- #1
USeptim
- 98
- 5
Hello,
I'm a bit stuck with a case in which the curl gives a vector that does not transform under rotation.
As an example, let's have a field with only [itex]\hat{x}[/itex] direction (but this does not mean that it's a scalar field!). The field has this expression:
F(x,y,z)= A*exp(-[itex]\sqrt{y^{2}+z^{2}}[/itex] [itex]\hat{x}[/itex]
Obviously, you have to take the positive value of the square root.
If you evaluate ∇x F in the x-axis you will get that the curl is:
∇x F = A (-[itex]\hat{y}[/itex]+[itex]\hat{z}[/itex]).
Now, it's straightforward to see that since F depends really on the radius ([itex]y^{2}+z^{2}[/itex]), if you rotate the YZ axis, the curl will not rotate but continue been the same.
I have heard that the curl is a pseudovector but I didn't expected to find that in some cases it does not rotates. Have I done something wrong?
Sergio
I'm a bit stuck with a case in which the curl gives a vector that does not transform under rotation.
As an example, let's have a field with only [itex]\hat{x}[/itex] direction (but this does not mean that it's a scalar field!). The field has this expression:
F(x,y,z)= A*exp(-[itex]\sqrt{y^{2}+z^{2}}[/itex] [itex]\hat{x}[/itex]
Obviously, you have to take the positive value of the square root.
If you evaluate ∇x F in the x-axis you will get that the curl is:
∇x F = A (-[itex]\hat{y}[/itex]+[itex]\hat{z}[/itex]).
Now, it's straightforward to see that since F depends really on the radius ([itex]y^{2}+z^{2}[/itex]), if you rotate the YZ axis, the curl will not rotate but continue been the same.
I have heard that the curl is a pseudovector but I didn't expected to find that in some cases it does not rotates. Have I done something wrong?
Sergio