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 $g(\Lambda^{-1}x,\Lambda^{-1}y)=g(x,y)$?

$\ $

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.

  • 2
    $\begingroup$ The idea is to distinguish between scalars and functions like density. If $\rho(x)$ is a density it will change under a Lorentz transformation! (The hint here is that rho is "really" a component of the stress-energy tensor. Nevertheless, the concept of "tensor densities" which behave this way is used in Landau's The Classical Theory of Fields towards the end with general relativity, if I recall correctly.) $\endgroup$
    – user12029
    Oct 24, 2016 at 18:47
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    $\begingroup$ So you have seen how $\phi '(x) \neq \phi (x)$. Compute the difference. Now check a plane wave, $e^{i k\cdot x}$ and compute the difference. $\endgroup$ Oct 24, 2016 at 22:57

2 Answers 2


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, $$\phi' (\mathbf{x'}) = \phi (\mathbf{\Lambda^{-1} \Lambda x}) = \phi (\mathbf{x})$$

  • $\begingroup$ What about constants that appear in $\phi(x)$, do they also transform accordingly? $\endgroup$
    – Tera
    Jul 3, 2021 at 10:01

You can also think of a scalar this way: it's a function $F$ that maps points of spacetime to numbers.

So, if $q$ is a point in spacetime, the representative of $F$ relative to a coordinate system $q \rightarrow x(p)$ is, say, $f$, defined by $f(x(q)) = F(q)$. If we have a second coordinate system $x' = \Lambda \cdot x$, then the representative of $F$ with respect to the new coordinate system is $f'(x'(q)) = F(q)$, so $f'(\Lambda \cdot x(q)) = f(x(q))$ or $f'\circ \Lambda = f$ or, finally, $f' = f \circ \Lambda^{-1}$.

The point is that while $F$ is independent of any coordinate system, its representatives w.r.t. different coordinate systems have to be related in this way since they're all tied to the same $F$.

  • $\begingroup$ This the best way I've ever seen it put. $\endgroup$ Mar 1, 2017 at 17:24

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