# I need to prove del cross f= 0

• Ashley1nOnly
In summary: So that's how to do it.It's not clear what you are trying to do.I think you need to learn some vector calculus before attempting this. You need to understand what the del operator is and how it works. The way you are writing things is confusing. You are writing a function ##f(r)## which depends on a vector ##r##, and trying to take the del operator of that. That's not how it works. The del operator operates on scalar functions (functions of position, not functions of position vectors) and vector functions.And you also need to understand the difference between the symbol ##\hat r## and the vector ##\vec r##. They are not the same thing. ##\vec r##
Ashley1nOnly

## Homework Statement

F(x) has the form F(x)=f(r)x where r=|x|

A.) prove that del cross f =0
B.) Now suppose also Del •F = O. What is the most general form allowed for f(r)?

## The Attempt at a Solution

I have done part b but what do I need for A

F(X)= r(hat)
Fr = 1, F(theta) = F(phi)= O.

Got this out of book.

Using equation from the book ( del cross F) r
I get zero.

But what is the proof behind it

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The formulas for ##\nabla\times## in polar coordinates make calculations with radial vectors easy because all the hard work has been done in deriving the formulas. If you have not been given those formulas, and hence are required to derive them if you use them, there is no point in using them because the work involved in deriving them is greater than the work involved in just doing the whole calc in Cartesian coordinates.

Given that, the simplest way is just to do the calc in Cartesian coordinates. Since the usual coordinate names in Cartesian are ##x,y,z## I suggest you instead call your point ##\vec r##. Then the coordinates of ##\vec F(\vec r)## are ##(f(r)x,f(r)y,f(r)z)##, writing ##r## for ##\|\vec r\|##.

Then use the Cartesian formula for ##\nabla\times##:
$$\nabla\times\vec u=(\partial_x,\partial_y,\partial_z)\times (u_x,u_y,u_z)=(\partial_y u_z-\partial_zu_y)\hat i +(\partial_z u_x-\partial_xu_z)\hat j +(\partial_x u_y-\partial_yu_x)\hat k$$
where ##\partial_x## denotes ##\frac{\partial}{\partial x}## and ##u_x## is the ##x## coordinate of ##\vec u##.

The calc is quite short.

Last edited:
Ashley1nOnly said:

## Homework Statement

F(x) has the form F(x)=f(r)x where r=|x|

A.) prove that del cross f =0
B.) Now suppose also Del •F = O. What is the most general form allowed for f(r)?

Your notation is terrible. It appears you are mixing up F and f in A. Apparently what you are calling x is what would normally be denoted ##\vec r = \langle x,y,z\rangle##. And in your work you use x as above and also as one of its components, which is very confusing. Also, your posting images instead of typing your work makes it difficult to edit and violates forum policy.

I wouldn't normally bother to answer such a post, but it's a slow day here so...
I think your question is, given ##\vec r = \langle x,y,z\rangle## and ##r = |\vec r| =\sqrt{x^2+y^2+z^2}##, and ##f(r)## is a differentiable function, calculate ##\nabla \times f(r)\vec r##. Write that out carefully and you should get the zero vector. If you get stuck and don't get that, type your work here for additional help.

Edit: I see Andrew posted while I was typing. Obviously, he and I agree about how to work it.

I must first ask, why does r=|r-hat|?

I understand that |r-hat| equals sqrt(rx^2+ry^2+rz^2)
Rx^2 is just r with respect to X.

Also f(r) is a scalar function.
When doing del X (f(r) (r-hat)
I know that you can't multiply Del with a scalar. You can only cross with a vector which is r-hat.

So in order to do this it would be
Del•(f(r)) X r-hat + f(r)•(Del X r-hat)

Wait, you can multiply Del with a scalar function but you can't cross Del with a scalar function

Of ##\phi## is a scalar and ##\vec V## is a vector, a standard formula (which you can prove yourself) is$$\nabla \times \phi \vec V = \nabla \phi \times \vec V + \phi ( \nabla \times \vec V)$$

## 1. What does "del cross f" mean?

"Del cross f" is a mathematical notation for the cross product between the del operator (∇) and a vector function (f). This operation results in a vector quantity.

## 2. Why do we need to prove del cross f = 0?

The equation del cross f = 0 is known as the curl-free condition. It is important in vector calculus because it represents a vector field that is conservative, meaning that the path integral of the field is independent of the path taken. Proving this condition allows us to simplify calculations and make certain conclusions about the behavior of the vector field.

## 3. How can we prove del cross f = 0?

There are several methods for proving del cross f = 0, depending on the context and available information. One approach is to use the properties of the del operator and the definition of the cross product to manipulate the equation and show that it simplifies to 0. Another approach is to use the divergence theorem, which states that the surface integral of a vector field is equal to the volume integral of its divergence. If the divergence of the field is 0, then the surface integral must also be 0, leading to del cross f = 0.

## 4. What are the implications of del cross f = 0?

The implications of del cross f = 0 depend on the specific problem or application. In general, it means that the vector field is conservative and has no rotational component. This can be useful in analyzing the behavior of a fluid flow, electromagnetic field, or other physical phenomenon.

## 5. Can del cross f ever be non-zero?

Yes, del cross f can be non-zero in certain cases. This occurs when the vector field has a rotational component, meaning that its direction and magnitude change as you move through space. In these cases, the curl of the field will not be equal to 0, and the del cross f equation will not hold. However, there are other conditions that can lead to a non-zero del cross f, such as a changing coordinate system or a non-uniform vector field.

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