- #1

Joggl

- 1

- 1

- 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##

This leads to:

##\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)##

##\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##

This leads to:

##\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)##