacuum solutions are again vacuum solutions). This linear space is infinite-dimensional because there are infinitely many independent vacuum solutions; for instance, all plane waves (6) with different null vectors kµ are linearly independent. A general vacuum solu(vac) tion Aµ (x) is an arbitrary superposition of plane waves with different amplitudes and different (null) wave vectors kµ ≡ (k, k), Z µ (vac) Aµ (x) = d3 k eikµ x ak , (9) where ak are arbitrary k-dependent amplitudes. To obtain a useful description of the space of vacuum solutions, we need to somehow reduce this huge space to a finitedimensional vector space. Let us consider the values of (vac) the vacuum solution Aµ (x0 ) at a fixed spacetime point (vac) x0 ≡ (t0 , x0 ). The set of values Aµ (x0 ) for all possible (vac) comprises a vector space because vacuum solutions Aµ electrodynamics obeys the superposition principle. This vector space is a subspace of the 4-dimensional vector space of all possible 4-vectors Aµ (t0 , x0 ). Of course, the dimension of this subspace does not depend on the chosen point (t0 , x0 ) because the equations of motion are invariant under spacetime translations. This motivates us to consider the following tentative definition. Definition 2: A field has n vacuum DOF if the values of all permissible vacuum solutions at a fixed point (t0 , x0 ) comprise a vector space of dimension n. In the next section we shall describe a more straightforward way to compute the number of DOF from this definition. The key ingredient will be to perform a Fourier transformation on the field. B.

Computing the vacuum DOF via the Fourier transform

where α, β, γ are constant coefficients (which are so far arbitrary). Furthermore, a gauge transformation of the form

The equations of motion (4) are linear differential equations with constant coefficients, so their Fourier transform can be written as

Aµ → Aµ + ∂µ [λ sin (kν xν )]

−kν k ν A˜µ (k) + kµ k ν A˜ν (k) = ˜jµ ,

(8)

(10)

3 where A˜µ (k) and ˜jµ (k) are the four-dimensional Fourier transforms of the 4-potential Aµ (t, x) and the 4-current jµ (t, x). The functions A˜µ (k) and ˜jµ (k) are usually called the Fourier modes of the field and the current. The vacuum equations (with ˜jµ = 0) can be written as
kµ k ν − k 2 δµν A˜ν ≡ Wµν A˜ν = 0,

(11)

where Wµν (k) is a “matrix” that depends on kµ and we write k 2 ≡ kµ k µ . It is clear that the solutions of the homogeneous equation (11) comprise a linear subspace S of the initial four-dimensional space of values of A˜µ , and that the dimensionality of the subspace S will depend on the 4-vector kµ . Therefore it will be convenient to analyze the solutions at a fixed kµ . In the physical space, these solutions correspond to plane waves with a fixed wave vector kµ . Note that A˜µ = λkµ (where λ is a scalar function of kµ ) is always a solution of Eq. (11), and thus the onedimensional subspace hkµ i spanned by the vector kµ is always a subspace of S. But note that the gauge transformations (2) are equivalent to A˜µ → A˜µ − ikµ α ˜. (12)   Since all solutions of the form A˜µ + λkµ are gauge equivalent, the space of physically relevant independent solutions are found as a factorspace of S by the one-dimensional subspace spanned by the vector kµ , i.e. Sphys = S/ hkµ i. The subspace hkµ i is thus the subspace of pure gauge solutions, i.e. solutions that are not physically different from A˜µ = 0. To analyze the space S of solutions of Eq. (11) at a fixed kµ , it suffices to consider two cases: k 2 6= 0 and k 2 = 0. If k 2 6= 0, Eq. (11) gives A˜ν k ν A˜µ = kµ 2 = akµ , k

(13)

where a is some constant. However, this solution is entirely within the gauge subspace and thus there are no nontrivial vacuum solutions for k 2 6= 0. In the remaining case k 2 = 0 (but kµ 6= 0), we have A˜µ k µ = 0 and thus A˜µ belongs to the three-dimensional subspace of 4vectors orthogonal to kµ . We have analyzed this subspace above and found its basis in the form (kµ , lµ , mµ ) with the decomposition (7). In other words, S = hkµ , lµ , mµ i and therefore the factorspace S/ hkµ i = hlµ , mµ i is twodimensional. Finally, we note that the mode with the zero wavenumber kµ = 0 (called the zero mode) corresponds to the (0) homogeneous solution Aµ (t, x) = Aµ = const, i.e. the 4-potential Aµ is constant in space and time. This solution can be reduced to Aµ = 0 by a gauge transformation of the form   ν (14) x Aµ → Aµ − ∂µ A(0) ν

and thus is not a physically nontrivial vacuum solution. (We call a solution physically trivial if it is gauge equivalent to the trivial solution Aµ = 0.) Our conclusion is that the space of vacuum solutions for electrodynamics is two-dimensional. We have obtained this result solely by analyzing the dimension of a certain finite-dimensional vector space that we constructed as the factor space of the vacuum solutions by the pure gauge subspace. C.

Vacuum DOF in general relativity

General relativity describes the dynamical geometry of the spacetime through the metric gµν (x). We shall consider the situation when the geometry is almost flat except for small gravitational perturbations, so that in appropriate coordinate systems we have gµν (x) = ηµν + hµν (x),

|hµν | ≪ 1.

(15)

The coordinate transformations xµ → xµ + ξ µ (x)

(16)

play the role of gauge transformations in this theory. If we consider only transformations with very small ξ, then to first order we may write the gauge transformation for hµν as hµν → hµν + ξµ,ν + ξν,µ .

(17)

Note that we do not need to employ covariant derivatives such as ξµ;ν because the difference between the covariant and the ordinary derivatives is of second order in the perturbations. The symmetric tensor hµν has 10 independent components, but not all of them are physically significant, and the number of independent components is further reduced by the equations of motion. In this section we shall compute the number of DOF in the gravitational perturbations. The vacuum equations of motion can be written as Gµν = 0. Since Rµν = Gµν − 21 Ggµν , we may also write the vacuum equations in the form Rµν = 0 which is a little simpler. A calculation shows that the Ricci tensor Rµν is expressed through the metric perturbation hµν as 2Rµν = η αβ (hµα,νβ + hνα,µβ ) − hµν − h,µν ,

(18)

where  = η αβ ∂α ∂β and h ≡ η αβ hαβ . It is convenient at this point to perform a Fourier transform in all spacetime ˜ µν and ξ˜µ . coordinates and consider the Fourier images h Then the equation of motion Rµν = 0 and the gauge transformation (17) are rewritten as   ˜ µν + hk ˜ µ kν = 0, ˜ µα kν + h ˜ να kµ k α + k 2 h (19) − h ˜ µν → h ˜ µν + kµ ξ˜ν + kν ξ˜µ . h

(20)

4 We have thus obtained a linear, homogeneous equa˜ µν and addition for the 10 independent components of h tionally a gauge equivalence relation containing four free parameters ξ˜µ . In principle we may now consider a 10dimensional linear space consisting of these independent components Qi , i = 1, ..., 10, and rewrite the equation of motion and the gauge equivalence relation in the matrix form, Fij Qi = 0,

Qi → Qi + Giµ ξµ ,

(21)

where Fij and Giµ are suitable 10×10 and 10×4 matrices. The solutions of the equation Fij Qi = 0 form a subspace S in the 10-dimensional space; for a given solution Qi(0) ∈ S, the solutions that are gauge equivalent to Qi(0) differ by a vector from the pure gauge subspace G ⊂ S. The problem of counting the DOF is then reduced to finding the dimensions of these subspaces S and G ⊂ S, which is a standard problem in finite-dimensional linear algebra. The number of DOF is then dim S − dim G. The calculations are thus made entirely straightforward but cumbersome. Rather than write the matrices Fij and Giµ explicitly, we shall simplify the equations ˜ µν in the tensor notation, use the gauge equivalence for h relations, and arrive to the answer more easily. ˜ αµ k α . Under a gauge First consider the 4-vector sµ ≡ h transformation, the vector sµ changes as sµ → sµ + ξ˜µ k 2 + ξ˜α k α kµ ,

(22)

˜ µν k µ k ν transforms as and the scalar product sµ k µ = h sµ k µ → sµ k µ + 2k 2 ξ˜α k α .

(23)

This transformation is trivial if k 2 = 0, so it appears that we need to consider the cases k 2 6= 0 and k 2 = 0 separately. If k 2 6= 0, then we can always find a vector ξ˜α such that µ

sµ k ξ˜α k α = − 2 , 2k

(24)

and then sµ k µ vanishes after the gauge transformation (23). After performing this gauge transformation, we still have the freedom to perform additional gauge transformations with vectors ξ˜α such that ξ˜α k α = 0. The vector sµ would change under such a transformation as sµ → sµ + k 2 ξ˜µ .

(25)

Therefore we may set sµ = 0 by choosing sµ ξ˜µ = − 2 ; k

(26)

note that this gauge transformation is allowed since we already have set sµ k µ = 0. After performing this last gauge transformation, we have no more remaining gauge freedom.

˜ µν k ν = 0, we may simplify Using the relation sµ ≡ h the equation of motion (19) to ˜ µν = 0. ˜ µ kν + k 2 h hk

(27)

Computing the scalar product of Eq. (27) with k µ , we ˜ µ k 2 = 0 and thus h ˜ = 0. Substituting this into find hk 2˜ ˜ µν = 0. Thus Eq. (27), we obtain k hµν = 0 and finally h there are no physically nontrivial solutions of the vacuum equations when k 2 6= 0. In the case k 2 = 0 (but kα 6= 0), we cannot perform the gauge transformations as above and need to take a ˜ changes under a gauge different approach. The trace h transformation as ˜→h ˜ + 2ξ˜α k α . h

(28)

˜ = 0 after a suitable gauge transforThus we may set h mation. There will remain the freedom to perform gauge transformations with vectors ξ˜α such that ξ˜α k α = 0. ˜ = 0, we simplify the equation of moAfter setting h tion (19) to sµ kν + sν kµ = 0,

˜ µν k ν . sµ ≡ h

(29)

Since the vector kµ 6= 0, the equation (29) has no other solutions except sµ = 0. [To verify this, one first computes a scalar product with a vector aµ such that aµ k µ = −1; the vector aµ exists as long as kµ 6= 0. Then one obtains sν = (sµ aµ ) kν and thus the vectors sµ and kµ are parallel, sµ = λkµ . Then the substitution into Eq. (29) gives λ = 0.] Thus we have reduced the equations of motion and the gauge equivalence relation to ˜h ≡ h ˜ µν η µν = 0, ˜hµν k ν = 0, ˜ µν → h ˜ µν + kµ ξ˜ν + kν ξ˜µ , ξ˜µ k µ = 0. h

(30) (31)

At this point we could introduce an explicit basis in the 4-dimensional space and find the remaining independent components, but we shall instead proceed without choosing a basis. The equations (30) represent 5 constraints on the 10 independent components of ˜hµν and thus we can expect the space of solutions to be 5-dimensional. The gauge equivalence relations are parametrized by the vector ξ˜µ which is constrained to remain orthogonal to k µ ; such vectors form a 3-dimensional space. Thus it appears that there are 5 − 3 = 2 independent DOF. The remaining task is to verify that the subspace of solutions of Eq. (30) and the pure gauge subspace are indeed 5dimensional and 3-dimensional respectively. The subspace S of solutions of the five constraints (30) will have more than 10 − 5 = 5 dimensions if the constraints were not independent, i.e. if there exists a vanishing linear combination of the constraints. This is equivalent to the existence of five coefficients A, B µ , not all equal to zero, such that Aη µν + B µ k ν = 0.

(32)

5 The scalar product of Eq. (32) with kν yields A = 0 and then the scalar product with any vector aµ such that aµ k µ 6= 0 yields B µ = 0. Therefore the constraints (30) are linearly independent and the space S has dimension five. Now we shall verify that the gauge equivalence reduces this five-dimensional space to a two-dimensional factor space. The gauge transformations (31) are compatible with the constraints (30) in the sense that the gaugetransformed solutions always satisfy the constraints. ˜ (0) Given a solution h µν of the constraints (30), we may consider the set of gauge-transformed solutions ˜ (0) + ξ˜µ kν + ξ˜ν kµ . h µν

(33)

This set is an (affine) space and we would like to show that this space has dimension three. The admissible vectors ξ˜µ must be orthogonal to k µ and thus form a threedimensional space. So the space of gauge-transformed solutions (33) cannot be more than three-dimensional, but it might be less than three-dimensional if some gauge˜ (0) transformed solutions coincide with h µν . This however cannot happen since the tensor ξ˜µ kν + ξ˜ν kµ is never zero for ξ˜µ 6= 0 and kµ 6= 0. Therefore the space of gaugeequivalent solutions is always a three-dimensional affine subspace of S, and the remaining physically independent space has dimension two. Finally, we shall consider the last remaining case kµ = 0 (the zero mode). This perturbation mode corresponds (0) to the space-independent perturbation hµν (x) = hµν = const. One can show that this perturbation is pure gauge, i.e. is gauge-equivalent to the zero perturbation. For instance, one can perform a gauge transformation hµν → hµν + ξµ,ν + ξν,µ

(34)

1 ξµ = − h(0) xν , 2 µν

(35)

with

h0µν

that the metric is approximately given by Eq. (15). Then the reasoning shown in this section will apply to all modes with sufficiently small wavelengths, and we shall again find two vacuum DOF. The number of DOF is defined as the dimension of space of solutions restricted to the point (t0 , x0 ) and is independent of the chosen neighborhood of that point (and of the choice of the point itself). In the next section we shall generalize this construction to an arbitrary nonlinear field theory and show that the number of vacuum DOF can be computed in essentially the same way in all such theories.

which will change hµν (x) = into zero. The conclusion is that small metric perturbations in general relativity have two vacuum DOF and are limited to modes with kα k α = 0, kα 6= 0. This result was obtained using the equations of motion for metric perturbations hµν on a flat background ηµν . A natural question is whether the number of vacuum DOF depends on the background: for instance, whether there are still two independent perturbation modes in the neighborhood of a black hole. The answer is that the number of vacuum DOF does not depend on the background. To understand this issue, let us consider a fixed spacetime point (t0 , x0 ). We expect that perturbations have the same number of DOF at all points. To describe the behavior of the perturbations, it is sufficient to impose the equations of motion in a small neigborhood of (t0 , x0 ). We may choose the coordinates in this neighborhood such

D.

Vacuum DOF in nonlinear field theories

The definition of vacuum DOF and the calculation in the previous sections relied on the fact that the Maxwell equations are linear in the fields and that the solutions form a vector space. For nonlinear field theories (e.g. general relativity) we therefore need to slightly modify our definition of DOF and the corresponding computational procedure. Consider a field theory with a field Qi having N realvalued components (i = 1, ..., N ); for instance, in general relativity the “field” is the metric gµν (x) which has 10 independent components. We shall assume that all independent components of all relevant fields are incorporated into the single “field” Qi . Suppose that the field Qi (x) satisfies some nonlinear equations of motion of the form
F ∂αβ Qi , ∂α Qi , Qi = 0, (36) where F (·, ·, ·) is some function. We have explicitly assumed that the equations of motion are second-order in derivatives of Qi , which is usually the case in field theories.1 The field theory may also admit gauge transformations of the form Qi (x) → Gi (∂α Qi (x), Qi (x), a(x)),

(37)

where Gi (·, ·, ·) is some function and a(x) is the parameter of the gauge transformation having an arbitrary dependence on x. At this moment we do not consider a specific form of the functions F and G and merely generalize the consideration of DOF to an arbitrary theory of this form. Later we shall consider the vacuum DOF in general relativity. Using the analogy with electrodynamics where electromagnetic waves are small perturbations of the vacuum Aµ = 0, one can expect that the “vacuum degrees of freedom” in the field theory (36) will be small perturbations of a “vacuum state.” Thus we are lead to assuming that

1

Below we shall consider fields with higher-order derivatives as well, but for now our considerations are valid only for fields with second-order equations of motion.

6 Eq. (36) admits a special “vacuum state,” i.e. a solution Qivac (x) which is physically interpreted as the “absence” of the field Qi . In electrodynamics the vacuum solution is Aµ = 0, but in a more general nonlinear field theory the vacuum solution may be a nontrivial function. For instance, the vacuum solution in general relativity is the flat Minkowski metric gµν (x) = ηµν . Small perturbations δQi (x) of the vacuum solution i Qvac (x) satisfy the linearized equation (36), namely
FIµνi ∂αβ Qivac , ∂α Qivac , Qivac ∂µν δQi
µi +FII ∂αβ Qivac , ∂α Qivac , Qivac ∂µ δQi
i +FIII ∂αβ Qivac , ∂α Qivac , Qivac δQi = 0, (38) µi i , and FIII where FIµνi , FII are the partial derivatives of the function F (·, ·, ·) with respect to its first, second, and third and we have neglected terms of order  arguments,  O (δQ)2 . The gauge transformation (37) induces the corresponding transformation of the perturbation δQi ,
∂Gi ∂α Qivac , Qivac , a i i δQ (x) → δQ (x) + a. (39) ∂a

However, we are now considering only infinitesimal perturbations δQi and so we cannot allow the vacuum solution Qivac (x) to be significantly changed by a gauge transformation. Thus we need to consider the gauge freedom only with respect to infinitesimal gauge transformations, where the gauge parameter a(x) is of the same order as the perturbation δQi . So we may approximate a ≈ 0 in the arguments of G in Eq. (39) because the error is second-order, and the gauge transformation of the perturbation δQi can be written as
δQi (x) → δQi (x) + aGiIII ∂α Qivac , Qivac , a = 0 . (40) Since Eq. (38) is linear in δQi and Eq. (40) is linear in δQi and a, the solutions form a linear space and we could now apply the construction developed in the previous sections to compute the number of the DOF, considering the small perturbations δQi as the “main field.” Namely, the values δQi (t0 , x0 ) at a fixed spacetime point (t0 , x0 ) the number of DOF is equal to the dimension of the space S modulo the gauge subspace. To compute the number of DOF, we need to apply the Fourier transform to Eqs. (38)-(40), but these are now differential equations with nonconstant coefficients such as FIiαβ (x). To overcome this difficulty, we note that the number of DOF is determined only by the behavior of the field Qi in an arbitrarily small neighborhood of an arbitrary point (t0 , x0 ). The coefficients FIiαβ (x), etc., entering Eqs. (38)-(40), can be treated as constants in an infinitesimal neighborhood of (t0 , x0 ), as long as the fields are not singular at that point. Since the number of DOF

does not depend on the choice of the point (t0 , x0 ), we may choose (t0 , x0 ) such that all coefficients in Eqs. (38)(40) evaluated at that point are finite and nonsingular. Thus we can replace the original system of equations (38)(40) by linear equations in δQi with constant coefficients, evaluated at the point (t0 , x0 ), µi i FIµνi (x0 )∂µν δQi + FII (x0 )∂µ δQi + FIII (x0 )δQi = 0,

δQi (x) → δQi (x) + GiIII (x0 )a. The number of DOF in the field δQi can be determined using this latter system of equations, since these equations describe a field theory which is equivalent to the original field theory in an infinitesimal neighborhood of the point x0 ≡ (t0 , x0 ). We now apply the Fourier transform and arrive to the equations h

i µj j ˜ j = 0, −FIµνj (x0 )kµ kν + ikµ FII (x0 ) + FIII (x0 ) δ Q

(41)

˜ i → δQ ˜ i + GiIII (x0 )˜ δQ a.

(42)

These are now treated as finite-dimensional algebraic equations depending on the parameter kµ and containing an arbitrary scalar a ˜. Since Eq. (41) is homogeneous, its solutions form a subspace S in the original N -dimensional ˜ i . The gauge transformation (42) field space of vectors δ Q generates a one-dimensional space P of “pure gauge” solutions which is a subspace (possibly empty) within S. The number of DOF is found as the dimension of the space S modulo the pure gauge subspace, dim(S/P ). The consideration is straightforwardly adapted to the case when the gauge transformation contains several scalar parameters aj (x), j = 1, 2, ... instead of a(x). Note that the dimensions of the spaces P and S depend on the wave vector kµ and thus we need to enumerate all the possibilities. Typically these possibilities will depend only on the value of kµ k µ ; for instance, in electrodynamics and in general relativity the dimension of P/S is nonzero only for null vectors kµ .

III.

SUMMARY

There are two rather different questions one asks about the DOF of a physical field: 1) How many different scalar-valued functions of spacetime do we need to fully describe the physical states of the field? (Definition 1) 2) How many different vacuum solutions does the field have? (Definition 2) In both cases one asks for the “degrees of freedom” but one means two quite different things.