# Is this a right approach to show that $\partial_{\mu} \phi \partial^{\mu} \phi$ is Lorentz Invariant?

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

• 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. Apr 18 '19 at 20:48

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 $$\Lambda^T \cdot \eta \cdot \Lambda = \eta$$.
• 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 '19 at 3:56
• 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 '19 at 4:12