QFT (derivative the covariant and contravariant fields) (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

7
0
Hi,
please help me ..

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:

vanhees71

Science Advisor
Insights Author
Gold Member
11,472
4,168
Usually what's meant by the symbols is
[tex]\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}.[/tex]
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!
 
7
0
Usually what's meant by the symbols is
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!

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

Fredrik

Staff Emeritus
Science Advisor
Gold Member
10,730
403
thanks vanhees71
I know that, but i ask can me consider these vectors as two multiplied functions (quadratic ∂ ø) and derivative becomes 2 ∂αø?
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.
 
7
0
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.

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?!
 

Fredrik

Staff Emeritus
Science Advisor
Gold Member
10,730
403
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.
 
7
0
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?
 
7
0
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:

Fredrik

Staff Emeritus
Science Advisor
Gold Member
10,730
403
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:
7
0
yess! i'm understand.
I am grateful to you
thanx Fredrik!
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top