I have the following definition of the functional derivative $ \frac{\delta S}{\delta\gamma}$, where $S$ is my functional and $\gamma$ is a curve:

$$\tag{1} \int^B_A \frac{\delta S}{\delta\gamma} h(x) dx = \left. \frac{d}{d\epsilon} S[\gamma + \delta\gamma]\; \right|_{\epsilon = 0} $$

where $h(x)$ is an arbitrary function having the same fixed endpoints as $y$, $\epsilon$ is a small constant, and $\delta \gamma = \epsilon h$ is a small variation of the curve $\gamma$.

According to my notes the first variation is

$$\delta S = \int_A^B \frac{\delta S }{ \delta \gamma}\;\; \delta \gamma\;\; dx \tag{2}$$

which is to my eyes is simply the LHS of $(1)$ multiplied by $\epsilon$ (because $\delta \gamma = \epsilon h$).

If that is correct then I could instead multiply the RHS of $(1)$ to find

$$ \delta S = \epsilon \; \cdot \left( \left. \frac{d}{d\epsilon}\; S[\gamma + \delta\gamma]\; \right|_{\epsilon = 0} \right) \tag{3}$$

However, wiki defines the first variation (using $y$ as the function defining $\gamma$) as

$$ \delta S(y,h) = \left.\frac{d}{d\varepsilon} S(y + \varepsilon h)\right|_{\varepsilon = 0} \tag{4}$$

i.e. without the factor of $\epsilon$. I understand that in practice if one were looking for a stationary curve, setting $\delta S = 0$ would mean $(3) = (4) = 0$ since $\epsilon \neq 0$. But shouldn't $\delta S$ have this factor of $\epsilon$?

Wiki article on first variation


1 Answer 1


In fact the mess of definitions of the variation in literature is unmatched by anything I've seen thus far.

Anyway I believe you cannot multiply by $\epsilon$ since it is a differentially small element. Also you define the Gateaux derivative below there. For me it is more clear to look at a general family of functions and use the general variation (without the affine approach $y=\gamma+h \epsilon$)!

But as far as I can tell, what confuses you is the same thing I strugled with having seen many definitions: a common approach is to define $$ \delta q := \lim_{\epsilon\to\epsilon_0} \cfrac{\tilde{q}(\epsilon)-\tilde{q}(\epsilon_0)}{\epsilon-\epsilon_0} $$ which in terms of a function derivative $$ \frac{dy}{dx} := \lim_{x\to x_0} \cfrac{y(x)-y(x_0)}{x-x_0} $$ is lacking the term $$ \delta \epsilon \;.$$

This is the case for the wiki definition (the Gateaux derivative). Your definition above however explicitly deals with this quantity. In the end it's a matter of definition. Another method I know includes the variation of the parameter in the definition (so it never has to be written down): $$ \delta q := \left. \frac{\partial Q(\epsilon)}{\partial \epsilon}\right|_{\epsilon=\epsilon_0} \delta\epsilon $$ where $Q$ is the family of testing functions. I hope this helps somewhat.

  • $\begingroup$ Thank you very much! I posted this quite some time ago, before my mathematical physics exam. Good to get a clear answer and I'm sure it will help others. In the end I turned to Hilbert and Courant's book, my lecturer being unable to help. I agree very strongly, there are so many bad treatments of the calculus of variations, despite it often being the very first chapter in Lagrangian mechanics texts. $\endgroup$
    – Potkin57
    Commented Jan 7, 2017 at 19:28

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