- #1
JD_PM
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- TL;DR Summary
- I want to prove certain properties related to Lorentz Transformations.
This exercise was proposed by samalkhaiat here
Given the defining property of Lorentz transformation [itex]\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}[/itex], prove the following identities
(i) [itex]\ (\Lambda k) \cdot (\Lambda x) = k \cdot x[/itex]
(ii) [itex]\ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x[/itex]
(iii) [itex]\ \left(\Lambda^{-1}k \right)^{2} = k^{2}[/itex]
Where ##\Lambda## is a linear transformation
My attempt
Alright, let us deal with the inner-product vector space ##(\Re, \Re^n, +, \langle \cdot , \cdot \rangle )## . Addition is defined as usual and the inner product is the standard inner product over ##\Re^n##
$$\langle \cdot , \cdot \rangle : \Re^n \times \Re^n \rightarrow \Re : (X=\Big(
\begin{pmatrix}
x^1 \\
. \\
. \\
x^n
\end{pmatrix}\Big), Y=
\Big(
\begin{pmatrix}
y^1 \\
. \\
. \\
y^n
\end{pmatrix}\Big)) \mapsto \sum_{i=1}^n x_i y_i = X^T \cdot Y$$
(i) Applying the definition of ##\langle \cdot, \cdot \rangle##
$$X^T \cdot Y=(\Lambda k)^T \cdot (\Lambda x)=(k^T \Lambda^T) \cdot \Lambda x $$
Well, I get the answer iff I assume that ##\Lambda## is an orthogonal matrix (i.e. ##\Lambda^{-1}=\Lambda^{T}##) and ##k## is a symmetric matrix (i.e. ##k^{T}=k##)
$$(k^T \Lambda^T) \cdot \Lambda x = (k^T \Lambda^{-1}) \cdot \Lambda x = k \cdot x$$
Mmm it doesn't smell good; too many restrictions I think.
(ii) Let's apply the definition of the given inner product to the RHS
$$X^T \cdot Y= (\Lambda^{-1}p)^{T} \cdot x$$
Assuming ##\Lambda## is an orthogonal matrix and ##p## is symmetric I indeed get the answer, but I've got the same issue as in (i).
(iii) I get it if I also assume that ##\Lambda## is involutory (i.e. when ##\Lambda## is raised to an even power yields the identity matrix and when is raised to an odd power yields ##\Lambda##; more properties here).
Is my reasoning correct? If yes, I think there may be a better way of proving them all (without that many assumptions).
Side note: ##\Re^4## is the most common space-time vector space; it is the four dimensional real vector space consisting of all 4-tuples (i.e. ##x=(x^0, x^1, x^2, x^3)##) and the inner-product here satisfies ##X^T \cdot Y = x^0y^0 - \vec x \cdot \vec y##
Any help is appreciated.
Thank you
Given the defining property of Lorentz transformation [itex]\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}[/itex], prove the following identities
(i) [itex]\ (\Lambda k) \cdot (\Lambda x) = k \cdot x[/itex]
(ii) [itex]\ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x[/itex]
(iii) [itex]\ \left(\Lambda^{-1}k \right)^{2} = k^{2}[/itex]
Where ##\Lambda## is a linear transformation
My attempt
Alright, let us deal with the inner-product vector space ##(\Re, \Re^n, +, \langle \cdot , \cdot \rangle )## . Addition is defined as usual and the inner product is the standard inner product over ##\Re^n##
$$\langle \cdot , \cdot \rangle : \Re^n \times \Re^n \rightarrow \Re : (X=\Big(
\begin{pmatrix}
x^1 \\
. \\
. \\
x^n
\end{pmatrix}\Big), Y=
\Big(
\begin{pmatrix}
y^1 \\
. \\
. \\
y^n
\end{pmatrix}\Big)) \mapsto \sum_{i=1}^n x_i y_i = X^T \cdot Y$$
(i) Applying the definition of ##\langle \cdot, \cdot \rangle##
$$X^T \cdot Y=(\Lambda k)^T \cdot (\Lambda x)=(k^T \Lambda^T) \cdot \Lambda x $$
Well, I get the answer iff I assume that ##\Lambda## is an orthogonal matrix (i.e. ##\Lambda^{-1}=\Lambda^{T}##) and ##k## is a symmetric matrix (i.e. ##k^{T}=k##)
$$(k^T \Lambda^T) \cdot \Lambda x = (k^T \Lambda^{-1}) \cdot \Lambda x = k \cdot x$$
Mmm it doesn't smell good; too many restrictions I think.
(ii) Let's apply the definition of the given inner product to the RHS
$$X^T \cdot Y= (\Lambda^{-1}p)^{T} \cdot x$$
Assuming ##\Lambda## is an orthogonal matrix and ##p## is symmetric I indeed get the answer, but I've got the same issue as in (i).
(iii) I get it if I also assume that ##\Lambda## is involutory (i.e. when ##\Lambda## is raised to an even power yields the identity matrix and when is raised to an odd power yields ##\Lambda##; more properties here).
Is my reasoning correct? If yes, I think there may be a better way of proving them all (without that many assumptions).
Side note: ##\Re^4## is the most common space-time vector space; it is the four dimensional real vector space consisting of all 4-tuples (i.e. ##x=(x^0, x^1, x^2, x^3)##) and the inner-product here satisfies ##X^T \cdot Y = x^0y^0 - \vec x \cdot \vec y##
Any help is appreciated.
Thank you