# Magnetic Field as the Rotational Component of Electromagnetic Field

Let (M, gab) be the background relativistic spacetime. We are assuming it is temporally orientable and endowed with a particular temporal orientation. Let ξa be a smooth, future-directed unit timelike vector field on M (or some open subset of M). We understand it to represent the four-velocity field of a fluid. Further, let hab be the spatial projection field determined by ξa. The rotation and expansion fields associated with ξa are defined as follows:

ωab = h[amhb]nmξn —– (1)

θab = h(amhb)nmξn —– (2)

They are smooth fields, orthogonal to ξa in both indices, and satisfy

aξb = ωab + θab + ξammξb) —– (3)

Let γ be an integral curve of ξa, and let p be a point on the image of γ. Further, let ηa be a vector field on the image of γ that is carried along by the flow of ξa and orthogonal to ξa at p. (It need not be orthogonal to ξa elsewhere.) We think of the image of γ as the worldline of a fluid element O, and think of ηa at p as a “connecting vector” that spans the distance between O and a neighboring fluid element N that is “infinitesimally close.” The instantaneous velocity of N relative to O at p is given by ξaaηb. But ξaaηb = ηaaξb. So, by equation (3) and the orthogonality of ξa with ηa at p, we have

ξaaηb = (ωab + θaba —– (4)

at the point. Here we have simply decomposed the relative velocity vector into two components. The first, (ωabηa), is orthogonal to ηa since ωab is anti-symmetric. It is naturally understood as the instantaneous rotational velocity of N with respect to O at p.

The angular velocity (or twist) vector ωa. It points in the direction of the instantaneous axis of rotation of the fluid. Its magnitude ∥ωa∥ is the instantaneous angular speed of the fluid about that axis. Here ηa connects the fluid element O to the “infinitesimally close” fluid element N. The rotational velocity of N relative to O is given by ωbaηb. The latter is orthogonal to ηa

In support of this interpretation, consider the instantaneous rate of change of the squared length (−ηbηb) of ηa at p. It follows from equation (4) that

ξaa(−ηbηb) = −2θabηaηb —– (5)

Thus the rate of change depends solely on θab. Suppose θab = 0. Then the instantaneous velocity of N with respect to O at p has a vanishing radial component. If ωab ≠ 0, N can still have non-zero velocity there with respect to O. But it can only be a rotational velocity. The two conditions (θab = 0 and ωab ≠ 0) jointly characterize “rigid rotation.”

The rotation tensor ωab at a point p determines both an (instantaneous) axis of rotation there, and an (instantaneous) speed of rotation. As we shall see, both pieces of information are built into the angular velocity (or twist) vector

ωa = 1/2 εabcd ξbωcd —– (6)

at p. (Here εabcd is a volume element defined on some open set containing p. Clearly, if we switched from the volume element εabcd to its negation, the result would be to replace ωa with −ωa.)

If follows from equation (6) (and the anti-symmetry of εabcd) that ωa is orthogonal to ξa. It further follows that

ωa = 1/2 εabcd ξbcξd —– (7)

ωab = εabcd ξcωd —– (8)

Hence, ωab = 0 iff ωa = 0.

a = εabcd ξbωcd = εabcd ξb h[crhd]srξs = εabcd ξbhcrhdsrξ

= εabcd ξbgcr gdsrξs = εabcd ξbcξd

The second equality follows from the anti-symmetry of εabcd, and the third from the fact that εabcdξb is orthogonal to ξa in all indices.) The equation (6) has exactly the same form as the definition of the magnetic field vector Ba determined relative to a Maxwell field Fab and four-velocity vector ξa (Ba = 1/2 εabcd ξb Fcd). It is for this reason that the magnetic field is sometimes described as the “rotational component of the electromagnetic field.”

# Dynamics of Point Particles: Orthogonality and Proportionality

Let γ be a smooth, future-directed, timelike curve with unit tangent field ξa in our background spacetime (M, gab). We suppose that some massive point particle O has (the image of) this curve as its worldline. Further, let p be a point on the image of γ and let λa be a vector at p. Then there is a natural decomposition of λa into components proportional to, and orthogonal to, ξa:

λa = (λbξba + (λa −(λbξba) —– (1)

Here, the first part of the sum is proportional to ξa, whereas the second one is orthogonal to ξa.

These are standardly interpreted, respectively, as the “temporal” and “spatial” components of λa relative to ξa (or relative to O). In particular, the three-dimensional vector space of vectors at p orthogonal to ξa is interpreted as the “infinitesimal” simultaneity slice of O at p. If we introduce the tangent and orthogonal projection operators

kab = ξa ξb —– (2)

hab = gab − ξa ξb —– (3)

then the decomposition can be expressed in the form

λa = kab λb + hab λb —– (4)

We can think of kab and hab as the relative temporal and spatial metrics determined by ξa. They are symmetric and satisfy

kabkbc = kac —– (5)

habhbc = hac —– (6)

Many standard textbook assertions concerning the kinematics and dynamics of point particles can be recovered using these decomposition formulas. For example, suppose that the worldline of a second particle O′ also passes through p and that its four-velocity at p is ξ′a. (Since ξa and ξ′a are both future-directed, they are co-oriented; i.e., ξa ξ′a > 0.) We compute the speed of O′ as determined by O. To do so, we take the spatial magnitude of ξ′a relative to O and divide by its temporal magnitude relative to O:

v = speed of O′ relative to O = ∥hab ξ′b∥ / ∥kab ξ′b∥ —– (7)

For any vector μa, ∥μa∥ is (μaμa)1/2 if μ is causal, and it is (−μaμa)1/2 otherwise.

We have from equations 2, 3, 5 and 6

∥kab ξ′b∥ = (kab ξ′b kac ξ′c)1/2 = (kbc ξ′bξ′c)1/2 = (ξ′bξb)

and

∥hab ξ′b∥ = (−hab ξ′b hac ξ′c)1/2 = (−hbc ξ′bξ′c)1/2 = ((ξ′bξb)2 − 1)1/2

so

v = ((ξ’bξb)2 − 1)1/2 / (ξ′bξb) < 1 —– (8)

Thus, as measured by O, no massive particle can ever attain the maximal speed 1. We note that equation (8) implies that

(ξ′bξb) = 1/√(1 – v2) —– (9)

It is a basic fact of relativistic life that there is associated with every point particle, at every event on its worldline, a four-momentum (or energy-momentum) vector Pa that is tangent to its worldline there. The length ∥Pa∥ of this vector is what we would otherwise call the mass (or inertial mass or rest mass) of the particle. So, in particular, if Pa is timelike, we can write it in the form Pa =mξa, where m = ∥Pa∥ > 0 and ξa is the four-velocity of the particle. No such decomposition is possible when Pa is null and m = ∥Pa∥ = 0.

Suppose a particle O with positive mass has four-velocity ξa at a point, and another particle O′ has four-momentum Pa there. The latter can either be a particle with positive mass or mass 0. We can recover the usual expressions for the energy and three-momentum of the second particle relative to O if we decompose Pa in terms of ξa. By equations (4) and (2), we have

Pa = (Pbξb) ξa + habPb —– (10)

the first part of the sum is the energy component, while the second is the three-momentum. The energy relative to O is the coefficient in the first term: E = Pbξb. If O′ has positive mass and Pa = mξ′a, this yields, by equation (9),

E = m (ξ′bξb) = m/√(1 − v2) —– (11)

(If we had not chosen units in which c = 1, the numerator in the final expression would have been mc2 and the denominator √(1 − (v2/c2)). The three−momentum relative to O is the second term habPb in the decomposition of Pa, i.e., the component of Pa orthogonal to ξa. It follows from equations (8) and (9) that it has magnitude

p = ∥hab mξ′b∥ = m((ξ′bξb)2 − 1)1/2 = mv/√(1 − v2) —– (12)

Interpretive principle asserts that the worldlines of free particles with positive mass are the images of timelike geodesics. It can be thought of as a relativistic version of Newton’s first law of motion. Now we consider acceleration and a relativistic version of the second law. Once again, let γ : I → M be a smooth, future-directed, timelike curve with unit tangent field ξa. Just as we understand ξa to be the four-velocity field of a massive point particle (that has the image of γ as its worldline), so we understand ξnnξa – the directional derivative of ξa in the direction ξa – to be its four-acceleration field (or just acceleration) field). The four-acceleration vector at any point is orthogonal to ξa. (This is, since ξannξa) = 1/2 ξnnaξa) = 1/2 ξnn (1) = 0). The magnitude ∥ξnnξa∥ of the four-acceleration vector at a point is just what we would otherwise describe as the curvature of γ there. It is a measure of the rate at which γ “changes direction.” (And γ is a geodesic precisely if its curvature vanishes everywhere).

The notion of spacetime acceleration requires attention. Consider an example. Suppose you decide to end it all and jump off the tower. What would your acceleration history be like during your final moments? One is accustomed in such cases to think in terms of acceleration relative to the earth. So one would say that you undergo acceleration between the time of your jump and your calamitous arrival. But on the present account, that description has things backwards. Between jump and arrival, you are not accelerating. You are in a state of free fall and moving (approximately) along a spacetime geodesic. But before the jump, and after the arrival, you are accelerating. The floor of the observation deck, and then later the sidewalk, push you away from a geodesic path. The all-important idea here is that we are incorporating the “gravitational field” into the geometric structure of spacetime, and particles traverse geodesics iff they are acted on by no forces “except gravity.”

The acceleration of our massive point particle – i.e., its deviation from a geodesic trajectory – is determined by the forces acting on it (other than “gravity”). If it has mass m, and if the vector field Fa on I represents the vector sum of the various (non-gravitational) forces acting on it, then the particle’s four-acceleration ξnnξa satisfies

Fa = mξnnξa —– (13)

This is Newton’s second law of motion. Consider an example. Electromagnetic fields are represented by smooth, anti-symmetric fields Fab. If a particle with mass m > 0, charge q, and four-velocity field ξa is present, the force exerted by the field on the particle at a point is given by qFabξb. If we use this expression for the left side of equation (13), we arrive at the Lorentz law of motion for charged particles in the presence of an electromagnetic field:

qFabξb = mξbbξa —– (14)

This equation makes geometric sense. The acceleration field on the right is orthogonal to ξa. But so is the force field on the left, since ξa(Fabξb) = ξaξbFab = ξaξbF(ab), and F(ab) = 0 by the anti-symmetry of Fab.

# Ruminations on Philosophy of Science: A Case of Volume Measure Respecting Orientation

Let M be an n–dimensional manifold (n ≥ 1). An s-form on M (s ≥ 1) is a covariant field αb1…bs that is anti-symmetric (i.e., anti-symmetric in each pair of indices). The case where s = n is of special interest.

Let αb1…bn be an n-form on M. Further, let ξi(i = 1,…,n) be a basis for the tangent space at a point in M with dual basis ηi(i=1,…,n). Then αb1…bn can be expressed there in the form

αb1…bn = k n! η1[b1…ηnbn] —– (1)

where

k = αb1…bnξ1b1…ξnbn

(To see this, observe that the two sides of equation (1) have the same action on any collection of n vectors from the set {ξ1b, . . . , ξnb}.) It follows that if αb1…bn and βb1…bn are any two smooth, non-vanishing n-forms on M, then

βb1…bn = f αb1…bn

for some smooth non-vanishing scalar field f. Smooth, non-vanishing n-forms always exist locally on M. (Suppose (U, φ) is a chart with coordinate vector fields (γ⃗1)a, . . . , (γ⃗n)a, and suppose ηib(i = 1, . . . , n) are dual fields. Then η1[b1…ηnbn] qualifies as a smooth, non-vanishing n-form on U.) But they do not necessarily exist globally. Suppose, for example, that M is the two-dimensional Möbius strip, and αab is any smooth two-form on M. We see that αab must vanish somewhere as follows.

A 2-form αab on the Möbius strip determines a “positive direction of rotation” at every point where it is non-zero. So there cannot be a smooth, non-vanishing 2-form on the Möbius strip.

Let p be any point on M at which αab ≠ 0, and let ξa be any non-zero vector at p. Consider the number αab ξa ρb as ρb rotates though the vectors in Mp. If ρb = ±ξb, the number is zero. If ρb ≠ ±ξb, the number is non-zero. Therefore, as ρb rotates between ξa and −ξa, it is always positive or always negative. Thus αab determines a “positive direction of rotation” away from ξa on Mp. αab must vanish somewhere because one cannot continuously choose positive rotation directions over the entire Möbius strip.

M is said to be orientable if it admits a (globally defined) smooth, non- vanishing n-form. So far we have made no mention of metric structure. Suppose now that our manifold M is endowed with a metric gab of signature (n+, n). We take a volume element on M (with respect to gab) to be a smooth n-form εb1…bn that satisfies the normalization condition

εb1…bn εb1…bn = (−1)nn! —– (2)

Suppose εb1…bn is a volume element on M, and ξi b (i = 1,…,n) is an orthonormal basis for the tangent space at a point in M. Then at that point we have, by equation (1),

εb1…bn = k n! ξ1[b1 …ξbn] —– (3)

where

k = εb1…bn ξ1b1…ξnbn

Hence, by the normalization condition (2),

(−1)nn! = (k n! ξ1[b1 …ξbn]) (k n! ξ1[b1 …ξbn])

= k2 n!2 1/n! (ξ1b1 ξ1b1) … (ξnbn ξnbn) = k2 (−1)n

So k2 = 1 and, therefore, equation (3) yields

εb1…bn ξ1b1…ξnbn = ±1 —– (4)

Clearly, if εb1…bn is a volume element on M, then so is −εb1…bn. It follows from the normalization condition (4) that there cannot be any others. Thus, there are only two possibilities. Either (M, gab) admits no volume elements (at all) or it admits exactly two, and these agree up to sign.

Condition (4) also suggests where the term “volume element” comes from. Given arbitrary vectors γ1a , . . . , γna at a point, we can think of εb1…bn γ1b1 … γnbn as the volume of the (possibly degenerate) parallelepiped determined by the vectors. Notice that, up to sign, εb1…bn is characterized by three properties.

(VE1) It is linear in each index.

(VE2) It is anti-symmetric.

(VE3) It assigns a volume V with |V | = 1 to each orthonormal parallelepiped.

These are conditions we would demand of any would-be volume measure (with respect to gab). If the length of one edge of a parallelepiped is multiplied by a factor k, then its volume should increase by that factor. And if a parallelepiped is sliced into two parts, with the slice parallel to one face, then its volume should be equal to the sum of the volumes of the parts. This leads to (VE1). Furthermore, if any two edges of the parallelepiped are coalligned (i.e., if it is a degenerate parallelepiped), then its volume should be zero. This leads to (VE2). (If for all vectors ξa, εb1…bn ξb1 ξb2 = 0, then it must be the case that εb1 …bn is anti-symmetric in indices (b1, b2). And similarly for all other pairs of indices.) Finally, if the edges of a parallelepiped are orthogonal, then its volume should be equal to the product of the lengths of the edges. This leads to (VE3). The only unusual thing about εb1…bn as a volume measure is that it respects orientation. If it assigns V to the ordered sequence γ1a , . . . , γna, then it assigns (−V) to γ2a, γ1a, γ3a,…,γna, and so forth.