Proving that curl of gradient of f=0 using Stokes' theorem

[kolahalb]

we are to prove that curl of gradient of f=0 using Stokes' theorem.

Applying Stokes' theorem we get-
LHS=cyclic int {grad f.dr}
Hence we have,
LHS=cyclic int d f=(f)|[upper limit and lower limit are the same]
=0
I need to be sure that I am correct.Please tell me if I went wrong in my logic.

Thank you.

 

[coros]

Re: vector problem

You are right. grad f is a potential field of f and an integral about any closed path is 0.

 

[pmonon]

Re: vector problem

we are to prove that curl of gradient of f=0 using Stokes' theorem.

Applying Stokes' theorem we get-
LHS=cyclic int {grad f.dr}
Hence we have,
LHS=cyclic int d f=(f)|[upper limit and lower limit are the same]
=0
I need to be sure that I am correct.Please tell me if I went wrong in my logic.

Thank you.

If f is a scaler, how do you even define ∫© df? Further more Stoke's theorem comes after knowing curl(grad f)=0. So my sol. is simply use the def. and evaluate curl (grad f). This will have 0 valued components for all if you work with the determinant formula for cross product.

 

[kolahalb]

Re: vector problem

The Fundamental Theorem of (Integral) Calculus is of the form:
(Lower Lt a,Upper Lt. b) ∫(df/dx)dx=f(b)-f(a)

where df/dx is often written as F(x).

So,when you use a cyclic integral,it means you are tking both the lower and upper limit to be the same...

We are bound to use an alternative of Stokes' theorem.The question asks for that.

 

[pmonon]

Re: vector problem

So,when you use a cyclic integral,it means you are tking both the lower and upper limit to be the same...
/quote]

No. The fundamental theorem of calculus is not applicable in this case. Because for a line integral it may depend on the path and the nature of the vector field (whether it is conservative or not)

 

[kolahalb]

Re: vector problem

cannot agree buddy!
Think of Gravitational force...whatever path you follow the line integral is always dependent on end points..

 

[pmonon]

Re: vector problem

cannot agree buddy!
Think of Gravitational force...whatever path you follow the line integral is always dependent on end points..

Gravitational force field is a conservative vector field. If all line integrals depend on just the end points then it is indpendent of path and no need to define a line integral, just take the end points and you are done. But there are fields for which it depends on the path not on the end points. See this info from Britannica and let's talk back:

The magnetic field B is an example of a vector field that cannot in general be described as the gradient of a scalar potential. There are no isolated poles to provide, as electric charges do, sources for the field lines. Instead, the field is generated by currents and forms vortex patterns around any current-carrying conductor. Figure 9 shows the field lines for a single straight wire. If one forms the line integral òB×dl around the closed path formed by any one of these field lines, each increment B×dl has the same sign and, obviously, the integral cannot vanish as it does for an electrostatic field.

 

[kolahalb]

Re: vector problem

Observe one thing-in case of magnetic field,you cannot in general put forward a scalar potential given by grad V.
Here it is given...
I have talked to 4-5 forums regarding this.None but you are still lingering with a problem that does not seem to be done in wrong method.
Ok,I may do mistake as well.Then,point with valid logic where I have gone wrong.

 

[coros]

Re: vector problem

If for some vector field F=[X,Y,Z] we can find a scalar function u so that F = grad u, then the field F is a scalar potential field. For these fields the line integral depends only on end points. rot(F)=0 is a condition that F is scalar potential field. It follows from the conditions

δu/δx = X, δu/δy = Y, δu/δz = Z

If F is not scalar potential field then line integral depends not only on end points but also on the path.

 

[pmonon]

Re: vector problem

We are to prove:

curl grad f = 0

LHS= | i j k |
|δ/δx δ/δy δ/δz |
|δf/δx δf/δy δf/δz |
= (δ2f/δyδz - δ2f/δzδy)i + ..........
= 0.i+ 0.j + 0.k
= 0=RHS

Simple proof... No need to worry about Stoke's theorem.

 

[kolahalb]

Re: vector problem

pmonon,proving was not the problem.
The question specified to use Stokes' theorem.