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

• praharmitra
In summary, the more general expression d (\delta f) means the variation of a definite integral, varying a function to which the integrand function depends, where f( x, y, z, t, ...) is a function of n variables.
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?

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"

malawi_glenn said:
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 )

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

praharmitra said:
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).

## What is the definition of "Prove d (\delta x) = \delta (d x)"?

The statement "Prove d (\delta x) = \delta (d x)" is a mathematical expression that represents the equality between the derivative of a function (d) and the delta function (δ) applied to the derivative of a variable (x).

## Why is proving the equality between d (\delta x) and \delta (d x) important?

This proof is important because it helps to establish the fundamental relationship between derivatives and delta functions, which are both essential concepts in calculus and mathematical analysis.

## What are the different methods for proving this equality?

There are several methods for proving the equality between d (\delta x) and \delta (d x). These include using the definition of the derivative, using the properties of delta functions, and using integration by parts.

## What are some common challenges when attempting to prove this equality?

One common challenge when trying to prove d (\delta x) = \delta (d x) is understanding and properly applying the properties and definitions of derivatives and delta functions. Another challenge can be manipulating and simplifying the complex mathematical expressions involved in the proof.

## Are there any tips or strategies for successfully proving this equality?

Some tips for successfully proving d (\delta x) = \delta (d x) include carefully reviewing the definitions and properties of derivatives and delta functions, using algebraic manipulation to simplify the expressions, and breaking the proof into smaller steps to make it more manageable.

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