Cauchy momentum equation

In summary, the equation expressed by taking the outer product of a vector j with itself results in a matrix, not a vector. However, taking the divergence of this outer product yields a vector, as the divergence operator can only be applied to vectors. To find out how to take the divergence of a 2nd order tensor, such as a dyadic, one can refer to literature such as Appendix A of "Transport Phenomena" by Bird, Stewart, and Lightfoot.
  • #1
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I found that the equation is expressed by
e3d8084bef8e3043efefed4ef511b1613eaa54a9


there is outer product ...what I really don't get it is if j is a vector then the outer product of j and j is is obtained by multiplying each element of j by the complex conjugate of each element of j which is basically a matrix not a vector
 
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  • #2
hassouna said:
I found that the equation is expressed by
e3d8084bef8e3043efefed4ef511b1613eaa54a9


there is outer product ...what I really don't get it is if j is a vector then the outer product of j and j is is obtained by multiplying each element of j by the complex conjugate of each element of j which is basically a matrix not a vector
But when you take the divergence of the outer product of j and j, this yields a vector.
 
  • #3
divergence is a vector operator we can't operate it on matrix can't we??
 
  • #4
hassouna said:
divergence is a vector operator we can't operate it on matrix can't we??
Yes. The divergence of a vector is a scalar. The divergence of a 2nd order tensor is a vector. You need to check the literature to see how to take the divergence of a tensor (basically a dyad). See Appendix A of Transport Phenomena by Bird, Stewart, and Lightfoot to see how to work with dyadics and other 2nd order tensors.
 
  • #5
thank you for your help :smile:
 

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