Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Einstein's big revelation was that time and space are inseparable components of the same fabric. Physical observation tells us that distant galaxies are moving away from us at an accelerated rate, and because of the difficulty (impossibility?) of defining a coordinate system where things have well defined coordinates while also moving away from each other without changing the metric on the space, we interpret this to mean that space itself is expanding.

Because space and time are so directly intertwined is it possible that time too is expanding? Or perhaps it could be contracting?

share|cite|improve this question
up vote 7 down vote accepted

The simple answer is that no, time is not expanding or contracting.

The complicated answer is that when we're describing the universe we start with the assumption that time isn't expanding or contracting. That is, we choose our coordinate system to make the time dimension non-changing.

You don't say whether you're at school or college or whatever, but I'm guessing you've heard of Pythagoras' theorem for calculating the distance, $s$, between two points $(0, 0, 0)$ and $(x, y, z)$:

$$ s^2 = x^2 + y^2 + z^2 $$

Well in special relativity we have to include time in the equation to get a spacetime distance:

$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$

and in general relativity the equation becomes even more complicated because we have to multiply the $dt^2$, $dx^2$, etc by factors determined by a quantity called the metric, and usually denoted by $g$:

$$ ds^2 = g_{00}dt^2 + g_{11}dx^2 + g_{22}dy^2 + ... etc $$

where the $... etc$ can include cross terms like $g_{01}dtdx$, so it can all get very hairy. To be able to do the calculations we normally look for ways to simplify the expression, and in the particular case of the expanding universe we assume that the equation has the form:

$$ ds^2 = -dt^2 + a(t)^2 d\Sigma^2 $$

where the $d\Sigma$ includes all the spatial terms. The function $a(t)$ is a scale factor i.e. it scales up or down the contribution from the $dx$, $dy$ and $dz$, and it's a function of time so the scale factor changes with time. And this is where we get the expanding universe. It's because when you solve the Einstein equations for a homogenous isotropic universe you can calculate $a(t)$ and you find it increases with time, and that's what we mean by the expansion.

However the $dt$ term is not scaled, so time is not expanding (or contracting).

share|cite|improve this answer
Impressive covering of heaps on information in a few paragraphs. I got my partner to read your explanation; she is a bright non physicist so a good model for an "educated lay reader" and she said it was crystal clear, so well done. – WetSavannaAnimal aka Rod Vance Nov 6 '13 at 13:15
My understanding is that you use a co-moving coordinate frame then fix all the points then define a metric on that space and define a rate of expansion for it. But if I'm understanding you correctly your saying that leaving the time factor out was done for simplicity, and you could conceivably construct a model where the scale factor applied to the dt term? – Loourr Nov 6 '13 at 15:51
You can choose any coordinate system you want, but some coordinate systems are more useful than others. The coordinates used in the FLRW metric mean there is a simple interpretation of the time - it's equal to the proper time time for a stationary observer (wrt the CMB) - and all such observers everywhere agree on the elapsed time since the Big Bang. – John Rennie Nov 6 '13 at 15:56
What kind of evidence will be seen to disprove this assumption? – Sujoy Gupta Jan 30 '14 at 8:51
@SujoyGupta: what assumption? If you have doubts about the procedure used for deriving the FLRW metric maybe you could post a new question. – John Rennie Jan 30 '14 at 8:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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