Index placement -- Lorentz transformation matrix

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  • #1
dyn
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Hi.
I came across the following statement , which seems wrong to me.
Λμρ = ( ΛT )ρμ
I have it on good authority (a previous post on this forum) that (ΛT)μν = Λνμ so I am hoping that the first equation is wrong ? It looks like the inverse not the transpose ?

The equation Λμρ η μνΛνσ = ηρσ is written in matrix form as ΛT η Λ = η
Why is one of the Lorentz matrices transposed ? It looks like it should be Λ η Λ = η

Thanks
 

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  • #2
Orodruin
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a previous post on this forum
Please do not make such statements without linking to that previous post.
 
  • #3
pervect
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Hi.
I came across the following statement , which seems wrong to me.
Λμρ = ( ΛT )ρμ
I have it on good authority (a previous post on this forum) that (ΛT)μν = Λνμ so I am hoping that the first equation is wrong ? It looks like the inverse not the transpose ?

The equation Λμρ η μνΛνσ = ηρσ is written in matrix form as ΛT η Λ = η
Why is one of the Lorentz matrices transposed ? It looks like it should be Λ η Λ = η

Thanks

There are apparently multiple conventions in use, but I use tensor notation and follow the approach in "Gravitation" (MTW) where the authors state the indices of a transformation matrix always go "from northwest to southeast".

My best guess is that you aren't even using tensor notation but some sort of matrix notation, and that ##\lambda^T## means the transpose of the matrix. There is no need for a transpose operator in tensor notation, so that's why I suspect you aren't using it.

Besides apparently not using tensor notation, you are using a different notation than my textbook does when you write ##(...)_\rho{}^\mu##, because the indices move from southwest to northeast, rather than northwest to southeast.
 
  • #4
dyn
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If you agree with the following equation Λμρ ημν Λνσ = ηρσ
can you tell me why in matrix form its written as ΛTηΛ = η and not just ΛηΛ = η ?
I don't understand why one of the Lorentz transformation matrices is transposed.
Thanks
 
  • #5
nrqed
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If you agree with the following equation Λμρ ημν Λνσ = ηρσ
can you tell me why in matrix form its written as ΛTηΛ = η and not just ΛηΛ = η ?
I don't understand why one of the Lorentz transformation matrices is transposed.
Thanks
Let's go back to ordinary metric multiplication. Writing out the indices, we have
## (M A B)_{ad} = M_{ab} A_{bc} B_{cd} ##
Note that the order of the indices, the second index of the M matrix is the first index of the A matrix. In the expression you wrote, the first index of the first ##\Lambda## (the ##\mu##) is the first index of the eta matrix. Now, note that we can write the product I just wrote
## (M A B)_{ad} = (M^T)_{ba} A_{bc} B_{cd} ## which is what you wanted.
 
  • #6
Ibix
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It's important to make a distinction between this tensor index notation and matrix notation, no matter how similar they are. Matrix notation tells you how to sum by ordering the matrices - so ##\vec u^T\mathbf{g}\vec v## and ##\mathbf{g}\vec v\vec u^T## are different things. But tensor notation assigns meaning to the index placement but not the order, so ##g_{ij}u^iv^j## and ##u^ig_{ij}v^j## are the same thing. This means that "transposing a tensor" is something you only have to do if you choose to represent it as a matrix and apply the rules of matrix multiplication instead of the index summation rules.

Second thing to note is that the Lorentz transform matrix is symmetric. So transposing it doesn't do anything anyway.

Final thing to note is that there seem to be multiple conventions for index placement on ##\Lambda##. Carroll uses northwest/southeast for Lorentz transforms and southeast/northwest for inverse Lorentz transforms. But he notes that Schutz, for example, prefers northwest/southeast always. He also points out that it doesn't matter, since both forward and reverse transforms are Lorentz transforms, and all that matters is that your indices are correctly associated with the upper and lower positions.

So assuming that whatever unreferenced (there's a reason for PF rules about references!) source @dyn saw is using Carroll's notation then I'd agree that ##\Lambda^\mu{}_\rho=(\Lambda^T)_\mu{}^\rho## is incorrect since Carroll would use that notation for inverse, not transpose. I can't think offhand of a notation where it would make sense, since transposing a Lorentz transform in tensor notation is doubly pointless (don't take that as any kind of authority on notation, though). I'd say that the rest of the OP is a good example of why you should be very careful mixing tensor and matrix notation.
 
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  • #7
dyn
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[QUOTE="

Second thing to note is that the Lorentz transform matrix is symmetric. So transposing it doesn't do anything anyway.[/QUOTE]

I know the LT matrix is symmetric for a standard boost in the x-direction but is every LT matrix symmetric for any combination of boosts ?
If yes , is this a direct consequence of the definition ΛTηΛ = η ?
 
  • #8
Orodruin
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I know the LT matrix is symmetric for a standard boost in the x-direction but is every LT matrix symmetric for any combination of boosts ?
No, it is not. For example, a standard spatial rotation is also in the Lorentz group and is not symmetric.
 
  • #9
Ibix
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I know the LT matrix is symmetric for a standard boost in the x-direction but is every LT matrix symmetric for any combination of boosts ?
An arbitrary boost, yes. If you throw rotations into the mix then no, to be fair.
 
  • #10
vanhees71
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Usually the notation is as follows. The Lorentz transformation matrix describes the transformation of contravariant tensor components. It's sufficient to consider the space-time four-vector:
$$x^{\prime \mu} = {\Lambda^{\mu}}_{\rho} x^{\rho}.$$
In order to be a Lorentz transformation you must have
$$\eta_{\mu \nu} x^{\prime \mu} x^{\prime \nu}=\eta_{\rho \sigma} x^{\rho} x^{\sigma},$$
where ##\eta_{\rho \sigma}## are the (purely covariant!) tensor components of the Minkowski fundamental form, i.e., in Minkowski orthonormal frames ##(\eta_{\rho \sigma})=\mathrm{diag}(1,-1,-1,-1)##. Since this should hold for any vector ##x##, the Lorentz-transformation matrix must fulfill
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}.$$
Sometimes it's convenient to work in matrix-vector notation (to be a bit evil you can also say in order to utmostly confuse the student you switch between the self-explaining Ricci calculus and the notationally less precise matrix-vector notaion all the time ;-)). Then one defines matrices
$$\hat{\Lambda}=({\Lambda^{\mu}}_{\rho}), \quad \hat{\eta}=(\eta_{\rho \sigma}).$$
As you see, then the Ricci calculus's most important property to clearly distinguish between co- and contravariant components of tensors and transformation matrices, is lost, but that's no problem as long as you keep in mind what kind of components you are dealing with. Just write down the matrices and column and row vectors and then translate the Ricci-calculus formulae to the matrix-vector multiplication calculus.

Obviously, using column vectors ##\overline{x}=(x^{\mu})## for the contravariant vector components, then the Lorentz transformations read
$$\overline{x}'=\hat{\Lambda} \overline{x}.$$
The Lorentz-transformation property of the matrix ##\hat{\Lambda}## translates to
$$\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta},$$
or since ##\hat{\eta}^2=1## after some simple matrix manipulation
$$\hat{\Lambda}^{-1} = \hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta}.$$
In the Ricci calculus this translates into
$${(\Lambda^{-1})^{\mu}}_{\nu}=\eta^{\mu \rho} \eta_{\nu \sigma} {\Lambda^{\sigma}}_{\rho}={\Lambda_{\nu}}^{\mu}.$$
This can be verified by the Ricci calculus easily, using the original Lorentz-transformation property:
$${\Lambda_{\nu}}^{\mu} {\Lambda^{\nu}}_{\rho}=g_{\nu \sigma} g^{\mu \alpha} {\Lambda^{\sigma}}_{\alpha} {\Lambda^{\nu}}_{\rho} = g^{\mu \alpha} g_{\alpha \rho} = {\delta^{\mu}}_{\rho}.$$
 
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