This is regards to the lorentz invariance of a classical scalar field theory. We assume that the action which is $S= \int d^4 x \mathcal{L}$, is invariant under a Lorentz transformation. How do you prove that the integration measure $d^4 x$ is Lorentz invariant.
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2 Answers
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It's invariant because the Lorentz group is $SO(3,1)$ and the letter "S" stands for "special" which mathematically means the condition $$\det M = +1.$$ But the determinant is exactly the coefficient by which the volume form gets multiplied when the coordinates are Lorentz-transformed: $$ x \to M\cdot x\quad \Rightarrow \quad d^4 x \to \det M \cdot d^4 x $$ (this determinant-based transformation rule may also be derived if one views the volume form as an antisymmetric tensor with 4 indices) so if the determinant is equal to $+1$, the measure doesn't change. Well, $d^4 x$ is usually interpreted as $|d^4 x|$, so it's actually invariant under the whole $O(3,1)$, including the metrices with $\det M =-1$. And the condition $\det M=\pm 1$ (with "OR") follows from the orthogonality condition itself, so the adjective "special" is really unnecessary when we're already focusing on pseudoorthogonal matrices.
answered Dec 25, 2012 at 15:28
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$\begingroup$ Thanks Lubos. Can you also tell me why $d^3k \delta(k^2+m^2) \theta(k)$, is also lorentz invariant, in the frequency domain, for the sine-gordon equation? $\endgroup$– user7757Dec 25, 2012 at 15:44
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$\begingroup$ Sorry. Here $\delta(x)$ is the dirac delta function, and $\theta(x)$ is the unit step function. $\endgroup$– user7757Dec 25, 2012 at 15:47
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$\begingroup$ @ramanujan_dirac:this integral, if non zero, is expressed via the invariant mass $m$, so it is the same in all reference frames. Technically $\delta(x)$ and $dx$ have inverse dimensions: $\delta(A\cdot x) = \delta(x)/|A|$. $\endgroup$ Dec 25, 2012 at 16:10
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$\begingroup$ Dear @ramanujan_dirac, your latest differential is strange because you're mixing 3D and 4D notations. If $k$ is a 3-momentum, $\delta(k^2+m^2)$ may only be nonzero for $\vec k = 0$. I doubt you meant it, it's a strange distribution. There exist similar Lorentz-invariant expressions that may be shown to be Lorentz-invariant if you rewrite them in terms of the 4D invariants. $\endgroup$ Dec 25, 2012 at 18:32
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$d^3 x$ is Lorentz contracted and $dt$ is Einstein dilated by the same factor, so these factors disappear in the new $d^4 x'$.
answered Dec 25, 2012 at 15:46