Sum of an infinitesimal variable

[Repetit]

Hey!

Can it be concluded generally that:

<br /> \sum_r dx_r = 0<br />

...because we are summing an infinitesimaly small variable a finite number of times, in contrast to an integral which is an infinite sum of infinitesimaly small variables? In one of my books a probability is given by:

<br /> p_r = \frac{1}{Z} Exp[-\beta E_r]<br />

... and in the next line they write that:

<br /> \sum_r dp_r = 0<br />

Does someone have an explanation to this?

 

[lalbatros]

Repetit,

No you are totally wrong. In general this is not zero.
It is zero in your example, because you are dealing with probabilities.

Remember that for any probability distribution we have:

\sum_r p_r = 1 (eq 1)

Assuming this sum extends over all possibilties, then this sum is a constant.
Therefore if you differentiate the probabilties, because for example a parameter of the problem changes (temperature for example for a velocity distribution in statistical thermodynamics), then the sum of all these variations must equal zero:

\sum_r dp_r = 0

This is a consequence of (eq 1).
For example, if you increase the temperature, the higher energy states get more populated, but this happens at the expense of lower energys states populations, because the total of the population must remain constant.

I assumed you were studying statistical mechanics, but this is true in any case.
In statistical mechanics,

\beta = \frac{1}{kT} is the reciprocal of the temperature T,

p_r is the probability

that the state of energy E_r is occupied,

and Z, the "partition function" is simply the normalisation factor of the distribution.

I hope you will enjoy,

Michel

 

[HallsofIvy]

Hey!

Can it be concluded generally that:

<br /> \sum_r dx_r = 0<br />

I have never heard of "summing" a finite number of "infinitesmals". I can't imagine any good reason for doing so. Are you sure you are understanding the text correctly. Unless you are dealing with "non-standard" analysis you shouldn't even be talking about "infinitesmals" except as a shorthand for derivatives and integrals!

...because we are summing an infinitesimaly small variable a finite number of times, in contrast to an integral which is an infinite sum of infinitesimaly small variables? In one of my books a probability is given by:

<br /> p_r = \frac{1}{Z} Exp[-\beta E_r]<br />

... and in the next line they write that:

<br /> \sum_r dp_r = 0<br />

Does someone have an explanation to this?

Sometimes probability text will do strange things but there should not be a "dp" without an integral connected!

 

[Repetit]

Thanks for the answers to both of you. I see now that the sum must be zero because p_r is a probability distribution. And lalbatros you were right, I am studying statistical mechanics.

HallsofIvy:
Yes it seems strange that summing differentials (what I called infinitesimals before, but isn't it the same?) can make any sense, but look at the following example:

<br /> E=\sum_r p_r E_r<br />

Where E is the mean energy. The differential element dE would then be given by:

<br /> dE=\sum_r p_r dE_r + \sum_r E_r dp_r<br />

I suppose that would be alright then? Sums of differentials are everywhere in the book and I'm pretty sure that I understand it correctly. It's the book Statistical Physics by F. Mandl page 84-85 if you want to check it for yourself.

 

[arildno]

By which "r" has now changed from being an index on a finite set, to an index on an infinite set.

 

[Repetit]

By which "r" has now changed from being an index on a finite set, to an index on an infinite set.

Could you elaborate on that? Isn't there still a finite number of terms in the sum, and is r not still the same index variable indexing the (finite) number of energy levels?