What is the difference between space-like four-momentum and time-like four-momentum? [closed]

Question

I was reading a wiki page on Tachyon, came across these terms? What i need is bit of a mathematical description to understand these terms?

Answer 1

Let's suppose you have a metric tensor $g_{\mu\nu}$ which defines your metric at any point in spacetime. Let's furthermore assume (only for simplicity; this is not necessary), that this spacetime is completely flat, so $g_{\mu\nu}$ becomes the Minkowski metric $\eta_{\mu\nu}$ which I choose to be

\begin{align} \left(\eta_{\mu\nu}\right) = \left(\begin{matrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{matrix}\right) \end{align}

This is either plus or minus $\eta$, depending on the convention you are using.

Now let's say you stand in point $A = (a^0, a^1, a^2, a^3)$ in spacetime. You may ask yourself which points in spacetime can be hit by a ray of light from there. These are all the points $X = (x^0, x^1, x^2, x^4)$ which satisfy:

\begin{align} (x^0-a^0)^2 = (x^1-a^1)^2 + (x^2-a^2)^2 + (x^3-a^3)^2 \end{align}

If we call $\Delta x^i = x^i-a^i$ and use our metric $\eta^{\mu\nu}$ we can write:

\begin{align} \sum_{\mu=0}^3\sum_{\nu=0}^3 \eta_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu} = 0 \end{align}

Using Einstein's sum convention, this gets even shorter:

\begin{align} \eta_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu} = 0 \end{align}

because we always sum over greek indices from 0 to 3. What we have now is the definition for a lighlike vector $X$, which lies directly on the so called light cone.

There are two more possibilities:

\begin{align} &\eta_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu} < 0\\ &\quad\text{and}\\ &\eta_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu} > 0 \end{align}

The first equation describes every vector inside the light cone.

Now we will understand this with velocity vectors. I will call it $\dot{x}^{\mu} = (\dot{x}^0, \dot{x}^1, \dot{x}^2, \dot{x}^3)$. Since $x^0 = ct$, we get $\dot{x}^0 = c$ with $c$ being the speed of light in a vacuum. Furthermore we can for simplicity assume, that $A = (0,0,0,0)$, so the equation becomes:

\begin{align} (\dot{x}^1)^2 + (\dot{x}^2)^2 + (\dot{x}^3)^2 &< c^2\\ \Leftrightarrow \sqrt{(\dot{x}^1)^2 + (\dot{x}^2)^2 + (\dot{x}^3)^2} &< c \end{align}

So the magnitude of the space velocity has to be smaller than the speed of light. This is called a timelike vector.

The second equation describes every vector outside the light cone. Using the same argumentation as above, we get:

\begin{align} \sqrt{(\dot{x}^1)^2 + (\dot{x}^2)^2 + (\dot{x}^3)^2} > c \end{align}

We call vectors, that satisfy this condition spacelike. An object with a spacelike velocity would move faster than light.

Momentum $p$ is velocity times mass, so $p^\mu = m\dot{x}^\mu$.