Prove d (\delta x) = \delta (d x): Solutions & Tips

[praharmitra]

Proof of something...

I posted this in the calculus section...but no reply... I came across this while studying the most general noether's theorem which I will later apply to quantum field theory. So I am posting this here, hoping that some one would have come across this while studying qft

I was learning noether's theorem and I came up with something that I can't seem to prove.

Prove -

$$d (\delta x) = \delta (d x)$$

I was trying to prove the more general expression

$$d (\delta f) = \delta (d f)$$

where f( x, y, z, t, ...) is a function of n variables.

If i could prove the first then the second is proved...any help?

 

[malawi_glenn]

I have never tried to prove it, I thought it should be taken for granted.

now to a mathematician, you have to specify what \delta is I think...

but this is how i think of it:

$$\delta f(x) = f(x+\delta x) - f(x)$$

$$d(\delta f(x)) = d(f(x+\delta x) - f(x)) = \dfrac{df(x+\delta x )}{dx}dx - \dfrac{df(x )}{dx}dx$$

$$\delta (df(x)) = d(f(x+\delta x) - f(x)) = \dfrac{df(x+\delta x )}{dx}dx - \dfrac{df(x )}{dx}dx$$

But as I said, this is how I see it, I don't think it is a "proof"

 

[AEM]

I have never tried to prove it, I thought it should be taken for granted.

now to a mathematician, you have to specify what \delta is I think...

but this is how i think of it:

$$\delta f(x) = f(x+\delta x) - f(x)$$

$$d(\delta f(x)) = d(f(x+\delta x) - f(x)) = \dfrac{df(x+\delta x )}{dx}dx - \dfrac{df(x )}{dx}dx$$

$$\delta (df(x)) = d(f(x+\delta x) - f(x)) = \dfrac{df(x+\delta x )}{dx}dx - \dfrac{df(x )}{dx}dx$$

But as I said, this is how I see it, I don't think it is a "proof"

You are right that one must be careful in specifying what $$\delta$$ is, particularly in the context of Nother's theorem where $$\delta L$$ is often used to denote the variational derivtaive of the Lagrangian. For what it's worth, on pages 89 and 90 of Anderson's Principles of Relativity Physics , he takes pains to use a bar over $$\delta$$to distinguish between the two uses. Furthermore, he states (without proof) the relationship you just derived. Personally, your argument convinces me (but then I'm a lousy mathematician )

 

[malawi_glenn]

Yeah, most physicsists are louse mathematicians, so am I, but this is the way I convince myself ;-)

 

[lightarrow]

I posted this in the calculus section...but no reply... I came across this while studying the most general noether's theorem which I will later apply to quantum field theory. So I am posting this here, hoping that some one would have come across this while studying qft

I was learning noether's theorem and I came up with something that I can't seem to prove.

Prove -

$$d (\delta x) = \delta (d x)$$

I was trying to prove the more general expression

$$d (\delta f) = \delta (d f)$$

where f( x, y, z, t, ...) is a function of n variables.

If i could prove the first then the second is proved...any help?

As they have written, it's necessary to specify what $$\delta f$$ means, but you also have to specify what $$df$$ means.

Usually, but not always, (or, if you prefer, just "sometimes") $$\delta f$$ means to compute the variation of a definite integral, varying a function to which the integrand function depends. Example, you have the action S:

$$S = \int_{t_1} ^{t_2} L[q(t),q'(t),t]\ dt$$

If you want to compute the variation of S when you varies the function q(t) for every value of t, without changing the extremes of integration, you will write:

$$\delta S = \delta \int_{t_1} ^{t_2} L[q(t),q'(t),t]\ dt\ =\ \int_{t_1} ^{t_2} \delta L[q(t),q'(t),t]\ dt\ =\ \int_{t_1} ^{t_2}\ [\frac{\partial\ L}{\partial\ q} q'(t)\ +\ \frac{\partial\ L}{\partial\ q'}q''(t)]\ dt$$

(If you wanted to prove the Euler-Lagrange equation, you have to fix the values $$q(t_1)$$ and $$q(t_2)$$ in the evaluation of that integral).

Now, what does dS mean, in this context? Of course nothing, since it's a number. It has a meaning if, for example, you write:

$$S(t) = \int_{t_1} ^{t} L[q(\tau),q'(\tau),\tau]\ d\tau$$

But then you evaluate $$\delta S$$ in a completely different way (I let it to you as exercise).