Why is the most probable energy different from the speed?

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SUMMARY

The discussion centers on the distinction between the most probable energy and the most probable speed in the context of the Maxwell-Boltzmann distribution. It is established that the most probable energy level, given by the formula $$E_{max} = \frac{k_BT}{2}$$, does not equal the most probable speed squared multiplied by mass, as indicated by $$v_{max} = \sqrt{\frac{2k_BT}{m}}$$. Participants clarify that while energy is related to speed, the distribution of particles across energy levels and speeds can lead to different most probable values. The conversation emphasizes the importance of understanding the mathematical relationships and distributions involved.

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  • #31
The complete solution is in #13. I still don't know, what else you need.
 
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  • #32
vanhees71 said:
The complete solution is in #13. I still don't know, what else you need.

Never mind, I figured it out, thanks. My calcuations in post #30 are wrong, I didn't pay attention when doing them. I was overlooking the fact that the relation between ##dE## and ##dv## is a function of ##v## while the probability itself is also a function of ##v##. If the relation between ##dE## and ##dv## were a constant, then then ##E_{max}## would correspond to ##v_{max}## .

Is there actually a mathematical way to explain why the maximum probability density in terms of energy ##E_{max}## is exactly half the energy that corresponds to ##v_{max}##?
 
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  • #33
JohnnyGui said:
I figured it out, thanks.

It doesn't look like it, since you come right back with the same question:

JohnnyGui said:
Is there a mathematical way to explain why the maximum probability density in terms of energy ##E_{max}## is exactly half the energy that corresponds to ##v_{max}##?

The math has already been discussed quite sufficiently in this thread. I think the discussion has run its course.

Thread closed.
 
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