| Topic: |
Science > Physics |
| User: |
"Matthew Brenneman" |
| Date: |
06 May 2005 06:29:29 AM |
| Object: |
Counter-Intuitive Collision Frequency Result |
Hi,
I have come up with a result that initially seemed counter-intuitive to
me, and I am wondering if there is a simple, intuitive way to make
sense of it.
I am considering a gas which is "ideal" in every sense: it is at
equilibrium, the only intermolecular interactions are elastic
collisions, and they are perfect spheres. I want to derive the
expression for the collision frequency of a single gas molecule (which
I'll call a "test particle") if its velocity is v. The derivation I use
seems straightforward: I transform the system to the rest frame of the
test particle, I use the fact that the collision frequency is the same
as the net flux of particles through the surface of the test particle,
and then do the appropriate integrations over the surface of the sphere
and then over the relative velocity distribution.
My result is suprising: it is the same as the standard formula for the
collision frequency of a single gas molecule (see Berry and Rice, pg
829). In other words, there does not seem to be _any_ dependence of the
collision frequency on the test particle's velocity. Initially, this
struck me as strange: I would think that the faster the test particle
was moving, the higher its collision rate would be (since it would make
more "head-on" collisions). However, it is also true that it would lose
collisions since slower particles would not be able to "catch up" with
the test particle and make collisions from behind.
Is there a simple intuitive argument to see why this result is wrong or
right?
TIA,
Matthew
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| User: "Baugh" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 11:10:40 AM |
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Matthew Brenneman wrote:
Hi,
I have come up with a result that initially seemed counter-intuitive to
me, and I am wondering if there is a simple, intuitive way to make
sense of it.
I am considering a gas which is "ideal" in every sense: it is at
equilibrium, the only intermolecular interactions are elastic
collisions, and they are perfect spheres. I want to derive the
expression for the collision frequency of a single gas molecule (which
I'll call a "test particle") if its velocity is v. The derivation I use
seems straightforward: I transform the system to the rest frame of the
test particle, I use the fact that the collision frequency is the same
as the net flux of particles through the surface of the test particle,
and then do the appropriate integrations over the surface of the sphere
and then over the relative velocity distribution.
My result is suprising: it is the same as the standard formula for the
collision frequency of a single gas molecule (see Berry and Rice, pg
829). In other words, there does not seem to be _any_ dependence of the
collision frequency on the test particle's velocity. Initially, this
struck me as strange: I would think that the faster the test particle
was moving, the higher its collision rate would be (since it would make
more "head-on" collisions). However, it is also true that it would lose
collisions since slower particles would not be able to "catch up" with
the test particle and make collisions from behind.
Is there a simple intuitive argument to see why this result is wrong or
right?
TIA,
Matthew
Did you by chance select the average speed for the velocity of
your test particle. If so then you may have already "integrated over
all velocities" in your calculation. I'd have to review the details
to be certain.
Hmmm...
You have a velocity distribution $f(v,T)$. (Maxwell?)
You select a particle at random with velocity between v' and v'+dv
with probability $f(v,T)$. You transform the distribution:
f'(v,T) = f(v-v',T).
The density won't change.
You should get something like the integral:
R(v') = \rho \int dv f(v-v',T)<v|dS>
integrated over all velocities v
which are **inward to the surface element dS** i.e
such that
<v|dS> > 0
then integrated and over the surface.
This should depend on v'. Are you asserting that it
in fact does not?
You would then average over the distribution of particle velocities
to get an average rate:
Ravg = \int dv' f(v') R(v')
Which of course is independent of v'.
Regards,
J. B.
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| User: "Matthew Brenneman" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 08:28:10 PM |
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You should get something like the integral:
R(v') = \rho \int dv f(v-v',T)<v|dS>
integrated over all velocities v
which are **inward to the surface element dS** i.e
such that
<v|dS> > 0
then integrated and over the surface.
This should depend on v'. Are you asserting that it
in fact does not?
Exactly.
The integral you need to compute is essentially the flux through the
test particle's surface.
Let:
v denote the relative velocity of the bath particles
p(v) denote the probability of a bath particle having velocity v
p denote total particle density (a number density, which is essentially
contstant)
n denote the outward normal to the surface
dA denote surface element of sphere
* denote dot-product
Then net flux is:
[ Integral over dvx dvy dvz (Integral over phi and theta of ( p p(v)
v*n dA) ]
The key point here is, as you noted, that I am essentially saying that
the integration over the spherical coords does not involve the relative
bath velocity magntiudes or orientations. How did I end up with this
result? My reasoning was to say that since the velocity integration is
on the _outside_, the integration I am doing over the surface of the
sphere is for a fixed v. Since dot products are invariant under
rotation, I can make the surface integration for term v*n easy by
choosing z to be along the direction of v. Now v*n = |v| cos(theta),
and the integral reduces easily with no dependence on the bath
particles velocity.
What you end up with after this integration is essentially the original
shifted Gaussian velocity distribution, which when you integrate over
the velocity coords just gives you unity.
If you want to see the details, I'll be more than happy to show you:
it's not even one page.
Thanks for the comments and if you have any others, please don't
hesitate to make further suggestions.
Sincere Thanks,
Matt
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| User: "Baugh" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
07 May 2005 11:38:31 AM |
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Mat,
I spent more time in this reply than I intended.
I think I've found you in error... see below:
Matthew Brenneman wrote:
You should get something like the integral:
R(v') = \rho \int dv f(v-v',T)<v|dS>
integrated over all velocities v
which are **inward to the surface element dS** i.e
such that
<v|dS> > 0
then integrated and over the surface.
This should depend on v'. Are you asserting that it
in fact does not?
Exactly.
The integral you need to compute is essentially the flux through the
test particle's surface.
Let:
v denote the relative velocity of the bath particles
p(v) denote the probability of a bath particle having velocity v
p denote total particle density (a number density, which is essentially
contstant)
n denote the outward normal to the surface
dA denote surface element of sphere
* denote dot-product
Then net flux is:
[ Integral over dvx dvy dvz (Integral over phi and theta of ( p p(v)
v*n dA) ]
The key point here is, as you noted, that I am essentially saying that
the integration over the spherical coords does not involve the relative
bath velocity magntiudes or orientations. How did I end up with this
result? My reasoning was to say that since the velocity integration is
on the _outside_, the integration I am doing over the surface of the
sphere is for a fixed v. Since dot products are invariant under
rotation, I can make the surface integration for term v*n easy by
choosing z to be along the direction of v. Now v*n = |v| cos(theta),
and the integral reduces easily with no dependence on the bath
particles velocity.
The domain of integration over v is
not spherically symmetric it depends on the point on the surface of the
particle and so the velocity integration must be on the *inside*!
This is to say we are integrating over velocities of particles
which may strike the original sphere at a given point on its surface.
(and then integrating over the surface.)
Since only particles in the half space tangent to this point
may strike it (we don't count collisions from inside the
test particle) our domain of integration depends on the surface
coordinates.
What is more, since v' is not zero the distribution f(v-v') is not
spherically symmetric over v. Thus the half plane integration
will depend on v'. The final integration over the surface should
remove any angular dependence on v' but the radius should still
contribute in general.
I left off last time with the rate integral:
R(v') = \int f(v-v')dv (v|dA) ...
(Notation: ( | ) is dot product,
I fixing T=temperature and scale to \rho = 1.)
.... integrated over the surface area of the test particle and
over all velocities v such that (v|dA) > 0.
This condition is critical! It puts a constraint on the domain
of integration w.r.t. v which depends on surface coordinates
relative to the direction of v'. You must integrate w.r.t.
v prior to integrating w.r.t. dA.
Let's change variables of integration v" = v-v', dv"=dv. Then the
requirement that (v|dA) > 0 becomes (v"+v'|dA) > 0.
(this effectively undoes change of frame to rest frame of test ptcl.)
Now lets get down to brass tacks.
Since f(v") is spherically symmetric we can as easily write f(v")=f(s").
{ f depends only on relative speed: s"^2 = (v"|v") }.
I choose spherical coordinates where z-axis is the direction of dA
and y-axis is perpendicular to both dA and v'.
This gives:
vz' = s'cos(\phi') and vz" = s" cos(\phi)
vx' = s'sin(\phi') and vx" = s" sin(\phi)cos(\theta)
vy' = 0 and vy" = s" sin(\phi)sin(\theta),
for \phi,\phi' azmuth angles of v",v resp.
and \theta the equitorial angle of v".
We have:
(v" + v'|dA) = (vz" + vz)da =
= [s' cos(\phi') + s" cos(\phi)] da
Then the domain of integration is defined by:
W = vz"+vz' = s'cos(\phi') + s"cos(\phi) > 0
as a condition on \phi and s".
Use the Heavyside step function:
H(r) = 1 for r>0 else H(r) = 0
to "internalize" the domain of integration:
R(s') = \int f(s") H(W) W da,
now integrated over all values of v".
R(s') = \int(ds" d\phi d\theta da) {s"^2 cos(\phi)f[s"] H[W] W }
Now we may express integration over the area by:
da = cos(\phi')d\phi' d\theta'.
That is we parameterize surface relative to v' by
varying the coordinates of v'/s' relative to the surface normal.
(may introduce change of sign ?)
R(s')= \int (d~)d\theta d\theta' {s"^2cos(\phi)cos(\phi')W f(s") H[W]}
with (d~) = ds" d\phi d\phi'
Since there is no argument dependence on \theta or \theta' we may
go ahead and integrate w.r.t these:
R(s')=(4\pi^2)\int (d~){ s"^2cos(\phi)cos(\phi')f(s")W H[W] }
You assert the above is a constant independent of s',
in the particular case where f(s") is Gaussian distribution?
I'm still not sure I believe it...
I spent some time on this but found errors so won't post yet.
I was trying to see if dR/d s' is necessarily zero.
I'll play with it some more and get back to you.
Regards,
James Baugh
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| User: "Matthew Brenneman" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
08 May 2005 10:51:23 AM |
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Thank you for the derivation Jim. I, too, need to study it more
(posting without a nice equation editor or LaTex is hard!).
I did the integral in the lab frame, which takes out the orientational
dependence of the pdf for the velocity of the particles striking the
test particle. In this case, the only angular dependence (before you
integrate over all of the velocities, so this is the flux integral for
fixed (vx,vy, and vz) comes from the dot product of the test particles
velocity and the unit normal (of a sphere).
I'll study your derivation, get my final result soon and hopefully, the
two will agree!
Thank you again,
Matt Brenneman
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| User: "Jan Panteltje" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 09:44:22 AM |
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On a sunny day (6 May 2005 04:29:29 -0700) it happened "Matthew Brenneman"
<pi2ovr6@netscape.net> wrote in
<1115378968.980542.228500@g14g2000cwa.googlegroups.com>:
Is there a simple intuitive argument to see why this result is wrong
It is wrong, because at speed 0 clearly no collisions take place.
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| User: "Baugh" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 11:14:00 AM |
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Jan Panteltje wrote:
On a sunny day (6 May 2005 04:29:29 -0700) it happened "Matthew Brenneman"
<pi2ovr6@netscape.net> wrote in
<1115378968.980542.228500@g14g2000cwa.googlegroups.com>:
Is there a simple intuitive argument to see why this result is wrong
It is wrong, because at speed 0 clearly no collisions take place.
Not all speeds are zero only his test particle's speed.
That particle if at rest must feel collisions, that is what causes
the pressure.
Regards,
JB
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| User: "Jan Panteltje" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 12:39:31 PM |
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On a sunny day (Fri, 06 May 2005 12:14:00 -0400) it happened Baugh
<baconbaugh@charter.net> wrote in <wJMee.13819$cZ6.11090@fe02.lga>:
Jan Panteltje wrote:
On a sunny day (6 May 2005 04:29:29 -0700) it happened "Matthew Brenneman"
<pi2ovr6@netscape.net> wrote in
<1115378968.980542.228500@g14g2000cwa.googlegroups.com>:
Is there a simple intuitive argument to see why this result is wrong
It is wrong, because at speed 0 clearly no collisions take place.
Not all speeds are zero only his test particle's speed.
That particle if at rest must feel collisions, that is what causes
the pressure.
Regards,
JB
OK OK, I got that part now!
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| User: "Matthew Brenneman" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 10:08:04 AM |
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No. Zero velocity here would refer to the velocity of a single
particle, the "test particle", not all of the particles in the thermal
bath. A particle simply sitting in space will defintely be hit by other
particles and have a non-zero collision frequency.
However, after some thought, it seems that perhaps this result is true
and there is a simple explanation for it. Consider a test particle
moving in one direction (say +x direction). The distribution of thermal
bath particles is now a Gaussian that has been shifted to the left. So,
the test particle collides with all of those particles having a
negative relative velocity to it and also with all of those particles
having a positive relative velocity to it. Although it collides with
more particles having a negative velocty, there are less of them (due
to the shifted velocity distbtn), while it gets hit by fewer particles
from behind, but there are more of them. The net fluxes, it seem, might
balance out.
Matt
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| User: "Jan Panteltje" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 10:47:49 AM |
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On a sunny day (6 May 2005 08:08:04 -0700) it happened "Matthew Brenneman"
<pi2ovr6@netscape.net> wrote in
<1115392084.103724.235660@o13g2000cwo.googlegroups.com>:
No. Zero velocity here would refer to the velocity of a single
particle, the "test particle", not all of the particles in the thermal
bath. A particle simply sitting in space will defintely be hit by other
particles and have a non-zero collision frequency.
However, after some thought, it seems that perhaps this result is true
and there is a simple explanation for it. Consider a test particle
moving in one direction (say +x direction). The distribution of thermal
bath particles is now a Gaussian that has been shifted to the left. So,
the test particle collides with all of those particles having a
negative relative velocity to it and also with all of those particles
having a positive relative velocity to it. Although it collides with
more particles having a negative velocty, there are less of them (due
to the shifted velocity distbtn), while it gets hit by fewer particles
from behind, but there are more of them. The net fluxes, it seem, might
balance out.
Matt
OK, maybe I misunderstood your question.
But in this case you describe, if the test particle has say a very high speed
to the right, it will in the same unit of time cross the path of other particles.
So, should hit more per unit of time?
Consider you, driving a race car on a highway, if you go 300mph (and it was
crash proof) you could hit more cars then if you did only 20 mph.
Is this right?
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| User: "Mark Fergerson" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 11:00:51 AM |
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Jan Panteltje wrote:
On a sunny day (6 May 2005 08:08:04 -0700) it happened "Matthew Brenneman"
<pi2ovr6@netscape.net> wrote in
<1115392084.103724.235660@o13g2000cwo.googlegroups.com>:
No. Zero velocity here would refer to the velocity of a single
particle, the "test particle", not all of the particles in the thermal
bath. A particle simply sitting in space will defintely be hit by other
particles and have a non-zero collision frequency.
However, after some thought, it seems that perhaps this result is true
and there is a simple explanation for it. Consider a test particle
moving in one direction (say +x direction). The distribution of thermal
bath particles is now a Gaussian that has been shifted to the left. So,
the test particle collides with all of those particles having a
negative relative velocity to it and also with all of those particles
having a positive relative velocity to it. Although it collides with
more particles having a negative velocty, there are less of them (due
to the shifted velocity distbtn), while it gets hit by fewer particles
from behind, but there are more of them. The net fluxes, it seem, might
balance out.
Matt
OK, maybe I misunderstood your question.
But in this case you describe, if the test particle has say a very high speed
to the right, it will in the same unit of time cross the path of other particles.
So, should hit more per unit of time?
Consider you, driving a race car on a highway, if you go 300mph (and it was
crash proof) you could hit more cars then if you did only 20 mph.
Is this right?
Free particles in a gas are not constrained the way cars on a
racetrack or highway are. Think destruction derby, with the average
velocities of the participants = temperature.
Mark L. Fergerson
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| User: "Matthew Brenneman" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 11:15:51 AM |
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OK, maybe I misunderstood your question.
But in this case you describe, if the test particle has say a very high
speed
to the right, it will in the same unit of time cross the path of other
particles.
So, should hit more per unit of time?
Consider you, driving a race car on a highway, if you go 300mph (and it
was
crash proof) you could hit more cars then if you did only 20 mph.
Is this right?
I like your car analogy. Let's say the cars are all traveling along a
straight road, and their speeds are Gaussian distributed about the
speed limit. When you speed up, you hit more cars in the front than you
would have if you went say the mean, but you get hit by less guys from
behind than you would have if you were traveling the mean speed. The
idea is that the increased number of hits from the front are
counterbalanced by the decreased number of hits from behind.
If you want to think about it in a more mathematical way, sketch a
particle moving along the +x axis with some velocity v. Now sketch the
distribution for the relative x-velocity component of the thermal bath
particles: it is a normal distribution shifted to the left by |v|. What
happens when the particle moves _really_ fast? Well, this shifts the
curve far to the left. This means that in the test particle's rest
frame, most of the particles are moving in the -x direction. It makes
more hits with these particles, yes, but notice that it makes less hits
with the particles from behind. Why? Because the pdf for these
particles having a positive speed has been reduced when the distbtn was
left shifted. The thing I want to argue is that the increased number of
hits from the front are counterbalanced by the decreased number of hits
from behind. Maybe someone clever can see a good, simple argument that
clearly shows why this is so.
Regards,
Matt
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| User: "Jan Panteltje" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 12:39:34 PM |
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On a sunny day (6 May 2005 09:15:51 -0700) it happened "Matthew Brenneman"
<pi2ovr6@netscape.net> wrote in
<1115396151.524308.131670@f14g2000cwb.googlegroups.com>:
OK, maybe I misunderstood your question.
But in this case you describe, if the test particle has say a very high
speed
to the right, it will in the same unit of time cross the path of other
particles.
So, should hit more per unit of time?
Consider you, driving a race car on a highway, if you go 300mph (and it
was
crash proof) you could hit more cars then if you did only 20 mph.
Is this right?
I like your car analogy. Let's say the cars are all traveling along a
straight road, and their speeds are Gaussian distributed about the
speed limit. When you speed up, you hit more cars in the front than you
would have if you went say the mean, but you get hit by less guys from
behind than you would have if you were traveling the mean speed. The
idea is that the increased number of hits from the front are
counterbalanced by the decreased number of hits from behind.
Nice try, but if you go faster then the traffic in your direction,
you increase hits from the ones coming at you from the opposite direction,
AND from the ones you overtake.
If you want to think about it in a more mathematical way, sketch a
Somebody just did that here.
particle moving along the +x axis with some velocity v. Now sketch the
distribution for the relative x-velocity component of the thermal bath
particles: it is a normal distribution shifted to the left by |v|. What
happens when the particle moves _really_ fast? Well, this shifts the
curve far to the left. This means that in the test particle's rest
frame, most of the particles are moving in the -x direction. It makes
more hits with these particles, yes, but notice that it makes less hits
with the particles from behind. Why? Because the pdf for these
No particles from behind, but new ones you 'overtake'.
Is this correct?
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| User: "Matthew Brenneman" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 08:36:03 PM |
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Nice try, but if you go faster then the traffic in your direction,
you increase hits from the ones coming at you from the opposite
direction,
AND from the ones you overtake.
This is a good point, and I'll have to give it some thought. I'll see
what I can come up with.
Thanks,
Matt
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| User: "Matthew Brenneman" |
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| Title: Re: Counter-Intuitive Collision Frequency Result |
06 May 2005 09:33:12 PM |
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Nice try, but if you go faster then the traffic in your direction,
you increase hits from the ones coming at you from the opposite
direction,
AND from the ones you overtake
You are right. I was able to take your highway analogy and come up with
a nice counter-example that convinces me that the collision frequency
does have a velocity dependence, and moreover, it shows where my error
probably crept in.
Let's consider a highway where the cars can have 8 speeds
{-4,-3,-2,-1,+1,2,3,4}. We'll make the prob of each speed follow a nice
symmetric distribution as follows:
p(-4) = 0.05
p(-3) = 0.1
p(-2) = 0.15
p(-1) =0.2
p(1) = 0.2
p(2) = 0.15
p(3) = 0.1
p(4) = 0.05
(The exact numbers don't matter - they just serve to illustrate that
the collsion rates are very different for radically differet velocities
and moreover, indicate what's going on).
Case 1 : Particle moves with speed of +2.
What is his collision rate?
It is the sum of the relative velocities times the prob of that
relative velocity.
He hits everybody going, as you say "the other way"
The contribution here is:
6*0.05 + 5*0.1 + 4*0.15 + 3*0.2 = 2
He hist verybody going in the same direction, but going slower than
him. Here only v=+1 cntributes, and its contribution is:
0.2*1 = 0.2
He gets hit by all those going faster than him. Here v=3 and 4
contribute:
1*0.1 + 2*0.05 = 0.2
The sum of these contributions is: 2.4
Case 2: Particle goes 4
OK, he hits everyone.
The guys going in the opposite direction are -4,-3,-2,-1.
The key point here is that his velocity relative to these particles is
greater, so he hits them more frequently.
The contribtion from these velocities is:
8*0.05 + 7*0.1 + 6*0.15 + 5*0.2 = 3
He also hits everyone going the same direction, but slower. The giys
that contribute here are 1,2,3
Theur contribution is:
3*0.2 + 2*0.15 + 1 * 0.1 = 1
The total sum is 4.
Definitely a difference. And from this example, I can begin to see
where my original assumtpion was wrong. If you have a shifted Gaussian,
the average of |x| is _not_ the same as the average of |x| when it is
not shifted.
Thanks for the example and the insight.
Matt
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