# Invariance of Lorentz measure

Let $$m > 0$$ be fixed. If $$x=(x_{0},x_{1},x_{2},x_{3})$$ and $$y = (y_{0},y_{1},y_{2},y_{3})$$ are elements of $$\mathbb{R}^{4}$$, we denote the Lorentz inner product by: $$x\cdot \tilde{y} := x_{0}y_{0}-x_{1}y_{1}-x_{2}y_{2}-x_{3}y_{3}$$ Let $$H_{m}$$ be defined by: $$H_{m} := \{x \in \mathbb{R}^{4}: \hspace{0.1cm} \mbox{x\cdot \tilde{x} = x_{0}^{2}-x_{1}^{2}-x_{2}^{2}-x_{3}^{2} = m^{2} and x_{0} > 0}\}$$ This is a measurable subset of $$\mathbb{R}^{4}$$. Hence, we can consider $$H_{m}$$ equipped with its Borel $$\sigma$$-algebra, which is simply the set of all Borel subsets of $$\mathbb{R}^{4}$$ which are also subsets of $$H_{m}$$. Next, define $$j_{m}: H_{m} \to \mathbb{R}^{3}$$ by $$j_{m}(x_{0},x_{1},x_{2},x_{3}) = (x_{1},x_{2},x_{3})$$. This is a projection map, so it's measurable. Finally, define a measure for each measurable subset $$E$$ of $$H_{m}$$ by: $$\Omega_{m}(E) := \int_{j_{m}(E)}\frac{d^{3}{\bf{x}}}{\sqrt{m^{2}+|{\bf{x}}|^{2}}}$$ where I denoted by $${\bf{x}} = (x_{1},x_{2},x_{3})$$ and $$|{\bf{x}}|^{2} = x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$$.

Let $$\Lambda \in \mathscr{P}_{+}^{\uparrow}$$ be a Lorentz measure, i.e. a linear transformation $$\Lambda: \mathbb{R}^{4} \to \mathbb{R}^{4}$$ with $$\operatorname{det} \Lambda = 1$$ and satisfying: $$(\Lambda x)\cdot \widetilde{(\Lambda y)} = x\cdot \tilde{y}$$ for every $$x,y \in \mathbb{R}^{4}$$.

By the above condition, $$H_{m}$$ is invariant under $$\mathscr{P}_{+}^{\uparrow}$$, that is, if $$\Lambda \in \mathscr{P}_{+}^{\uparrow}$$ then $$\Lambda(H_{m}) \subseteq H_{m}$$. Additionally, it is known that $$\Omega_{m}$$ is also invariant under $$\mathscr{P}_{+}^{\uparrow}$$, i.e. $$\Omega_{m}(\Lambda E) = \Omega_{m}(E)$$ holds for every $$\Lambda \in \mathscr{P}_{+}^{\uparrow}$$ and $$E$$ measurable. There is a proof of this result in Reed & Simon - Vol II. In their proof, they consider the measure: $$\Omega_{m}^{f}(E) = \int_{h(E)} \frac{f(y)d{\bf{x}}dy}{\sqrt{m^{2}+|{\bf{x}}|^{2}}}$$ where $$h: (x_{0},{\bf{x}}) \mapsto ({\bf{x}}, x\cdot \tilde{x})$$ and construct a sequence $$f_{n}$$ in $$C_{0}^{\infty}(0,\infty)$$ which converges to $$\delta(y-m)$$.

Question: Simon and Reed mention that an alternative proof of the invariance of $$\Omega_{m}$$ under $$\mathscr{P}_{+}^{\uparrow}$$ can be obtained in a straightfoward way by computing the action of $$\Lambda \in \mathscr{P}_{+}^{\uparrow}$$ on $$H_{m}$$. It seems reasonable to me that a more straightfoward proof exists, but I couldn't find it myself and I don't understand exactly what Reed & Simon meant with their statement. Do you know what is this straightfoward approach to prove the invariance of $$\Omega_{m}$$?

I don't think this is quite what you had in mind, but I recommend the following approach, using differential geometry. The surface $$H_m$$ is a submanifold of $$\mathbb{R}^4$$, and the Lorentz metric $$g$$ naturally restricts (pullback) to a (negative-definite) Riemannian metric on $$H_m$$, call it $$g_m$$. Any Riemannian manifold has a natural measure associated with the metric, whose density in any coordinate system is $$\sqrt{\det g}$$. The measure $$\Omega_m$$, is, up to a constant, precisely this measure associated to $$g_m$$. From this point of view, $$\Omega_m$$ is manifestly invariant under the Lorentz group, because the Lorentz group preserves $$g$$. Moreover, the pullback $$g_m$$ and its determinant may be explicitly computed using standard methods.
On $$H_m$$ we have the constraint $$x_0=\sqrt{m^2 + |\mathbf{x}|^2}$$ Thus $$dx_0 = \frac{x_1dx_1 + x_2dx_2 + x_3dx_3}{\sqrt{m^2 + |\mathbf{x}|^2}}$$ So $$g_m=dx_0^2 - dx_1^2 - dx_2^2 - dx_3^2 = \frac{(x_1dx_1 + x_2dx_2 + x_3dx_3)^2}{m^2+|\mathbf{x}|^2}- dx_1^2 - dx_2^2 - dx_3^2$$ After some computation, one finds $$\sqrt{-\det g_m}=\frac{m}{\sqrt{m^2 + |\mathbf{x}|^2}}$$ which is $$m$$ times the density cited in the question.
• @idrinehart, thanks for your answer! It seems a very nice approach and I can get the idea behind it but, to be sincere, I have almost no background on differential geometry and it is hard to me to follow your answer 100%; I can appreciate the elegance of the approach, however. I was thinking more like some sort of "change of variables"; I mean, $\Omega_{m}(\Lambda E)$ is the integral over $j_{m}(\Lambda E)$ so we maybe we can redefine ${\bf{x}}$ or something like that to get back to $j_{m}(E)$. It is not clear to me yet, tho. Commented Jun 4, 2021 at 22:59