- #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]

[tex]divA^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}[/tex]

scalar (invariant)

[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]

[tex]j_{\mu}=(-\vec{j},c\rho)[/tex]

[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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