I finally realized Mathematica didn't do all the simplifications! :smile:
By using Simplify command it comes up that curl(E)=(0,0,0) even in cartesian coordinates, as it should be.
Thank you, anyway!
Here is a quick computation I made with Mathematica regarding this problem.
As you can clearly see, in one case the curl is 0, in the second one is different from 0.
Homework Statement
Since the electrostatic field is conservative, show that it is irrotational for an electric dipole, whose dipole momentum is p .Homework Equations
\nabla \times \mathbf{E} = 0 The Attempt at a Solution
I know that the components of the electric field in spherical...
On the other hand, one can think to use the definition of canonical trasformation:
A time-independent transformation Q = Q(q,p) , and P = P(q,p) is called canonical if and only if there exists a function F(q,p) such that:
dF(q,p) = p_i dq_i - P_i(q,p) dQ_i(q,p)
In other words...
I know that:
[Q_j, P_k] = \delta_{jk}
so, in this case, since Q = Q_1 and P = P_1 , we have:
[Q_1, P_1] = 1 = [Q, P]
These are the calculations I made:
[Q,P]_{q,p} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial...
Hi, I tried to solve this problem, but I was unsuccessful
Here is the problem:
Given the transformation:
\left \{ \begin{array}{l} Q = p^\gamma \cos(\beta q) \\ P = p^\alpha \sin(\beta q) \end{array} \right.
a) Determine the values of the constants \alpha , \beta and \gamma...
Is someone able to proove the invariance under Galilean transformations of F=dp/dt within a system of variable mass? In particular is the momentum invariant? i.e. p=p', as Goldstein states? Please answer me! :wink: