What is a "Time-like" direction?

Question

In a question to do with time and energy's relationship someone said that

"energy is the component of momentum in a time-like direction."

I was wondering if someone could go into a bit more detail about what a "timelike" direction is.

Answer 1

I would flip it and say momentum in energy in the space-like direction. When at rest (four velocity $u = (c, {\bf 0})$), your are moving entirely in the time direction. Four momentum is:

$$ p_{\mu} = mu_{\mu} = (mc^2, {\bf 0}) $$

For a observer boosted to velocity $\bf v$:

$$ p_{\mu} = mu_{\mu} =\gamma(mc^2, m{\bf v}c)=(mc^2+T, {\bf p}c)$$

your rest mass (energy) now appears as momentum (and the timelike term has kinetic energy added to it).

That last equality may make it clear that when in motion (c=1):

$$ m \rightarrow m+T$$ $$ {\bf 0}\rightarrow {\bf 0} + {\bf p}$$

which could be interpreted as "Kinetic energy is momentum in the time-like direction".

Answer 2

This is very simple. In general relativity (and any other fields of physics that use Einstein tensor notation), we represent most physical quantity as tensors. A subset of this form of representation includes those things we represent as 4-vectors. I'm sure you're used to the concept of the 3-vector; such as some position vector: $$\vec d=(a,b,c)=a\hat x+b\hat y+c\hat z$$

A 3-vector has 3 components; one for each spatial dimension. The simplest way to understand a 4-vector is to think of it like we are extending it to include the 4th dimension; time. In this system, our position vector would resemble the following: $$\vec d=(s,a,b,c)=s\hat t+a\hat x+b\hat y+c\hat z$$ Of course, nothing is ever as simple as it first appears; the way in which 4-vectors are actually used would make most physicists grind their teeth at me over my portrayal above, but I think it's a good enough representation.

The source of your quote was clearly referring to the fact that the 4-vector for momentum is extended from the 3-vector in the following way. If the 3-vector is: $$\vec p=(p_x,p_y,p_z)$$ then the 4-vector is written as $$\vec p=(E,p_x,p_y,p_z)$$ where $E$ is the energy of the system in question. You see, energy fills the spot associated to the temporal dimension.

Let me give you some additional background as to why this makes sense and fills physicists with a warm glowy feeling when we see others doing this. Momentum is a conserved quantity, as is energy. The mathematical description for the conservation of momentum comes from a spatial translation invariance symmetry. That is, the laws of physics remain the same everywhere between where you are and, say, three meters to your right. That means momentum is conserved everywhere along that path. Similarly, we can show the same thing for three meters in front of you and three meters above you (and the mere fact that you are thinking that it's unnecessary to have to show it separately for all three directions means that angular momentum is conserved as well, but let's not get into that).

This is great, there is an invariance in the x-direction, so $p_x$ is conserved. There's an invariance in the y- and z-directions, so $p_y$ and $p_z$ are conserved respectively. There also happens to be a time translation symmetry; the laws of physics remain unchanged between now and 5 minutes from now (hopefully). This leads to the conservation of energy all throughout those 5 minutes. So when we treat energy as the temporal component of the momentum 4-vector, it really feels right to us. After all, a translation symmetry in each spatial dimension leads to momentum conservation and a translation symmetry in the temporal dimension leads to energy conservation. It may not be a rigorous proof why it is acceptable, but the similarity is very comforting nonetheless.

This is all why someone might refer to energy as "the component of momentum in a time-like direction".

Answer 3

For a description of time-like, suppose an observer measures two events as being separated in time by $Δ t$ and a spatial distance $Δ x$. Then the spacetime interval ( $Δ s$ ) between the two events that are separated by a distance $Δ x$ in space and by $Δ c t = c Δ t$ in the $ct$ coordinate is:

$$ (\Delta s)^{2}= (\Delta ct)^2-(\Delta x)^{2}$$

or for three space dimensions:

$$(\Delta s)^{2}=(\Delta ct)^2− ( Δ x )^ 2 − ( Δ y )^2 − ( Δ z )^ 2 $$

Because of the minus sign, the spacetime interval between two distinct events can be zero. If $(\Delta s)^{2}$ is positive, the spacetime interval is timelike, meaning that two events are separated by more time than space. If $(\Delta s)^2$ is negative, the spacetime interval is spacelike, meaning that two events are separated by more space than time.

Spacetime intervals are zero when $x = ± c t$ . In other words, the spacetime interval between two events on the world line of something moving at the speed of light is zero. Such an interval is termed lightlike or null.

See https://en.wikipedia.org/wiki/Spacetime

Answer 4

What must be meant by this statement is that energy and momentum form a four-vector, $(E,\vec{P})$, and that this vector under Lorentz transformations transforms in the same way as $(t,\vec{r})$.

Answer 5

In relativity theory 'time-like' denotes a vector or direction that points towards the future light-cone from its origin. That is, it is an allowed velocity for a material particle. Light-like or null vectors point along the light cone, space-like vectors outside the cone.

Technically, it is a vector with $g_{ab}x^ax^b<0$ where $g$ is the metric tensor and $x$ is the vector ($a,b$ are the components); in SR this assumes the tensor signature $(-,+,+,+)$.

See also this answer.