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When trying to convince myself that $\partial_{\mu} \phi \partial^{\mu} \phi $ is Lorentz Invariant, I stumbled upon this approach:

enter image description here

The last equation should read - $\partial_{i} \phi \partial^{i} \phi = \partial_{i^{'}} \phi^{'} \partial^{i^{'}} \phi^{'} $

Here since $C_1$, $C_2$ and $C_3$ are just scalars, it permits us do something like $\frac{C_1 C_2}{C_3}$. And this shows that $\partial_{\mu} \phi \partial^{\mu} \phi $ is Lorentz Invariant.

Does this seem logical to do this and prove it this way? I understand that there exist other better methods to show the same, I am just wondering if this method is consistent.

** $dx$ and $dx'$ are related by the Lorentz Transform.

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    $\begingroup$ In the future, please try to type up photos using mathjax, as it is much easier to read for others, and easier for search engines to parse. $\endgroup$ Apr 18 '19 at 20:48
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It's not really consistent. You're manipulating symbols in a way that doesn't make sense, especially when you bring together $c_1$, $c_2$ and $c_3$ in $c_1 c_2/c_3$. Just start by showing how $$\frac{\partial}{\partial x^\mu} \phi$$ transforms under a Lorentz transformation $x \to x' = \Lambda \cdot x$. Finally use a key fact you know about the matrix $\Lambda$, namely $\Lambda^T \cdot \eta \cdot \Lambda = \eta$.

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  • $\begingroup$ 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 $C_1 C_2$ ? $\endgroup$
    – 256ABC
    Apr 17 '19 at 3:56
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    $\begingroup$ 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. $\endgroup$ Apr 17 '19 at 4:12

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