Infinitesimal Lorentz transform and its inverse, tensors

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  • #1
fluidistic
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Homework Statement


The problem can be found in Jackson's book.
An infinitesimal Lorentz transform and its inverse can be written under the form ##x^{'\alpha}=(\eta ^{\alpha \beta}+\epsilon ^{\alpha \beta})x_{\beta}## and ##x^\alpha = (\eta ^{\alpha \beta}+\epsilon ^{'\alpha \beta}) x^{'}_\beta## where ##\eta _{\alpha \beta}## is Minkowski's metric and the epsilons are infinitesimals.
1)Demonstrate, using the definition of the inverse, that ##\epsilon ^{'\alpha \beta}=-\epsilon ^{\alpha \beta}##.
2)Demonstrate, using the conservation of the norm, that ##\epsilon ^{\alpha \beta}=-\epsilon ^{\beta \alpha}##


Homework Equations


Not really sure, but I used some eq. found on some page earlier in the book: ##\epsilon ^{'\alpha \beta}=\frac{\partial x^{'\alpha }}{\partial x^\alpha} \frac{\partial x^{'\beta}}{\partial x^\beta} \epsilon ^{\alpha \beta}##.


The Attempt at a Solution


1)I used the relevant equation and wrote that it's equal to ##\frac{\partial x^{'\alpha }}{\partial x^\beta} \frac{\partial x^{'\beta}}{\partial x^\alpha}\epsilon ^{\alpha \beta}##. Then I calculated the partial derivatives using the 2 equations given in the problem statement, I made an approximation (depreciated terms with epsilons multiplied together because they are "infinitesimals") and I reached that ##\epsilon ^{'\alpha \beta}\approx \frac{\eta^{\alpha \beta}}{\eta ^{\alpha \beta}+\epsilon ^{'\alpha \beta}}\cdot \epsilon ^{\alpha \beta}##. I don't see how the first term can be worth -1 here... So I guess my approach is wrong. Or if it's right, I still don't see how I can show that the first term is worth -1. Thanks for any comment.
 

Answers and Replies

  • #2
vela
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For #1, apply the transformation and followed by its inverse and use the fact that that product should yield the identity.
 
  • #3
fluidistic
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For #1, apply the transformation and followed by its inverse and use the fact that that product should yield the identity.
I see... do you mean that I should perform ##x^{'\alpha}x^{\alpha}=\text{Identity}##?
Also I don't see what's wrong with what I did in my attempt.
By the way I don't understand how tensors "work" yet, I am self studying this topic right now.

Edit: Nevermind my first question, the answer is no, the expression I wrote makes no sense...
 
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  • #4
vela
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There's a problem with the equation you started with. A dummy index should only appear twice in each product. You wrote
$$\epsilon ^{'\alpha \beta}=\frac{\partial x^{'\alpha }}{\partial x^\alpha} \frac{\partial x^{'\beta}}{\partial x^\beta} \epsilon ^{\alpha \beta}.$$ The indices ##\alpha## and ##\beta## appear only once on the lefthand side, so they should appear only once on the righthand side. What you probably meant was something like
$$\epsilon ^{'\alpha \beta}=\frac{\partial x^{'\alpha }}{\partial x^\gamma} \frac{\partial x^{'\beta}}{\partial x^\delta} \epsilon ^{\gamma \delta}.$$ Notice how ##\gamma## and ##\delta## appear in pairs, which implies a summation over those indices.

This relationship, however, doesn't apply here. It relates the components of a tensor in one frame to the components of the same tensor in a different frame. In this problem, ##\epsilon## and ##\epsilon'## aren't the same tensor. One generates a Lorentz transformation, and the other, the inverse Lorentz transformation.
 
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  • #5
fluidistic
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I see... thanks. Yes this is what I've done and I replaced gamma and delta by alpha and beta respectively... Ok I didn't know this could not be done.
 
  • #6
vanhees71
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I'd start from the Lorentz-transformation property of the representing matrices:
[tex]\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}.[/tex]
Now write
[tex]{\Lambda^{\mu}}_{\rho}=\delta_{\rho}^{\mu}+{\epsilon^{\mu}}_{\rho},[/tex]
plug this in the above defining equation for LT matrices and expand up to the 1st-order terms in [itex]{\epsilon^{\mu}}_{\rho}[/itex]. Finally note that by definition one applies the index-dragging rule not only to tensor components but also to Lorentz matrices. Note that a Lorentz matrix does not form tensor components but the infintesimal generators do, but that doesn't matter for your problem here. The only additional thing you need to know is that
[tex]\epsilon_{\rho \sigma}=\eta_{\rho \mu} {\epsilon^{\mu}}_{\sigma}.[/tex]
 
  • #7
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1)I used the relevant equation and wrote that it's equal to ##\frac{\partial x^{'\alpha }}{\partial x^\beta} \frac{\partial x^{'\beta}}{\partial x^\alpha}\epsilon ^{\alpha \beta}##. Then I calculated the partial derivatives using the 2 equations given in the problem statement, I made an approximation (depreciated terms with epsilons multiplied together because they are "infinitesimals") and I reached that ##\epsilon ^{'\alpha \beta}\approx \frac{\eta^{\alpha \beta}}{\eta ^{\alpha \beta}+\epsilon ^{'\alpha \beta}}\cdot \epsilon ^{\alpha \beta}##. I don't see how the first term can be worth -1 here... So I guess my approach is wrong. Or if it's right, I still don't see how I can show that the first term is worth -1.

I will say you will not get anywhere using this.Just follow the idea jackson gave to you.Like when he say use conservation of norm then use xαxα=x'αx'α.You will have to leave some second order terms here to show εαβ=-εβα.First one is simple manipulation.
 
  • #8
WannabeNewton
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Hi there fluidistic! For the first part, note that all you have to do is a simple substitution. By definition we have ##x'_{\beta} = (\eta_{\beta \gamma} + \epsilon_{\beta \gamma}) x^{\gamma}## so plug this into ##x^{\alpha} = (\eta^{\alpha \beta} + \epsilon'^{\alpha\beta})x'_{\beta}## to get the quadratic ##x^{\alpha} = (\eta^{\alpha \beta} + \epsilon'^{\alpha\beta})(\eta_{\beta \gamma} + \epsilon_{\beta \gamma}) x^{\gamma}##. Expand the quadratic to ##O(\epsilon^2)## and at the every end of the calculation use the fact that ##x^{\alpha}## is arbitrary. The other parts should be very similar.
 
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  • #9
fluidistic
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Thanks guys for all the help. Unfortunately I'm way too sloppy with tensors in order to manipulate them.
Here's what I reached using WBN's suggestion: ##x^\alpha \approx (\delta ^\alpha _\gamma +\eta ^{\alpha \beta}\epsilon _{\beta \gamma}+\epsilon ' ^{\alpha \beta}\eta _{\beta \gamma})x^\gamma##. By looking at this equation I have a feeling that gamma must be worth alpha and that what is in parenthesis must be worth the identity. This would imply that ##\eta ^{\alpha \beta}\epsilon _{\beta \gamma}=-\epsilon ' ^{\alpha \beta}\eta _{\beta \gamma}##. And I guess that it's by working on that equation that I will get the desired result.
 
  • #10
WannabeNewton
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You're pretty much there. Recall that ##\eta^{\alpha\beta}## converts covariant indices into contravariant indices, and vice versa, upon contraction (more precisely, it is what we call a musical isomorphism and serves as a map between elements of the tangent space and its dual). So ##\eta^{\alpha \beta}\epsilon_{\beta \gamma} = \epsilon^{\alpha}{}{}_{\gamma}## and similarly ##\epsilon'^{\alpha \beta}\eta_{\beta \gamma} = \epsilon'^{\alpha}{}{}_{\gamma}##.

So now you have ##\delta^{\alpha}{}{}_{\gamma}x^{\gamma} + \epsilon'^{\alpha}{}{}_{\gamma}x^{\gamma} + \epsilon^{\alpha}{}{}_{\gamma}x^{\gamma} + O(\epsilon^2) = x^{\alpha}##. The result should then be immediate.
 
  • #11
fluidistic
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You're pretty much there. Recall that ##\eta^{\alpha\beta}## converts covariant indices into contravariant indices, and vice versa, upon contraction (more precisely, it is what we call a musical isomorphism and serves as a map between elements of the tangent space and its dual). So ##\eta^{\alpha \beta}\epsilon_{\beta \gamma} = \epsilon^{\alpha}{}{}_{\gamma}## and similarly ##\epsilon'^{\alpha \beta}\eta_{\beta \gamma} = \epsilon'^{\alpha}{}{}_{\gamma}##.

So now you have ##\delta^{\alpha}{}{}_{\gamma}x^{\gamma} + \epsilon'^{\alpha}{}{}_{\gamma}x^{\gamma} + \epsilon^{\alpha}{}{}_{\gamma}x^{\gamma} + O(\epsilon^2) = x^{\alpha}##. The result should then be immediate.
I see.
So I reach that ##(\delta ^\alpha _\gamma +\epsilon ^\alpha _\gamma+\epsilon '^\alpha _\gamma )x^\gamma \approx x^\alpha##. The only way for both sides to be approximately equal requires that ##\gamma =\alpha## and that what is in parenthesis is worth the identity right?
In this case I reach that ##\epsilon '^\alpha _\alpha =-\epsilon ^\alpha _\alpha##. But I still don't see how to reach the final result. I guess I must contract these tensors as to get only upper scripts with alpha and beta.
If I multiply the last expression by ##\eta ^{\beta \alpha}##, I reach that ##\epsilon '^{\beta \alpha}=-\epsilon ^{\beta \alpha}##. Now if this tensor is symmetric then I reach the desired result but how do I know whether it is symmetric?
I'm sure I made some error(s)...


Edit: Nevermind, I think I reach the final result, if I multiply both sides of ##\epsilon '^\alpha _\alpha =-\epsilon ^\alpha _\alpha## by ##\eta ^{\alpha \beta}##, but a multiplication by the RIGHT and not the left. This yields the result. Is that correct?
 
  • #12
vela
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I see.
So I reach that ##(\delta^\alpha{}_\gamma +\epsilon^\alpha{}_\gamma + \epsilon'^\alpha{}_\gamma) x^\gamma \approx x^\alpha##. The only way for both sides to be approximately equal requires that ##\gamma =\alpha## and that what is in parenthesis is worth the identity right?
Remember that you're summing over repeated indices. So you really have
$$\sum_\gamma \delta^\alpha{}_\gamma x^\gamma + \sum_\gamma (\epsilon^\alpha{}_\gamma + \epsilon'^\alpha{}_\gamma) x^\gamma = x^\alpha.$$ The first term just leaves you with ##x^\alpha## after the summation, so you end up with
$$\sum_\gamma (\epsilon^\alpha{}_\gamma + \epsilon'^\alpha{}_\gamma) x^\gamma = 0.$$ Now use the fact that this has to hold for arbitrary ##x^\mu##.
 
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  • #13
fluidistic
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Remember that you're summing over repeated indices. So you really have
$$\sum_\gamma \delta^\alpha{}_\gamma x^\gamma + \sum_\gamma (\epsilon^\alpha{}_\gamma + \epsilon'^\alpha{}_\gamma) x^\gamma = x^\alpha.$$ The first term just leaves you with ##x^\alpha## after the summation, so you end up with
$$\sum_\gamma (\epsilon^\alpha{}_\gamma + \epsilon'^\alpha{}_\gamma) x^\gamma = 0.$$ Now use the fact that this has to hold for arbitrary ##x^\mu##.
Ah right, thanks!
This leaves me with ##\epsilon ^\alpha _\gamma=-\epsilon ' ^\alpha _\gamma##. By multiplying each side from the right by ##\eta ^{\gamma \beta}##, I do reach the final result.
 
  • #14
fluidistic
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I attempted part 2) but I get some non sense.
As andrien pointed out, conservation of norm is ##x^\alpha x_\alpha=x'^\alpha x'_\alpha##.
What I did:
##x'^\alpha x'_\alpha=(\eta^{\alpha \beta }+ \epsilon ^{\alpha \beta})x_\beta x'_\alpha=(\eta^{\alpha \beta }+ \epsilon ^{\alpha \beta}) x_\beta \eta _{\alpha \gamma}x'^\gamma =(\eta^{\alpha \beta }+ \epsilon ^{\alpha \beta}) x_\beta \eta _{\alpha \gamma}(\eta^{\gamma \omega }+ \epsilon ^{\gamma \omega} ) x_\omega## ## = (\eta^{\alpha \beta } x_\beta + \epsilon ^{\alpha \beta} x_\beta )(\delta ^\omega _\alpha x_\omega + \epsilon ^\omega _\alpha x_\omega )=\eta ^{\alpha \beta}x_\beta \delta ^\omega _\alpha x_\omega +\eta ^{\alpha \beta}x_\beta \epsilon ^\omega _\alpha x_\omega +\epsilon ^{\alpha \beta} x_\beta \delta ^\omega _\alpha x_\omega +O(\epsilon ^2)## ##\approx x^\alpha \delta ^\omega _\alpha x_\omega + x ^\alpha \epsilon ^\omega _\alpha x_\omega + \epsilon ^{\alpha \beta} x_\beta \delta ^\omega _\alpha x_\omega##.
Here on my draft I rewrote that last expression with sums (2 double sums, 1 triple sum) just to simplicate the Kronecker's delta's.
Then I got rid once again of the sums. And I reached that it's worth ##x^\alpha x_\alpha +x^\alpha \epsilon ^\omega _\alpha x_\omega +\epsilon ^{\alpha \beta}x_\beta x_\alpha##. Now using the conservation of the norm, I got that ##x^\alpha \epsilon ^\omega _\alpha x_\omega =-\epsilon ^{\alpha \beta}x_\beta x_\alpha##.
But here I realized that this is a non sense: for if ##x_\beta## is a 1x4 covector, epsilon a 4x3 matrix, then ##x _\omega## sould be a 3x1 vector... but it is a 1x3 covector. Therefore the left hand side doesn't make any sense.
I don't see where I went wrong though.
 
  • #15
WannabeNewton
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##\epsilon^{\alpha}{}{}_{\beta}## has a 4x4 matrix representation, not 4x3.
 
  • #16
fluidistic
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##\epsilon^{\alpha}{}{}_{\beta}## has a 4x4 matrix representation, not 4x3.
Oh you're right.
But I still have the same problem, the left hand side of the last equation would be (1x4)x(4x4)x(1x4) where the colored digits indicate that something doesn't match. I seem to have a covector multiplied by a covector instead of a covector multiplied by a vector.
 
  • #17
WannabeNewton
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What you have is (1x4)x(4x4)x(4x1) = 1x1 on the left hand side, which is fine. The right hand side is the same thing once you raise the index of one of the ##x_{\alpha}## and lower the corresponding index on ##\epsilon^{\alpha\beta}##. If you want to represent ##\epsilon^{\alpha\beta}## as a matrix then it has to be in the form ##\epsilon^{\alpha}{}{}_{\beta}##.
 
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  • #18
fluidistic
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What you have is (1x4)x(4x4)x(4x1) = 1x1 on the left hand side, which is fine. The right hand side is the same thing once you raise the index of one of the ##x_{\alpha}## and lower the corresponding index on ##\epsilon^{\alpha\beta}##. If you want to represent ##\epsilon^{\alpha\beta}## as a matrix then it has to be in the form ##\epsilon^{\alpha}{}{}_{\beta}##.
I see thank you. I'll need to digest this. I'll come back tomorrow on this.
 
  • #19
vanhees71
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I don't understand the trouble you make with this problem. Just expand
[tex]\eta_{\mu \nu} \left (\delta^{\mu}_{\rho} + {\epsilon^{\mu}}_{\rho} \right) \left (\delta^{\nu}_{\sigma} + {\epsilon^{\nu}}_{\sigma} \right) \stackrel{!}{=}\eta_{\rho \sigma}.[/tex]
up to first order in [itex]\epsilon[/itex], and you'll find that [itex]\epsilon_{\mu \nu}=-\epsilon_{\nu \mu}[/itex].
 
  • #20
fluidistic
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I don't understand the trouble you make with this problem. Just expand
[tex]\eta_{\mu \nu} \left (\delta^{\mu}_{\rho} + {\epsilon^{\mu}}_{\rho} \right) \left (\delta^{\nu}_{\sigma} + {\epsilon^{\nu}}_{\sigma} \right) \stackrel{!}{=}\eta_{\rho \sigma}.[/tex]
up to first order in [itex]\epsilon[/itex], and you'll find that [itex]\epsilon_{\mu \nu}=-\epsilon_{\nu \mu}[/itex].
Hmm but I wouldn't be using the conservation of the norm by doing so, or I'm wrong?
 
  • #21
vanhees71
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But this is the conservation of the "norm"!
 
  • #22
vela
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Conservation of the norm says
$$\eta_{\mu\nu} x^\mu x^\nu = \eta_{\rho\sigma}x'^\rho x'^\sigma.$$ Again, write x' in terms of x to get
$$\eta_{\mu\nu} x^\mu x^\nu = \eta_{\rho\sigma}[(\delta^\rho_\mu + \epsilon^\rho{}_\mu) x^\mu] [(\delta^\sigma_\nu + \epsilon^\sigma{}_\nu)x^\nu] = [\eta_{\rho\sigma}(\delta^\rho_\mu + \epsilon^\rho{}_\mu) (\delta^\sigma_\nu + \epsilon^\sigma{}_\nu)] x^\mu x^\nu.$$ Comparing the two sides of the equation, you should be able to see you must have
$$\eta_{\mu\nu} = \eta_{\rho\sigma}(\delta^\rho_\mu + \epsilon^\rho{}_\mu) (\delta^\sigma_\nu + \epsilon^\sigma{}_\nu).$$ Without the x's cluttering things up, it's a little easier to see where you're headed.
 

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