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Let us be in $\mathbb{R}^3$, with the wave equation $u_{tt} - \Delta_x u =0$ satisfying the initial conditions $u(0,x)=f(x)$, $u_t(0,x)=g(x)$. We define the spherical average of $u$, $U(t,r,x) = \frac{1}{4\pi r^2} \int_{\partial B(x,r)} u(t,\sigma) d \sigma$.

My question is, why is ($B(0,1)$ is the ball centered at $0$ with radius $1$) $$U(t,r,x) = \frac{1}{4\pi r^2} \int_{\partial B(x,r)} u(t,\sigma) d \sigma = \frac{1}{4\pi} \int_{\partial B(0,1)} u(t,x + r \omega) d \omega?$$ Clearly we are doing change of variables, but I don't understand the part where we scale by $r^2$.

Albert
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  • We're not scaling by $r^2$. When we plug $r=1$ into $\frac{1}{4\pi r^2}$ you just get $\frac{1}{4\pi}$. – K.defaoite Oct 06 '20 at 11:45
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    As you said you're doing change of variables. If you calculated the Jacobian for $\omega\mapsto x+r\omega$ on $\mathbb{R}^3$ (i.e. $\omega\in\mathbb{R}^3$) you'd get $r^3$ but you are doing change of variables on the 2-dimensional sphere (i.e. $\omega\in\mathbb{S}^2$) which is why you are missing one power. This might be helpful: https://math.stackexchange.com/questions/673526/how-to-change-variables-in-a-surface-integral-without-parametrizing – Sven Pistre Oct 06 '20 at 11:48
  • @SvenPistre The link that you attached is very helpful. Thanks! Maybe I should learn differential geometry sometime. – Albert Oct 08 '20 at 02:23
  • Definitely do! ;) – Sven Pistre Oct 19 '20 at 15:05

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