Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Electromagnetism
Boltzmann vs Maxwell distribution?
Reply to thread
Message
[QUOTE="Toby_phys, post: 5788212, member: 591646"] So I worked through the Boltzmann distribution and got: $$ P\propto e^{\frac{-E}{k_BT}} $$ Where $E$ is the energy. So surely this means the kinetic energy (and therefore speed) of particles is distributed over a Boltzmann distribution. Or in equation: $$ P\propto e^{\frac{-mv^2}{2k_BT}} $$ However, this is not what my book says. The books says velocity is distributed in this way. They appear to have split up the components of energy Speed is distributed as follows, as a Maxwell distribution. $$ P\propto v^2e^{\frac{-mv^2}{2k_BT}} $$ I understand that the probability of the speed is the sum of all the different degrees of freedom and such. However, why can you split up the components of the kinetic energy is this way? Note, I am not asking how to get from Boltzmann to Maxwell, I am asking why velocity, and not kinetic energy, is Boltzmann distributed? [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Electromagnetism
Boltzmann vs Maxwell distribution?
Back
Top