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1

No, not really. A gradient is the derivative of a scalar. It is not actually a vector, but a dual vector or 1-form. http://en.wikipedia.org/wiki/Gradient Vectors and 1-forms have different transformation properties, and used to be called contra-variant and co-variant vectors, but the language of exterior calculus makes this much cleaner. Intuitively, a ...

5

It's purely notation. Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) ...

2

The answer is no. Just as in the case without a gauge field, it is just a product of two derivatives of the field $\phi$. You might be interested in the chapter "Scalar Electrodynamics" in Srednicki's book.

1

I don't think that you really understand integration. Let me clear this up for you. In that question there is a rod of length l. You know how to calculate gravitational force between two point masses but not in continuous mass bodies. If you apply the formula to find the gravitational force you don't know what to take the distance as because it is ...

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