# Commutation of Curl and the partial time derivative?

1. Nov 16, 2011

### ksmith1281

I am curious if there are any issue with commuting the curl of a vector with the partial time derivative?

For example if we take Faraday's law:

Curl(E)-dB/dt=0

And I take the curl of both sides:

Curl(Curl(E))-Curl(dB/dt)=0

Is

Curl(dB/dt)=d/dt(Curl(B))

I assume this is only true for a certain condition but I cant think of anything?

If you have some insight please let me know

2. Nov 16, 2011

### ZapperZ

Staff Emeritus
If I have a function of two variables, say f(x,y), does it matter if I take the derivative with respect to one variable first over the other?

The curl is a derivative with respect to the spatial variable, while the time derivative is with respect to... well, what else, but time! If you can answer the first question, you have your answer here.

Zz.

3. Nov 16, 2011

### ksmith1281

Thanks!

I agree with you comparison. The funny thing about physics is it can be approached from a physics or math approach. I guess to refine my question: Does the ability to commute derivatives have any implication on the physics. I trust you can commute (I have already assumed and done so in my calculations) however I am wondering if there is an assumption implicit in "allowing" commutation between these two derivative.

Perhaps this is more of a math questions than anything else.

4. Nov 18, 2011

### aesir

If you search for "Symmetry of second derivatives" in wikepedia you will find the intro to the mathematical answer.
In physics you usually look for functions which are either $C^2$ or a tempered distributions

5. Oct 16, 2012

### Trifis

Hmmm this makes me wonder... if curl and time derivative are commutable, then why the curl of the derivative of a position vector DOES NOT vanish? Symbolically:
$\nabla$$\times$$\vec{r}$(x,y,z)=0 but $\nabla$$\times$$\frac{d$\vec{r}$(x,y,z)}{dt}$$\neq$0
In the second case the time-dependency of the position vector is exposed and we can no longer assert that it is only position dependent. However given the $\vec{r}$ without further analysis, is it safe to say that it is a conservative field, since it is also a central field.
That is why function domains are a huge benefit not only to mathematics...

Last edited: Oct 16, 2012
6. Oct 18, 2012

### Trifis

All in all curl and time derivative DO NOT commute...

7. Oct 18, 2012

### Muphrid

If $r=r(t)$ and $v = dr/dt$, then how exactly would you compute $\nabla \times v$?

8. Oct 18, 2012

### Trifis

You can't. That's why they could not commute in the first place (answer to the OP) and that's why there is no time-dependent conservative field!

9. Oct 18, 2012

### Muphrid

No, I pointing out to you that you've conflated a couple things and put them together in a way that makes no sense, drawing an erroneous conclusion as a result.

Furthermore, let's say I extrapolate from what you're saying to talk about continuous vector fields. You state without proof that the curl of the velocity is nonzero. Can you give an example of a time-dependent field that would obey this relation?

10. Oct 18, 2012

### Trifis

If you are reffering to post 5, then it is just a physical intuition, due to the fact that energy conservation arises from time invariance. I do not think that curl is definable in this case. If it is the $\neq$ relation that bothers you, let us say it was my not-so-rigorously-mathematic way, to make my point clear...

11. Oct 19, 2012

### Staff: Mentor

In the left equation, you use r as position in space? The position is not time-dependent, so both equations are 0. If r is some arbitrary vector field, the curl can be non-zero.

12. Oct 19, 2012

### Trifis

Ok let's make this clear...
$\vec{r} (x,y,z): ℝ^{3}\rightarrowℝ^{3}$, $φ(t): [α,β]\rightarrowℝ^{3}$
$\frac{d\vec{r}}{dt}$ implies $\frac{d\vec{r}(φ(t))}{dt}$

Exactly that confusion, I wanted to avoid, by saying that curl and time derivative do not always commute, for if we are not given the domains some things could not make sense/be defined. Maybe I haven't expressed myself correctly so far

13. Oct 19, 2012

### Staff: Mentor

What do you mean by "implies"? If r is not a function of t, the time-derivative is meaningless.

For any $\vec{r} (x,y,z,t): ℝ^{4}\rightarrowℝ^{3}$ with reasonable properties ($C^2$):

$\nabla \times \frac{d\vec{r}}{dt} = \frac{d}{dt} (\nabla \times \vec{r})$

14. Oct 19, 2012

### Trifis

My whole point was to draw attention to the fact that it is preferable to state explicitly the function domains, cause I've seen misconceptions like the following many a time in the past:
"Determine if the following force is conservative or not: $\vec{F}=-k\vec{r}+α\dot{\vec{r}}$", where students commuted the curl operation with the time derivative and concluded that F is conservative...
Of course from ℝ4 to ℝ3, what you wrote does hold.

*Implies means that if we are to write the derivative of a field, that field'd better be time dependent.