# Inverse transformations

1. Oct 7, 2013

### Mentz114

1. The problem statement, all variables and given/known data

${\Lambda_c}^b$ 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 ?

2. Relevant equations

3. The attempt at a solution

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

Wrong.

Last edited: Oct 7, 2013
2. Oct 7, 2013

### WannabeNewton

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

3. Oct 8, 2013

### Mentz114

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

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

Last edited: Oct 8, 2013
4. Oct 8, 2013

### 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.

Last edited: Oct 8, 2013
5. Oct 8, 2013

### Mentz114

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. Oct 8, 2013

### WannabeNewton

Is this an exercise from a textbook?

7. Oct 8, 2013

### Mentz114

No. It comes from

\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}
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. Oct 8, 2013

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

9. Oct 8, 2013

### Mentz114

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.