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I want to show the Lorentz invariance of $d^3 p/E$ (Eq. 8.11 of Mandl-Shaw), where $E$ is the relativistic energy. Peskin-Schroeder gives sort-of, a proof in section 2.3 which I am convinced of. But my independent line of reasoning leads me to a contradiction.

$d^4$ p is Lorentz invariant $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$(of course)

$\implies d^3p~dp^0$ is Lorentz invariant

$\implies d^3p~E$ is Lorentz invariant because $E$ and $p^0$ are both time-components of 4-vectors $(E, p^1, p^2,p^3)$ and $(dp^0, dp^1, dp^2,dp^3)$ respectively.

So finally, I arrive at the conclusion that it's $d^3p~E$ which is Lorentz invariant, not $ d^3p/E$. What did I do wrong above?

Sashwat Tanay
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  • Possible duplicates: https://physics.stackexchange.com/q/83260/2451 , https://physics.stackexchange.com/q/53534/2451 and links therein. – Qmechanic Dec 03 '18 at 20:46
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    I think that your assertion "...$d^3p~E$ is Lorentz invariant because $E$ and $p^0$ are both time-components of 4-vectors..." is wrong. – Frobenius Dec 03 '18 at 22:15
  • @Frobenius Please tell me where exactly I am going wrong. Does $E$ not transform as the time-component? If it does not, then why? If it does, then why is $dp^3 E$ not invariant? I need to pinpoint the problem exactly. – Sashwat Tanay Dec 03 '18 at 23:19
  • $d^3p\ dE$ is Lorentz invariant. $d^3p\ E$ is not. $E$ and $dE$ transform differently under Lorentz transformations. – AccidentalFourierTransform Dec 04 '18 at 00:02

1 Answers1

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I think that your assertion "...$d^3p~E$ is Lorentz invariant because $E$ and $p_0$ are both time-components of 4-vectors..." is wrong.

Replace the 4-vector $\mathbf p=\left(p^0,p^1,p^2,p^3\right)$ by the space-time position 4-vector $\mathbf x=\left(x^0\equiv t,x^1,x^2,x^3\right)$. The infinitesimal 4-volume $\mathrm dt\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3$ is Lorentz invariant \begin{equation} \mathrm dt'\,\mathrm d x'^1\,\mathrm d x'^2\,\mathrm d x'^3=\mathrm dt\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3 \tag{01}\label{01} \end{equation} but I don't think you could prove that $t\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3$ is Lorentz invariant \begin{equation} t'\,\mathrm d x'^1\,\mathrm d x'^2\,\mathrm d x'^3=t\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3 \tag{02}\label{02} \end{equation}

It's not permitted to mix the components of various 4-vectors in infinitesimal Lorentz invariant scalars like $\mathrm dx^0\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3$. The relation between the infinitesimal 4-volumes in Minkowski space is \begin{equation} dx'^{1}dx'^{2}dx'^{3}dx'^{0} =\begin{vmatrix} \dfrac{\partial x'_{1}}{\partial x_{1}}& \dfrac{\partial x'_{1}}{\partial x_{2}}&\dfrac{\partial x'_{1}}{\partial x_{3}}&\dfrac{\partial x'_{1}}{\partial x_{0}}\\ \dfrac{\partial x'_{2}}{\partial x_{1}}& \dfrac{\partial x'_{2}}{\partial x_{2}}&\dfrac{\partial x'_{2}}{\partial x_{3}}&\dfrac{\partial x'_{2}}{\partial x_{0}}\\ \dfrac{\partial x'_{3}}{\partial x_{1}}& \dfrac{\partial x'_{3}}{\partial x_{2}}&\dfrac{\partial x'_{3}}{\partial x_{3}}&\dfrac{\partial x'_{3}}{\partial x_{0}}\\ \dfrac{\partial x'_{0}}{\partial x_{1}}& \dfrac{\partial x'_{0}}{\partial x_{2}}&\dfrac{\partial x'_{0}}{\partial x_{3}}&\dfrac{\partial x'_{0}}{\partial x_{0}} \end{vmatrix} dx^{1}dx^{2}dx^{3}dx^{0}=\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{0}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{0}\right)}\right\vert dx^{1}dx^{2}dx^{3}dx^{0} \tag{03}\label{03} \end{equation} where $\:\left\vert\partial\left(x'^{1},x'^{2},x'^{3},x'^{0}\right)/\partial\left(x^{1},x^{2},x^{3},x^{0}\right)\right\vert\:$ the Jacobian, that is determinant of the Jacobi matrix. But the Jacobi matrix is the Lorentz matrix $\:\Lambda\:$ with $\:\det(\Lambda)=+1$, that is \begin{equation} \left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{0}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{0}\right)}\right\vert=\det(\Lambda)=+1 \tag{04}\label{04} \end{equation} so \begin{equation} dx'^{1}dx'^{2}dx'^{3}dx'^{0} =dx^{1}dx^{2}dx^{3}dx^{0}=\text{scalar invariant} \tag{05}\label{05} \end{equation} There is no similar way to prove that \begin{equation} x'^{0} dx'^{1}dx'^{2}dx'^{3} =x^{0}dx^{1}dx^{2}dx^{3}=\text{scalar invariant} \tag{06}\label{06} \end{equation} because simply it's incorrect.

Frobenius
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