If time is treated as a fourth dimension of spacetime, what is relation between length and time units?

Or in other words, how can I convert time units to length units, for instance seconds to meters?


The length of one second in meters is the distance travelled by light in one second.

$1\ \mathrm s=c\times1\ \mathrm s= 299\,792\,458\ \mathrm m$

The reason we use the same units for time and distance is special relativity, whose foundation rests on the speed of light (in vacuum) being constant in all inertial frames of reference. Its universality allows us to use the same units for both time and distance.

  • 2
    $\begingroup$ +1: Your answers are quite charming, I see :-) $\endgroup$ Oct 22 '12 at 3:23
  • $\begingroup$ constant in all frames of reference - I thought the speed of light was only constant in a vacuum? $\endgroup$
    – mowwwalker
    Oct 22 '12 at 3:26
  • 5
    $\begingroup$ @Walkerneo when Prathyush says "the speed of light being constant in all frames in reference" this is not about the literal speed at which light waves travel. It means there is a particular speed which is invariant across all reference frames. This speed happens to be called "the speed of light" for historical reasons (because light, in a vacuum, was the first thing found to travel at that speed), even if you are in a situation where light does not travel at that speed. $\endgroup$
    – David Z
    Oct 22 '12 at 8:14
  • 2
    $\begingroup$ edit: added speed of light in vacuum and inertial frames of reference. $\endgroup$
    – Prathyush
    Oct 22 '12 at 8:40
  • 2
    $\begingroup$ Just to put this number in relation: 1 ligsecond is the distance earth-moon. $\endgroup$
    – lalala
    Aug 23 '17 at 20:04

In special theory, the space time geometry contains 4 dimensions (3 space + 1 time) which look like $(x, y, z, ct)$. The distance between two points in this space is not given by the usual euclidean geometry $$d_{12}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2+c^2(t_1-t_2)^2}$$ rather is given by $$d_{12}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2-c^2(t_1-t_2)^2}$$ This space is called a Minkowski space. This is the relation between length and time.

I feel that as such you can not define a relation between to convert time units to length units. I think Prathyush has assumed $d_{12}=0$ and ${(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}=1$

  • 3
    $\begingroup$ My explanation is clear and is correct. Yes I assumed D12=0 which is true for light rays. Just like in quantum mechanics Energy and Frequency have the same units, in special relativity time and space have the same units. setting c=1 which is done often in theoretical physics means precisely this. $\endgroup$
    – Prathyush
    Oct 29 '12 at 20:33

Prathyush's answer is the most useful one, but time is just another dimension -- you could use any units you like, same as the other dimensions. For example another useful unit is $c$ itself.


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.