How would a Maxwell-Boltzmann speed distribution change for a moving observer? Imagine I have a container of gas molecules whose speeds are given by the Maxwell-Boltzmann distribution while I'm stationary with respect to it. If I observe this system while moving with a velocity u relative to it, how would the distribution change? Would this be simply analogous to increasing the temperature, if so, can you give some explanation why? It feels to me like there would be a more substantial change.
Edit:
Just to rephrase my question: I am looking for some function P(v) which gives me the probability density function for the speeds of the particles in the system in my new frame of reference. Just by considering the cases of very high velocities, it is obvious that there would be some change at least.
The function in terms of individual velocity components would be
$$ P(v_x,v_y,v_z) = \left( \frac{m}{2\pi kT} \right)^{3/2} \exp(-\frac{m}{2kT}((v_x-u)^2+v_y^2+v_z^2))    $$
Then, as encouraged by these lecture notes, I assumed that I needed to convert the expression into spherical coordinates (v,$\theta,\phi$) and integrate the expression with respect to the angles to obtain an integral with dv, such that what's inside the integral would be the distribution function. I couldn't make sense of it in Cartesian coordinates, and spherical conversion could have made it more difficult considering our linear motion and the lack of spherical symmetry.
I am not confident in the the following working at all and would appreciate it if someone can help.
I have rewritten the function as
$$ P(v_x,v_y,v_z) = \left( \frac{m}{2\pi kT} \right)^{3/2} \exp(-\frac{m}{2kT}((v_x^2+v_y^2+v_z^2 +u^2 - 2uv_x))    $$
Then, by $ v^2 = v_x^2+v_y^2+v_z^2 $ (if it still is true in our new frame of reference) and by $ v_x = v \sin\theta \cos\phi $ (angle with the z-axis is $\theta$), I got
$$ P(v,\theta,\phi) = \left( \frac{m}{2\pi kT} \right)^{3/2} \exp(-\frac{m}{2kT}((v^2 +u^2 - 2uv \sin\theta \cos\phi))    $$
And then
$$ \int\int_0^{2\pi}\int_0^{\pi} \left( \frac{m}{2\pi kT} \right)^{3/2} \exp(-\frac{m}{2kT}((v^2 +u^2 - 2uv \sin\theta \cos\phi)) \; v^2 \sin\theta \; d\theta \: d\phi \: dv  $$
which is where I am stuck. The exponential term appears to be non-integrable. So either my question is unsolvable, or it is unsolvable in spherical coordinates, or it is me who can't see the answer, or I made a mistake. It would be interesting to see what a simulation would say.
 A: The 3-dimensional Maxwell velocity distribution function in the rest frame of the container is
$$f(v_x,v_y,v_z) = \left(\frac{m}{2 \pi kT}\right)^{3/2}\,  exp(-\frac{m(v_x^2+v_y^2+v_z^2)}{2kT})$$
If the container is moving with velocity $V$ say in the x-direction, this becomes
$$f(v_x,v_y,v_z) = \left(\frac{m}{2 \pi kT}\right)^{3/2}\,  exp(-\frac{m((v_x-V)^2+v_y^2+v_z^2)}{2kT})$$
You can see that the maximum of the distribution in the $x$-direction changes from $v_x=0$ to $v_x=V$ but the temperature $T$ is not changed. The temperature is related to the random velocities of the molecules within the container not to any systematic velocity of the molecules as a whole.
The Maxell distribution for the speed is
$$f (v) =
\left(\frac{2}{\pi} \right)^{1/2}
\left(\frac{m}{ kT} \right)^{3/2}
v^2
\exp \left[-
\frac{mv^2}{2kT}
\right]$$
where in the container's reference frame
$$v^2=v_x^2+v_y^2+v_z^2$$
In the moving reference frame you would have to replace this with
$$v^2=(v_x-V)^2+v_y^2+v_z^2$$
A: From the Hyperphysics website:
The Maxwell-Boltzmann distribution is the classical distribution function for distribution of an amount of energy between identical but distinguishable particles
$$f(E)=\frac{1}{Ae^{E/kT}}$$
Where $f(E)$ is the probability of  that a particle will have energy E, A is a constant, $k$ is Boltzmann’s constant, and $T$ is the absolute kinetic temperature.
So if the absolute temperature does not change, then the probability that  the particle will have energy $E$ will  not change. So the question boils down to, is the kinetic temperature of the gas a function of the velocity of the container of gas? And the answer to that is no.
The temperature of the gas is based on the average random translational kinetic energy about the center of mass of the contents of the container (the gas molecules themselves). This is different then the translational kinetic energy of the center of mass of the system as a whole based on the velocity of the container.
In short, motion of the container at constant velocity does not change the distribution of the velocities of the individual molecules where all the molecules move at constant velocity.
Incidentally, if the temperature did change, it would violate the law of special relativity, which states
“The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of co-ordinates in uniform translatory motion”
If we were able to measure a difference in temperature between the container of gas moving at constant velocity and the container of gas at rest, we would have a means of determining the existence of absolute motion.
Hope this helps.
A: I think I understand your problem, I've been stuck at it too and your question actually proved helpful pushing me in the right direction.
The solution is:
$$
P(v) = \sqrt{\frac{2}{\pi}} \sqrt{\frac{m}{kT}} \frac{v}{u} \exp{\left(-\frac{m \cdot (v^2 + u^2)}{2kT}\right)} \sinh{\left(\frac{v u \cdot m}{kT}\right)}.
$$
You were actually on the right track, but your integral in the end is solvable.
First do the $\phi$ integral which gives according to mathematica
$$
\mathrm{const.} \cdot 2\pi \cdot \int_{0}^{\pi} d\theta \sin{\theta} \cdot I_{0}\left( \frac{vu \cdot m}{kT} \cdot \sin{\theta} \right)
$$
where $I_{0}(x)$ is the zeroth-order modified Bessel function of the first kind. Integrating this, again via mathematica, leads to
$$
\mathrm{const.} \cdot \frac{2kT \cdot \sinh{\frac{vu \cdot m}{kT}}}{vu \cdot m}
$$
if $\frac{vu \cdot m}{kT}$ is real and $\geq 0$ (which it is, otherwise just define $u$ the other way round). Since $P(v)$ is supposed to be a probability density you do not want to integrate over $v$, which means we're done and obtained the solution presented above.
If you don't trust me and my mathematica skills, have a look at
this publication. Their Eq.(25) is exactly what you are looking for. (When cross-checking be aware that they are missing a factor $\pi$ though, as can be seen from comparison with their Eq. (24) in the limit $\lim_{u \rightarrow 0}$.
