Divergence of a vector field is a scalar field?

  • Thread starter merrypark3
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  • #1
merrypark3
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Hello.

How can I show the Divergence of a vector field is a scalar field(in [tex]E^{3}[/tex]) ?
Should I show that Div is invariant under rotation?

[tex]x^{i'}=a^{ij}x^{j},V^{'}_{i}(\stackrel{\rightarrow}{x})=a_{ij}v_{j}(\stackrel{\rightarrow}{x})[/tex]

then

[tex]\frac{\partial V^{'}_{i}(\stackrel{\rightarrow}{x})}{\partial x^{'i}}=\frac{\partial(a_{ij}V_{j} (\stackrel{\rightarrow}{x})) }{\partial(a^{ij} x^{j} ) } = \frac{\partial V_{i} (\stackrel{\rightarrow}{x})}{\partial x^{i}} [/tex]

How can I prove the last equality?
 

Answers and Replies

  • #2
chrisk
288
1
Start with the dot product of the del operator with the vector field; this is the expression for divergence.
 
  • #3
vela
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Hello.

How can I show the Divergence of a vector field is a scalar field(in [tex]E^{3}[/tex]) ?
Should I show that Div is invariant under rotation?

[tex]x^{i'}=a^{ij}x^{j},V^{'}_{i}(\stackrel{\rightarrow}{x})=a_{ij}v_{j}(\stackrel{\rightarrow}{x})[/tex]

then

[tex]\frac{\partial V^{'}_{i}(\stackrel{\rightarrow}{x})}{\partial x^{'i}}=\frac{\partial(a_{ij}V_{j} (\stackrel{\rightarrow}{x})) }{\partial(a^{ij} x^{j} ) } = \frac{\partial V_{i} (\stackrel{\rightarrow}{x})}{\partial x^{i}} [/tex]

How can I prove the last equality?
Use the chain rule to express the derivative with respect to x'i in terms of the derivatives with respect to the unprimed coordinates.
 

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