Vector field question

A vector field V is not irrotational.Show that it is always possible to find f such that fV is irrotational.

curl [fV]=f curlV-Vxgrad f
I have to equate the LHS to zero.But then,how can I extract f out of the resulting equation?
Please help

TPT curl fv=fcurlv-v cross gradf=0

now, if curlv=0, v≡gradk(k any scaler)

So, the RHS becomes: 0 - v cross grad k(f/k)=(f/k)v cross v=0
=> LHS = 0.

Hope you are done!

You are simply WRONG.V is a vector field and by question,the curl is non-zero.

Added after 16 minutes:

I mean if the question is correct,then you are wrong.But if the question is wrong then you are correct.

Added after 22 minutes:

So, the RHS becomes: 0 - v cross grad k(f/k)=(f/k)v cross v=0

Click to expand...

It is not convincing.You have missed a term...(f/k) is not a constant.

rot(fV) = grad(f) x V + f rot(V)

The last term is 0 because of the assumption rot(V) = 0, so rot(fV) ≠ 0 iff grad(f) x V ≠ 0 iff f ≠ const and grad(f) ≠ V.

kolahalb said:

You are simply WRONG.V is a vector field and by question,the curl is non-zero.

Added after 16 minutes:

I mean if the question is correct,then you are wrong.But if the question is wrong then you are correct.

Added after 22 minutes:

So, the RHS becomes: 0 - v cross grad k(f/k)=(f/k)v cross v=0

Click to expand...

It is not convincing.You have missed a term...(f/k) is not a constant.

Click to expand...

I am 100% correct because v is a rotational vector field. So curl v=0. And f/k is a scaler so it can come out of the grad operator.

Basic silly conceptual mistake.
rotational vector field means curl V is non-zero.
It can also be understood by commonsense.Curl measures the amount of rotation of a vector field.So,irrotational means curl V=0 and rotational means curl is non-zero.

And f/k is a scaler so it can come out of the grad operator.

Click to expand...

k was also scalar when you extracted (f/k) from grad [k(f/k)].Why did not you bring it outside?

coros,you read my question first.I said the vector field is not irrotational.

kolahalb said:

Basic silly conceptual mistake.
rotational vector field means curl V is non-zero.
It can also be understood by commonsense.Curl measures the amount of rotation of a vector field.So,irrotational means curl V=0 and rotational means curl is non-zero.

And f/k is a scaler so it can come out of the grad operator.

Click to expand...

k was also scalar when you extracted (f/k) from grad [k(f/k)].Why did not you bring it outside?

coros,you read my question first.I said the vector field is not irrotational.

Click to expand...

Ok then let's think from the start...
If V is rotational, then how can you find a scalar so that fV is not rotational? That makes me doubtful about the proposition you are trying to prove...

This is indeed a doubtful question and I am still thinking if it is correctly printed...
However,given this is correct,I am trying this...

Ok, so we must have

rot(fV) = 0, f scalar field, V vector field V = [ Vx, Vy, Vz], the equation rot(fV) = 0 we can split into set of 3 differential equations accordingly to the 3 coordinates

| i j k |
| δ/δx δ/δy δ/δz | = 0
| fVx fVy fVz |

Vz • δf/δy - Vy • δf/δz + f • δVz/δy - f • δVy/δz = 0

Vx • δf/δz - Vz • δf/δx + f • δVx/δz - f • δVz/δx = 0

Vy • δf/δx - Vx • δf/δy + f • δVy/δx - f • δVx/δy = 0

after ordering

Vz • δf/δy - Vy • δf/δz = -f • (δVz/δy - δVy/δz)

Vx • δf/δz - Vz • δf/δx = -f • (δVx/δz - δVz/δx)

Vy • δf/δx - Vx • δf/δy = -f • (δVy/δx - δVx/δy)

The derivatives on the right are the components of rot(V) multiplied by f, so they are given functions
we can call them f1, f2, f3

Vz • δf/δy - Vy • δf/δz = -f • f1

Vx • δf/δz - Vz • δf/δx = -f • f2

Vy • δf/δx - Vx • δf/δy = -f • f3

now let's multiply the first by Vx/Vz and add it to the third

Vy • δf/δx - Vx•Vy/Vz • δf/δz = -f • f3 - Vx/Vz • f • f1

we multiply the second by Vy/Vz and add it

Vy • δf/δx - Vy • δf/δx = -f • f3 - Vx/Vz • f • f1 - Vy/Vz • f • f2

finally

-f • (f3 + Vx/Vz • f1 + Vy/Vz • f2) = 0, so f = 0 or f3 + Vx/Vz • f1 + Vy/Vz • f2 = 0

I hope you are satisfied.