Hi all I'm a little confused

So I have that in special relativity time is included as a coordinate so that in 1 spatial dimension we have 2 space time coordinates. The most basic metric is the Minkowski metric given by $\left[ {\begin{array}{*{20}{c}} { - 1}&0 \\ 0&1 \end{array}} \right]$ so now if I have a vector $\left[ {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right]$ and I use Einstein Summation Convention with the metric I get $-1 + 1 =0$ ... so what does this tell me about the length squared of the vector.. why is the $0$ significant ?

EDIT: I found out that if the length squared = $0$ the interval is then light-like and light-like particles have world-lines confined to the light cone and the square of the separation of any two points on a light-like world line is zero. Is this correct ?

Help is very much appreciated

Thank you =)


We assume that we are working in units where the speed of light $c=1$. Given a vector with time and space components $A=(A^t, A^x)$, the Minkowski square of the vector is \begin{align} A^2 = -(A^t)^2+(A^x)^2 \end{align} We can classify the vector $A$ according to the sign of its square length. In particular, in the given signature (with the minus sign in the time coordinate) we use the term timelike to refer to any vector with $A^2<0$, lightlike to refer to any vector with $A^2=0$, and spacelike to refer to any vector with $A^2>0$.

Therefore, the vector you wrote down is said to be lightlike.

One way of attaching physical significance to these terms is by considering trajectories of particles moving in spacetime. If a massive particle moves in spacetime, then the tangent vector to its trajectory will be timelike, while when a massless particle moves through spacetime, the tangent to its trajectory will be lightlike. In order for the tangent vector to a trajectory to be spacelike, it would have to be moving faster than light.

So the vector you wrote down could be the tangent vector to the path of a massless particle moving through spacetime, like a photon.

  • $\begingroup$ Thank you, follows the information I found in special relativity =) Should be sufficient.. I'll go make sense of it. Thank you once again $\endgroup$ – KennyB Sep 3 '13 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.