Derivation of the average translational kinetic energy of a molecule

AI Thread Summary
The discussion centers on the derivation of the average translational kinetic energy of a molecule, specifically questioning the addition of velocity components. The user is confused about the textbook's assertion that the average of the squared velocities in three dimensions can be simplified to three times the average of one component's squared velocity. Clarification is provided that while the square of the magnitude of the velocity vector is indeed the sum of the squares of its components, when averaging, the equality of the averages allows for simplification. The key point is that the average squared velocity can be expressed in terms of one component due to their equal contributions in a three-dimensional isotropic system. Understanding this concept resolves the confusion regarding the addition of vector components in the context of kinetic energy.
NeuronalMan
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Homework Statement



Hello, this is not actually a homework problem. I just can't seem to understand the derivation of the average translational kinetic energy of a molecule. I am startled by the way the velocities are added.


Homework Equations




My undergraduate level textbook says that (vx^2)av = (vy^2)av = (vz^2)av, but then it says (v^2)av = (vx^2)av + (vy^2)av + (vz^2)av = 3(vx^2)av

How can velocities being vectors be added this way? I am surely missing something here.


The Attempt at a Solution

 
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I thinking you are missing the importance of subscripts. It is certainly true that the square of the magnitude of the 3-d velocity vector is always

v2=vx2+vy2+vz2

what does this become when you replace the components with their averages which are all equal to each other?
 

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