In my GR course, the teacher first introduced the flat metric:

$$ d \tau^2 = dt^2-dx^2-dy^2-dz^2=dt^2-dr^2-r^2 d\theta^2-r^2 sin(\theta)^2 d\phi^2. \tag{1}$$

Then he said "universe in expansion, homotethy depending on time":

$$ d \tau^2 = dt^2-a^2(t) (dr'^2+r'^2 d\theta^2+r'^2 sin(\theta)^2 d\phi^2). \tag{2}$$

But for me there is on physic in this last equation, we just did a change of variable where we wrote:

$$r(t) = a(t)r'.\tag{3} $$

First question: Do you agree with me that this last metric where he commented "universe in expansion" is just a change of variable from the flat metric, and thus we don't really have the information of a "universe in expansion" here ? I don't understand why he wrote universe in expansion here if it is only a change of coordinates.

Also, after that, he wrote the Robertson and Walker metric :

$$ d\tau^2=dt^2-a^2(t)(\frac{dr^2}{1-kr^2}+r^2 d\theta^2 + r^2 sin(\theta)^2 d\phi^2) \tag{4}$$

He said that with $k>0$ we have finite universe. $k=0$ : flat universe. $k<0$ : 3-hyperboloïd geometry.

Second question : Just to understand : does that mean that if $k \neq 0$ I can't find a change of variable that would lead me from the $$ d \tau^2 = dt^2-dx^2-dy^2-dz^2\tag{5}$$ to this metric? So, indeed the discussion about topology makes sense (it is not just a change of coordinates, there is "absolute" information with thoose choice of $k$).

PS: I'm a huge beginner in GR.

  • 2
    $\begingroup$ No, the change of coordinates (3) does not change the metric (1) into the metric (2). $\endgroup$ – Qmechanic Nov 28 '17 at 16:26
  • $\begingroup$ @Qmechanic Really ? I don't understand why. If I write $r=a(t) r'$, then I will have $dr=a(t) dr'$. Thus : $dr^2=a(t)^2 dr'^2$, the $d\theta$ and $d\phi$ don't change, but because $r^2=a(t)^2 r'^2$, I will end up with : $ d\tau^2=dt^2-a(t)^2(dr'^2-r'^2 d\theta^2 -r'^2 sin(\theta)^2 d\phi^2)$ ? (If it was a confusion between $a(t)$ and $1/a(t)$ I just edited my message to make my notations more clear. Was it only this problem ?? $\endgroup$ – StarBucK Nov 28 '17 at 16:34
  • 1
    $\begingroup$ Aaah no it's false because of the time dependance of $a$. I think it is what you meant... Thus in conclusion there is really "physical meaning" in the new metric with the $a(t)$ right ? (Because the new metric is not deduced from the other one from change of coordinates, thus "physical assumption" behind) $\endgroup$ – StarBucK Nov 28 '17 at 16:36
  • 1
    $\begingroup$ If $r=a(t)r'$, then $$dr = \frac{\partial r}{\partial t}dt+\frac{\partial r}{\partial r'}dr' = r'\frac{da}{dt}dt +a(t)dr'$$ $\endgroup$ – Colin MacLaurin Feb 10 '18 at 9:59

Your Answer

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

Browse other questions tagged or ask your own question.