Understanding the Invariance of dl in Biot-Savart Law

[erece]

In the expression of Biot-Savart law
B = (µo/4π) ∫ (I dl x r^)/r2
why dl does not depend on the coordinate systems ?
in books they are using del X dl = 0

 

[Muphrid]

It's natural d\ell doesn't depend on the coordinate system; it's a vector.

Still, \nabla \times d\ell doesn't make a whole lot of sense. Could you give some more context for this question?

 

[erece]

i am attaching a file in which the derivation of biot-savart law is given. Now after equation 6-29 they used del X dl' = 0. i want to know the reason

 

[Muphrid]

It says the reason right in the screenshot:

Now, since the unprimed and primed coordinates are independent, \nabla \times d\ell' = 0

Here, d\ell and d\ell' are two different things. The first depends on "unprimed" coordinates (i.e. x, y, z) and the second depends on "primed" coordinates x', y', z'). The operator \nabla differentiates only with respect to the unprimed coordinates. Hence, \nabla \times d\ell' is zero because x', y', z' are not functions of x,y, z; they can't be differentiated with respect to the unprimed coordinates.

This use of primed and unprimed variables is pretty common in proofs of integral theorems, particularly when you have a function on the left that is, say, X(r), generally the variable of integration on the right is r'. For example,

E(r) = \int \frac{\rho(r')/\epsilon_0}{4\pi |r - r'|^2} \; dV'

Here, r' is the dummy variable of integration, and dV' is the associated volume element for that variable.

 

[erece]

now i got it. thanks a lot