How to calculate redshift for a very large object? I've been scratching my head on this concept of redshift. I can understand redshift for a pointlike object that is either coming towards an observer or going away. However, I am confused on what color the object will look like if it is very long instead of pointlike. 
For example, imagine an infinitely long horizontal bar moving away from you very quickly. Looking directly in front of you and in the middle you will see the usual redshift for pointlike objects. Will you see the same value for redshift (as in the middle of the bar) if looking far to the right or left? I've tried drawing right triangles for this, but I don't know how to relate light's speed, one fast-moving massive object, and redshift.
 A: Let's just start with classical Doppler. A person moving through a medium claps at some time $t$, then again at some time $t + \delta t$, and those two claps travel towards you, and you are stationary with respect to the medium, at the origin of our coordinates.
If their position is $\vec r(t)$ then the first clap reaches you at time $t + |\vec r(t)|/c$ and the second reaches you at time $t + \delta t + |\vec r(t + \delta t)|/c,$ where $c$ is the speed of sound in the medium. You hear the time between the claps as therefore $$\delta t' = \delta t + \frac{1}{c} \big(|\vec r(t + \delta t)| - |\vec r(t)|\big).$$ As you might imagine, this means that there is a definite limit for ${\delta t'}/{\delta t}$, which is $$\alpha = \lim_{\delta t\to0}\frac{\delta t'}{\delta t} =1 + \frac{1}{c} \frac{d}{dt} \sqrt{\vec r\cdot \vec r} = 1 + \frac1c~\frac{\vec v \cdot \vec r + \vec r \cdot \vec v}{2\sqrt{\vec r \cdot \vec r}} = 1 + \hat r \cdot \frac{\vec v}{c}~.$$When something is moving precisely away from you $\vec v \propto \hat r$and this formula is $1 + v/c$, when it is moving precisely towards you $\vec v \propto -\hat r$ and this formula is $1 - v/c.$
Now if those "claps" are wavefronts and they are high-enough frequency that we no longer hear them as discrete (which happens at about 100ms or less), then $f = 1/\delta t$ and so we have that $\alpha \approx f/f'$ as well, hence the usual formula $f' = f_0/(1 \pm v/c).$
Now let's talk about light. Two effects happen: first, usually, we describe an object in terms of its normalized velocity $\vec \beta = \vec v/c$. Second the only thing we relativistically need to add is a time-dilation effect at the source as perceived by the receiver, lowering its frequency by a factor of $\gamma=1/\sqrt{1-\beta^2}$: $$f' = \frac{f_0}{1 + \hat r \cdot \vec\beta}~\sqrt{1 - \beta^2}.$$In fact the $1/\gamma$ factor that you see here factorizes nicely into $\sqrt{(1 + \beta)(1-\beta)}$ and therefore when an object is moving straight toward/away from you the factor is $f' = f_0\sqrt{1+\beta\over1-\beta}$ (blueshift) or $ f' = f_0 \sqrt{1-\beta\over1+\beta}$ (redshift). You can then choose to memorize only one of these as you get the other one from the simple switch $\beta \mapsto -\beta.$ 
But, in general, if you do not have those effects then you will need the above equation rather than either of these specializations. Note that $\hat r$ still points, in the coordinates where the observer is stationary, from the observer to the point where the light was emitted. If you're doing a high-quality visualization you will also probably want to incorporate the real time delay that this light takes to reach the observer.
