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)


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


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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s