- #1
HJ Farnsworth
- 126
- 1
Greetings,
On page 7 of Landau and Lifschitz Vol. 1 3rd. Ed, it says
L(v'^2)=L(v^2+2\bf{v}\cdot\bf{ε}+\bf{ε}^2).
They then Taylor expand in powers of ε, getting (ignoring second order terms and higher)
L(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}2\bf{v}\cdot\bf{ε}.
The \frac{\partial L}{\partial v^2} confuses me. We effectively have a function of the form
f(g(x))=f(a+bx+cx^2), which, Taylor expanding around x=0, would give
f(a+bx+cx^2)=f(x=0)+\frac{df}{dg}(x=0)\frac{dg}{dx}(x=0)x=\frac{df}{dg}(x=0)bx.
So, where I have \frac{df}{dg}(x=0), Landau has \frac{\partial f}{\partial a}.
How do I go from what I have to what Laundau has?
Thanks.
-HJ Farnsworth
On page 7 of Landau and Lifschitz Vol. 1 3rd. Ed, it says
L(v'^2)=L(v^2+2\bf{v}\cdot\bf{ε}+\bf{ε}^2).
They then Taylor expand in powers of ε, getting (ignoring second order terms and higher)
L(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}2\bf{v}\cdot\bf{ε}.
The \frac{\partial L}{\partial v^2} confuses me. We effectively have a function of the form
f(g(x))=f(a+bx+cx^2), which, Taylor expanding around x=0, would give
f(a+bx+cx^2)=f(x=0)+\frac{df}{dg}(x=0)\frac{dg}{dx}(x=0)x=\frac{df}{dg}(x=0)bx.
So, where I have \frac{df}{dg}(x=0), Landau has \frac{\partial f}{\partial a}.
How do I go from what I have to what Laundau has?
Thanks.
-HJ Farnsworth