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Feb 2, 2016 at 7:00 history edited user36790 CC BY-SA 3.0
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Nov 22, 2015 at 13:10 comment added TheQuantumMan @grizzlyadam so in the end, they are not independent but can be treated as much because the mathematics prove it, right?
Jan 14, 2011 at 15:57 comment added wsc "Not a word of Landau, not a thought of Lifshitz."
Nov 17, 2010 at 19:55 comment added Sklivvz Landau is a great mathematical physicist but he's not known as a simple writer :-)
Nov 17, 2010 at 17:02 comment added grizzly adam Yes, the notation is confusing. That's another problem.
Nov 17, 2010 at 9:34 comment added Greg Graviton ... and to express the variations $\delta u$ and $\delta v$ in terms of $\delta x$ afterwards.
Nov 17, 2010 at 9:33 comment added Greg Graviton The notation for the arguments $L$ is somewhat confusing, in which case it is instructive to consider the following example: take $F(x,2x-y)$ and vary $F(x+\delta x,2(x+\delta x)-y) = \frac{\partial F}{\partial x}\delta x + \frac{\partial F}{\partial (2x-y)}2\delta x$. You might say that the arguments of $F$ are varied independently, but that sounds weird. If anything, it's just that the notation for the partial derivatives of $F$ is bad; it is much better to write $F(u,v)$ and $(u,v)=(x,2x-y)$ to obtain $\delta F = \frac{\partial F}{\partial u}\delta u + \frac{\partial F}{\partial v}\delta v$
Nov 17, 2010 at 9:30 comment added Greg Graviton Well, instead of saying "$q$ and $\dot q$ are varied independently", you could also say "$q$ and $\dot q$ are varied (perhaps independently, perhaps not)" and later note that the variation $\delta\dot q$ is given by $\delta\dot q = \frac d{dt} \delta q$.
Nov 17, 2010 at 6:25 history edited grizzly adam CC BY-SA 2.5
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Nov 17, 2010 at 6:19 history answered grizzly adam CC BY-SA 2.5