Pathria page 2 -- total energy for mutually interacting particles

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SUMMARY

The discussion centers on the total energy of mutually interacting particles as described in Pathria's 'Statistical Mechanics.' For noninteracting particles, the total energy is simply the sum of individual energies, expressed as ##E = \sum_{i} n_{i}\epsilon_{i}##. However, for mutually interacting particles, additional internal energy from interactions must be considered, which cannot be captured by this summation. An example provided involves small charged dust particles, where electrostatic repulsion adds complexity to the total energy calculation.

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In page 2 of Pathria's textbook on 'Statistical Mechanics,' it is mentioned that

Consider the total energy ##E## of the system. If the particles comprising the system could be regarded as noninteracting, the total energy ##E## would be equal to the sum of the energies ##\epsilon_i## of the individual particles:

##E = \sum_{i} n_{i}\epsilon_{i},##

where ##n_i## denotes the number of particles each with energy ##\epsilon_i##.

If the particles were mutually interacting, the total energy ##E##
cannot be written in the form above.

Why can't the total energy be written in the form above for mutually interacting particles?
 
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failexam said:
Why can't the total energy be written in the above for mutually interacting particles?
Hi failexam:

Since no one has replied to your question for a while, I will try to be helpful, although I am very rusty about this topic.

The interaction of the particles implies that there is additional internal energy in these interactions that is not part of the energy of the individual particles. Here is an example. Suppose the particles are small dust particles, each with a very small electric charge, say all negative. Suppose the charge on each particle is so small that the electrostatic repulsive force between particle is about half the gravitational attractive force. The total energy is the sum of the energy of all the particles, plus additional energy in the electrostatic field.

The details in this scenario may not be quite right, but perhaps it will help you find some helpful and more accurate information on the internet.

Regards,
Buzz
 
If you try it, you will find you are double counting!
 

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