- #1
Nishant Garg
- 4
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Hello, new to this website, but one question that's been killing me is how can curl of a gradient of a scalar field be null vector when mixed partial derivatives are not always equal??
consider Φ(x,y,z) a scalar function
consider the determinant [(i,j,k),(∂/∂x,∂/∂y,∂/∂z),(∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z)] (this is from ∇×(∇Φ))
When you expand this you will get
[(∂^2Φ/∂y∂z)-(∂^2Φ/∂z∂y)]i-[(∂^2Φ/∂x∂z)-(∂^2Φ/∂z∂x)]j+[(∂^2Φ/∂x∂y)-(∂^2Φ/∂y∂x)]k
Now this can only be null vector when individual components are 0, and that's only when mixed partials are equal, but they are not always equal now are they?
consider Φ(x,y,z) a scalar function
consider the determinant [(i,j,k),(∂/∂x,∂/∂y,∂/∂z),(∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z)] (this is from ∇×(∇Φ))
When you expand this you will get
[(∂^2Φ/∂y∂z)-(∂^2Φ/∂z∂y)]i-[(∂^2Φ/∂x∂z)-(∂^2Φ/∂z∂x)]j+[(∂^2Φ/∂x∂y)-(∂^2Φ/∂y∂x)]k
Now this can only be null vector when individual components are 0, and that's only when mixed partials are equal, but they are not always equal now are they?