What sort of vector is this?

[pellman]

What sort of "vector" is this?

A contravariant transforms as $$A^\rho \rightarrow {\Lambda^\rho}_\sigma A^\sigma$$.

A covariant vector $$A_\mu$$ can be built from $$A^\rho$$ by $$A_\mu=\eta_{\mu\rho}A^\rho$$. Then it transforms according to

$$A_\mu\rightarrow\eta_{\mu\rho}{\Lambda^\rho}_\sigma A^\sigma=\Lambda_{\mu\sigma}A^\sigma={\Lambda_\mu}^\sigma A_\sigma$$.

(Indeed, $${\Lambda_\mu}^\sigma {\Lambda^\rho}_\sigma=\delta^\rho_\mu$$ can be considered the definition of a Lorentz transformation, that is, one which leaves $$A^\mu A_\mu$$ invariant.)

But are there (covariant?) vectors whose components transform according to

$$B_\sigma \rightarrow {\Lambda^\rho}_\sigma B_\rho$$

?

What sort of a "vector" is B? It is not dual to a contravariant vector in the ordinary sense. And (why I put this in relativity forum and not a math forum) what does this type of vector signify within relativity theory? Are there any physical quantities which transform like this?Edit:
Ok. I think I get it. B is a covariant vector and the transformation being questioned represents transforming it by the inverse of the given Lorentz transform.

But please correct if I am wrong. Thanks!

 

[Hurkyl]

Just in case you haven't noticed it:

A covariant vector $$A_\mu$$ ... transforms according to

$$A_\mu\rightarrow {\Lambda_\mu}^\sigma A_\sigma$$.

are there (covariant?) vectors whose components transform according to

$$B_\sigma \rightarrow {\Lambda^\rho}_\sigma B_\rho$$

Those two expressions are the same expression, just with different variable names.

 

[HallsofIvy]

A contravariant transforms as $$A^\rho \rightarrow {\Lambda^\rho}_\sigma A^\sigma$$.

A covariant vector $$A_\mu$$ can be built from $$A^\rho$$ by $$A_\mu=\eta_{\mu\rho}A^\rho$$. Then it transforms according to

$$A_\mu\rightarrow\eta_{\mu\rho}{\Lambda^\rho}_\sigma A^\sigma=\Lambda_{\mu\sigma}A^\sigma={\Lambda_\mu}^\sigma A_\sigma$$.

(Indeed, $${\Lambda_\mu}^\sigma {\Lambda^\rho}_\sigma=\delta^\rho_\mu$$ can be considered the definition of a Lorentz transformation, that is, one which leaves $$A^\mu A_\mu$$ invariant.)

But are there (covariant?) vectors whose components transform according to

$$B_\sigma \rightarrow {\Lambda^\rho}_\sigma B_\rho$$

?

What sort of a "vector" is B? It is not dual to a contravariant vector in the ordinary sense.

Well, yes, it is. Just as you can construct a covariant vector that is dual to a contravariant vector (some texts would say "covariant" and "contravariant" components of the same vector), you can construct a covariant vector that is dual to a contravariant vector (that is basically what "dual" means- it works both ways). The dual to [tex]B_\sigma[/itex] is $$B^\gamma= \nu^{\gamma\sigma}B_\sigma$$.<br /> Assuming that your $\nu_{\mu\rho}$ is the metric tensor, $\mu^{\gamma\sigma}$ is the tensor such that $\nu_{\mu\rho}\nu^{\mu\gamma}= \nu^{\mu\gamma}\nu_{\mu\rho}= \delta^\gamma_\rho$. Algebraically, $\nu^{\gamma\sigma}$ is represented by the matrix that is the inverse to $\nu_{\gamma\sigma}$.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> And (why I put this in relativity forum and not a math forum) what does this type of vector signify within relativity theory? Are there any physical quantities which transform like this?<br /> <br /> <br /> Edit:<br /> Ok. I think I get it. B is a covariant vector and the transformation being questioned represents transforming it by the inverse of the given Lorentz transform. <br /> <br /> But please correct if I am wrong. Thanks! </div> </div> </blockquote> Yes, your edit is correct.[/tex]

 

[pellman]

Thanks!

 

[Fredrik]

Those two expressions are the same expression, just with different variable names.

Actually, they're not, because we have

$${\Lambda_\mu}^\sigma=(\Lambda^{-1})^\sigma{}_\mu$$

 

[Fredrik]

Covariant transformation from (say) frame S to frame S':

$$A'_\mu=\Lambda_\mu{}^\nu A_\nu$$

Multiply by [itex]\Lambda^\mu{}_\rho[/tex].<br /> <br /> $$\Lambda^\mu{}_\rho A'_\mu=\Lambda^\mu{}_\rho\Lambda_\mu{}^\nu A_\nu=\Lambda^\mu{}_\rho(\Lambda^{-1})^\nu{}_\mu A_\nu=\delta^\nu_\rho A_\nu=A_\rho$$<br /> <br /> This is clearly just a covariant transformation from S' to S.[/itex]