Evaluating an integral for an expanding, charged sphere

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1. Jan 30, 2015

ELB27

1. The problem statement, all variables and given/known data
An expanding sphere, radius $R(t) = vt$ ($t>0$, constant $v$) carries a charge $Q$, uniformly distributed over its volume. Evaluate the integral $$Q_{eff} = \int \rho(\vec{r},t_r) d\tau$$ with respect to the center. ($t_r$ is the retarded time and $d\tau$ is an infinitesimal volume element). Show that $Q_{eff}≈Q(1-\frac{3v}{4c})$, if $v<<c$.

2. Relevant equations
The definition of the retarded time: $t_r ≡ t-\frac{d}{c}$ where $d$ is the distance between observation point and charge.

3. The attempt at a solution
At any time $t$, the volume charge density is constant: $\rho(t) = \frac{Q}{\frac{4}{3}\pi (vt)^3}$. However, I believe that for the observer at the center, the density will not seem uniform because the retarded time depends on separation distance (and hence, signals emitted from points near the center will arrive simultaneously with earlier signals from far points). At a certain radius $r<R$, the retarded time is $t_r = t-\frac{r}{c}$ and at this retarded time, the density at $r$ is $rho(r,t_r) = \frac{Q}{\frac{4}{3}\pi (vt_r)^3} = \frac{3Q}{4\pi}\frac{c^3}{v^3}\frac{1}{(tc-r)^3}$. Integrating over the entire sphere at time $t$ and noting that the (spherical) $\theta$ and $\phi$ integrals together give $4\pi$, $$Q_{eff} = 3Q\frac{c^3}{v^3}\int_0^{vt}\frac{r^2}{(tc-r)^3}dr = 3Q\frac{c^3}{v^3}\left[\ln\left(\frac{c}{c-v}\right) +\frac{3v^2-2cv}{2(c-v)^2}\right]$$
The problem is that if I expand the above expression in Taylor series about $v=0$, I get $Q_{eff}≈Q(1+\frac{9v}{4c})$ in disagreement with the problem statement.
I have been thinking about this problem for quite some time and tried different approaches but none led to the correct approximation in the end and now I'm at a loss. Where is my argument wrong?
By the way, the author of my book argued before (when discussing potentials of a moving point charge) that the apparent volume in an integral such as the above is altered by the factor $$\frac{1}{1-\hat{d}\cdot\vec{v}/c}$$ where $\hat{d}$ is a unit vector pointing from source point to observation point. If so, the apparent volume (and hence, the effective charge) is bigger, not smaller as the approximation of the problem suggests.
Any comments/corrections will be greatly appreciated!

2. Jan 30, 2015

DEvens

What is the limit on the integral?

3. Jan 30, 2015

ELB27

Ah, I see... The largest radius from which the signal has just arrived at the time of evaluation: $R = vt_r=c(t-t_r)$. From this, $t_r=\frac{ct}{c+v}$ and $R=vt_r=\frac{vct}{v+c}$ and this last expression constitutes the upper limit of the integral. With this, $$Q_{eff}=3Q\frac{c^3}{v^3}\int_0^{\frac{vct}{v+c}}\frac{r^2}{(tc-r)^3} dr = 3Q\frac{c^3}{v^3}\left[\ln\left(\frac{c+v}{c}\right)+\frac{v^2-2cv}{2c^2}\right]$$ If $v<<c$, this indeed gives $Q_{eff}≈Q(1-\frac{3v}{4c})$.

Thank you very much! (Such a long OP and such a short answer )