1
$\begingroup$

In Minkowski space, coordinates which satisfy $\Delta s^2 = \Delta t^2 - \Delta ^2 > 0$ are in the region of spacetime that is time-like. If it's $\Delta s^2 = \Delta t^2 - \Delta x^2 < 0$, the region is space-like. But if $\Delta s^2 = \Delta t^2 - \Delta x^2 = 0$, then it's the "trajectory of light-like particles".

I have understood the first two points about time-like and space-like regions but I could not get the third one about "light-like particles".

Why just light-like particles? There are many other particles at quantum level.

Improve this question
$\endgroup$
2
  • $\begingroup$ In the future, please use MathJax, not HTML markup, to display math. Thanks. $\endgroup$ Commented Apr 6, 2019 at 22:38
  • $\begingroup$ What is $X$...? $\endgroup$ Commented Aug 12, 2024 at 22:45

2 Answers 2

Reset to default
7
$\begingroup$

My confusion is about why just light like particles? there are many other particles at quantum level.

You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.

An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.

Improve this answer
$\endgroup$
3
$\begingroup$

Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.

The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=\pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.