Flux of molecules

1. Dec 4, 2017

Pushoam

1. The problem statement, all variables and given/known data
Derive the flux of molecules i.e. no. of molecules striking a surface per unit area per unit time.

2. Relevant equations

3. The attempt at a solution

Let's say that there are n molecules per unit volume.

The no. of molecules which will touch the surface in the +ve x direction in time $d t$ is
dN = n A dx = n A <vx> dt .
So, the flux should be $\phi = n <v_x>$.
$< v_x> = < v_y> = < v_z> \neq \frac {<v>} 3$
The probability of getting a particle with speed between $v_x$ and $v_x + dv_x$ is given by
$g(v_x) dv_x$ $\propto$ $e ^ \left( - \frac {m{v_x}^2}{2 k_B T} \right)$.
Then,
$<v_x> = \int_{0 }^{\infty } v_x C$ $e^{ \left( - \frac { m{v_x}^2 } {2 k_BT } \right) }$ $dv_x$
Where C is the appropriate constant.
I have to express $< v_x >$ into $<v>$. For this I should express $<v_x>$ as $v \sin { \theta } \cos { \phi }$ and then take integration.

Is this correct?

Last edited: Dec 4, 2017
2. Dec 4, 2017

haruspex

Where dx is what, exactly?

3. Dec 4, 2017

Pushoam

The particle is moving in the x - direction. So, dx is the distace it moves in the x - direction in time dt before hitting the surface with area A.

4. Dec 4, 2017

haruspex

Ok.
I don't think you need to do quite such a messy integral. If a particle has speed v but arbitrary direction, what, on average, is its speed in the x direction?

5. Dec 4, 2017

Pushoam

Since the body has speed v in any arbitrary direction, then the body will have speed v when it is moving in the x - direction only and speed v $\cos \theta$ when it is moving with speed v along a direction whose projection upon x is $\cos \theta$.
Using $< f(x)> = \int_{x_i}^{x_f} f(x) P(x) dx$
$<v_x> = \int_0 ^{\frac {\pi}{2} }v \cos \theta P(\theta) d \theta$
Now, should I take $P(\theta) d\theta = \frac{d \theta}{2 \pi}$?

6. Dec 4, 2017

haruspex

No. Consider the area on a sphere that would be hit by leaving the centre in a direction forming an angle to to the x axis between θ and θ+dθ.

7. Dec 4, 2017

Pushoam

$P(\theta ) d \theta$ is the probability that the particle is moving in the direction making an angle $\theta~ and~ \theta + d \theta$ with x – axis.

Since $\theta$ goes from $\frac { - \pi } {2 }$ to $\frac { \pi } {2 }$ and each value is equally probable, the probability that the particle is moving in the direction making an angle $\theta ~and ~\theta + d \theta$ with x – axis is $\frac {d\theta } { \pi }$. Is this correct?

Another way is:

$P(\theta ) d \theta$ = $\frac { \text { The area between the angle } \theta ~and~ \theta + d \theta } {\text { The area between the angle } \frac { - \pi } {2 } ~ to ~ \frac { \pi } {2 } } = \frac { 2 \pi R^2 \sin \theta d\theta } {4 \pi R^2 } = \frac {\sin \theta d\theta } { 2 }$

How to decide which one is correct?

8. Dec 4, 2017

haruspex

Not so. If you consider a particle going in an arbitrary direction from the centre of a sphere the probability distribution for where it hits the surface should, by spherical symmetry, be uniform.

9. Dec 5, 2017

Pushoam

Since there is no reason why I should take the probability of hitting at one place more than the other place on the spherical surface, I have to take the probability of hitting any place on the surface to be uniform. This I understand.

In a similar way, I have no reason to take the probability of velocity in a particular direction more than other. So, the probability of a particle moving in any direction should also be uniform. This gives me $P(\theta) d\theta = \frac{ d \theta} {\pi}$. What is wrong with this logic? I didn't get it.

10. Dec 5, 2017

haruspex

But that is not treating all directions equally. It is treating all angular deviations from a given direction equally, which is not the same thing. The choice of that reference direction creates a bias.

11. Dec 5, 2017

Pushoam

From now on I will write x as s (cylindrical coordinate ) as I am considering spherical coordinate system. Then $\theta$is the angle which the velocity vector makes with z- axis.

And I have to consider the particles which hit the y- x surface. So, I have to calculate $<v_z>$.

So, the probability that the particle's velocity will be in a direction between $\theta ~ and ~ \theta + d\theta$ from z- axis is P($\theta$) = $\frac {d \theta } { \pi }$ in s-z plane.

But, here s- axis is the projection of velocity vector upon x – y plane. Since, the velocity vector is changing, this axis, too, is changing. So, I unknowingly assumed that velocity vector is changing in such a way that its projection upon x-y plane remains constant in direction. This is where I am wrong. Right?

12. Dec 5, 2017

Pushoam

The above tells me the probability of the particle hitting the surface.
It doesn't tell me the probability of the particle moving in a certain direction.

Another way is:

$P(\theta ) d \theta$ = $\frac { \text { The area between the angle } \theta ~and~ \theta + d \theta } {\text { The area between the angle } \frac { - \pi } {2 } ~ to ~ \frac { \pi } {2 } } = \frac { 2 \pi R^2 \sin \theta d\theta } {4 \pi R^2 } = \frac {\sin \theta d\theta } { 2 }$

I don't see the connection between the above and the idea that the above is the way one can treat all directions equally.

13. Dec 5, 2017

haruspex

As I wrote, if you consider a particle emitted from the centre of a sphere, all directions equally likely must be the same as the probability of hitting a given patch of the surface being proportional to the area of the patch.

14. Dec 5, 2017

Pushoam

Is this an experimental fact which could not be proved but verified experimentally?

15. Dec 6, 2017

haruspex

I would say it is the definition of "all directions equally likely". Actually, that is not a proper statement anyway since there are infinitely many directions and each by itself has probability zero. It needs to be expressed in terms probability densities, and I do not see any way to do that other than referring to solid angles, and solid angles are defined in terms of areas projected from a point.