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 $\Omega_{m}$?


1 Answer 1


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.

  • $\begingroup$ @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. $\endgroup$
    – IamWill
    Jun 4, 2021 at 22:59

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