Transforming Contra & Covariant Vectors

[dyn]

Hi.
The book I am using gives the following equations for the the Lorentz transformations of contravariant and covariant vectors
x^/μ = Λ^μ_ν x^ν ( 1 )

x_μ^/ = Λ_μ^ν x_v ( 2 )

where the 2 Lorentz transformation matrices are the inverses of each other. I am trying to get equation 2 from equation 1 but if I lower the index on the LHS of (1) using the metric g_ρμ and apply it to both sides of (1) I get

x_ρ^/ = Λ_ρν x ^ν

Then I'm stuck because how can I lower the ν on x^ν as ν is already repeated twice on the RHS and so I can't use a metric with ν in it ?
Thanks

 

[Orodruin]

The relation between the contravariant and covariant components is $x^\nu = g^{\nu\mu}x_\mu$.

 

[stevendaryl]

Hi.
The book I am using gives the following equations for the the Lorentz transformations of contravariant and covariant vectors
x^/μ = Λ^μ_ν x^ν ( 1 )

x_μ^/ = Λ_μ^ν x_v ( 2 )

where the 2 Lorentz transformation matrices are the inverses of each other. I am trying to get equation 2 from equation 1 but if I lower the index on the LHS of (1) using the metric g_ρμ and apply it to both sides of (1) I get

x_ρ^/ = Λ_ρν x ^ν

Then I'm stuck because how can I lower the ν on x^ν as ν is already repeated twice on the RHS and so I can't use a metric with ν in it ?
Thanks

You have

$x'^\mu = \Lambda^\mu_\nu x^\nu \Rightarrow g^{\mu \rho} x'_\rho = \Lambda^\mu_\nu g^{\nu \lambda} x_\lambda$

Now, you operate on both sides with the $g$ to get:

$x'_\rho = g_{\mu \rho} \Lambda^\mu_\nu g^{\nu \lambda} x_\lambda$

So the transformation matrix for the lowered components is $g_{\mu \rho} \Lambda^\mu_\nu g^{\nu \lambda}$

The final step is to realize that $g_{\mu \rho} \Lambda^\mu_\nu g^{\nu \lambda} = \Lambda^\lambda_\rho$. That might seem obvious, but it's actually not, because $\Lambda$ is not a tensor; the two indices refer to different coordinate systems. So it's not immediately obvious that you can raise and lower indices the way you could with a tensor.

 

[dyn]

You have

$x'^\mu = \Lambda^\mu_\nu x^\nu \Rightarrow g^{\mu \rho} x'_\rho = \Lambda^\mu_\nu g^{\nu \lambda} x_\lambda$

Now, you operate on both sides with the $g$ to get:

$x'_\rho = g_{\mu \rho} \Lambda^\mu_\nu g^{\nu \lambda} x_\lambda$
.

Thanks for your reply. Can you explicitly explain this step for me ? What do I multiply each side of the equation by in terms of indices ?

 

[Orodruin]

You are not to multiply each side with anything. You are to insert the relation given in #2, which @stevendaryl did for you explicitly in the first line of #3.

 

[dyn]

I don't understand that step

 

[Orodruin]

It is just an insertion of a known relation.

 

[dyn]

It is just an insertion of a known relation.

I presume you mean the relation g^αβg_αγ = δ^β_γ ?
So that means I multiply each side of the equation by g_μα ?

 

[stevendaryl]

Thanks for your reply. Can you explicitly explain this step for me ? What do I multiply each side of the equation by in terms of indices ?

Let me write it without any indices. I think you will find that there is only one way to put in indices so that it makes sense.

1. $x' = \Lambda x$
2. Let's write: $x = g^{-1} g x$
3. Substituting this expression for $x$ into equation 1: $x' = \Lambda g^{-1} g x$
4. Operate on 3 using $g$: $g x' = g \Lambda g^{-1} g x$
5. Now, let's define the combination: $\widetilde{x} \equiv g x$
6. Also, $\widetilde{x'} \equiv g x'$
7. And $\widetilde{\Lambda} \equiv g \Lambda g^{-1}$
8. So the contravariant transformation law is: $\widetilde{x'} = \widetilde{\Lambda} \widetilde{x}$

There is really only one way that you can insert indices into the equations 1-8 that makes sense.

 

[dyn]

Thanks for that but its the placement of indices in each step that confuses me. Was the following statement correct ?

I presume you mean the relation g^αβg_αγ = δ^β_γ ?
So that means I multiply each side of the equation by g_μα ?

 

[stevendaryl]

Thanks for that but its the placement of indices in each step that confuses me. Was the following statement correct ?

As I said, there really is only one way to do the indices so that it makes sense.

But yes, if you start with $x'^\mu = \Lambda^\mu_\nu x^\nu$, and multiply both sides by $g_{\mu \alpha}$ (and sum over $\mu$). Then you rewrite $x^\nu$ as $g^{\nu \lambda} x_\lambda$

 

[dyn]

The final step is to realize that $g_{\mu \rho} \Lambda^\mu_\nu g^{\nu \lambda} = \Lambda^\lambda_\rho$. That might seem obvious, but it's actually not, because $\Lambda$ is not a tensor; the two indices refer to different coordinate systems. So it's not immediately obvious that you can raise and lower indices the way you could with a tensor.

I always see Λ as a 4x4 matrix and I always thought a matrix was a tensor of rank 2. I see Λ^μ_ν as the entry on row μ and column ν of that matrix but I'm unsure how the rows and columns relate to the inverse Λ_μ^ν

 

[Nugatory]

I always thought a matrix was a tensor of rank 2

The individual components of a rank-2 tensor can be represented as a matrix, but not all matrices are representations of rank-2 tensors.

The components of a tensor transform in a particular way when you change coordinate systems; not all matrices have that property. Also when you represent a tensor as a matrix, you lose the distinction between contravariant and covariant components.