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I followed the method u suggested only but still getting the same expression. Here are equations cocerning the relevant terms: dr/d# =-rsin#[email protected] ex [email protected]# ey + 0 ez |dr/d#| = [email protected] which gives, e# = -sin# ex +cos# ey And the matrix is, [Ax] [[email protected]# [email protected]# -sin#] [Ar] [Ay]...

Yeah checked it again from a book where only the matrix is given. The transformation matrix is actually same as that comes from the above expression.

I was actually studying this inorder to find the spherical coordinates representation of the 'laplacian' of a vector( scalar electric potential)...how to reach into that form for here? laplacian ∇2V = Vxx +Vyy + Vzz must be converted to ∇2V = 1/r2 ∂/∂r(r2∂V/∂r) + ...

So here is what i got after grouping the terms acc to cartesian unit vectors...whats next? A = ([email protected]# + [email protected]@cos# - A#sin#) ex + ([email protected]# + [email protected]@sin# + A#cos#) ey + ([email protected] [email protected]@) ez Edit: Thanks i was able to complete the solution myself. Thanks for providing me hints and making me...

Ya thanks. So here is what i have done, Taking r = [email protected]# ex + [email protected]# ey + [email protected] ez I calculated the partials with respect to r, @ and # from these. ∂r/∂r = [email protected]# ex + [email protected]# ey + [email protected] ez |∂r/∂r| = √[email protected]# +sin2sin2# + [email protected] Likewise, i have arrived at the partials and magnitude...

Sorry, Iam not able to reach a such relations. Can u please tell what the relations are?

Which relation to use to find the partial derivatives for finding each unit vectors? Is it r^2 = x^2 + y^2 + z^2 ?

Iam having trouble understanding how one arrives at the transformation matrix for spherical to rectangular coordinates. I understand till getting the (x,y,z) from (r,th ie., z = [email protected] y = [email protected]# x = [email protected]# Note: @ - theta (vertical angle) # - phi (horizontal angle) Please show me how...