# What sort of vector is this?

1. Jul 7, 2009

### 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!

Last edited: Jul 7, 2009
2. Jul 7, 2009

### Hurkyl

Staff Emeritus
Re: What sort of "vector" is this?

Just in case you haven't noticed it:

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

3. Jul 7, 2009

### HallsofIvy

Re: What sort of "vector" is this?

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 $$B_\sigma[/itex] is $$B^\gamma= \nu^{\gamma\sigma}B_\sigma$$. Assuming that your $\nu_{\mu\rho}$ is the metric tensor, $\mu^{\gamma\sigma}$ is the tensor such that $\nu_{\mu\rho}\nu^{\mu\gamma}= \nu^{\mu\gamma}\nu_{\mu\rho}= \delta^\gamma_\rho$. Algebraically, $\nu^{\gamma\sigma}$ is represented by the matrix that is the inverse to $\nu_{\gamma\sigma}$. Yes, your edit is correct. 4. Jul 7, 2009 ### pellman Re: What sort of "vector" is this? Thanks! 5. Jul 7, 2009 ### Fredrik Staff Emeritus Re: What sort of "vector" is this? Actually, they're not, because we have $${\Lambda_\mu}^\sigma=(\Lambda^{-1})^\sigma{}_\mu$$ 6. Jul 7, 2009 ### Fredrik Staff Emeritus Re: What sort of "vector" is this? Covariant transformation from (say) frame S to frame S': $$A'_\mu=\Lambda_\mu{}^\nu A_\nu$$ Multiply by [itex]\Lambda^\mu{}_\rho$$.

$$\Lambda^\mu{}_\rho A'_\mu=\Lambda^\mu{}_\rho\Lambda_\mu{}^\nu A_\nu=\Lambda^\mu{}_\rho(\Lambda^{-1})^\nu{}_\mu A_\nu=\delta^\nu_\rho A_\nu=A_\rho$$

This is clearly just a covariant transformation from S' to S.