# Timelike and Spacelike Four-vectors

## Main Question or Discussion Point

Hi,

I've been reading about four-vectors and I have two quick questions:

If the position 4-vector is timelike, are the corresponding momentum and acceleration vectors also timelike? If the position 4-vector is spacelike, are the corresponding momentum and acceleration vectors also spacelike? Why?

Also, I understand the physical meaning of timelike and spacelike intervals, i.e. that if an interval between two events is timelike the events can have a causal link. But what's the physical meaning of timelike and spacelike vectors?

Thanks in advance! :-)

## Answers and Replies

Related Special and General Relativity News on Phys.org
WannabeNewton
Since you're talking about position vectors, I assume you're strictly speaking about Minkowski space-time (special relativity). Keep in mind the concept has no meaning whatsoever in general relativity.

The answer to your first question is no. Consider the path of an observer uniformly accelerating in the ##+x## direction of an inertial frame with associated global inertial coordinates ##(t,x)##; such an observer is also known as a Rindler observer. The path can be represented in this frame as ##x^{\mu} = (g^{-1}\sinh g\tau, g^{-1}\cosh g\tau)## where ##g > 0## is the magnitude of the acceleration and ##\tau## is the proper time along the Rindler observer's worldline. We have ##x^{\mu}x_{\mu} = g^{-2}(-\sinh^2 g\tau + \cosh^2 g\tau) = g^{-2} > 0 ## but ##u^{\mu}u_{\mu} = - 1<0## where ##u^{\mu}## is the 4-velocity of the Rindler observer. Also ##a^{\mu}a_{\mu} = g^2 > 0##. The convention in relativity is to always normalize the 4-velocity of a time-like observer to unity, in which case the 4-acceleration is always orthogonal to the 4-velocity. Since the 4-velocity is always time-like, the 4-acceleration will always be space-like.

As for your second question, think about it in terms of light cones. At any given point in space-time, we can consider the set of all possible light rays which pass through that point. Taking the set of all tangent vectors to all of these light rays (these tangent vectors are all null vectors of course) forms the light cone at that point. If a vector is time-like then it will lie inside of this cone and if a vector is space-like it will lie outside of this cone; the vectors lying on the cone are as stated the tangent vectors to the light rays and these are null vectors.

This framework codifies the (local) causal structure of space-time. For example, time-like particles cannot propagate faster than light so if you draw the worldline of a time-like particle then you will trace out a curve in space-time which, at each point in space-time, lies inside of the lightcone at that point; this is the same thing as saying that the tangent vector to the worldline at each point lies inside of the light cone at that point.

Also, I understand the physical meaning of timelike and spacelike intervals, i.e. that if an interval between two events is timelike the events can have a causal link. But what's the physical meaning of timelike and spacelike vectors?
This question occurred to me during a recent discussion on electromagnetism in SR. First consider the four velocity ##U = (u^0,u^1,0,0) ##. Using the +--- convention, the magnitude of the four velocity vector is always c for a real particle. This means that we are restricted in our choices of ##u^0## and ##u^1## to values where ##u^1<u^0## and where ##(u^0)^2=(c)^2+(u^1)^2##.

Now if we consider the electromagnetic four vector, ##J = (j^0,j^1,0,0)## where ##j^0## is the charge density and ##j^1## is the current, it appears we are unrestricted in our choices of ##j^0## and ##j^1##. What does J represent and what is the physical significance of J being timelike or spacelike?

Dale
Mentor
If the position 4-vector is timelike, are the corresponding momentum and acceleration vectors also timelike? If the position 4-vector is spacelike, are the corresponding momentum and acceleration vectors also spacelike? Why?
The only physical significance of timelike vs spacelike nature of a position 4-vector is whether or not it is causally connected to the origin. Since you are free to pick the origin you can always pick an origin where a given position is timelike or spacelike.

Essentially, position vectors (both the position 4- and 3-vectors) are not physical since there is no physically identifiable point which uniquely serves as a physical origin. I.e. positions are not vectors, they are better represented as an affine space.

Also, I understand the physical meaning of timelike and spacelike intervals, i.e. that if an interval between two events is timelike the events can have a causal link. But what's the physical meaning of timelike and spacelike vectors?
That depends on the physical meaning of the specific vector. It is a little like asking what the physical nature of 3-vectors is. You can say it has a magnitude and a direction, etc. but those are mathematical properties. Similarly with 4 vectors.

So, a general answer to your question would simply be a listing of mathematical properties of 4-vectors. But here are a few specific 4-vectors and their physical interpretation:
4-momentum: timelike → massive particle, lightlike → massless particle, spacelike → doesn't exist
4-current: timelike → stationary charge in some frame, spacelike → uncharged current in some frame
4-acceleration: always spacelike
4-force: always spacelike
4-velocity: timelike → massive particle, lightlike → massless particle, spacelike → doesn't exist

1 person
WannabeNewton
Now if we consider the electromagnetic four vector, ##J = (j^0,j^1,0,0)## where ##j^0## is the charge density and ##j^1## is the current, it appears we are unrestricted in our choices of ##j^0## and ##j^1##. What does J represent and what is the physical significance of J being timelike or spacelike?
I'm not sure what you mean by "we are unrestricted in our choices...". The 4-current density has to still satisfy ##\nabla_{\mu}J^{\mu} = 0## and ##\nabla_{\mu}F^{\mu\nu} = J^{\nu}##.

By the way, say we have a charged dust field with 4-velocity field ##u^{\mu}##. Then the 4-current density is given by ##J^{\mu} = \sigma u^{\mu}## where ##\sigma = J^{\mu}u_{\mu}## is the charge density as measured by a comoving observer. So ##J^{\mu}## here is time-like and the physical interpretation is clear: relative to a comoving observer, the 3-current density vanishes so the 4-current density only has a time-like component i.e. ##J^{\mu} = \sigma \delta ^{\mu}_0 = \sigma u^{\mu}## but this is a covariant expression meaning the 4-current density must be parallel to the 4-velocity field of the charged fluid in any frame hence it must itself be time-like.

This question occurred to me during a recent discussion on electromagnetism in SR. First consider the four velocity ##U = (u^0,u^1,0,0) ##. Using the +--- convention, the magnitude of the four velocity vector is always c for a real particle. This means that we are restricted in our choices of ##u^0## and ##u^1## to values where ##u^1<u^0## and where ##(u^0)^2=(c)^2+(u^1)^2##.

Now if we consider the electromagnetic four vector, ##J = (j^0,j^1,0,0)## where ##j^0## is the charge density and ##j^1## is the current, it appears we are unrestricted in our choices of ##j^0## and ##j^1##. What does J represent and what is the physical significance of J being timelike or spacelike?

No, J is not unrestricted. Just like the 4-velocity, the 4-current (of a particle) cannot be spacelike. You must keep in mind that the 4-current of a particle is proportional to its 4-velocity.

Dale
Mentor
No, J is not unrestricted. Just like the 4-velocity, the 4-current (of a particle) cannot be spacelike. You must keep in mind that the 4-current of a particle is proportional to its 4-velocity.
J is not restricted to individual particles. In most circuits J will be spacelike and will consist of the net 4 current due to all charge carriers.

No, J is not unrestricted. Just like the 4-velocity, the 4-current (of a particle) cannot be spacelike. You must keep in mind that the 4-current of a particle is proportional to its 4-velocity.
But we can for example, analyse a system with a static charge and zero current in one reference frame that transforms in an invariant way, can we not?

BruceW
Homework Helper
J is not restricted to individual particles. In most circuits J will be spacelike and will consist of the net 4 current due to all charge carriers.
If we wrote down the 'true' 4 current (i.e. a bunch of Dirac-Delta's for each particle), then I guess we always have a timelike 4 current? Since there are no magnetic monopoles. But then, we don't really care about the 'true' 4 current when we are talking about circuits, so we allow spacelike 4 current. Is that about right?

If we wrote down the 'true' 4 current (i.e. a bunch of Dirac-Delta's for each particle), then I guess we always have a timelike 4 current? Since there are no magnetic monopoles. But then, we don't really care about the 'true' 4 current when we are talking about circuits, so we allow spacelike 4 current. Is that about right?
DaleSpam's (correct) point is that it is possible to add two (or more) time-like vectors and obtain a space-like vector so even though the 4-current of each individual particle is time-like, the total 4-current may not be. That's why I emphasized that I was talking about a single particle.

BruceW
Homework Helper
yeah. But if we assume that (in 'reality') we have only point charges, then they will not be in the same place, so the total 4-current at any position will only be due to one point charge. If we add up the 4-currents from nearby point charges in a wire, then we are making a coarse-grained model really. There's no problem with this. I was just checking that if we assume only point charges, then the 'true' 4-current will always be timelike. (that is what I'd guess).

edit: sorry, this is off-topic really.

BruceW
Homework Helper
If we wrote down the 'true' 4 current (i.e. a bunch of Dirac-Delta's for each particle), then I guess we always have a timelike 4 current? Since there are no magnetic monopoles. But then, we don't really care about the 'true' 4 current when we are talking about circuits, so we allow spacelike 4 current. Is that about right?
hmm. That's not quite right. The existence of magnetic monopoles is not relevant, I think. If we assume point charges, and that each is massive, it must have timelike 4-velocity. So therefore, the 4-current must also be timelike. I guess only if the charge was imaginary, we would get spacelike 4-current. (but that's not allowed, charge needs to be real).

Dale
Mentor
If we wrote down the 'true' 4 current (i.e. a bunch of Dirac-Delta's for each particle), then I guess we always have a timelike 4 current?
Either you are in the classical regime where you have a continuous charge density, or you are in the quantum regime where you don't have dirac deltas but rather overlapping wave functions. Either way I think you legitimately wind up with spacelike 4 currents, but at a minimum I certainly would avoid a calling a dirac delta the true 4 current.

When we transform to a different reference frame, the charge density of the single electron changes. If the electron is considered a point particle (and the charge of the electron is invariant), what is the physical significance of the electron's charge density changing?

Last edited:
WannabeNewton
The 4-current density can be time-like even for a charged fluid such as charged dust (as explained above); it doesn't have to be only for a single particle. On the other hand, if we are say in the interior of a homogenous Ohmic material containing a steady current then we will have a space-like 4-current density.

A charged point particle has a Dirac delta distribution for its charge density so the charge moving would simply mean that the Dirac delta distribution takes on a time dependence i.e. instead of ##\rho = q\delta(r - r')## for fixed ##r'## we will have ##\rho = q\delta(r - r'(t))##.

BruceW
Homework Helper
Either you are in the classical regime where you have a continuous charge density, or you are in the quantum regime where you don't have dirac deltas but rather overlapping wave functions. Either way I think you legitimately wind up with spacelike 4 currents, but at a minimum I certainly would avoid a calling a dirac delta the true 4 current.
ah, ok. We never see the situation of a bunch of dirac delta, since quantum effects will become important. So restricting the 4 current to only be timelike is not necessary.

yuiop said:
When we transform to a different reference frame, the charge density of the single electron changes. If the electron is considered a point particle (and the charge of the electron is invariant), what is the physical significance of the electron's charge density changing?
the number density of electrons would change, since the length between each electron would change in the direction of relative velocity between the two reference frames.

edit: er, I mean the charge of a single electron is Lorentz invariant, so it does not change when you change reference frame. But if you have a bunch of electrons, the distance between them will change when you change reference frame. So the coarse-grained charge density will change when you change reference frames.

Last edited: