Proving Lorentz Transformation identities

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Summary:

I want to prove certain properties related to Lorentz Transformations.

Main Question or Discussion Point

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 :smile:
 

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  • #2
PeroK
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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]
I think you have gone wrong conceptually. ##\Lambda## here is a Lorentz transformation. I would have:
$$k' = \Lambda k, \ \ \text{where} \ \ k'^{\mu} = \Lambda^{\mu}{}_{\rho} k^{\rho}$$
And:
$$x' = \Lambda x, \ \ \text{where} \ \ x'^{\nu} = \Lambda^{\nu}{}_{\sigma} x^{\sigma}$$
Then:
$$k' \cdot x' = \eta_{\mu \nu} k'^{\mu}x'^{\nu} = \eta_{\mu \nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} k^{\rho} x^{\sigma} = \eta_{\rho \sigma} k^{\rho} x^{\sigma} = k \cdot x$$

PS ##\Lambda## can't just be any linear transformation.
 
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Ahhh I see the approach thanks.

(ii) [itex]\ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x[/itex]

We now have

$$p' = \Lambda p, \ \ \text{where} \ \ p'^{\rho} = \Lambda^{\rho}{}_{\lambda} p^{\lambda}$$

$$x' = \Lambda x, \ \ \text{where} \ \ x'^{\nu} = \Lambda^{\nu}{}_{\sigma} x^{\sigma}$$

We need to solve for ##p##

$$p^{\rho}= \Lambda_{\lambda}{}^\rho p^{'\lambda}$$

Thus

$$p \cdot x' = \eta_{\rho \nu} p^{\rho}x'^{\nu}=\eta_{\rho \nu} \Lambda_{\lambda}{}^\rho \Lambda^{\nu}{}_{\sigma} p^{'\lambda} x^{\sigma} =\Lambda_{\lambda}{}^\rho p^{'\lambda} x_{\rho} = (\Lambda^{-1} p') \cdot x$$

Do you agree so far? :biggrin:

Working on (iii)...
 
  • #4
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(iii) [itex]\ \left(\Lambda^{-1}k \right)^{2} = k^{2}[/itex]
We have

$$k' = \Lambda k, \ \ \text{where} \ \ k'^{\mu} = \Lambda^{\mu}{}_{\rho} k^{\rho}$$

We solve for ##k##

$$k^{\mu}= \Lambda_{\rho}{}^\mu k^{'\rho}$$

Thus we get

$$k \cdot k = \eta_{\mu \nu} k^{\mu} k^{\nu} = \eta_{\mu \nu} \Lambda_{\rho}{}^\mu \Lambda_{\sigma}{}^\nu k^{\rho} k^{\sigma}$$

Mmm but here I do not see how to show that

$$\eta_{\mu \nu} \Lambda_{\rho}{}^\mu \Lambda_{\sigma}{}^\nu k^{\rho} k^{\sigma} = (\Lambda^{-1} k)^2$$
 
  • #5
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Ahhh I see the approach thanks.




We now have

$$p' = \Lambda p, \ \ \text{where} \ \ p'^{\rho} = \Lambda^{\rho}{}_{\lambda} p^{\lambda}$$

$$x' = \Lambda x, \ \ \text{where} \ \ x'^{\nu} = \Lambda^{\nu}{}_{\sigma} x^{\sigma}$$

We need to solve for ##p##

$$p^{\rho}= \Lambda_{\lambda}{}^\rho p^{'\lambda}$$

Thus

$$p \cdot x' = \eta_{\rho \nu} p^{\rho}x'^{\nu}=\eta_{\rho \nu} \Lambda_{\lambda}{}^\rho \Lambda^{\nu}{}_{\sigma} p^{'\lambda} x^{\sigma} =\Lambda_{\lambda}{}^\rho p^{'\lambda} x_{\rho} = (\Lambda^{-1} p') \cdot x$$

Do you agree so far? :biggrin:

Working on (iii)...
Hmm. I think I did all the work for (i). All you need for (ii) is to let ##k = \Lambda^{-1} p##.
 
  • #6
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Hmm. I think I did all the work for (i). All you need for (ii) is to let ##k = \Lambda^{-1} p##.
I am afraid I do not understand. You don’t agree with my # 3?
 
  • #7
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I am afraid I do not understand. You don’t agree with my # 3?
How are you getting the components of ##\Lambda^{-1}##?
 
  • #8
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How are you getting the components of ##\Lambda^{-1}##?
Based on

$$p' = \Lambda p, \ \ \text{where} \ \ p'^{\rho} = \Lambda^{\rho}{}_{\lambda} p^{\lambda}$$

I solved for ##p## and got

$$p^{\rho}= \Lambda_{\lambda}{}^\rho p^{'\lambda}$$

I've read that it is standard notation to write the inverse of ##\Lambda^{\rho}{}_{\lambda}## as ##\Lambda_{\lambda}{}^\rho##

I have to say I do not see how to justify that ##\Lambda_{\lambda}{}^\rho p^{'\lambda} x_{\rho} = (\Lambda^{-1} p') \cdot x## and ##\eta_{\mu \nu} \Lambda_{\rho}{}^\mu \Lambda_{\sigma}{}^\nu k^{\rho} k^{\sigma} = (\Lambda^{-1} k)^2##.

Hmm. I think I did all the work for (i).
I agree you've already been really helpful solving (i), thank you.
 
  • #9
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(ii) and (iii) are immediate consequences of (i). There was no more work to be done.
 
  • #10
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(ii) and (iii) are immediate consequences of (i). There was no more work to be done.
Once I get how to prove (ii) and (iii) I'll post it.
 
  • #11
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Once I get how to prove (ii) and (iii) I'll post it.
There's nothing to prove. They are immediate corollaries of (i).

What you could prove is that ##\Lambda^{-1}## exists and is also a Lorentz Transformation. But, I don't think you can get a simple formula for the components. You will have the usual matrix inverse relationship.
 
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  • #12
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PS You could also prove that the set of Lorentz Transformations, as defined, forms a group.
 
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  • #13
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Hmm. I think I did all the work for (i). All you need for (ii) is to let ##k = \Lambda^{-1} p##.
Alright I see now how to show (ii):

Assuming ##p = \Lambda k## and using (i)

$$ (\Lambda k) \cdot (\Lambda x) = (\Lambda \Lambda^{-1} p) \cdot (\Lambda x) = p \cdot (\Lambda x) = k \cdot x = (\Lambda^{-1} p) \cdot x$$

For (iii) I am still having difficulties;

Assuming ##p = \Lambda k##, ##k=\Lambda x## and (ii)

$$\ p \cdot (\Lambda x) = (\Lambda k) \cdot (\Lambda x) = k \cdot (\Lambda^{-1} k) \neq (\Lambda^{-1} k)^2$$

I guess I am missing something basic here.
 
  • #14
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Alright I see now how to show (ii):

Assuming ##p = \Lambda k## and using (i)

$$ (\Lambda k) \cdot (\Lambda x) = (\Lambda \Lambda^{-1} p) \cdot (\Lambda x) = p \cdot (\Lambda x) = k \cdot x = (\Lambda^{-1} p) \cdot x$$

For (iii) I am still having difficulties;

Assuming ##p = \Lambda k##, ##k=\Lambda x## and (ii)

$$\ p \cdot (\Lambda x) = (\Lambda k) \cdot (\Lambda x) = k \cdot (\Lambda^{-1} k) \neq (\Lambda^{-1} k)^2$$

I guess I am missing something basic here.
##\Lambda^{-1}## is a Lorentz Transformation. We know from (i) that for any Lorentz Transformation ##L##:
$$(Lk)^2 = Lk \cdot Lk = k \cdot k$$
This is true with ##L = \Lambda^{-1}##.

And, a proof of (ii) is simply:
$$(\Lambda^{-1} p) \cdot x = (\Lambda \Lambda^{-1} p) \cdot \Lambda x \ \ (\text{from (i)}) \\ = p \cdot \Lambda x$$
 
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  • #15
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What you could prove is that ##\Lambda^{-1}## exists and is also a Lorentz Transformation.
Good point, I've been assuming so far ##\Lambda^{-1}## exists.

I've come across this proof:

Given the Lorentz Transformation condition ##\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}##, we multiply both sides by ##\eta^{\sigma \tau}## to get

$$\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} \eta^{\sigma \tau} = \eta_{\rho \sigma} \eta^{\sigma \tau}=\delta_{\rho}^{\tau}$$

Then we can rearrange at will because, as indices are written explicitly, we can treat matrices as numbers. Thus we get

$$\eta_{\mu\nu} \eta^{\sigma \tau} \Lambda^{\nu}{}_{\sigma} \Lambda^{\mu}{}_{\rho}=\delta_{\rho}^{\tau}$$

Contracting we get

$$\Lambda_{\mu}{}^\tau \Lambda^{\mu}{}_{\rho}=\delta_{\rho}^{\tau}$$

Thus we can conclude that:

$$\Big(\Lambda^{-1}\Big)^{\tau}{}_{\mu}=\eta_{\mu\nu} \eta^{\sigma \tau} \Lambda^{\nu}{}_{\sigma}=\Lambda_{\mu}{}^\tau$$

QED.
 
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  • #16
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There's a neater proof on that page, which is to write the defining relation in matrix form:
$$\Lambda^{T} \eta \Lambda = \eta$$
And since ##\det(\eta) = -1##, we have ##\det(\Lambda) = \pm 1##, hence ##\Lambda## is invertible.

And, as ##\eta^{-1} = \eta##, we have:
$$\Lambda^{-1} = \eta \Lambda^T \eta$$

To prove that these Lorentz Transformations form a group, it's easier to stay in matrix form. Note that:
$$(LM)^T = M^TL^T, \ \ \text{and} \ \ LL^{-1} = I \ \Rightarrow \ (L^{-1})^TL^T = I \ \Rightarrow \ (L^T)^{-1} = (L^{-1})^T$$
To show that ##L^{-1}## is also a Lorentz Transformation, we have:
$$L^T \eta L = \eta \ \Rightarrow \ \eta = (L^T)^{-1}\eta L^{-1} = (L^{-1})^T \eta L^{-1}$$
You can also show that if ##L, M## are Lorentz transformations then so is ##LM##. And the identity is also a Lorentz transformation. Therefore, these matrices form a group.

That said, doing all this in index notation must be good practice.
 
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  • #17
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Ahhh I was clearly overcomplicating it!

For (iii) I am still having difficulties;
##\Lambda^{-1}## is a Lorentz Transformation. We know from (i) that for any Lorentz Transformation ##L##:
$$(Lk)^2 = Lk \cdot Lk = k \cdot k$$
This is true with ##L = \Lambda^{-1}##.

You proved (iii) using (i), which is perfectly fine.

I came up with the following:

Using (ii) and assuming that ##k = \Lambda x##.

$$k \cdot (\Lambda x) = (\Lambda^{-1} k) \cdot (\Lambda^{-1} k)$$

Then we use (i) to prove the RHS of (iii)

$$(\Lambda^{-1} k) \cdot (\Lambda^{-1} k)=k \cdot k$$

Good! But there's something that makes me think; I've assumed (in my proofs of (ii) and (iii)) that ##p=\Lambda k## and ##k = \Lambda x##. Is that OK or should I try to avoid such assumptions and find a neater route?
 
  • #18
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PS You could also prove that the set of Lorentz Transformations, as defined, forms a group.
To prove that these Lorentz Transformations form a group, it's easier to stay in matrix form. Note that:
$$(LM)^T = M^TL^T, \ \ \text{and} \ \ LL^{-1} = I \ \Rightarrow \ (L^{-1})^TL^T = I \ \Rightarrow \ (L^T)^{-1} = (L^{-1})^T$$
To show that ##L^{-1}## is also a Lorentz Transformation, we have:
$$L^T \eta L = \eta \ \Rightarrow \ \eta = (L^T)^{-1}\eta L^{-1} = (L^{-1})^T \eta L^{-1}$$
You can also show that if ##L, M## are Lorentz transformations then so is ##LM##. And the identity is also a Lorentz transformation. Therefore, these matrices form a group.

That said, doing all this in index notation must be good practice.
If I am not mistaken, to prove that the Lorentz Transformations ##\Lambda## form a group (called Lorentz Group), we need:

1) To equip the set of LTs with a binary operation: the inner-product.

2) To show closure (i.e. the inner product of two LTs yields a LT; I think this is what you suggested at #16).

3) To show the existance of the identity element (i.e. ##\Lambda I = \Lambda##).

4) To show invertibility (i.e. ##\Lambda^{-1}\Lambda=\Lambda\Lambda^{-1}=I##, which I think we've already done).

And that'd do the proof.

PS: I cannot wait to dive into this!! 😍 Mmm I see that the full Lorentz Group is made of the union of four disjoint sets

Captura de pantalla (1051).png


But first things first!
 
  • #19
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When we say the Lorentz group is a group, it is assumed that the binary operation is matrix multiplication. The set of all invertible matrices is a group, so we are really looking to show that the Lorentz group is a subgroup of that, which is why we have only a few things to prove. Although, actually, it's not that much less to do.

Note that the binary operation is not the inner product. A group is an algebraic structure. Some groups also have an analytic or topological structure. This requires some definition of magnitude or distance - metric, norm, inner product - or, more abstractly, some concept of open sets (which is the more abstract topology). This is the basis of continuity, differentiability and calculus.

For example, the group of boosts and rotations is a normal subgroup of the Lorentz Group. Normality is an algebraic property. It is also a connected subgroup. Connectedness is an analytic or topological property.
 
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  • #20
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When we say the Lorentz group is a group, it is assumed that the binary operation is matrix multiplication. The set of all invertible matrices is a group, so we are really looking to show that the Lorentz group is a subgroup of that
Oh I see.

Let me label the Lorentz Group G; To prove that G is a subgroup of all invertible matrices, we need to show the following

1) ##G \neq \emptyset ##

2) G is closed under matrix multiplication (i.e. show that the product of two Lorentz-Transformation matrices is also a Lorentz Transformation).

3) G is closed under inverses (i.e. show that ##\Lambda^{-1}## is a Lorentz Transformation)


1) Here something's going wrong. The zero matrix doesn't belong to the Lorentz Group, as it doesn't satisfy the Lorentz condition ##\eta = L^{T} \eta L## (i.e. plugging the zero matrix into it yields ##\eta=0##, which is not true).

2) Let's assume L and M are LTs. Thus we have

$$\eta = L^T \eta L, \ \ \ \ \eta = M^T \eta M$$

$$(LM)^T \eta LM = M^T L^T \eta L M = M^T \eta M = \eta$$

Thus the product of two Lorentz-Transformation matrices is also a Lorentz Transformation.

3)

To show that ##L^{-1}## is also a Lorentz Transformation, we have:
$$L^T \eta L = \eta \ \Rightarrow \ \eta = (L^T)^{-1}\eta L^{-1} = (L^{-1})^T \eta L^{-1}$$
So I guess that to conclude the proof we only have to solve 1)
 
  • #21
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1) ##G \neq \emptyset ##

So I guess that to conclude the proof we only have to solve 1)
It's the multiplicative identity (##I##) you need. That also takes care of the set not being empty. Note that the subgroup of boosts and rotations is the (connected) subgroup that contains the identity element.
 

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