Are the following 2 equations lorentz invariant?

[Dreak]

[itex]\partial[/itex]_μ[itex]\phi[/itex][itex]\partial[/itex]^μ[itex]\phi[/itex]

and

[itex]\partial[/itex]_μ[itex]\partial[/itex]^μ[itex]\phi[/itex]

with [itex]\phi[/itex](x) a scalar field

 

[Mentz114]

Suppose ##\partial_{\mu'}=\Lambda^\mu_{\mu'}\partial_{\mu}## and ##\partial^{\mu'}=\Lambda^{\mu'}_{\mu}\partial^{\mu}##.
Then
##\partial_{\mu'}\partial^{\mu'}\phi=\Lambda^\mu_{\mu'}\partial_{\mu} \Lambda^{\mu'}_{\mu}\partial^{\mu}\phi=\Lambda^\mu_{\mu'} \Lambda^{\mu'}_{\mu}\partial_{\mu}\partial^{\mu}\phi##.

So the scalar is invariant if ##\Lambda^\mu_{\mu'}## is the inverse of ## \Lambda^{\mu'}_{\mu}##.

Now, ##\Lambda^\mu_{\mu'}## is not a tensor but a transformation matrix. It so happens that the inverse of the LT is found by reversing the sign of the boost velocity. This operation ##\Lambda_b^{b'}=\eta_{ba}\Lambda_{a'}^a\eta^{a'b'}## also reverses the sign. So for the Lorentz transformation ##\Lambda^\mu_{\mu'}## is the inverse of ## \Lambda^{\mu'}_{\mu}##.

 

[Dreak]

So the answer is yes?!
I actually thought the second one was lorentzinvariant because:

[itex]\partial[/itex]_μ[itex]\partial[/itex]^μ is the d'Alembertian which I thought was Lorentzinvariant, while a scalar field is always Lorentzinvariant?

And what about the other equation? I'm not sure what the influence of the second scalar field in the equation yields.

 

[Mentz114]

So the answer is yes?!
I actually thought the second one was lorentzinvariant because:

[itex]\partial[/itex]_μ[itex]\partial[/itex]^μ is the d'Alembertian which I thought was Lorentz invariant, while a scalar field is always Lorentz invariant?

And what about the other equation? I'm not sure what the influence of the second scalar field in the equation yields.

The answer is 'yes' for both equations. If you write out the first expression in terms of the transformed coords, there is another cancellation of ##\Lambda^\mu_{\mu'}\Lambda_\mu^{\mu'}##.

Rank-0 tensors ( ie no indexes) are always invariant. Differentiating a rank-0 tensor creates a new tensor with rank-1. Scalars formed by contractions of tensors are invariant.

 

[Dreak]

Thank you very much! :)

 

[vanhees71]

One should, however, be a bit more careful with the indices in the Lorentz transformations and write them also with proper order. Let
[tex]{\Lambda^{\mu}}_{\nu}[/tex]
the Lorentz-transformation matrix for contravariant vector/tensor components, i.e.,
[tex]x'^{\mu}={\Lambda^{\mu}}_{\nu} x^{\nu}.[/tex]
A Lorentz transform obeys
[tex]\eta_{\mu \nu} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma}=\eta_{\rho \sigma},[/tex]
or
[tex]\eta_{\mu \nu} \eta^{\rho \alpha} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.[/tex]
This implies
[tex]\eta^{\rho \alpha} \eta_{\mu \nu} {L^{\mu}}_{\rho}={(L^{-1})^{\alpha}}_{\nu}={L_{\mu}}^{\alpha}.[/tex]

 

[Mentz114]

One should, however, be a bit more careful with the indices in the Lorentz transformations and write them also with proper order. Let
[tex]{\Lambda^{\mu}}_{\nu}[/tex]

...

Thanks for clarifying that.