Proof of Lorentz invariance of Klein-Gordon equation

  • #1
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Main Question or Discussion Point

I would like to prove the Lorentz invariance of the Klein-Gordon equation by proving the invariance of the action ##\mathcal{S} = \int d^{4}x\ \mathcal{L}_{KG}## under a Lorentz tranformation.

I would like to do this by first proving the Lorentz invariance of the ##\mathcal{L}_{KG}## and then by proving the Lorentz invariance of the spacetime volume element ##d^{4}x##.

Firstly, under a Lorentz transformation ##\Lambda##,

##\mathcal{L}_{KG} = \frac{1}{2}(\partial_{\mu}\phi)(x)(\partial^{\mu}\phi)(x)-\frac{1}{2}m^{2}\phi^{2}(x) = \frac{1}{2}\eta^{\mu\nu}(\partial_{\mu}\phi)(x)(\partial_{\nu}\phi)(x)-\frac{1}{2}m^{2}\phi^{2}(x)##

##\qquad \rightarrow \frac{1}{2}\eta^{\mu\nu}{(\Lambda^{-1})^{\rho}}_{\mu}{(\Lambda^{-1})^{\sigma}}_{\nu}(\partial_{\rho}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

##\qquad = \frac{1}{2}{(\Lambda^{-1})^{\rho}}_{\mu}{(\Lambda^{-1})^{\sigma}}_{\nu}\eta^{\nu\mu}(\partial_{\rho}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

##\qquad = \frac{1}{2}{(\Lambda^{-1})^{\rho}}_{\mu}(\Lambda^{-1})^{\sigma\mu}(\partial_{\rho}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

##\qquad = \frac{1}{2}{(\Lambda^{-1})^{\rho}}_{\mu}(\Lambda)^{\mu\sigma}(\partial_{\rho}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

##\qquad = \frac{1}{2}\eta^{\rho\sigma}(\partial_{\rho}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

##\qquad = \frac{1}{2}\eta^{\sigma\rho}(\partial_{\rho}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

##\qquad = \frac{1}{2}(\partial^{\sigma}\phi)(\Lambda^{-1}x)(\partial_{\sigma}\phi)(\Lambda^{-1}x)-\frac{1}{2}m^{2}\phi^{2}(\Lambda^{-1}x)##

Are my steps to show that the Klein-Gordon Lagrangian ##\mathcal{L}_{KG}## is Lorentz invariant all correct?
 

Answers and Replies

  • #2
18,091
7,510
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
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There's nothing new to add to my post.

If someone checks my working and gives me a thumbs up, I will proceed to show that the spacetime volume element ##d^{4}x## is also Lorentz invariant, thus proving that the Klein-Gordon equation is Lorentz invariant (since the Klein-Gordon equation is derived from the variation of a soon-to-be-proved Lorentz invariant Klein-Gordon action).

I'm new to quantum field theory and general relativity, and so I'm practicing my technical skills and, in the process, learning the subjects.
 
  • #4
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As my query falls under the subject of classical (relativistic) field theory, I initially thought that the 'Classical Physics' subforum was better suited for this thread.
 
  • #5
Paul Colby
Gold Member
1,034
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Your steps look correct to me. The transformation ##\phi \rightarrow \phi(\Lambda^{-1} x)## permutes the points of ##\mathbb{R}^4## and since you're summing over them all the action is invariant. The invariance of ##dx^4## I believe follows from ##\det(\Lambda)=1##.
 
  • #6
1,344
32
Thanks!

Let me now prove the Lorentz invariance of the spacetime volume element ##d^{4}x##.

Under a Lorentz transformation ##\Lambda##,

##d^{4}x \rightarrow d^{4}x \Big|\frac{\partial x^{\mu}}{\partial x^{\mu'}}\Big|##, where ##\Big|\frac{\partial x^{\mu}}{\partial x^{\mu'}}\Big|## is the Jacobian of the Lorentz transformation.

The Jacobian of the Lorentz transformation = 1, so the spacetime volume element ##d^{4}x## is Lorentz invariant.
 

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