If I have a function $\ f(x)$, what does it mean for it to be Lorentz invariant? I believe it is that $\ f( \Lambda^{-1}x ) = f(x)$, but I think I'm missing something here.
Furthermore, if $g(x,y)$ is Lorentz invariant, does this means that $g(\Lambda^{-1}x,\Lambda^{-1}y)=g(x,y)$?
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EDIT: Allow me to explain the source of my confusion….I've been looking at resources online, where it says that the definition should be $\phi^{\prime}(x^{\prime})=\phi(x)$. But $\phi^{\prime}(x)=\phi(\Lambda^{-1}x)$ under a Lorentz transformation $x^{\prime}=\Lambda x$. So then I get:
$\phi^{\prime}(x^{\prime})=\phi(\Lambda^{-1}x^{\prime})=\phi(\Lambda^{-1}\Lambda x)=\phi(x)$
This seems completely trivial, and doesn't look like a condition I would need to check on a case-by-case basic.
Best Answer
It may look trivial, but that's what it is.
A Lorentz scalar is an element of the 0-dimensional vector space considered as a representation space of the trivial $(0,0)$ representation of the Lorentz group. In other words, a scalar $s$ transforms trivially:
$$s \rightarrow s' = s$$
A Lorentz vector is an element of the 4-dimensional vector space considered as a representation space of the standard $(\frac{1}{2}, \frac{1}{2})$ representation of the Lorentz group. In other words, a vector $\mathbf{v}$ transforms as:
$$\mathbf{v} \rightarrow \mathbf{v}' = \mathbf{\Lambda} \mathbf{v}$$
$\mathbf{\Lambda}$ is the Lorentz transformation matrix.
A scalar function, such as $\phi(\mathbf{x})$ maps a Lorentz vector to a Lorentz scalar, i.e. $$\mathbf{x} \mapsto \phi(\mathbf{x})$$
Consequently, it transforms to $$ \mathbf{\Lambda x} \mapsto \phi'(\mathbf{\Lambda x}) = \phi(\mathbf{x}) $$
Therefore, $$\phi' (\mathbf{x}) = \phi (\mathbf{\Lambda^{-1} x})$$ Indeed, $$\phi' (\mathbf{x'}) = \phi (\mathbf{\Lambda^{-1} \Lambda x}) = \phi (\mathbf{x})$$