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

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

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

<br /> \delta x(t)<br />

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...