Transforming Contra & Covariant Vectors

In summary: I do not understand why this step is necessary.Can you explain why this step is necessary?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.
  • #1
dyn
773
61
Hi.
The book I am using gives the following equations for the the Lorentz transformations of contravariant and covariant vectors
x = Λμν xν ( 1 )

xμ/ = Λμν xv ( 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 gρμ and apply it to both sides of (1) I get

xρ/ = Λρν x ν

Then I'm stuck because how can I lower the ν on xν as ν is already repeated twice on the RHS and so I can't use a metric with ν in it ?
Thanks
 
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  • #2
The relation between the contravariant and covariant components is ##x^\nu = g^{\nu\mu}x_\mu##.
 
  • #3
dyn said:
Hi.
The book I am using gives the following equations for the the Lorentz transformations of contravariant and covariant vectors
x = Λμν xν ( 1 )

xμ/ = Λμν xv ( 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 gρμ and apply it to both sides of (1) I get

xρ/ = Λρν x ν

Then I'm stuck because how can I lower the ν on xν 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.
 
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  • #4
stevendaryl said:
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 ?
 
  • #5
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.
 
  • #6
I don't understand that step
 
  • #7
It is just an insertion of a known relation.
 
  • #8
Orodruin said:
It is just an insertion of a known relation.

I presume you mean the relation gαβgαγ = δβγ ?
So that means I multiply each side of the equation by gμα ?
 
  • #9
dyn said:
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.
 
  • #10
Thanks for that but its the placement of indices in each step that confuses me. Was the following statement correct ?
dyn said:
I presume you mean the relation gαβgαγ = δβγ ?
So that means I multiply each side of the equation by gμα ?
 
  • #11
dyn said:
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##
 
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  • #12
stevendaryl said:
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 Λμν 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 Λμν
 
  • #13
dyn said:
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.
 
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Related to Transforming Contra & Covariant Vectors

1. What are covariant and contravariant vectors?

Covariant and contravariant vectors are two types of vectors used in the study of linear algebra and tensor calculus. Covariant vectors are defined by their transformation properties under a change of basis, while contravariant vectors are defined by their transformation properties under a change of coordinates.

2. How do you transform covariant and contravariant vectors?

To transform a covariant vector, you use the transformation matrix associated with the change of basis. For a contravariant vector, you use the inverse of the transformation matrix associated with the change of coordinates. This ensures that the vector remains in the same space and maintains its properties.

3. What is the significance of transforming covariant and contravariant vectors?

Transforming covariant and contravariant vectors is essential in understanding the relationship between different coordinate systems and their corresponding vector spaces. It allows for calculations and measurements to be made in one coordinate system and then easily transformed to another, making it a crucial tool in many areas of science and engineering.

4. Can covariant and contravariant vectors be transformed independently?

Yes, covariant and contravariant vectors can be transformed independently. This is because they have different transformation properties and require different transformation matrices. However, they are still closely related and often appear together in equations and calculations.

5. Are there any real-life applications of transforming covariant and contravariant vectors?

Yes, there are many real-life applications of transforming covariant and contravariant vectors. For example, in physics, these transformations are used to understand the behavior of physical quantities in different reference frames. In computer graphics, they are used to rotate and scale objects in three-dimensional space. They are also used in engineering, specifically in the study of stress and strain in materials.

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