Notational problem in tensor calculus

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The discussion centers on the application of the Einstein summation convention in tensor calculus, specifically regarding the divergence of the expression \nabla \bullet (x_{i}\vec{a}). Participants clarify that the notation x_{i}\vec{a} refers to the i-th component of a tensor multiplied by a constant vector. The divergence is computed as \partial_{j} (a_{j}x_{i}), leading to the conclusion that since the components of vector a are constants, this simplifies to a_j \partial_{j} x_{i}. The conversation emphasizes the need for clarity in notation and understanding of the underlying calculus principles.
jacobrhcp
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Using the Einstein convention, is this about right? (indexes run from 1 to 3):

\nabla\bullet(x_{i}a)=div(x_{i}a)=\partial_{j}a_{j}x_{i}=3x_{i}
 
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The partial derivatives are take wrt what variable ? x or a ?
 
it's said x above the exercises \partial_{j}=\frac{\partial}{\partial x_{j}}}

and the question is:

calculate explicitly: \nabla \bullet (x_{i}\vec{a}), where a is a constant vector...

my attempt is way off, but I don't feel at home in these new symbols enough to get the right answer.
 
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x_i{\vec a} is not clear notation. Does it mean the i component of a tensor
[xa], or does it mean {\vec\hat i}\cdot{\vec r}{\vec a}?
 
jacobrhcp said:
Using the Einstein convention, is this about right? (indexes run from 1 to 3):

\nabla\bullet(x_{i}a)=div(x_{i}a)=\partial_{j}a_{j}x_{i}=3x_{i}

You are right that it means \partial_{j} (a_{j}x_{i}) But the components of a are constants so this is equal to a_j \partial_{j} x_{i} Now, what does \partial_{j} x_{i}give ?
 

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