# What does it mean when one says that a vector field is spacelike, timelike, or null separated?

In Goldstein Classical Mechanics Chapter 7 (3rd edition, page 287), the authors classify vector fields as follows:

Name Time Portion Space Portion (Magnitude$$)^2$$ Type
Coordinate $$ct$$ $$\mathbf{r}$$ $$c^2t^2 - r^2$$ spacelike, null, or timelike
Velocity $$\gamma c$$ $$\gamma \mathbf{v}$$ $$c^2$$ timelike
Momentum $$\displaystyle \frac{E}{c}$$ $$\mathbf{p}$$ $$m^2c^2$$ timelike
Force $$\displaystyle \frac{\gamma}{c} \frac{\mathrm{d}E}{\mathrm{d}t}$$ $$\displaystyle \gamma \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = \gamma \mathbf{F}$$ $$-\left(\mathbf{F}_{\mathrm{Newtonian}}\right)^2$$ spacelike
Current density $$\gamma pc$$ $$\gamma \mathbf{J}$$ $$\rho^2 c^2$$ timelike

where $$\gamma = \displaystyle \frac{1}{\sqrt{1 - \beta^2}}$$ and $$\beta = v/c$$. Other symbols have their usual meanings.

Now, I understand what it means to have a spacelike, timelike, or null separation between coordinates. But what does it mean when the authors use this term for a vector field? Apart from the mathematical notion of the norm of the vectors of these fields being of a certain sign everywhere what does it physically mean to say that the momentum field is always timelike and the force field is always spacelike?

• Where exactly in Goldstein are you looking? Or maybe which edition? I don't find that characterization on quick glance. Dec 29 '20 at 19:37
• @Buzz You've edited the question into an inconsistent form as the "field" still remains in the question at the end. I'm inclined to think that this edit is more of an answer than an edit to the question. Chapter 7 does talk about true fields though, which is why I asked about the specific location for the information presented in the table by the OP. Dec 29 '20 at 20:12
• @Brick I updated my edit. I am also working on an answer, which will include clarification of the edit.
– Buzz
Dec 29 '20 at 20:14
• Sorry @Buzz I am not comfortable with the edit. Although I understand the answer would be the same but my question is for vector fields and not for an individual vector. In a vector field there are different vectors at different points in spacetime. When the authors say the vector field is spacelike I assume that they say that all vectors in that field are spacelike. But the question then becomes --what does this say about the field in general? Dec 29 '20 at 20:34
• See my answer. Most of the things in that table are not vector fields at all.
– Buzz
Dec 29 '20 at 20:38

The "canonical" meaning of a vector in general, in a geometric sense, is a displacement: if you have a point $$P$$ in a suitably homogeneous space, such as Euclidean three-dimensional space, then

$$P + \mathbf{v}$$

is a point displaced in the direction and through the distance encoded within $$\mathbf{v}$$. Likewise, if $$Q$$ and $$P$$ are two points, then

$$Q - P$$

is the vector that represents the displacement from $$P$$ to $$Q$$.

The same applies to space-time vectors: we can use them to represent displacements between space-time points, just as we can use them to represent displacements between purely spatial points. Thus, it is natural to suggest that if we can classify the separations between points into those categories, that we should also analogously be able to classify vectors producing those separations when added to one or the other point, into the same categories.

Hence, we can define that a vector is spacelike, timelike, or lightlike if it produces, respectively, a space-time point (or event) that is spacelike, timelike, or lightlike separated from another such point when added to that given point. That is, $$\mathbf{v}$$ is space/time/light-like if $$P$$ and $$P + \mathbf{v}$$ are space/time/light-like separated, respectively.

From this, you can prove that, in turn, in terms of the length of the vector,

$$||\mathbf{v}|| = \sqrt{v_t^2 - v_x^2 - v_y^2 - v_z^2}$$

that

• $$\mathbf{v}$$ is timelike if $$||\mathbf{v}||$$ is nonzero and real,
• $$\mathbf{v}$$ is lightlike if $$||\mathbf{v}||$$ is zero,
• $$\mathbf{v}$$ is spacelike if $$||\mathbf{v}||$$ is nonzero and imaginary.

And this likewise can, then, be extended to define analogous notions for other vectors that are not dimensionally suitable to represent displacements directly. And we can then also likewise analogously extend it further to vector fields: a vector field $$\mathbf{F}$$ is (s/t/l)-like if at every space-time point $$P$$, $$\mathbf{F}(P)$$ is (s/t/l)-like as we have just defined and generalized.

Finally, physically, the 4-momentum is never spacelike (not "always timelike" - the 4-momentum of photons is lightlike) because it, in effect, represents the displacement within space-time that a physical object is undergoing while it is moving. And a displacement in a space-like direction would represent a motion faster than light. And as far as we have discovered, nothing travels faster than light.

• Thanks. I can only accept one answer. So I will give you a +1 instead. Both answers are good and yours corrects the textbook by pointing out that the momentum and velocity fields can be nulllike (in case of light). Dec 29 '20 at 20:58

That a vector $$V$$ is timelike, lightlike/null, or spacelike indicates solely that $$V^{\mu}V_{\mu}=V_{0}^{2}-V_{1}^{2}-V_{2}^{2}-V_{3}^{2}=V_{0}^{2}-{\bf V}^{2}$$ is positive, zero, or negative. What that means in different specific case varies somewhat.

For the four-momentum $$p^{\mu}$$: Since the momentum components are the canonical conjugates to the position variables,* $$p^{\mu}$$ is the generator of spacetime translation. This is automatically true quantum mechanically, and it is also true classically, when the theory is interpreted with Poisson brackets implementing transformations or through Hamilton's Principle, with the trajectory of a particle being the configuration of Least Action. Qualitatively, that $$p^{\mu}$$ must be timelike (or lightlike for a massless particle) means that the spacetime trajectory followed by a particle must also be timelike (or lightlike**).

The situation for the four-velocity is the same, since the four-velocity is just defined by rescaling the four-momentum by $$m^{-1}$$. The current density is slightly more complicated, but if $$J^{\mu}$$ is viewed as generated by the motion of a (large) collection of pointlike charges, $${\bf J}=\sum_{n}q_{n}{\bf v}_{n}\delta^{3}[{\bf r}-{\bf r}_{n}(t)]$$, then its timelike nature also descends from that the timelike nature of $$p^{\mu}$$.

The requirement that the four-force be spacelike is a consequence of the fact that a particle of mass $$m$$ must, both before and after it experiences a four-impulse $$\Delta I^{\mu}=F^{\mu}\Delta t$$ satisfy the same energy-momentum relation, $$E^{2}-{\bf p}^{2}c^{2}=m^{2}c^{4}$$. Because $$|\partial E/\partial{\bf p}|=|{\bf v}|, the energy must change less than the momentum under the impulse (modulo factors of $$c$$), meaning the four-impulse (and hence four-force) must have a larger $$|{\bf F}|$$ than $$F_{0}$$; in other words, the four-force is spacelike.

*This is not true formally, for the time components (since the time $$t$$ is not a dynamical variable), but the same qualitative statements still apply to the time components.

**Barring reflections, which change the momentum from one lightlike vector to another.

(I have edited the question to refer to "vectors," rather than "vector fields," since only the coordinate and current density are vector fields, rather than single vectors. A vector field would be a vector-valued function of position, whereas the momentum of a particle is just a function of time, having no meaning except at the instantaneous location of the particle.)

• Thanks. While I disagree with you on the distinction between a vector and a vector field, that distinction is not hugely relevant to this question. I have thus reverted your edits but accepted your answer. Dec 29 '20 at 20:56