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In page 24 of Srednicki's QFT textbook, he says that $\mathrm{d}^4x$ is a Lorentz scalar. I understand that the determinant of a Lorentz matrix is always $\pm 1$. So in an improper Lorentz transformation, shouldn't the Jacobian be negative?

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1: $\int_{-\infty}^{\infty} \mathrm{d}x f(x) \rightarrow \int_{\infty}^{-\infty} \mathrm{d}(-x) f(-x) = \int_{-\infty}^{\infty} \mathrm{d}x f(-x)$ since you interchange the limits as well. 2: What do you mean exactly? That the equations of motion are invariant but not the action itself? That would be okay in a classical theory, I guess, since the action is just a tool to derive equations of motion. But in a quantum theory the off-shell behaviour of the action matters as well, so the action actually has to be a Lorentz scalar. –  Michael Brown Jan 19 '13 at 15:06
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Srednicki does say a couple of pages earlier that when we talk of something being Lorentz invariant, we generally mean under the proper orthochronous Lorentz group. In this case his statement is OK. (talking about first question here) –  twistor59 Jan 19 '13 at 15:08
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Actually, the integration measure $\mathrm{d}^n x$ (which is an $n$-form in $n$ dimensions) transforms with the absolute value of the Jacobian, so also for improper Lorentz transformations the measure is Lorentz invariant. –  Vibert Jan 19 '13 at 17:29
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Hi Nabil, and welcome to Physics Stack Exchange! We prefer that you ask each individual question on its own, so I've edited out the second of your questions. Feel free to make a separate post for it. –  David Z Jan 20 '13 at 0:16

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