Why is the trace of jacobian=the divergence?

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The discussion explores the theoretical connection between the trace of the Jacobian and the divergence of a vector field, emphasizing that the trace represents the sum of the partial derivatives of the vector field components. It establishes that the trace of the Jacobian matrix at a point equals the divergence, which is mathematically expressed as the sum of the derivatives of each component of the vector field. The conversation highlights the importance of invariants like the trace and determinant, which are coordinate-independent and thus provide a consistent representation of the function. Additionally, the participants clarify terminology regarding coordinate independence and the potential confusion arising from coordinate-dependent derivatives. This understanding is crucial for accurately interpreting mathematical relationships in vector calculus.
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What is the theoretical connection intuitively justifying that the trace of the jacobian=the divergence of a vector field? I know that this also equals the volume flow rate/original volume in the vector field but leaving that aside, what is the mathematical background behind establishing this equality?
 
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Are you asking for a proof? The proof is trivial (it follows immediately from the form of the jacobian). Let ##U\subseteq \mathbb{R}^{n}## be open and let ##F:U\rightarrow \mathbb{R}^{m}## be differentiable at ##a\in U##. Then the matrix representation of the total derivative of ##F## at ##a##, in the standard basis ##S##, is given by ##(DF(a))_{S} = (\frac{\partial F^j}{\partial x^i}(a))##. We call this matrix representation the jacobian of such a map. Hence ##Tr(DF(a))_{S} = \sum\frac{\partial F^{i}}{\partial x^{i}}(a)##. Thus if ##X:U\subseteq \mathbb{R}^{3}\rightarrow \mathbb{R}^{3}## is a vector field and ##a\in U##, ##Tr(DX)_{S}(a) = \frac{\partial X^{1}}{\partial x^1}(a) + \frac{\partial X^{2}}{\partial x^2}(a) + \frac{\partial X^{3}}{\partial x^3}(a) = (\nabla\cdot X)(a)##.
 
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Well the Jacobian it the derivative of the function in the f:V->V senses. So any other derivative can be found from it, in particular we like invariants of the Jacobian including the trace and determinant as they are coordinate independent and as such represent the function without interference.
 
lurflurf said:
in particular we like invariants of the Jacobian including the trace and determinant as they are coordinate independent and as such represent the function without interference.

I think you meant "the trace and determinant are independent of the coordinate system".

Could you please explain the meaning of "represent the function without interference"?
 
coordinate independent and independent of the coordinate system mean the same thing, but coordinate independent is shorter and sounds better. By "represent the function without interference" I mean that sometimes one uses coordinate dependent, but when one does she must always consider what is caused by the function and what is caused by the coordinate system. A condition must be considered in terms of the coordinates. A simple example if we define a derivative (x^2f)' the condition f'=0 becomes ((x^2f)'-2xf)/x^2=0 when might cause confusion.
 
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