# Why is $d^4p$ "obviously" Lorentz invariant?

Is there a mathematical and a physical reason? I believe it follows from the jacobian of the Lorentz transformation. But, if that is so, I don't see why it would be "obvious" unless you can do such calculations instantly in your head. What is the physical meaning of the invariance of Lorentz measure? Is it just conservation of 4-momentum?

• See the answer below, but it's "obvious" because the one of the basic properties of the Lorentz transformations is they are unitary. Commented Oct 26, 2015 at 13:26
• @levitopher But what is the physical meaning of the measure itself? Commented Oct 26, 2015 at 13:40
• Well, $d^4p$ is the volume element in momentum space with coordinates $(p^0,p^1,p^2,p^3)$. The physical meaning of it being invariant is just that if you picked a different set of coordinates $(p'^0,p'^1,p'^2,p'^3)$ that was related to the first by Lorentz boosts and translations, you would get the same measure. Commented Oct 26, 2015 at 13:45

Under a Lorentz transformation $p \to p' = \Lambda p$, we have $$d^4 p' = d^4 p | \det \Lambda | = d^4 p$$ since $\det \Lambda = \pm 1$.

• Thanks. I'm also asking for a physical interpretation of dp. Commented Oct 26, 2015 at 16:26

The defining property of the Lorentz transform is $$\Lambda^T\eta\Lambda = \eta$$ where $\eta_{\mu \nu}$ is the flat space-time metric $diag(-+++)$. Since the determinant is linear and the transformations are non-degenerate (as is also the metric itself), we have $$\mathrm{det}\Lambda^T \, \mathrm{det}\eta \;\mathrm{det}\Lambda=\mathrm{det}\eta\,,\implies \mathrm{det}\Lambda^T \mathrm{det}\Lambda= (\mathrm{det}\Lambda)^2=1$$ I.e., very straightforwardly from the defining properties of the Lorentz transform, the determinant of it can be only $\pm 1$.

One way to intuitively expect this is from the understanding that the Lorentz transform is "orthogonal" or a "space-time rotation" in a certain sense and thus it is natural that it's determinant is $\pm 1$ similarly to the spatial rotation.

Another way to see this is the equivalence principle. By Lorentz-transforming into another frame, you should get a fully equal and emancipated set of coordinates. But $\mathrm{det} \Lambda < 1$ would mean that you are somehow "loosing information per coordinate unit". You could then repeat a series of $\mathrm{det} \Lambda < 1$ transformations to reach a set of degenerate coordinates, or conversely by $\mathrm{det} \Lambda > 1$, a blown-up set of coordinates. This is simply against the spirit of special relativity, the transformation should not prefer the "initial" or "latter" reference frame, they should all represent an equivalent description of the physical situation.

The transformation is linear, so the Jacobian is just $\Lambda$ itself, so the $\mathrm{d}^4 p$ (where we mean the momenta in Cartesian coordinates!) is trivially Lorentz-invariant.

• Thanks. I also wanted to relate this invariant to conservation of 4-momentum directly somehow. It relates indirectly because lorentz vectors are defined such that their dot product is invariant. Commented Oct 26, 2015 at 16:30
• Well, the conservation of four-momentum is in fact a consequence of invariance of the theory with respect to the full Poincaré group, i.e. Lorentz plus translations. The four-momentum is simply invariant under translations, so this is usually not even commented upon. But the invariance of momentum is not the reason for conservation, you can see the four-momentum conservation by applying Noether's theorem to translation symmetry of the Lagrangian of a free relativistic particle.
– Void
Commented Oct 26, 2015 at 18:10