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I'm teaching a course using D. V. Schroeder, An Introduction to Thermal Physics, and there is a "derivation" in the book that is making me cringe a bit. I would like the opinion of mathematicians on the subject.
Take a (continuous) degree of freedom ##q## from which you can get the energy ##E(q) = cq^2##. Then artificially take ##q## to be discrete, with equally-spaced intervals ##\Delta q##. This allows you to write the partition function as the sum
$$
Z = \sum_q e^{-\beta E(q)} = \sum_q e^{-\beta c q^2},
$$
which you multiply by ##\Delta q / \Delta q##,
$$
Z = \frac{1}{\Delta q} \sum_q e^{-\beta c q^2} \Delta q.
$$
Then take ##\Delta q## to be small, such that
$$
Z = \frac{1}{\Delta q} \int_{-\infty}^{\infty} e^{-\beta c q^2} \, dq.
$$
It is this last step that I don't like at all. I don't like how ##\Delta q## is made infinitesimally small at one place but not the other.
Also, since ##q## is continuous, I would have taken directly the partition function as
$$
Z = \int_{-\infty}^{\infty} e^{-\beta c q^2} \, dq,
$$
i.e., without the ##1 / \Delta q##. In the end, that factor cancels out, so the answer is the same.
Any pointers on how to correctly transform a summation into an integral will be appreciated.
Take a (continuous) degree of freedom ##q## from which you can get the energy ##E(q) = cq^2##. Then artificially take ##q## to be discrete, with equally-spaced intervals ##\Delta q##. This allows you to write the partition function as the sum
$$
Z = \sum_q e^{-\beta E(q)} = \sum_q e^{-\beta c q^2},
$$
which you multiply by ##\Delta q / \Delta q##,
$$
Z = \frac{1}{\Delta q} \sum_q e^{-\beta c q^2} \Delta q.
$$
Then take ##\Delta q## to be small, such that
$$
Z = \frac{1}{\Delta q} \int_{-\infty}^{\infty} e^{-\beta c q^2} \, dq.
$$
It is this last step that I don't like at all. I don't like how ##\Delta q## is made infinitesimally small at one place but not the other.
Also, since ##q## is continuous, I would have taken directly the partition function as
$$
Z = \int_{-\infty}^{\infty} e^{-\beta c q^2} \, dq,
$$
i.e., without the ##1 / \Delta q##. In the end, that factor cancels out, so the answer is the same.
Any pointers on how to correctly transform a summation into an integral will be appreciated.