Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
    
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? –  ramanujan_dirac Dec 25 '12 at 15:44
    
Sorry. Here $\delta(x)$ is the dirac delta function, and $\theta(x)$ is the unit step function. –  ramanujan_dirac Dec 25 '12 at 15:47
    
@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: $\delta(A\cdot x) = \delta(x)/|A|$. –  Vladimir Kalitvianski Dec 25 '12 at 16:10
    
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. –  Luboš Motl Dec 25 '12 at 18:32
add comment

$d^3 x$ is Lorentz contracted and $dt$ is Einstein dilated by the same factor, so these factors disappear in the new $d^4 x'$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.