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Before the question, I need to mention some necessary definitions.

The rapidity is defined as: $$y=\frac{1}{2}\ln\frac{E+p_z}{E-p_z}=\frac{1}{2}\ln\frac{1+v_z}{1-v_z}=\tanh^{-1}(v_z)$$ where $v_z=p_z/E$ is the velocity along $z$ direction. $v_z=\tanh y$

We have defined the transverse mass $m_T$ and the longitudinal boost factor $\gamma_z$: $$m^2_T=m^2+p^2_T$$ $$\gamma_z=\frac{1}{\sqrt{1-v^2_z}}=\frac{E}{\sqrt{E^2-p^2_z}}=\frac{E}{\sqrt{m^2+p^2_T}}=\frac{E}{m_T}=\cosh y$$ It is easy to show that: $$E=m_T\gamma_z=m_T\cosh y$$ $$p_z=m_T\gamma_zv_z=m_T\sinh y$$ We note that under longitudinal boost, both $p_T$ and $m_T$ remain constant.

In high energy physics, one usually uses the Lorentz-invariant particle spectrum $EdN/d^3p$. $$p_z = m_T\sinh y \Rightarrow dp_z=m_T\cosh ydy=Edy\Rightarrow\frac{dp_z}{E}=dy$$ Therefore, $$\frac{d^3p}{E}=\frac{dp_zd^2p_T}{E}=dyd^2p_T=dyp_Tdp_Td\phi_p$$ The above Lorentz-invariant spectrum is often written as $$E\frac{dN}{d^3p}=\frac{dN}{dyd^2p_T}=\frac{dN}{dyp_Tdp_Td\phi_p}=\frac{dN}{dym_Tdm_Td\phi_p}$$ One can see that under Lorentz boost, $d^2p_T$ and $dy$ remain invariant, therefore, $d^3p/E = d^2p_Tdy$ is Lorentz invariant quantity.

My question is that is there another different method to show that $E\frac{dN}{d^3p}$ is Lorentz-invariant?

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$$\int\!d^4p\,\delta(p^2-m^2) =\\ \int\!d^4p\left(\frac{1}{2\sqrt{\vec{p}^2+m^2}|}\delta(p^0+\sqrt{\vec{p}^2+m^2})+\frac{1}{2\sqrt{\vec{p}^2+m^2}}\delta(p^0-\sqrt{\vec{p}^2+m^2})\right)\\ =\int\!d^4p\frac{1}{2\sqrt{\vec{p}^2+m^2}|}\left(\delta(p^0+\sqrt{\vec{p}^2+m^2})+\delta(p^0-\sqrt{\vec{p}^2+m^2})\right)\\ =\int\!d^3p\frac{1}{2\sqrt{\vec{p}^2+m^2}}\Big|_{p^0=+\sqrt{\vec{p}^2+m^2}} + \int\!d^3p\frac{1}{2\sqrt{\vec{p}^2+m^2}}\Big|_{p^0=-\sqrt{\vec{p}^2+m^2}}$$

The thing your wrote $d^3p/E$ is the first term and is invariant under proper orthochronous Lorentz transformations that do not change the sign of $p^0$.

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