Let γ be a smooth, future-directed, timelike curve with unit tangent field ξ^{a} in our background spacetime (M, g_{ab}). 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 λ^{a} be a vector at p. Then there is a natural decomposition of λ^{a} into components proportional to, and orthogonal to, ξ^{a}:

λ^{a} = (λ^{b}ξ_{b})ξ^{a} + (λ^{a} −(λ^{b}ξ_{b})ξ^{a}) —– (1)

Here, the first part of the sum is proportional to ξ^{a}, whereas the second one is orthogonal to ξ^{a}.

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

k_{ab} = ξ_{a} ξ_{b} —– (2)

h_{ab} = g_{ab} − ξ_{a} ξ_{b} —– (3)

then the decomposition can be expressed in the form

λ^{a} = k^{a}_{b} λ^{b} + h^{a}_{b} λ^{b} —– (4)

We can think of k_{ab} and h_{ab} as the relative temporal and spatial metrics determined by ξ^{a}. They are symmetric and satisfy

k^{a}_{b}k^{b}_{c} = k^{a}_{c} —– (5)

h^{a}_{b}h^{b}_{c} = h^{a}_{c} —– (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 ξ^{a} and ξ′^{a} are both future-directed, they are co-oriented; i.e., ξ^{a} 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 = ∥h^{a}_{b} ξ′^{b}∥ / ∥k^{a}_{b} ξ′^{b}∥ —– (7)

For any vector μ^{a}, ∥μ^{a}∥ is (μ^{a}μ_{a})^{1/2} if μ is causal, and it is (−μ^{a}μ_{a})^{1/2} otherwise.

We have from equations 2, 3, 5 and 6

∥k^{a}_{b} ξ′^{b}∥ = (k^{a}_{b} ξ′^{b} k_{ac} ξ′^{c})^{1/2} = (k_{bc} ξ′^{b}ξ′^{c})^{1/2} = (ξ′^{b}ξ_{b})

and

∥h^{a}_{b} ξ′^{b}∥ = (−h^{a}_{b} ξ′^{b} h_{ac} ξ′^{c})^{1/2} = (−h_{bc} ξ′^{b}ξ′^{c})^{1/2} = ((ξ′^{b}ξ_{b})^{2} − 1)^{1/2}

so

v = ((ξ’^{b}ξ_{b})^{2} − 1)^{1/2} / (ξ′^{b}ξ_{b}) < 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′^{b}ξ_{b}) = 1/√(1 – v^{2}) —– (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 P^{a} that is tangent to its worldline there. The length ∥P^{a}∥ of this vector is what we would otherwise call the mass (or inertial mass or rest mass) of the particle. So, in particular, if P^{a} is timelike, we can write it in the form P^{a} =mξ^{a}, where m = ∥P^{a}∥ > 0 and ξ^{a} is the four-velocity of the particle. No such decomposition is possible when P^{a} is null and m = ∥P^{a}∥ = 0.

Suppose a particle O with positive mass has four-velocity ξ^{a} at a point, and another particle O′ has four-momentum P^{a} 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 P^{a} in terms of ξ^{a}. By equations (4) and (2), we have

P^{a} = (P^{b}ξ_{b}) ξ^{a} + h^{a}_{b}P^{b} —– (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 = P^{b}ξ_{b}. If O′ has positive mass and P^{a} = mξ′^{a}, this yields, by equation (9),

E = m (ξ′^{b}ξ_{b}) = m/√(1 − v^{2}) —– (11)

(If we had not chosen units in which c = 1, the numerator in the final expression would have been mc^{2} and the denominator √(1 − (v2/c2)). The three−momentum relative to O is the second term h^{a}_{b}P^{b} in the decomposition of P^{a}, i.e., the component of P^{a} orthogonal to ξ^{a}. It follows from equations (8) and (9) that it has magnitude

p = ∥h^{a}_{b} mξ′^{b}∥ = m((ξ′^{b}ξ_{b})^{2} − 1)^{1/2} = mv/√(1 − v^{2}) —– (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 ξ^{a}. Just as we understand ξ^{a} to be the four-velocity field of a massive point particle (that has the image of γ as its worldline), so we understand ξ^{n}∇_{n}ξ^{a} – the directional derivative of ξ^{a} in the direction ξ^{a} – to be its four-acceleration field (or just acceleration) field). The four-acceleration vector at any point is orthogonal to ξ^{a}. (This is, since ξ^{a}(ξ^{n}∇_{n}ξ_{a}) = 1/2 ξ^{n}∇_{n}(ξ^{a}ξ_{a}) = 1/2 ξ^{n}∇_{n} (1) = 0). The magnitude ∥ξ^{n}∇_{n}ξ^{a}∥ 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 F^{a} on I represents the vector sum of the various (non-gravitational) forces acting on it, then the particle’s four-acceleration ξ^{n}∇_{n}ξ^{a} satisfies

F^{a} = mξ^{n}∇_{n}ξ^{a} —– (13)

This is Newton’s second law of motion. Consider an example. Electromagnetic fields are represented by smooth, anti-symmetric fields F_{ab}. If a particle with mass m > 0, charge q, and four-velocity field ξ^{a} is present, the force exerted by the field on the particle at a point is given by qF^{a}_{b}ξ^{b}. 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:

qF^{a}_{b}ξ^{b} = mξ^{b}∇_{b}ξ^{a} —– (14)

This equation makes geometric sense. The acceleration field on the right is orthogonal to ξ^{a}. But so is the force field on the left, since ξ_{a}(F^{a}_{b}ξ^{b}) = ξ^{a}ξ^{b}F_{ab} = ξ^{a}ξ^{b}F_{(ab)}, and F_{(ab)} = 0 by the anti-symmetry of F_{ab}.