I Is The Curl Operator Linear?

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1. May 20, 2016

Sheldon Cooper

Hi,

I stumbled upon thinking that "Is curl operator a linear operator" ?
I was reading EM Theory and studied that the electromagnetic field satisfies the curl relations of E and B. But if the operator was not linear then how can a non linear operator give rise to a linear solution. Thus it becomes apparent that curl is linear but how can we prove it mathematically?

2. May 20, 2016

ShayanJ

Use the definition $\displaystyle (\vec \nabla \times \vec F)\cdot \hat n \equiv \lim_{A_n\to 0} \frac{1}{A_n} \oint_{\partial A_n} \vec F \cdot \vec{dr}$ (where $A_n$ is a surface with unit normal $\hat n$ and $\partial A_n$ is its boundary curve.)!

3. May 21, 2016

houlahound

Does an operator have to be linear to generate a linear solution to an arbitrary equation, I did not know that??

A proof for the general case?

4. May 21, 2016

ShayanJ

The phrase "linear solution" seems meaningless to me. Linearity is a property of an operator/equation.

5. May 21, 2016

houlahound

I can't get the question, linear has multiple meanings in math.

6. May 21, 2016

ShayanJ

The OP has realized that Maxwell's equations are linear equations and so the curl operation is linear too. He just wants to prove that it actually is.

7. May 21, 2016

Staff: Mentor

Which are? You can have linearity on the objects $O$ or in the arguments $A$, but it's always the same condition:

8. May 21, 2016

houlahound

Linear can mean supper position, straight line, type of vector space , something raised to power one, a type of DE...

9. May 21, 2016

Staff: Mentor

Which are all together just linear combinations of certain objects, or linear in the argument in case of scalar multiplication.
However, I have not the slightest idea what "type of vector space" means. Vector spaces are per definition a set of linear combinations.
There is no "multiple meaning" of linearity in math.

10. May 21, 2016

houlahound

Correction type of space eg vector space.

How is a linear equation eg y=2x+1the same as a linear first order DE or a linear operator?

Linearity has context.

11. May 21, 2016

Staff: Mentor

This is no linear function. It's called an affine transformation. To call it "linear" isn't exact. $y(0) ≠ 0$

It is linearity on the considered objects, here differential operators which are a special case of linear operators.

Wrong. The objects or arguments it applies to have a context, e.g. the derivatives in the example above.
Linearity itself is a property independent of what you regard in a special case, it simply says that you have a rule for addition and for scalar multiplication - whether it's applied to a superposition or an operator. The meaning of "linear" does not change.

12. May 21, 2016

houlahound

I learned something, thanks.

13. May 21, 2016

Sheldon Cooper

Hey Shyan, tried your procedure to prove the linearity problem but and am stuck can u please add a few more steps(if possible).

14. May 22, 2016

ShayanJ

Just put $\vec F=\vec G+b \vec H$ in the definition!

15. May 22, 2016

Sheldon Cooper

ok got it thanks!