Thermal Equilibrium: Exchange of Kinetic Energy

[boletoms]

as we know that atomic collisions are perfectly ellastic
perfectly ellastic collisions are given by
v1=u1(m1-m2)+2(m2)u2 / m1+m2
and v2=u2(m2-m1)+2(m1)u1 / m1+m2
when m1=m2
v1=u2
v2=u1

we know heat is stored in the form of vibrational energy(kinetic energy)
so to exchange energy they should collide
according to ellasticity the should exchange their energies
T1° will become T2° and T1° will become T2°
the how will they come to equillibrium

 

[Avodyne]

Your formulas apply only in one dimension. In one dimension, it is true that colliding particles will not thermalize.

 

[boletoms]

this is just a doubt in ellasticity and not thermodynamics
in 2d collisions
when m1=m2
v1=(u2+u1)cos(x/2)
v2=(u2-u1)sin(x/2) [scource : http://en.wikipedia.org/wiki/Elastic_collision ]
initial total energy(T.E)=1/2(m^2)(u1^2 +u2^2)
final T.E=1/2(m^2)((u2-u1)^2)

according to physics laws initial T.E should be equal to final T.E

hence , 1/2(m^2)(u1^2 +u2^2)=1/2(m^2)((u2-u1)^2)

(u1^2 +u2^2)=((u2-u1)^2)

which is only possible when u1 or u2 is equal to zero
but when u1 & u2 are not equal to zero they violate the law of conservation of energy

why is that so?

 

[Bill_K]

Please move this to the Classical Physics forum. This has nothing to do with Quantum Mechanics.

 

[boletoms]

Please move this to the Classical Physics forum. This has nothing to do with Quantum Mechanics.

could you tell me how to do it
i'm new to PF

 

[Avodyne]

Your formulas do not appear in the wikipedia article that you cite, and they are not correct.

 

[boletoms]

Your formulas do not appear in the wikipedia article that you cite, and they are not correct.

look for 2d collisions i just modified the relative initial velocity as u1+u2

and before that please substitute m1=m2

 

[Avodyne]

Your formulas are from the 1d section.