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Flux of molecules

  1. Dec 4, 2017 #1
    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. jcsd
  3. Dec 4, 2017 #2

    haruspex

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    Where dx is what, exactly?
     
  4. Dec 4, 2017 #3
    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.
     
  5. Dec 4, 2017 #4

    haruspex

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    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?
     
  6. Dec 4, 2017 #5
    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}##?
     
  7. Dec 4, 2017 #6

    haruspex

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    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.
     
  8. Dec 4, 2017 #7
    ##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?
     
  9. Dec 4, 2017 #8

    haruspex

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    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.
     
  10. Dec 5, 2017 #9
    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.
     
  11. Dec 5, 2017 #10

    haruspex

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    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.
     
  12. Dec 5, 2017 #11
    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?
     
  13. Dec 5, 2017 #12
    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.
     
  14. Dec 5, 2017 #13

    haruspex

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    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.
     
  15. Dec 5, 2017 #14
    Is this an experimental fact which could not be proved but verified experimentally?
     
  16. Dec 6, 2017 #15

    haruspex

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    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.
     
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