I have to obtain a time-dependent metric for a disk which satisfies a simple differential equation but I am stuck with making sure that the physics is correct.

To be explicit, I'll first describe a shrinking disk:

Suppose that we are working in a $4D$ Minkowski spacetime $(\mathcal{M},\eta_{ab})$ where the metric has signature $\text{sign}(\eta)=\text{diag}(-,+,+,+)$ and $\mathcal{M}=\mathbb{R}_t\times\mathbb{R}_{\Sigma}^3$. Consider embedding a compact disk $\Omega_t\hookrightarrow\mathbb{R}^3_{\Sigma}$ which is defined at each time slice by $$\Omega_t=\left\{(x,y)\in\mathbb{R}^2: x(t)^2+y(t)^2\le R(t)^2 \right\}$$ where the radius of the disk solves the differential equation $$\frac{\text{d}R(t)}{\text{d}t} = -\gamma R(t)$$ with the initial and final conditions $R(t_i)=R_i$, $R(t_f)=R_f$, and $R_f<R_i$. The disk starts at some initial radius $(t_i,R_i)$ and shrinks until it reaches the smaller radius $(t_f,R_f)$. The spacetime diagram looks like a cone.

I can solve the differential equation to find that the radius has the explicit form $$R(t) = R_i \exp(-\gamma (t-t_i)) = R_i\exp\left(\frac{t\,- t_i}{t_f-t_i}\log\left[\frac{R_f}{R_i}\right]\right).$$

I'm now stuck trying to construct a metric for the disk. My initial guess was that it should look like FLRW since it looks like a shrinking universe; $$\text{d}s^2 \overset{?}{=} -\text{d}t^2 + \exp(-2\gamma(t-t_i))\left(\text{d}r^2+ R_i^2\text{d}\theta^2\right)$$ but I am unsure how to derive this metric, assuming that it is in the correct form. Could I get some help with deriving a metric for this disk model? Where is a good starting point?

  • $\begingroup$ Your final equation is a metric of 3D de Sitter space. What is the point of working in 4D if you are interested in a 3D metric? $\endgroup$
    – A.V.S.
    Commented Mar 17, 2019 at 5:37
  • $\begingroup$ @A.V.S. I had to consider an embedded disk for the following reason: After I get the metric of the $2+1D$ disk, I need to look at the symmetries of the disk in the full 3+1D Minkowski spacetime. I should have made that clearer, thanks! $\endgroup$
    – Dhuality
    Commented Mar 17, 2019 at 12:25

1 Answer 1


The metric



$$x^2+y^2=R^2\,\quad \Rightarrow \quad y=\sqrt{R(t)^2-x^2}$$ $$dy=-{\frac {x{\it dx}}{\sqrt { \left( R \left( t \right) \right) ^{2}-{x }^{2}}}}+{\frac {R \left( t \right) \left( {\frac {d}{dt}}R \left( t \right) \right) {\it dt}}{\sqrt { \left( R \left( t \right) \right) ^{2}-{x}^{2}}}} $$


$$ds^2= \left( -1+{\frac { \left( R \left( t \right) \right) ^{2} \left( { \frac {d}{dt}}R \left( t \right) \right) ^{2}}{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}} \right) {{\it dt}}^{2}-2\,{\frac {{ \it dx}\,R \left( t \right) \left( {\frac {d}{dt}}R \left( t \right) \right) x{\it dt}}{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}} +{{\it dx}}^{2} \left( 1+{\frac {{x}^{2}}{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}} \right) $$

$$g= \left[ \begin {array}{cc} -1+{\frac { \left( R \left( t \right) \right) ^{2} \left( {\frac {d}{dt}}R \left( t \right) \right) ^{2}}{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}}&-{\frac {R \left( t \right) \left( {\frac {d}{dt}}R \left( t \right) \right) x}{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}} \\ -{\frac {R \left( t \right) \left( {\frac {d}{dt }}R \left( t \right) \right) x}{ \left( R \left( t \right) \right) ^ {2}-{x}^{2}}}&1+{\frac {{x}^{2}}{ \left( R \left( t \right) \right) ^ {2}-{x}^{2}}}\end {array} \right]$$

$$\det(g)={\frac { \left( R \left( t \right) \right) ^{2} \left( -1+ \left( { \frac {d}{dt}}R \left( t \right) \right) ^{2} \right) }{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}} $$


Metrix with polar coordinate

with $x=R(t)\,\cos(\phi)\quad, y=R(t)\,\sin(\phi)$ this satisfy the constraint equation $x^2+y^2=R(t)^2$

you get the metric:

$$ds^2=-dt^2+dx^2+dy^2=-dt^2+\left( {\frac {d}{dt}}R \left( t \right) \right) ^{2}{{\it dt}}^{2}+ \left( R \left( t \right) \right) ^{2}{d\phi }^{2}=\left( -1+ \left( {\frac {d}{dt}}R \left( t \right) \right) ^{2} \right) {{\it dt}}^{2}+ \left( R \left( t \right) \right) ^{2}{d \phi }^{2} $$

  • $\begingroup$ Could you elaborate why you chose to begin with a flat metric for the disk? $\endgroup$
    – Dhuality
    Commented Mar 17, 2019 at 12:47
  • $\begingroup$ No special reason for it, but why not?. For the time coordinate you can choose arbitrary Faktor $a(t,x,y)$ but I can’t tel you if it make any sense? $\endgroup$
    – Eli
    Commented Mar 17, 2019 at 13:06
  • $\begingroup$ Okay, I'm also unsure why the angle $y(t)/x(t)=\tan(\theta(t))$ does not appear. I'll have to think about this. $\endgroup$
    – Dhuality
    Commented Mar 17, 2019 at 13:09
  • $\begingroup$ @Aditya new matric with polar coordinate,but you get a singularity for the $t$ coordinate ? $\endgroup$
    – Eli
    Commented Mar 17, 2019 at 13:50
  • 1
    $\begingroup$ after thinking about this for a while, I believe you've answered the original question. Thank you. $\endgroup$
    – Dhuality
    Commented Jan 31, 2020 at 20:54

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