How did my professor get this integral

[grandpa2390]

Homework Statement

derive maxwell distribution function in case of 1-d and 2-d classical gas

Homework Equations

The Attempt at a Solution

[/B]
The constant K can be solved from normalization.
$\int_{-∞}^{∞} F(V_x)dV_x = 1$

substituting $F(V_x)=Ke^{+/- kV_x^2}$

$1 = K\int_{-∞}^{∞} e^{-kV_x^2}dV_x$
which equals:
$K(\frac{π}{k})^{1/2}$

how did that integral become that result?

 

[gneill]

I suspect that he used the definition of the Gauss error function and its limiting values at +/- ∞. At least, that's what I would do

 

[Daniel Gallimore]

This is the integral of a Gaussian (a bell curve essentially). There are plenty of resources online that give you the general form of this integral. One such resource is http://www.umich.edu/~chem461/Gaussian Integrals.pdf. They also include a very handy derivation.

 

[grandpa2390]

This is the integral of a Gaussian (a bell curve essentially). There are plenty of resources online that give you the general form of this integral. One such resource is http://www.umich.edu/~chem461/Gaussian Integrals.pdf. They also include a very handy derivation.

I followed that. And it assuming I did it correctly... I got the same result.
$K = \frac{k}{π}^\frac{1}{2}$ in 1-D
and
$K^2 = \frac{k}{π}$ which is the same thing for 2-D

is that right?

 

[Daniel Gallimore]

I followed that. And it assuming I did it correctly... I got the same result.
$K = \frac{k}{π}^\frac{1}{2}$ in 1-D
and
$K^2 = \frac{k}{π}$ which is the same thing for 2-D

is that right?

For 1D, make sure to put some parenthesis around you fraction so we can tell you're setting the whole fraction the the 1/2 power.

For 2D, I believe your formula generalizes to $$1 = K\int_{-∞}^{∞}\int_{-∞}^{∞} e^{-k(V_x^2+V_y^2)}dV_x \, dV_y$$

 

[grandpa2390]

For 1D, make sure to put some parenthesis around you fraction so we can tell you're setting the whole fraction the the 1/2 power.

For 2D, I believe your formula generalizes to $$1 = K\int_{-∞}^{∞}\int_{-∞}^{∞} e^{-k(V_x^2+V_y^2)}dV_x \, dV_y$$

well I thought it was $F(V_x)F(V_y) = Ke^{(..)}Ke^{(...)}$ and so K^2

 

[Daniel Gallimore]

well I thought it was $F(V_x)F(V_y) = Ke^{(..)}Ke^{(...)}$ and so K^2

A one-dimensional Gaussian has the form $$F(x)=Ke^{-\alpha x^2}$$ A two-dimensional Gaussian has the form $$F(x,y)=Ke^{-\alpha (x^2+y^2)}$$ For an ideal gas in two dimensions, the Maxwell velocity distribution is a two-dimensional Gaussian (I'm assuming $V_i$ represents the speed in the $i$th direction). This is a consequence of the equipartition theorem.