I want to learn how to check if any scalar is a Lorentz scalar, so what is the definition of being invariant under Lorentz transformation? Is it correct to say that $\phi$ is invarant under Lorentz transformation if and only if:

$\phi(x^\nu) = \phi({\Lambda^\nu}_\mu x^\mu)$ for all $x^\mu \in \mathbb{R}^4$ and Lorentz transformtions $\Lambda$?

Or is the definition something different? A short answer that is very clear, would be best.


In case of non-field quantity that has one value for the whole inertial system, like net electric charge of a body, it means its value is the same in all inertial systems. For example, electron has the same charge in all inertial systems. Therefore it is Lorentz invariant.

In case of a field quantity like $E^2 - c^2B^2$, the value depends on position and time (event). In one inertial system the value of this quantity for event $x^\mu$ is

$$ E^2(x^\mu) - c^2B^2(x^\mu) $$

In another system where the same event has coordinates $x'^{\mu}$ and the electric and magnetic fields are given by functions $\mathbf E',\mathbf B'$, the value is

$$ E'^2(x'^\mu) - c^2B'^2(x'^\mu). $$

It can be shown that $$ E^2(x^\mu) - c^2B^2(x^\mu) = E'^2(x'^\mu) - c^2B'^2(x'^\mu). $$

It is this property that is meant when saying $E^2-c^2B^2$ is Lorentz invariant. In general case, a field $\phi(x^\mu)$ is Lorentz invariant if its evaluation in two inertial systems, connected via Lorentz transformation, leads to the same value:

$$ \phi(x^\mu) = \phi'(x'^\mu). $$

  • $\begingroup$ Unfortunately, I didn't understand this. Are you confirming that what I proposed is the definition? $\endgroup$ – Mikkel Rev Aug 28 '16 at 16:20
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    $\begingroup$ @MariusJonsson No yours is wrong. Think of temperature. If I rotate around the room I may measure different temperature at different points in the room (the room is not isotropic). That's what your formula describes. If I rotate around but still measure the original point from my new point of view then I measure the same temperature - this is scalar. Now extend from 3d space to 4d space-time. $\endgroup$ – Bruce Greetham Aug 28 '16 at 16:35

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