What is the meaning of 'delta' in variational calculations?

[jostpuur]

I always like to do the variational calculations in rigor way for example like this

$$0 = D_{\alpha} \Big(\int dt\; L(x(t)+\alpha y(t))\Big)\Big|_{\alpha=0} = \cdots$$

because this way I understand what is happening. However in literature I keep seeing the quantity

$$\delta x(t)$$

being used most of the time. What does this delta mean really? Does it have a rigor meaning? It seems to be same kind of mystical* quantity as the $df$, but this time an... infinite dimensional differential?

*: mystical in the way, that even if the rigor meaning exists, it is not easily available, and the concept is usually used in non-rigor way.

 

[Ben Niehoff]

According to my copy of Goldstein's Classical Mechanics (3rd ed., p.38), the variation is defined as:

$$\delta y = \left \frac{\partial y}{\partial \alpha} \right|_{\alpha = 0} d\alpha$$

where

$$y(x,\alpha) = y(x,0) + \alpha \eta(x)$$

for some arbitrary function $\eta$.

I share your sentiment that the mathematics of physics ought to be done in a rigorous way...for my own edification, I went through the variational calculus bits of Goldstein's and proved them all myself. It's actually not too hard, despite the fact that function space is infinite-dimensional.

 

[jostpuur]

Now $\delta y$ depends on the chosen $\eta$. If a notation $\delta y_{\eta}$ was given, would that be analogous to $dx_i$ where i=1,2,3?

 

[Ben Niehoff]

No, because $\eta(x)$ is not really an "index". $\eta(x)$ is just an arbitrary function; it is simply left undefined exactly what sort of function it is. You never "choose" it. When you work through the details of the math, you get to a point where the definition "$\eta(x)$ is arbitrary" can be used to advance the proof. Specifically, you will get to a point where you have

$$\int_{x_1}^{x_2} M(x)\eta(x)\, dx = 0$$

which, if it is true for all functions $\eta(x)$, must therefore imply

$$M(x) = 0$$

And, setting M(x)=0 allows you to derive the Euler equations (which are generic to the calculus of variations, and have nothing specific to do with the Lagrangian formalism).

 

[jostpuur]

There was a mistake with my analogy. In three dimension we can take a gradient of a function with expression $u\cdot\nabla f$, where $u=(u_1,u_2,u_3)$ is some vector. The mapping $x\mapsto \eta(x)$ should be considered analogous to the mapping $i\mapsto u_i$. So $\eta$ alone, is analogous to $u$, and not to the indexes.

Anyway there are several things that don't make sense to me. What is the $d\alpha$?

If $\eta$ is arbitrary, then $\delta y$, whatever it is, at least seems to be arbitrary too.

What did the "arbitrary" mean? I would like to have some kind of parameter interpretation for it. For example we might say that x is arbitrary in f(x), but we don't really want to mean an arbitrary number f(x), but instead a mapping $x\mapsto f(x)$.

 

[Ben Niehoff]

Anyway there are several things that don't make sense to me. What is the $d\alpha$?

$\alpha$ is just a parameter, so $d\alpha$ is just the (ordinary) differential of $\alpha$. It ends up going under an integral sign.

If $\eta$ is arbitrary, then $\delta y$, whatever it is, at least seems to be arbitrary too.

What did the "arbitrary" mean? I would like to have some kind of parameter interpretation for it. For example we might say that x is arbitrary in f(x), but we don't really want to mean an arbitrary number f(x), but instead a mapping $x\mapsto f(x)$.

$\eta$ is arbitrary in the same sense that a constant of integration is arbitrary:

$$\int f(x) dx = F(x) + C$$

Here, C is arbitrary, meaning that you can choose any number to be C, and the equation is true.

The only difference is that $\eta(x)$ is a function. So, "arbitrary" in this case means that $\eta(x)$ can be any function. (Actually, there is a slight restriction, in that $\eta(x_1) = 0$ and $\eta(x_2) = 0$, but $\eta(x)$ can take on any value whatsoever at other points).

 

[jostpuur]

$\alpha$ is just a parameter, so $d\alpha$ is just the (ordinary) differential of $\alpha$. It ends up going under an integral sign.

But we are not integrating over alpha anywhere in the calculation.

 

[jostpuur]

This started very confusingly. It could be smarter to not try to understand everything that got already said...