- #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)##