Question

in a system of n particles are distributed in the x direction with the speed of the particle vx.and if distribution function with speed between vx and vx + dvx expressed as:
then how i find the probability particles that often appears??
and how i find constans C

Answer 1

function in here:

Answer 2

how i found the most probability velocity in the function above?
and how i found constant of C?
and kinetic energy??

Answer 3

You can find the most probable velocity in the above function by finding its maximum. The constant C can be found by normalizing the function, i.e. integrating it over the whole domain of interest and setting it equal to one. I'm not sure what you're asking about the kinetic energy, I apologize. Do you mean you want the expectation value of the kinetic energy?

Answer 4

i means that,with how much kinetic energy the particle will move....
can you show me how normalizing function?

Answer 5

You are entering the realm of statistical physics, in which you are describing not one particle but a large collection of them. You therefore cannot say that a particular particle has a particular velocity. Instead, you talk about the distribution of velocities.

You can therefore only make calculations such as "What is the most probable speed of a particle?" or "What is the average kinetic energy of the particles?".

Answer 6

Your function is the following:
\[ f(v) = C e^{\frac{-mv^2}{2kT}} \]
We can find that constant by demanding that
\[ \int_{-\infty}^\infty f(v) dv = 1 \]
That yields
\[ \int_{-\infty}^\infty C e^{\frac{-mv^2}{2kT}}dv = C \sqrt{\frac{2\pi kT}{m}} = 1\]
so
\[ C = \sqrt{\frac{m}{2\pi kT}} \]
and
\[f(v) = \sqrt{\frac{m}{2\pi k T}}e^{\frac{-mv^2}{2kT}} \]

Answer 7

for kinetic energy,i found this:

Answer 8

You need to be very careful here. We need to transform the function as follows:

\[f(v)dv = f(E)dE \]
Very quickly, is this system completely restricted to one dimension?

Answer 9

yes,..
f(v)dv=f(E)dE the same means f(E)dE= f(v)dv

Answer 10

Yes, but watch:
\[ dv = \frac{dv}{dE} dE\]
since
\[ E = \frac{mv^2}{2} \]
\[ \frac{dE}{dv} = mv = m\sqrt{\frac{2E}{m}}= \sqrt{2mE}\]
Substituting
\[v = \sqrt{\frac{2E}{m}} \]
into f(v)dv, we obtain
\[f(E)dE = \sqrt{\frac{m}{2\pi kT}}e^{\frac{-E}{kT}}\cdot\frac{1}{\sqrt{2mE}} \space dE \]
so we get as a result
\[f(E) = \frac{1}{2}\sqrt{\frac{1}{\pi E k T}}e^{\frac{-E}{kT}} \]
Please bear in mind that this is for a system that is restricted to one dimension. This is very different from a system that occupies three spatial dimensions, as I will show now:

Answer 11

When the system is 3 dimensional rather than one dimensional, we need to substitute
\[dv \rightarrow 4\pi v^2 dv\]
due to integration in spherical polar coordinates. Our normalization factor must therefore change:
\[\int_{0}^\infty C e^{\frac{-mv^2}{2kT}}\cdot 4\pi v^2 dv = C\cdot \left(\frac{\pi k T}{m}\right)^{\frac{3}{2}}= 1\]
so
\[C = \left(\frac{m}{2\pi k T}\right)^{\frac{3}{2}} \]
and
\[f(v) = \left(\frac{m}{2\pi k T}\right)^{\frac{3}{2}} e^{\frac{-mv^2}{2kT}} \]

Furthermore, when we transform,
\[f(v) \cdot 4\pi v^2 dv = f(E) dE\]
since
\[v^2 = \frac{2E}{m}\]
and \[\frac{dv}{dE} = \frac{1}{\sqrt{2mE}} \]
we get
\[f(E) dE = \left(\frac{m}{2\pi k T}\right)^{\frac{3}{2}} e^{\frac{-E}{kT}}\cdot 4\pi \frac{\sqrt{2E}}{m^\frac{3}{2}} dE = 2 \sqrt{\frac{E}{\pi}}\left(\frac{1}{\pi k T}\right)^{\frac{3}{2}}e^\frac{-E}{kT} \space dE \]
so
\[f(E) = 2 \sqrt{\frac{E}{\pi}}\left(\frac{1}{ k T}\right)^{\frac{3}{2}}e^\frac{-E}{kT} \]

So you see the physics fundamentally changes when the spatial dimensions are altered or restricted.

Answer 12

Oops, I made a typo in the second to last line.... it should be
\[f(E) dE = 2 \sqrt{\frac{E}{\pi}}\left(\frac{1}{kT}\right)^{\frac{3}{2}} e^\frac{-E}{kT} \space dE \]

Answer 13

The physical implications of this are very important and not necessarily intuitive. The so-called "density of states" , which is loosely speaking the number of available ways to divide up the energy amongst the particles in your system, increases in 3-D like
\[\sqrt{E}\]
Which means that higher energies yield more available configurations. However, when we are restricted to one dimension, the f(E) actually DECREASES like
\[\frac{1}{\sqrt{E}} \]
A strange result. As may be guessed from this, the density of states in two dimensions actually is independent of the energy.

I'm sure that's more than you wanted to know, but I can tell you nothing but the truth :)

Answer 14

yes, thxs myfriend...
by theway are you study or teach??

Answer 15

I will be entering Graduate School in the fall, so I'm not a professor but I do teach a little on the side.

Answer 16

where are you from and your school? sorry i ask this...

Answer 17

Not at all. I graduated from the University of Alabama in the United States, and I'm originally from the state of Pennsylvania. I'm not sure what graduate school I will be attending yet, though. What about you?

Answer 18

I graduated from state universities of Jakarta in Indonesia, majoring in physics... I will be entering Graduate School university education Indonesia.....

Answer 19

how i solved this:
how i find standard deviation (\[\sigma\]) and variance?
how i find laplace constanta?

Answer 20

You should start a new question for this

Answer 21

ok