From summation to integral

In summary, the derivation in the book is causing a mathematicians to cringe because they don't like how the Δq factor is introduced. However, if you don't mind using a different method of summation, then the order of growth of the function is still correctly represented.
  • #1
DrClaude
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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.
 
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  • #2
What's your definition of an integral? If you're fine with Riemann integration, then it's already a shorthand for a limit of discrete summations.
 
  • #3
economicsnerd said:
What's your definition of an integral? If you're fine with Riemann integration, then it's already a shorthand for a limit of discrete summations.

I have no problem with a sum becoming an integral. I am only unsure about the proper way of doing it.

With a Riemann sum, there is already a factor ##\Delta x## that becomes ##dx## when going to the Riemann integral. What we have here is a normal sum, which is taken over a continuous variable. Do you simply replace the sum with an integral and the ##dx## that comes with it, or do you have to introduce it in a contorted way as above?
 
  • #4
DrClaude said:
Take a (continuous) degree of freedom ##q## from which you can get the energy ##E(q) = cq^2##.

What would that mean, mathematically? If [itex] q [/itex] is a variable then how can the expected value of [itex] q [/itex] be a function of [itex] q [/itex]? If the derivation is to make mathematical sense, shouldn't this be the first thing to straighten out? Or is [itex] E(q) =c q^2 [/itex] intended only as the definition of the function [itex] E() [/itex] ?
 
  • #5
It is going to be true that if [itex] \delta q [/itex] is very small, then
[tex] \frac{1}{\delta q} \sum_{q} e^{-c\ beta q^2} \delta q \approx \frac{1}{\delta q} \int_{-\infty}^{\infty} e^{-c\beta q^2} dq [/tex]
where the approximation is saying that the relative error is going to zero, i.e. the ratio of the two is going to 1. Since you are dividing/multiplying this by something else that has a [itex]\delta q[/itex] in it, then you are only really interested in the ratio going to 1, not that they are absolutely converging to each other.

Alternatively, you can simply not take the limit until after you have canceled the extra [itex] \delta q[/itex] factors.

Your third choice is to start off by not writing down something which goes to infinity as [itex] \delta q[/itex] goes to zero, but that might be harder to motivate (I don't know too much about the subject).
 
  • #6
When 1/Δq and Δq are introduced, it is not immediately clear that the summation is well behaved. Soon it is clear that the summation with the Δq factor is approaching a limit, L, (the integral value). Then, if you are only interested in the order of growth of the function as Δq goes to zero, you are free to go ahead and replace the summation with it's limit value, L. So we end up with a good representation of the order of growth of the function when Δq goes to zero. Apparently that is what they need.
 

1. What is the purpose of going from summation to integral?

The purpose of going from summation to integral is to approximate the area under a curve by breaking it into smaller rectangles or trapezoids, and then taking the limit as the width of these rectangles approaches zero. This allows us to solve more complex problems and calculate exact values rather than just approximations.

2. How is the summation symbol related to the integral symbol?

The summation symbol (∑) and the integral symbol (∫) are both used to represent the addition of many small values. The main difference is that the summation symbol is used when the values being added are discrete, while the integral symbol is used when the values being added are continuous.

3. What is the relationship between Riemann sums and integrals?

Riemann sums are a way to approximate the area under a curve by dividing it into smaller rectangles or trapezoids. As the number of rectangles increases, the Riemann sum approaches the exact value of the integral. In essence, integrals are the limit of Riemann sums as the width of the rectangles approaches zero.

4. Can any function be represented by an integral?

Yes, any continuous function can be represented by an integral. The integral is a fundamental tool in calculus that allows us to find the area under a curve and solve a variety of problems. However, some functions may have more complex integrals that cannot be easily calculated.

5. What are some real-world applications of going from summation to integral?

Going from summation to integral has many real-world applications, including calculating the distance traveled by a moving object, the volume of a 3D shape, the work done by a variable force, and the population growth of a species. It is also used in economics, physics, engineering, and many other fields to solve complex problems and make accurate predictions.

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