# 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.