# Derivation of Lagrangian in the calculus of variations

• Joggl
In summary, the conversation discusses the application of the chain rule in calculating the derivative of a function involving variables ##q## and ##\epsilon##. The conversation also addresses the evaluation of the arguments of the operator ##\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)}## at ##\epsilon = 0## and suggests using the Taylor-Maclaurin series expansion to write ##L(q+\epsilon \psi)## in terms of derivatives of ##L## with respect to ##q##.
Joggl
Homework Statement
##\frac {d} {d\epsilon}\left. L(q+\epsilon \psi) \right|_{\epsilon = 0} ##

where the index ##\epsilon = 0## means the derivative of the function evaluated at this point.
Relevant Equations
##q=q(t)##
##\psi = \psi (q(t), \dot{q}(t), t)##
Hello. In a chapter of a book I just read it is given that

##\frac {d} {d\epsilon}\left. L(q+\epsilon \psi) \right|_{\epsilon = 0} = \frac {\partial L} {\partial q} \psi ##

While trying to get to this conclusion myself I've stumbled over some problem.
First I apply the chain rule:

##\left.(\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)} \frac {d (q+\epsilon \psi)} {d \epsilon}) \right|_{\epsilon = 0} ##

The second part of the product can be evaluated since ##q## and ##\psi## do not depend on ##\epsilon##:

##\frac {d (q+\epsilon \psi)} {d \epsilon} = \psi##

##\frac {d} {d\epsilon}\left. L(q+\epsilon \psi) \right|_{\epsilon = 0} = \left.(\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)} \frac {d (q+\epsilon \psi)} {d \epsilon}) \right|_{\epsilon = 0} = \left.(\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)} \psi) \right|_{\epsilon = 0}##

And now, I don't know how to continue since I think it is not allowed to evaluate the arguments of the Operator ##\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)}## at ##\epsilon = 0##
In my opinion this evaluation has to be done after the derivation which I can't calculate since it's not clear how ##L## depends on ##(q+\epsilon \psi)##

Delta2
As you write it, the Lagrangian is a function only of the one variable. You shouldn't be writing a partial derivative. But to answer your question the two functions ##L'(q+\epsilon\psi)## and ##L'(q)## agree at ##\epsilon=0## where ##L'(y)=\frac{dL(y)}{dy}##.

Delta2
Can you think of a way to write ##L(q+\epsilon \psi)## in terms of derivatives of ##L## with respect to ##q##, given that ##\epsilon \psi## is very small?

Yes... look up Taylor-Maclaurian series expansion of a function in your Calculus 1 textbook.

## 1. What is the calculus of variations?

The calculus of variations is a branch of mathematics that deals with finding the optimal solution to a problem involving a functional. It involves finding the function that minimizes or maximizes a certain quantity, such as the length of a curve or the time taken for a particle to move between two points.

## 2. What is the Lagrangian in the calculus of variations?

The Lagrangian is a function that is used to represent the energy of a system in classical mechanics. In the calculus of variations, it is used to derive the equations of motion for a system by minimizing the action, which is the integral of the Lagrangian over time.

## 3. How is the Lagrangian derived in the calculus of variations?

The Lagrangian is derived by first defining the functional for the system, which is the integral of the system's energy over time. Then, by using the Euler-Lagrange equation, which is a necessary condition for an extremum, the Lagrangian can be found by taking the derivative of the functional with respect to the system's variables.

## 4. What is the significance of the Lagrangian in the calculus of variations?

The Lagrangian is significant because it allows for the derivation of the equations of motion for a system by minimizing the action. This is a powerful tool in classical mechanics and is used to solve a wide range of problems, from simple mechanics to complex systems in quantum field theory.

## 5. How is the calculus of variations used in other fields of science?

The calculus of variations has applications in various fields of science, including physics, engineering, economics, and biology. It is used to optimize systems and find the most efficient or optimal solutions. It is also used to model and analyze complex systems, such as in fluid dynamics and control theory.

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