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Inverse transformations

  • Thread starter Mentz114
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Homework Statement



##{\Lambda_c}^b## is a Lorentz transformation and ##{\Lambda^c}_b## is its inverse, so ##{\Lambda_c}^b {\Lambda^c}_b## gives an identity matrix.

How can I write this, assuming it's possible, in terms of ##\delta##'s ?


Homework Equations




The Attempt at a Solution



##{\Lambda_a}^b {\Lambda_c}^a =\delta^b_c ##

Wrong.
 
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Answers and Replies

  • #2
WannabeNewton
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##\Lambda^{a}{}{}_{c}\Lambda^{c}{}{}_{b} = \delta^{a}{}{}_{b}##; it's just regular matrix multiplication.
 
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  • #3
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##\Lambda^{a}{}{}_{c}\Lambda^{c}{}{}_{b} = \delta^{a}{}{}_{b}##; it's just regular matrix multiplication.
OK, thanks. I wasn't too far off.
Can I use a ##\delta## to relabel these indexes ? Like

##{\Lambda_c}^b {\Lambda^d}_a {\delta_d}^c = {\delta^b}_a\ \Rightarrow\ {\Lambda_c}^b {\Lambda^d}_a {\delta_d}^c{\delta^d}_c = {\delta^b}_a {\delta^d}_c ##
 
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  • #4
WannabeNewton
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I can't parse what you wrote down because the indices are being summed over more than twice. But if you're asking if something like ##\Lambda^{a}{}{}_{c}\delta_{ab} = \Lambda_{bc}## is true then the answer is yes. Keep in mind that this is technically not index raising and lowering but rather just applying the definition of ##\delta_{ab}##. The two are different mathematical operations in spaces that aren't Euclidean.
 
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  • #5
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I can't parse what you wrote down because the indices are being summed over more than twice. But if you're asking if something like ##\Lambda^{a}{}{}_{c}\delta_{ab} = \Lambda_{bc}## is true then the answer is yes. Keep in mind that this is technically not index raising and lowering but rather just applying the definition of ##\delta_{ab}##. The two are different mathematical operations in spaces that aren't Euclidean.
Yes, I edited while you were writing and took out the raising/lowering.

I can also see now that what I wrote is not sensible. I'm actually trying to establish if for a geodesic ##u^a##

##{\Lambda_c}^b {\Lambda^d}_a u^c \nabla_b u_d = 0##

Messing ##\delta##s looks like a red-herring.
 
  • #6
WannabeNewton
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Is this an exercise from a textbook?
 
  • #7
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Is this an exercise from a textbook?
No. It comes from

[tex]
\begin{align}
\dot{v}_a &=\left({\Lambda_c}^b u^c \right)\nabla_b \left( {\Lambda^d}_a u_d \right)\\
&= {\Lambda_c}^b {\Lambda^d}_a u^c \nabla_b u_d + {\Lambda_c}^b u^c u_d \nabla_b {\Lambda^d}_a
\end{align}
[/tex]
I reason that if ##\Lambda## is an inertial boost then both terms must be zero. But the first term does not depend in any way on this distinction ( no derivatives of ##\Lambda##) so it must be identically zero. Am I wrong ?
 
  • #8
WannabeNewton
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I really can't understand what you're doing. What is ##\dot{v}_a##? I assume since you're using ##\nabla## that this is a general curved space-time in which case ##\Lambda## must be a local Lorentz transformation which is different from a global Lorentz transformation of ##(\mathbb{R}^{4},\eta)##.
 
  • #9
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I really can't understand what you're doing. What is ##\dot{v}_a##? I assume since you're using ##\nabla## that this is a general curved space-time in which case ##\Lambda## must be a local Lorentz transformation which is different from a global Lorentz transformation of ##(\mathbb{R}^{4},\eta)##.
Sorry, ##v^a = {\Lambda_b}^a u^b## and ##\dot{v}_a## is its proper acceleration.

So, this is not the correct way to boost in curved spacetime ? I suppose I could take this into the frame field of ##u^\mu## in which case I can boost the basis, which is legitimate. Hmm.
 

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