Timeline for Is this a right approach to show that $\partial_{\mu} \phi \partial^{\mu} \phi $ is Lorentz Invariant?
Current License: CC BY-SA 4.0
4 events
when toggle format | what | by | license | comment | |
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Apr 17, 2019 at 6:30 | vote | accept | 256ABC | ||
Apr 17, 2019 at 4:12 | comment | added | Hans Moleman | Of course the multiplication $c_1 c_2$ is technically fine. The weird part is dividing by $dx^i dx_i$ and turning $dx^\sigma/dx_i$ into a Jacobian. It's not! More generally there's no need to introduce line elements $dx^i$ or $dx^\sigma$ in this proof at all. When you prove that an inner product in $\mathbb{R}^3$ is invariant under $SO(3)$ you don't need it either. | |
Apr 17, 2019 at 3:56 | comment | added | 256ABC | Would you say that if $\vec{a}.{b} = \vec{c}.\vec{d}$ and $\vec{a_1}.{b_1} = \vec{c_1}.\vec{d_1}$, then $(\vec{a}.{b})(\vec{a_1}.{b_1}) = (\vec{c}.\vec{d})(\vec{c_1}.\vec{d_1})$ is true ? And if so, I am just wondering why that doesn't translate when we are talking about $C_1 C_2$ ? | |
Apr 17, 2019 at 2:04 | history | answered | Hans Moleman | CC BY-SA 4.0 |