Can you find out what's wrong ?

I'll present a proof that \(0 = 1\). Of-course there is an error in the proof. Can you find it?

Proof\large \textbf{Proof}

Let us start with a vector field v=1r2r^\overrightarrow{v} = \dfrac{1}{r^2} \hat{r}.

Let us find the Divergence of this field. .v=1r2r(r21r2)=0 \overrightarrow{\nabla} . \overrightarrow{v} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \dfrac{1}{r^2} \right) = \boxed{0}

Now let us calculate the flux of this field. Sv.da=4π\oint_{S}{ \overrightarrow{v}. \overrightarrow{da}} = 4\pi

Green’s Theorem state that V(.v)dτ=Sv.da\int_{V} {\left (\overrightarrow{\nabla} . \overrightarrow{v} \right ) d\tau} = \oint_{S}{ \overrightarrow{v}. \overrightarrow{da}}

Applying it we see L.H.S = 0 since .v=0\overrightarrow{\nabla} . \overrightarrow{v} = 0 and R.H.S = 4π4\pi.

This implies 0=4π 0 = 4\pi.
Dividing by 4π4\pi we get
0=1\boxed{ 0 = 1 }

Hence proved.

Note by Rajdeep Dhingra
5 years, 1 month ago

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1 vote

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Green's theorem does not hold as the divergence is not continuous inside a ball containing the origin.

Abhishek Sinha - 5 years, 1 month ago

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Yes , correct. To correct this error we have to use a Diract Delta Function.

Rajdeep Dhingra - 4 years ago

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Hint 1: There is some thing fishy with the divergence.

Rajdeep Dhingra - 5 years, 1 month ago

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@Rajdeep Dhingra is nishant abhangi better than u??

Meera Somani - 4 years, 1 month ago

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