# Still having difficulty in my introduction to fields

1. Jun 15, 2014

### grzz

1. The problem statement, all variables and given/known data
I was reading in a book that,

" For any function f(x) whose functional form changes from f(x) to f'(x), we can write
$\bar{\delta}$f(x) = f'(x) - f(x)
= {f'(x') - f(x)} - {f'(x') - f'(x)}
= $\delta$f(x) - $\partial$$_{μ}$f(x)$\delta$x$^{μ}$ "

I am understanding that the $\bar{\delta}$ represents only the change in the functional form at the same value of x and $\delta$ represents the total change.

My difficulty is in the term $\partial$$_{μ}$f(x)$\delta$x$^{μ}$.

2. Relevant equations

3. The attempt at a solution

If I expand {f'(x') - f'(x)} I get

f'(x$^{μ}$ + $\delta$x$^{μ}$) - f'(x$^{μ}$)
= f'(x$^{μ}$) + $\partial$$_{μ}$f'(x)$\delta$x$^{μ}$ - f'(x$^{μ}$)

= $\partial$$_{μ}$f'(x)$\delta$x$^{μ}$ .

Hence I get the the $\partial_{μ}$ of f'(x) and not of f(x) as the book says.

Can somebody tell me what am I missing?

2. Jun 15, 2014

### maajdl

Tell us which book this is, or at least what is the domain or topic of this book or chapter.
Is that about calculus or functional programming?
What is the meaning of this sentense:

"any function f(x) whose functional form changes from f(x) to f'(x)"

What is the meaning of f'? A derivative or just another function.
What does that mean: a "functional form"?

I don't know what you are missing.
But I do know what I am missing: a clear question.
I have no doubt the answer would be a piece of cake if the question was clear.

3. Jun 15, 2014

### maajdl

The difference is a second order term and it is neglected.

4. Jun 16, 2014

### grzz

Maybe the following will make it more clear.

The topic is on classical field theory which then leads to quantum field theory.

f(x) is any function of spacetime, e.g. the potential A$_{μ}$ of the electromagnetic field.

f' is not to be taken as any derivative of f.

The phrase “ …any function f(x) whose functional form changes from f(x) to f'(x) …” means that only the type of function is changing, i.e. from f to f’, but the old function f and of the new function f’ are evaluated at the same spacetime point x.

Any further help is much appreciated.

5. Jun 16, 2014

### vanhees71

If this is field theory in the usual sense, and $f$ is a scalar (relativistic or non-relativistic) field, then the behavior under translations ${x'}^{\mu}=x^{\mu}-\delta x^{\mu}$, $\delta x^{\mu}=\text{const}.$, is defined (!) to be
$$f'(x')=f(x)=f(x'+\delta x)=f(x')+\delta x^{\mu} \partial_{\mu} f(x').$$

6. Jun 16, 2014

### CAF123

So you have $$f'(x') = f'(x + \delta x) \approx f'(x) + \delta x^{\mu}\partial_{\mu} f(x)$$ and want to know what happened to the prime on the f at the end?

I have the same question, and have seen the above in other texts such as in Di Francesco's book on CFT. I thought it might be simply the case of having two functions f and f' that are numerically equal just different functions (one in the primed system, the other unprimed). But that is not the case here since we are only deciding between f(x) and f'(x).
Vanhees71, would you have any ideas?

7. Jun 16, 2014

### grzz

I am imagining that f(x) is a constant temperature field where f changes to f' just because the origin of the coordinates is changed. With that in my mind I understand the post of vanhees.

Thanks vanhees71.

But my difficulty is rather what CAF123 is hinting at.

My original quotation is on p22 from the book "A first book on qft", 2nd edition by Lahiri and Pal.

Last edited: Jun 16, 2014
8. Jun 16, 2014

### maajdl

I think you need to remember the basis of the calculus of variations.
See for example:

http://home.comcast.net/~peter.m.brown/am/calc_var.htm [Broken]

http://home.comcast.net/~peter.m.brown/am/image_gif/am02-img-01.gif [Broken]

On this picture, you can see three different functions.
These functions have been chosen to have some common features, but that is not fundamental here.
Let's call them f1(x), f2(x), f3(x) .
You can consider the difference between two of these functions like this:

df12(x) = f1(x) - f2(x)

This difference accounts only for taking two different functions.

If you wish, you can also define a "combined difference" where both the functions and the x values are different:

Df12(x,x') = f1(x) - f2(x')

It looks like your question is about relating these two differences, assuming f1 and f2 are close as well as x and x'.
Using Taylor expansion, this is quite straightforward.
You will get the claimed result by eliminating any second-order term.

Last edited by a moderator: May 6, 2017
9. Jun 16, 2014

### grzz

Let me use the notation of maajdi.

My original quotation from the book would then read as follows:

df12 = f1(x) - f2(x)

= {f1(x) - f2(x')} - {f2(x) - f2(x')}

= Df12(x,x') - {f2(x) - f2(x')}

= Df12(x,x') - {f2(x' + ε) - f2(x')} where x = x' + ε

= Df12(x,x') - {f2(x') + ε$^{μ}$∂$_{μ}$f2(x') - f2(x')}

= Df12(x,x') - ε$^{μ}$∂$_{μ}$f2(x').

Last edited: Jun 16, 2014
10. Jun 16, 2014

### king vitamin

Well, how are you defining $f'(x)$? As vanhees was saying, for a scalar field, it's natural to define (!*) the transformation law for the field as $f'(x')=f(x)$ (how else?). Then the steps vanhees made follow trivially.

* I put an exclamation point since vanhees did and I don't think you paid enough attention to it

11. Jun 17, 2014

### maajdl

I just understood what the topic was.
If I guess correctly, x and x' are two different system of coordinates.
And the question then is: how does the scalar field change when changing the coordinates! (Vanhees!)

This illustrates the need to state the problem very clearly.
In the current case, the reference to the book would have helped a lot.

Well, I am still interested to read a few page from this book, since I am still not totally sure about what the question is.

12. Jun 17, 2014

### vanhees71

The question also is what you want to describe physically with scalar fields. As an example take the temperature field, $T(\vec{x})$ that describes the temperature as measured at any point $\vec{x}=(x_1,x_2,x_3)$. Now the temperature is a physical fact about the air in that room and doesn't care about which coordinate system we choose. Thus, we we simply change the description by shifting the origin of our coordinate system such that we use new coordinates $\vec{x}'=\vec{x}-\vec{a}$, the temperature at a given location doesn't change (why should it?). So the correct description is to introduce a new function $T'(\vec{x}')$ that describes the temperature at the same location using different coordinates as being independent of this choice of description, i.e., by
$T'(\vec{x}')=T(\vec{x})=T(\vec{x}'+\vec{a}),$
and this leads to the general definition of how a scalar field transforms under spatial translations. It's dictated by the physical meaning of the description.

13. Jun 17, 2014

### grzz

Again I thank vanhees for his explanation.

I have no problem in understanding what vanhees says in his last post. That is what I meant when in one of my posts I said that

"I am imagining that f(x) is a constant temperature field where f changes to f' just because the origin of the coordinates is changed."

For any possible further discussion I am giving the quotation form the book: please see attachment.

In the last line the book has the partial derivative of f(x) and I managed to get the partial derivative of f'(x).

#### Attached Files:

• ###### quotation from book.pdf
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14. Jun 17, 2014

### vanhees71

Hm, this book doesn't make it very clear. I think what they wanted to do is to distinguish between transformations concerning space time like Lorentz transformations or translations (or both together forming the Poincare group) from intrinsic transformations. One prominent example of the latter is to multiply with a phase factor (global U(1) invariance). There you have
$$\phi(x)=\exp(\mathrm{i} \alpha) \phi(x).$$
Here you have the the space-time coordinates unchanged.

For transformations involving the space-time coordinates one always defines the transformed field as function of the new space-time variables. E.g., for a vector field you have the following behavior under Lorentz transformations:
$$x'=\Lambda x, \quad A'(x')=\Lambda A(x)=\Lambda A(\Lambda^{-1} x').$$

There is a little difficulty of understanding when it comes to quantized fields and the corresponding unitary transformations. Then the unitary operation is understood as acting as follows (again for a vector field in a local QFT)
$$\hat{U}(\Lambda) \hat{A}(x) \hat{U}^{\dagger}(\Lambda)=\Lambda \hat{A}(\Lambda^{-1} x).$$
Here on the left-hand side we act with the unitary representation of the Lorentz group (in Hilbert space). On the right-handside the $\Lambda$ is the corresponding Lorentz matrix with c-number entries acting in the usual way on the operator valued components of the vector field.

I hope, now the issue becomes clearer.

15. Jun 17, 2014

### CAF123

Yes, and in fact that is exactly what I was alluding to in my first post:
However, at the case at hand, we are wondering where the prime went on the last expression for f. Consider the single variable case as an illustration, $f(x+\delta x) \approx f(x) + \delta x df/dx$. That is fine, so shouldn't $f'(x +\delta x) \approx f'(x) + \delta x^{\mu} \partial_{\mu} f'(x)$

All I did there is send f → f' really and put in indices because of the different dimension of the space.

16. Jun 17, 2014

### vanhees71

Once more. I don't know, how to explain it differently. By definition we have
$$x'=x-\delta a, \quad f'(x')=f(x)=f(x'+\delta \alpha)=f(x')+\delta \alpha \mathrm{d}_x f(x') + \mathcal{O}(\delta a^2).$$
Note that here the prime doesn't mean a derivative!

17. Jun 17, 2014

### bloby

2 lines above the quotation you post in the book they do the same thing(dropping the prime) arguing "as the difference is of higher order in $\delta x^\lambda$"

18. Jun 17, 2014

### TSny

Yes, I think that's the key. The expression $\partial$$_{μ}$f '(x)$\delta$x$^{μ}$ is of first order due to the factor $\delta$x$^{μ}$. The expression will remain accurate to first order if you replace f '(x) by f(x), since f '(x) ≈ f(x) to "zeroth" order.

19. Jun 18, 2014

### CAF123

Ok, that makes sense. The transformation is an infinitesimal offset $\delta x^{\mu}$ so the new field at the same position in the unprimed system is approximately that of the old field (before the transformation) at the same position x.

20. Jun 18, 2014

### grzz

I thank everybody for the help given.