Physics, asked by varunmittal17951, 1 year ago

Determination of most probable velocity from maxwell equation

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Answered by bhavikbhatia20p9ckc2
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We can estimate the typical speed of a particle of mass m in thermal equilibrium with its surroundings. In thermal equilibrium, a particle is excited thermally. We can write its kinetic energy as:

 \frac{1}{2} m v^2 = \frac{3}{2} kT  \,\!

Thus, a typical particle has speeds of

 v \approx \sqrt {\frac{3 kT}{m}} \,\!

Velocity Distribution

Now we will derive f(\vec{v}), the distribution of velocities of a particle in thermal equilibrium. Since we are in thermal equilibrium, we can use Maxwell-Boltzmann statistics. Recall that for Maxwell-Boltzmann statistics, the probability of an energy state εi at temperature T is given by:

P(\epsilon _ i) = \frac{\exp (-\epsilon _ i/kT)}{\sum _ j \exp (-\epsilon _ j/kT)}  \,\!

The denominator is the normalization factor which is calculated by summing over all possible energy states. Now, consider a velocity \vec{v} = (v_ x, v_ y, v_ z). to calculate the probability of a particle of having that particular velocity.

 f(\vec{v}) = C \exp \left(-\frac{m|\vec{v}|^2}{2kT}\right) = C\exp \left(-\frac{m}{2kT} (v_ x^2+v_ y^2+v_ z^2) \right)  \,\!

To get C, the normalizing factor, we need to sum over all possible velocities.

 1/C = \sum _ i \exp \left(-\frac{m\vec{v_ i}^2}{2kT}\right)  \,\!

Because velocities are continuous, the sum then becomes an integral. And since velocity is in three dimensions, we need to integrate over dvxdvydvz.

 1/C = \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \exp \left(-\frac{m}{2kT} (v_ x^2+v_ y^2+v_ z^2)\right) dv_ xdv_ ydv_ z  \,\!

We note that we can break this up into three Gaussian integrals:

 1/C = \int _{-\infty }^{\infty }\exp \left(-\frac{m v_ x^2}{2kT}\right) dv_ x \int _{-\infty }^{\infty } \exp \left(-\frac{m v_ y^2}{2kT}\right) dv_ y \int _{-\infty }^{\infty } \exp \left(-\frac{m v_ z^2}{2kT}\right) dv_ z  \,\!

We will define \tilde{v}^2 = 2kT/m and integrate a Gaussian.

 1/C = ( \tilde{v} \sqrt {\pi })^3 = \left( \frac{ 2 \pi kT}{m} \right)^{3/2}  \,\!

Plugging it all in, the velocity distribution is simply a Gaussian centered around zero with a typical fluctuation in the velocity of \sqrt {2kT/m}:

 f(\vec{v}) = \left( \frac{m}{ 2 \pi kT} \right)^{3/2} \exp \left(-\frac{mv^2}{2kT}\right)  \,\!

where v = |\vec{v}| = \sqrt {v_ x^2+v_ y^2+v_ z^2}. Note that f(\vec{v}) has units of [cm/s] − 3 since to obtain a probability you need to integrate over all three velocity directions.

Maxwellian Speed Distributions

The Maxwellian Speed Distribution is a bit more interesting than the velocity distribution (and used more often). What we want to find is f(v) which is different than f(\vec{v}). This is because there are more ways to combine vx, vy, vz to make larger v’s so there are more available states at higher speeds. For a given speed v, any combination of velocity components that satisfy v_ x^2+v_ y^2+v_ z^2=v^2 will work. This is analogous to the surface area of a sphere and likewise we find that there are 4πv2 states at any given v. Thus, we can modify the Maxwell velocity distribution with the number of states at a given speed:

 f(v) = 4\pi \left( \frac{m}{ 2 \pi kT} \right)^{3/2} v^2 \exp \left(-\frac{mv^2}{2kT}\right)  \,\!

This is the Maxwell speed distribution and note that it has units of [cm/s] − 1.

Typical Speeds Revisited


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