energy distribution and Maxwell speed distributionby r4nd0m Tags: distribution, energy, maxwell, speed 

#1
Jan606, 06:27 AM

P: 96

I am in the first semester and I'm taking the course Molecular physics, which is an introduction to statistical physics and thermodynamics. It is not very advanced  it doesn't involve much of the higher mathematics.
But my problem is: The probability density function of energy distribution is an exponential function which in general has the form: [tex]w(\epsilon) = \frac{\exp(\beta\epsilon)}{\int\exp(\beta\epsilon)d\Gamma}[/tex] that means that it is a decreasing function with the maximum in E = 0  so the highest probability is that the molecules have an energy around zero. If the gas is not in a force field, then the whole energy is just knietic energy. that means that the highest probability is that the velocity of molecules is around zero. On the other hand the Maxwell speed distribution tells us that velocities around zero are not very probable. What is wrong in this speculation? 



#2
Jan606, 09:45 AM

P: 49

The problem with your assumption is that you've assumed that the maximum is at E = 0, thereby implying that almost all particles in a system are near the ground state. This would mean that we would see BoseEinstein condensation everywhere, and that, in the classical limit, the vast majority of the air would be right next to the ground. Clearly that assumption must be incorrect, so you need to try to figure out something else that would give you the most probable energy, and it should be a function of temperature.




#3
Jan606, 11:19 AM

P: 96

So what is incorrect? The function is OK (at least that's what the textbook says).
What the function is saying is that the probability that the energy is in the interval of (0,dE) }(where dE is 1 for example) is higher than the probability that the energy is in the interval (E,E+dE) (where E>0). But the maxwell speed distribution on the other hand says that the probability that the speed (and therefore also the kinetic energy) is in the interval (0,dv) is lower than the probability tha the speed is in the interval (v, v+dv) (where v>0 and v< mean value). 



#4
Jan1006, 06:41 AM

P: 143

energy distribution and Maxwell speed distribution
There is only one state with energy zero, but in three dimensions there is a "sphere" of states with energies larger than zero. The surface of this sphere increases with energy.




#5
Jan1106, 05:32 AM

Sci Advisor
HW Helper
P: 11,863

I hope you mean [itex] \exp\left(\beta \epsilon\right) [/itex] with [itex] \beta=\frac{1}{kT} [/itex], right...?
Daniel. 



#6
Jan1106, 06:17 AM

Sci Advisor
HW Helper
P: 2,004

Expanding on what Pieter said... First, let's consider the difference between speed and velocity. Velocity is a vector which has magnitude and direction. Speed is a number with no direction. It is the magnitude of the velocity. If no external field exists, then the MaxwellBoltzmann distribution becomes the simpler Maxwell distribution: [tex]f(\vec v)=C \exp(\frac{1}{2}mv^2 \beta)[/tex] with C some constant which can be calculated and [itex]\beta =1/kT[/itex]. This f gives the probability of picking a particle with velocity components between [itex]v_x,v_y,v_z[/itex] and [itex]v_x+dv_x,v_y+dv_y,v_z+dv_z[/itex] respectively (think of this range of velocities as a volume element in velocityspace). Also [itex]f(\vec v)[/itex] is the number of particles with velocities in the range [itex]d\vec v[/itex] about [itex]\vec v[/itex]. For a large number of particles there's no difference. As you can see, the probability of getting zero velocity ([itex]\vec v = \vec 0[/itex]) is the highest. But now we want to know how the speed is distributed. Let's call the speed distribution F (F=F(v)). Since the velocity distribution is spherically symmetrical (all directions equal) the points with equal speed will be a spherical surface. So the number of particles with a particular speed is between v and v+dv is [itex]F(v)dv=f(\vec v)dv \cdot 4\pi v^2[/itex]. This will give you the speed distribution: [tex]F(v)=C' v^2 \exp(\frac{1}{2}mv^2 \beta)[/tex] which is zero for v=0, but now you see where it comes from. 


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