- #1
Petar Mali
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[tex]\frac{\partial \rho}{\partial t}+div\vec{j}=0[/tex]
In Deckart coordinate system
[tex]\frac{\partial j_x}{\partial x}+\frac{\partial j_y}{\partial _y}+\frac{\partial j_z}{\partial _z}+\frac{\partial (c\rho)}{\partial (ct)}=0[/tex]
definition
[tex]divA^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}[/tex]
scalar (invariant)
Why I define divergence like that? Is there some certain rules for that?
[tex]j^{\mu}=(j_x,j_y,j_z,c\rho)=(\vec{j},c\rho)[/tex]
[tex]\frac{\partial j^{\mu}}{\partial x^{\mu}}=0[/tex]
[tex]divj^{\mu}=0[/tex]
Now is satisfied
[tex]j_{\mu}=(-\vec{j},c\rho)[/tex]
Can I interprate this like time inversion. Changing od indeces, think of that?
What is with
[tex]divj_{\mu}=?[/tex]
In Deckart coordinate system
[tex]\frac{\partial j_x}{\partial x}+\frac{\partial j_y}{\partial _y}+\frac{\partial j_z}{\partial _z}+\frac{\partial (c\rho)}{\partial (ct)}=0[/tex]
definition
[tex]divA^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}[/tex]
scalar (invariant)
Why I define divergence like that? Is there some certain rules for that?
[tex]j^{\mu}=(j_x,j_y,j_z,c\rho)=(\vec{j},c\rho)[/tex]
[tex]\frac{\partial j^{\mu}}{\partial x^{\mu}}=0[/tex]
[tex]divj^{\mu}=0[/tex]
Now is satisfied
[tex]j_{\mu}=(-\vec{j},c\rho)[/tex]
Can I interprate this like time inversion. Changing od indeces, think of that?
What is with
[tex]divj_{\mu}=?[/tex]
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