The question is asking me to show that the line-of-sight intensity of radiation is invariant when there is no emission or absorption.
Starting with the radiative transfer equation:
$$ \frac{d}{dz}I_{\nu}=-\kappa_{\nu}I_{\nu}+j_{\nu} $$
I'm assuimg therefore that $\kappa_{\nu},j_{\nu}=0$? i.e., that there is no emission and no absorption? Or should it be 1? Not sure which is correct?
And would it be invariant if $I_{\nu} =$ constant? Which is really simple given $\frac{d}{dz}I_{\nu}=0$ ??
So, my question is how would I show that the line-of-sight velocity is invariant and is what I have thus far, correct?
You are right: if both the radiation $\kappa_{\nu}$ and absorption $j_{\nu}$ are equal to zero, then the given differential equation becomes simply $$\frac{d}{dz}I_{\nu}= 0$$ and we know from calculus that a function with zero derivative is constant. (The precise reason is the Mean Value Theorem that relates the change of function on an interval to the value of its derivative there.)