Conservation of linear momentum with mass defect Suppose we have an insulating container, like a perfect black body, which absorbs all the radiation coming from a radioactive element placed in the center (or some equivalent process like matter annihilation). Assuming a spherically symmetric emission. It will transform all radiation into thermal energy until reaching thermal equilibrium.
If the $\Delta$m, due to mass defect, doesn't "pop out" from the body, like the example of the rocket, how is momentum preserved?
Does the body accelerate????
$\sum_{i} F_i = \frac{dp}{dt}=\frac{d(vm)}{dt}$

 A: From the spherical symmetry it is clear that the body cannot accelerate. (Which direction would it go, everything is symmetric). However, let’s look in a little more detail. 
As you say, the center radiates and therefore loses mass. The mass lost is equal to the energy of the radiation. By the problem setup the container absorbs all of the radiated energy and therefore the container gains the same mass that was lost by the center. So again, there is no acceleration, but let’s look in a little more detail. 
Specifically, let’s consider the situation in an inertial reference frame where the container and center are moving. In this frame, the mass of the center is decreasing so it is losing momentum and the mass of the container is increasing so it is gaining momentum. Since the mass decrease of the center is equal to the mass increase of the container there is no violation of the conservation of momentum, but how does the momentum transfer without causing acceleration? 
Although the radiation is spherically symmetric in the rest frame, in this frame the forward traveling radiation is blue-shifted due to the Doppler effect, and the backward radiation is redshifted. The overall radiation therefore carries momentum from the center to the container. The amount of momentum thus carried is exactly equal to the momentum lost due to the mass decrease of the center and the mass gain of the container. Thus the center and the container do not accelerate because all of the momentum transferred by the radiation is already accounted for in the mass change. 
A: With a simple calculation, it can be shown that there is no acceleration.
Let's assume $dm/dt$ is constant we call $\delta$. We then have
\begin{equation}
\frac{dv}{dt}=-\delta v
\end{equation}
Then:
\begin{equation}
\frac{dv}{v} = -\delta dt
\end{equation}
By integrating:
\begin{equation}
\ln (v) = -\delta t + A
\end{equation}
Where $A$ is an integration constant, dependent on initial conditions as we shall see.
\begin{equation}
v=B e^{-\delta t}
\end{equation}
Where $B=exp(A)$. But at $t=0$ we have $v=B$. But $v=0$ at $t=0$, then $B=0$ ($A\rightarrow - \infty$). So now:
\begin{equation}
v=0 ; \forall t\geq 0
\end{equation}
Thus:
\begin{equation}
a=\frac{dv}{dt}=0
\end{equation}
No acceleration!
(The case for $dm/dt$ not constant which is realistic is not solvable for most realistic states. Should try for $dm/dt=Cexp(-ct)$ like a radioactive material, but in my knowledge this isn't solvable analytically, I'm not that good.)
