# Difficulty with Lagrange multipliers in Kardar's Statistical Physics book

• AndreasC
In summary, the author is trying to solve the Euler Lagrange equation of the variational principle, but gets stuck because he cannot understand how to get the result. He starts by understanding the equation and then tries to solve it using the variational principle, but runs into problems because he cannot understand the variation of the entropy. He eventually solves it by taking the variation of the entropy directly.

#### AndreasC

Gold Member
Homework Statement
You are given an entropy function in terms of a probability density function, and you are asked to maximize said entropy function in terms of the density function, under the constraint of constant energy. The details aren't very relevant, what I don't understand is a leap in the mathematics of the solution.
Relevant Equations
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Alright, so I did some progress and then I got stuck. After some time I went to check the solution. Up to some point, it's all well and good:

I understand everything that is happening up to the point where he takes the partial derivative of S wrt ρ(Γ). I don't understand how he gets the result he does. I imagine it's just a calculus thing that I am failing really bad at right now. Damn book always demanding you to know how to do things >:(

I think it's the Euler-Lagrange equation from the calculus of variations,
$$\frac{d}{d\Gamma}\left(\frac{\partial S}{\partial \frac{d\rho}{d\Gamma}}\right) - \frac{\partial S}{\partial \rho} = 0$$ where the integrand is independent of $\frac{d\rho}{d\Gamma}$; this is combined with an abuse of notation which identifies $S$ with the integrand $$\rho (-\ln \rho - \alpha - \beta \mathcal{H}(\Gamma))$$ which is easily differentiated with respect to $\rho$.

AndreasC
pasmith said:
I think it's the Euler-Lagrange equation from the calculus of variations,
$$\frac{d}{d\Gamma}\left(\frac{\partial S}{\partial \frac{d\rho}{d\Gamma}}\right) - \frac{\partial S}{\partial \rho} = 0$$ where the integrand is independent of $\frac{d\rho}{d\Gamma}$; this is combined with an abuse of notation which identifies $S$ with the integrand $$\rho (-\ln \rho - \alpha - \beta \mathcal{H}(\Gamma))$$ which is easily differentiated with respect to $\rho$.
Abuse of notation is what I thought too... But I'm not quite sure. And it doesn't make sense any other way. You have a valid point about the Euler-Lagrange equation, I didn't think of that.

Let ##F[\rho (\Gamma)] = \rho (\Gamma) \ln \rho (\Gamma) + \alpha \rho (\Gamma) + \beta \mathcal H(\Gamma) \rho (\Gamma)##

so

##S = \alpha +\beta E-\int d\Gamma F[\rho (\Gamma)]##

When ##S## is a maximum, ##S## will not change to first order for arbitrary variations of ##\rho (\Gamma)##. That is, ##\delta S = 0## to first order in ##\delta \rho## when ##\rho (\Gamma) = \rho_{\rm max} (\Gamma)##.

Note that to first order in ##\delta \rho(\Gamma)##, ##\delta S = -\int d\Gamma \large \frac {\partial F}{\partial \rho (\Gamma)} \normalsize\delta \rho (\Gamma)##.

For this to be zero for arbitrary ##\delta \rho(\Gamma)##, we must have ##\large \frac {\partial F}{\partial \rho (\Gamma)} \normalsize \bigg|_{\rho = \rho_{max}}= 0##.

What do you get for ##\large \frac {\partial F}{\partial \rho (\Gamma)} \normalsize ?##

Abhishek11235, vanhees71 and AndreasC
It's not an abuse of notation, it's plain wrong. What you have to take is the variation of the entropy, which is a functional (not function!) of ##\rho##. What you then get is the "Euler Lagrange equation" of the variational principle. Here, however, it's easier to just take the variation wrt. ##\rho## directly. First of all, it should be, including the Lagrange multipliers for the constraints
$$S=\int \mathrm{d} \Gamma \rho [-\ln \rho-\alpha - \beta \mathcal{H}].$$
I don't know what the additional terms are good for.

Then, just take the variation of this functional of ##S## wrt. ##\rho##. This gives
$$S=\int \mathrm{d} \Gamma \left \{ \delta \rho [-\ln \rho - \alpha -\beta \mathcal{H}]-\rho \frac{1}{\rho} \delta \rho \right \} = \int \mathrm{d} \Gamma \delta \rho [-\ln \rho - \alpha -\beta \mathcal{H}-1].$$
Now thanks to the Lagrange multipliers to get the maximum entropy the bracket must vanish, because you can treat ##\rho## as an independent function, because you can adjust the constraints by choosing the right Lagrange multipliers. This leads to
$$-\ln \rho - (\alpha+1)-\beta \mathcal{H}=0$$
or solved for ##\rho##
$$\rho = \exp(-\alpha-1) \exp(-\beta \mathcal{H}).$$
From the first constraint you get
$$\exp(-\alpha-1)=\frac{1}{Z} \quad \text{with} \quad Z=\int \mathrm{d} \Gamma \exp(-\beta \mathcal{H}).$$
So finally you get the canonical distribution
$$\rho=\frac{1}{Z} \exp(-\beta \mathcal{H}).$$

atyy

## 1. What are Lagrange multipliers and why are they important in Kardar's Statistical Physics book?

Lagrange multipliers are mathematical tools used to optimize a function subject to a set of constraints. In Kardar's Statistical Physics book, they are used to find the extrema of a thermodynamic potential while satisfying the constraints of the system.

## 2. Why do many students struggle with understanding Lagrange multipliers in Kardar's book?

Many students struggle with understanding Lagrange multipliers in Kardar's book because it requires a strong understanding of calculus and mathematical optimization techniques. Additionally, the concept of constraints can be difficult to grasp for some students.

## 3. How can I improve my understanding of Lagrange multipliers in Kardar's book?

To improve your understanding of Lagrange multipliers in Kardar's book, it is important to have a strong foundation in calculus and mathematical optimization. It may also be helpful to work through practice problems and seek clarification from a tutor or instructor if needed.

## 4. Are there any real-world applications of Lagrange multipliers discussed in Kardar's book?

Yes, Lagrange multipliers have many real-world applications in fields such as economics, physics, and engineering. In Kardar's book, they are used to solve problems related to thermodynamics and statistical mechanics.

## 5. Can I skip over the section on Lagrange multipliers in Kardar's book and still understand the rest of the material?

No, it is not recommended to skip over the section on Lagrange multipliers in Kardar's book. They are an important concept in statistical physics and are used throughout the book to solve various problems. It is important to have a solid understanding of Lagrange multipliers in order to fully grasp the material in the rest of the book.