- #1
HJ Farnsworth
- 128
- 1
Greetings,
On page 7 of Landau and Lifschitz Vol. 1 3rd. Ed, it says
[itex]L(v'^2)=L(v^2+2\bf{v}\cdot\bf{ε}+\bf{ε}^2)[/itex].
They then Taylor expand in powers of ε, getting (ignoring second order terms and higher)
[itex]L(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}2\bf{v}\cdot\bf{ε}[/itex].
The [itex]\frac{\partial L}{\partial v^2}[/itex] confuses me. We effectively have a function of the form
[itex]f(g(x))=f(a+bx+cx^2)[/itex], which, Taylor expanding around [itex]x=0[/itex], would give
[itex]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[/itex].
So, where I have [itex]\frac{df}{dg}(x=0)[/itex], Landau has [itex]\frac{\partial f}{\partial a}[/itex].
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
[itex]L(v'^2)=L(v^2+2\bf{v}\cdot\bf{ε}+\bf{ε}^2)[/itex].
They then Taylor expand in powers of ε, getting (ignoring second order terms and higher)
[itex]L(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}2\bf{v}\cdot\bf{ε}[/itex].
The [itex]\frac{\partial L}{\partial v^2}[/itex] confuses me. We effectively have a function of the form
[itex]f(g(x))=f(a+bx+cx^2)[/itex], which, Taylor expanding around [itex]x=0[/itex], would give
[itex]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[/itex].
So, where I have [itex]\frac{df}{dg}(x=0)[/itex], Landau has [itex]\frac{\partial f}{\partial a}[/itex].
How do I go from what I have to what Laundau has?
Thanks.
-HJ Farnsworth