# QFT (derivative the covariant and contravariant fields)

1. Mar 10, 2014

### aams

Hi,

How can I derivative covariant and contravariant fields?

as in the attached picture

Thanks..

[PLAIN]http://www.gulfup.com/?tNXcaN [Broken][/PLAIN]

w.r.t alpha

Last edited by a moderator: May 6, 2017
2. Mar 10, 2014

### vanhees71

Usually what's meant by the symbols is
$$\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}.$$
It's important to keep in mind that the derivative with respect to contravariant vector components (upper indices) gives gives an operator that transforms like a covariant component (lower indices). This is, of course, only true in flat space-time!

3. Mar 10, 2014

### aams

thanks vanhees71
I know that, but i ask can me consider these vectors as two multiplied functions (quadratic ∂ ø) and derivative becomes 2 ∂αø?

4. Mar 10, 2014

### aams

5. Mar 10, 2014

### Fredrik

Staff Emeritus
You have to use the product rule for derivatives. I would also like to suggest that since the alphas in the first factor are dummy variables, you change those two to $\gamma$ before you begin taking the derivative with respect to $\phi^\alpha$. Either that, or write out exactly what the notation $\partial_\alpha\phi^\alpha\partial_\beta\phi^\beta$ means before you start.

6. Mar 10, 2014

### aams

aah, thanx Fredrik. please just tell me the steps for method to choose the dummy variables. why must be using it? what is the idea of choose it? and how can me use the dummy variables?!

7. Mar 11, 2014

### Fredrik

Staff Emeritus
I don't want to do the problem for you, but I'll do another one that illustrates the idea. Suppose that you're asked to compute the ith partial derivative of the function $f:\mathbb R^3\to\mathbb R^3$ defined by $f(\mathbf x)=|\mathbf{x}|^2$ for all $\mathbf x\in\mathbb R^3$, would you know how? Sure you do. (If not, you will have to open up a calculus books and study partial derivatives again). I'll use the notation $\mathbf x=(x_1,x_2,x_3)$.
\begin{align}
\frac{\partial f(\mathbf x)}{\partial x_i} &=\frac{\partial}{\partial x_i}|\mathbf x|^2 =\frac{\partial}{\partial x_i} \sum_{j=1}^3 x_j x_j = \frac{\partial}{\partial x_i} \left( x_1{}^2+x_2{}^2+x_3{}^2\right)\\
&=\frac{\partial}{\partial x_i}x_1{}^2 + \frac{\partial}{\partial x_i}x_2{}^2 + \frac{\partial}{\partial x_i}x_3{}^2.
\end{align}
If for example i=1, then the last two terms are obviously zero, and the first one is
$$\frac{\partial}{\partial x_1}x_1{}^2=2x_1.$$ If i=2 instead, you get the result $2x_2$ in a similar way, and if i=3, you end up with $2x_3$. So regardless of what i is, you end up with $2x_i$. This means that for all $i\in\{1,2,3\}$, we have
$$\frac{\partial f(\mathbf x)}{\partial x_i}=2x_i.$$ That wasn't so hard, was it? You will find that your example isn't any harder, if you just do the summation explicitly the way I did here. The only difficulty in your problem is that the notation makes it hard to see that your problem is essentially the same as this one.

If we use a Kronecker delta ($\delta_{ij}$ is 1 when i=j and 0 when i≠j), we can do the calculation without doing the summation explicitly (as I did in the last equality on the first line).
$$\frac{\partial}{\partial x_i} \sum_{j=1}^3 x_j x_j = \sum_{j=1}^3 \frac{\partial}{\partial x_i}\left(x_j x_j\right) =\sum_{j=1}^3 \left(\left(\frac{\partial}{\partial x_i}x_j\right)x_j + x_j\left( \frac{\partial}{\partial x_i}x_j\right)\right) =\sum_{j=1}^3 \left(\delta_{ij}x_j +x_j\delta_{ij}\right) =\sum_{j=1}^3 2\delta_{ij}x_j =2x_i.$$ When you use the summation convention, you can drop all the summation sigmas. What I'm saying is that when you do, it's going to be confusing if the dummy variable used for summation is the same symbol as the (not dummy) variable that indicates which partial derivative you're interested in.

8. Mar 11, 2014

### aams

thank you very much .. i'm understand ,but the last question, is not to need conversion between contra and covariant ..? How does contra and covariant affect the differentiation?

9. Mar 11, 2014

### aams

sorry, I mean you explain where (x sub i) what about upper indices? what is the difference?!

http://www.gulfup.com/?dSTVRF [Broken]
??!!

Last edited by a moderator: May 6, 2017
10. Mar 11, 2014

### Fredrik

Staff Emeritus
I used an example where the indices would cause as little confusion as possible, because I wanted you to see that the only thing you're supposed to do is to take partial derivatives of a multiple-variable polynomial. This is something you already know how to do. If the notation confuses you, then you should rewrite the expression in a way that doesn't confuse you. In particular, if you're asked to evaluate something like $\frac{\partial}{\partial x^\alpha}\left(x_\beta x^\beta\right)$, and you don't immediately see how, you can rewrite it like this:
$$\frac{\partial}{\partial x^\alpha}\left(x_\beta x^\beta\right) = \frac{\partial}{\partial x^\alpha}\left(g_{\beta\gamma} x^\gamma x^\beta\right) = \frac{\partial}{\partial x^\alpha}\left(g_{00}x^0 x^0 +g_{11}x^1x^1 +g_{22}x^2x^2 + g_{33}x^3x^3\right).$$
Edit: Also, don't write things like $\frac{\partial}{\partial x^\alpha}\left(x_\alpha x^\alpha\right)$. This expression doesn't make sense, since the summation convention is defined only for indices that occur exactly twice. If you're asked to apply $\frac{\partial}{\partial x^\alpha}$ to $x_\alpha x^\alpha$, you have to realize that the alphas in $x_\alpha x^\alpha$ are just there to tell you that you're dealing with the sum $x_0x^0+x_1x^1+x_2x^2+x_3x^3$, which can also be written as $x_\beta x^\beta$. So what you actually write down is $\frac{\partial}{\partial x^\alpha}\left(x_\beta x^\beta\right)$.

Last edited: Mar 11, 2014
11. Mar 11, 2014

### aams

yess! i'm understand.
I am grateful to you
thanx Fredrik!