How Is the Inverse Lorentz Transformation Represented with Kronecker Deltas?

[Mentz114]

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.

 

[WannabeNewton]

$\Lambda^{a}{}{}_{c}\Lambda^{c}{}{}_{b} = \delta^{a}{}{}_{b}$; it's just regular matrix multiplication.

 

[Mentz114]

$\Lambda^{a}{}{}_{c}\Lambda^{c}{}{}_{b} = \delta^{a}{}{}_{b}$; it's just regular matrix multiplication.

OK, thanks. I wasn't too far off.
Can I use a $\delta$ to relabel these indexes ? Like

${\Lambda_c}^b {\Lambda^d}_a {\delta_d}^c = {\delta^b}_a\ \Rightarrow\ {\Lambda_c}^b {\Lambda^d}_a {\delta_d}^c{\delta^d}_c = {\delta^b}_a {\delta^d}_c$

 

[WannabeNewton]

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.

 

[Mentz114]

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.

 

[WannabeNewton]

Is this an exercise from a textbook?

 

[Mentz114]

Is this an exercise from a textbook?

No. It comes from

$$\begin{align}<br /> \dot{v}_a &=\left({\Lambda_c}^b u^c \right)\nabla_b \left( {\Lambda^d}_a u_d \right)\\<br /> &= {\Lambda_c}^b {\Lambda^d}_a u^c \nabla_b u_d + {\Lambda_c}^b u^c u_d \nabla_b {\Lambda^d}_a<br /> \end{align}$$
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 ?

 

[WannabeNewton]

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 $(\mathbb{R}^{4},\eta)$.

 

[Mentz114]

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.