Transpose and Inverse of Lorentz Transform Matrix

[cianfa72]

That's a shame, because you'd be correct.
You can of course view it either actively or passively.

Surely, my point was that even if do not 'attach' any physical quantity/property to it from the active point of view it is actually a (1,1) tensor, though.

In fact using the identification of a vector space $V$ with its double dual $V^{**}$ we get an isomorphism between Linear transformations of $V$ and (1,1) type tensors.

 

[ergospherical]

Surely, my point was that even if do not 'attach' any physical quantity/property to it from the active point of view it is actually a (1,1) tensor, though.

That is a mistake. The elements ${\Lambda^{\mu}}_{\nu}$ are not the elements of a tensor. Nonetheless you may still raise and lower its indices with the metric $\eta_{\mu \nu}$.

 

[cianfa72]

That is a mistake. The elements ${\Lambda^{\mu}}_{\nu}$ are not the elements of a tensor.

I'm really struggling with it

From the point of view of active linear transformation of $V$ into itself, ${\Lambda^{\mu}}_{\nu}$ is actually its representation in a chosen basis of $V$. Furthermore as linear transformation it is isomorphic to a specific (1,1) tensor.

Then upon a change of basis in $V$ (and implicitly in the dual $V^{*}$) the entries in ${\Lambda^{\mu}}_{\nu}$ change in the same way the components of the (1,1) tensor isomorphic to it, dont' you ?

 

[ergospherical]

Not every object with two indices is a rank-2 tensor! [Recall that a tensor is specifically a multilinear map of vectors and covectors to some field e.g. $\mathbb{R}$, and whose components in a given basis transform according to the tensor transformation law.]

Rather a Lorentz transformation is an element of the Lorentz group [its elements are endomorphisms of $\mathbb{R}^4$ which map one orthonomal basis to another orthonormal basis], and each has a matrix representation with respect to a given basis. That a Lorentz transformation is an isometry implies that, for arbitrary vectors $\mathbf{u},\mathbf{v} \in \mathbb{R}^4$,\begin{align*}
\eta(L(\mathbf{u}), L(\mathbf{v})) = \eta_{ab} [L(\mathbf{u})]^a [L(\mathbf{v})]^b &\overset{!}{=} \eta_{cd} u^c v^d = \eta(\mathbf{u},\mathbf{v})\\

\eta_{ab} {\Lambda^a}_c u^c {\Lambda^b}_d v^d &= \eta_{cd} u^c v^d \\

\eta_{ab} {\Lambda^a}_c {\Lambda^b}_d &= \eta_{cd} \\

{(\Lambda^T)_c}^a \eta_{ab}{\Lambda^b}_d &= \eta_{cd} \\

\Lambda^T \eta \Lambda &= \eta
\end{align*}which is the defining property of a Lorentz matrix $\Lambda$. Taking the determinant of both sides yields $|\Lambda|^2 = 1$; the sub-group of $O(3,1)$ with determinant $+1$ is called the proper Lorentz group and is denoted $SO(3,1)$. Meanwhile an orthochronous Lorentz transformation is one for which, given any timelike vector $\mathbf{u}$ [$\eta(\mathbf{u},\mathbf{u}) < 0$], we have $\eta(L(\mathbf{u}), \mathbf{u}) < 0$. You can show that a Lorentz matrix is orthochronous if ${\Lambda^0}_0 \geq 1$ [exercise!]. The group of Lorentz transformations which are both proper and orthochronous [called restricted Lorentz transformations] is denoted $SO^{+}(3,1)$; any Lorentz transformation can in fact be reduced to a restricted one.

 

[cianfa72]

Rather a Lorentz transformation is an element of the Lorentz group [its elements are endomorphisms of $\mathbb{R}^4$ which map one orthonomal basis to another orthonormal basis], and each has a matrix representation with respect to a given basis.

Take an element of the Lorentz group (i.e. a Lorentz transformation) and consider its matrix representation in a given not orthonormal basis. Is actually this matrix a 'Lorentz matrix' ?

 

[ergospherical]

I'd say no, a Lorentz matrix $\Lambda$ is specifically a representation of a Lorentz transformation in an orthonormal basis of $\mathbb{R}^4$, i.e. a matrix for which there exists a map $L : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ and an orthonormal basis $\{\hat{\mathbf{e}}_a \}$ such that $L(\hat{\mathbf{e}}_a) = {\Lambda^b}_a \hat{\mathbf{e}}_b$. That's to say the columns of $\Lambda$ are the images $L(\hat{\mathbf{e}}_a)$.

[But of course, you can represent $L$ in terms of any basis you wish!]

 

[cianfa72]

[But of course, you can represent $L$ in terms of any basis you wish!]

Maybe this was the missing point to me.
Hence a Lorentz transformation is isomorphic to a (1,1) tensor instead a Lorentz matrix (since restricted to represent a Lorentz transformation in an orthonormal basis) is not.

 

[ergospherical]

a Lorentz transformation is isomorphic to a (1,1) tensor

You should drop this idea, it's misleading! Lorentz transformations don't really have anything to do with tensors.

 

[cianfa72]

You should drop this idea, it's misleading! Lorentz transformations don't really have anything to do with tensors.

Sorry, surely I am wrong but...is a Lorentz transformation actually a linear transformation of $\mathbb {R}^4$ in itself ?
Is any linear transformation isomorphic to a (1,1) tensor ?

Hence...

 

[ergospherical]

I think @vanhees71 already said this; Lorentz transformations are endomorphisms $L: \mathbb{R}^4 \rightarrow \mathbb{R}^4$, whilst tensors are maps $T: {V^*}^p \times {V}^q \rightarrow \mathbb{F}$, i.e. they're different types of objects.

However, it is a result of linear algebra that vector spaces of equal dimension are isomorphic. So yeah, it's certainly possible to write down isomorphisms between $(1,1)$ tensors and endomorphisms on $\mathbb{R}^4$, although I'm not sure if that's helpful when dealing with Lorentz transformations...

 

[cianfa72]

So yeah, it's certainly possible to write down isomorphisms between $(1,1)$ tensors and endomorphisms on $\mathbb{R}^4$, although I'm not sure if that's helpful when dealing with Lorentz transformations...

The point to be highlighted is that actually there exist a natural (or canonical) isomorphism between the two that requires no choice about the basis.

See for example the first answer here https://math.stackexchange.com/questions/1108842/why-is-a-linear-transformation-a-1-1-tensor

 

[ergospherical]

Ah cool, yeah that's one worked example! Though I'm not sure how helpful it is to put a Lorentz transformation into correspondence with a $(1,1)$ tensor. The isomorphism between linear transformations and matrices (e.g. taking $L$ to $\Lambda$) seems like the only important one.

Maybe someone else can think of an example, though?

 

[cianfa72]

The isomorphism between linear transformations and matrices (e.g. taking $L$ to $\Lambda$) seems like the only important one.

Btw that isomorphism is not natural because the selection of a basis is needed.

 

[vanhees71]

That's a shame, because you'd be correct.
You can of course view it either actively or passively.

The active point of view: A Lorentz transformation $L : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ is an isometry (read: preserves the scalar square of vectors) of spacetime and the Lorentz matrix $\Lambda = [{\Lambda^{\mu}}_{\nu}]$ of this transformation in some orthonormal basis $\{ \hat{\mathbf{e}}_{\mu} \}$ is the matrix that satisfies $L(\hat{\mathbf{e}}_{\nu}) = {\Lambda^{\mu}}_{\nu} \hat{\mathbf{e}}_{\mu}$.

[The set of all these transformations $L$ (along with a binary operation $\circ$) forms the Lorentz group $O(3,1)$]

Another way to see that a Lorentz matrix are not tensor components is

I think @vanhees71 already said this; Lorentz transformations are endomorphisms $L: \mathbb{R}^4 \rightarrow \mathbb{R}^4$, whilst tensors are maps $T: {V^*}^p \times {V}^q \rightarrow \mathbb{F}$, i.e. they're different types of objects.

However, it is a result of linear algebra that vector spaces of equal dimension are isomorphic. So yeah, it's certainly possible to write down isomorphisms between $(1,1)$ tensors and endomorphisms on $\mathbb{R}^4$, although I'm not sure if that's helpful when dealing with Lorentz transformations...

It's not an isomorphism, but you can look at the (proper orthochronous) Lorentz group as a Lie group. Then the infinitesimal generators indeed behave like tensor components. Note, though, that you cannot write all Lorentz transformations as a single exponential of a generator. So the map is not surjective.

 

[cianfa72]

It's not an isomorphism

Which is not an isomorphism ?

 

[vanhees71]

There is no isomorphism between the Lie algebra of the proper orthochronous Lorentz transformation and the Lie group.

 

[cianfa72]

There is no isomorphism between the Lie algebra of the proper orthochronous Lorentz transformation and the Lie group.

Surely, in fact we were talking about the canonical isomorphism between Linear transformation (such as Lorentz transformations in $\mathbb{R}^4$) and (1,1) tensors.

 

[PeterDonis]

The point to be highlighted is that actually there exist a natural (or canonical) isomorphism between the two that requires no choice about the basis.

See for example the first answer here https://math.stackexchange.com/questions/1108842/why-is-a-linear-transformation-a-1-1-tensor

the canonical isomorphism between Linear transformation (such as Lorentz transformations in $\mathbb{R}^4$) and (1,1) tensors.

The isomorphism referred to requires you to view the reals as a vector space. But from the point of view of tensor calculus, the reals--the numbers you get when you contract tensors with vectors and covectors--are not a vector space. They are just a field.

 

[ergospherical]

Every field is a vector space over itself.

 

[PeterDonis]

Every field is a vector space over itself.

As a matter of math, yes, this is true. But as a matter of physics, sometimes the vector space structure of a field is part of the physical model, and sometimes it's not.

 

[ergospherical]

As a matter of math, yes, this is true. But as a matter of physics, sometimes the vector space structure of a field is part of the physical model, and sometimes it's not.

I don't understand the distinction, could you explain? Any field $(F, \times, \cdot)$ is automatically a vector space over itself because all the vector space axioms are contained within the field axioms. It doesn't have anything to do with Physics?

 

[cianfa72]

The isomorphism referred to requires you to view the reals as a vector space. But from the point of view of tensor calculus, the reals--the numbers you get when you contract tensors with vectors and covectors--are not a vector space. They are just a field.

Do you mean $\mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} =\mathbb{R}^4$ as vector space ?

 

[PeterDonis]

Do you mean $\mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} =\mathbb{R}^4$ as vector space ?

For the case of Lorentz transformations in 4D spacetime, yes, you would have to consider $\mathbb{R}^4$ as a vector space. But note that it is not the same vector space as Minkowski spacetime; i.e., it is not the vector space that the Lorentz transformations are normally viewed as acting on.

Note also that, for the case of tensors, as defined by how they contract with other tensors to form scalars, the scalars are just reals; they are elements of $\mathbb{R}$, not $\mathbb{R}^N$ (where $N$ is the number of dimensions of the space or spacetime being considered).

 

[PeterDonis]

Any field $(F, \times, \cdot)$ is automatically a vector space over itself because all the vector space axioms are contained within the field axioms.

But $\mathbb{R}$ is not the only possible vector space over the field $\mathbb{R}$. And in the case of the physics we are discussing, the vector space in question is Minkowski spacetime, which is not $\mathbb{R}$, and not even $\mathbb{R}^4$ (which is not the same vector space as $\mathbb{R}$, but considered as a vector space over a field, is a vector space over the field $\mathbb{R}$), but is a different vector space over the field $\mathbb{R}$ (since the scalars you get when you contract Minkowski vectors and covectors are reals--this is the same point I made to @cianfa72 in my previous post just now). So the vector space structure being used to model the physics in this case is not the vector space structure of $\mathbb{R}$ (even though it automatically has one, as you point out), it's the vector space structure of Minkowski spacetime.

 

[ergospherical]

Minkowski spacetime, which is ... not even $\mathbb{R}^4$

This is a bit of a digression... but Minkowski spacetime is $M = (\mathbb{R}^4, \eta)$ with $\eta$ a symmetric, non-degenerate bilinear form of signature (-+++), and where the Lorentz group acts on the tangent space $T_{p}M$ at every point.

 

[PeterDonis]

This is a bit of a digression... but Minkowski spacetime is $M = (\mathbb{R}^4, \eta)$ with $\eta$ a symmetric, non-degenerate bilinear form of signature (-+++), and where the Lorentz group acts on the tangent space $T_{p}M$ at every point.

As a manifold with metric, yes. But we're talking about Minkowski spacetime as a vector space.

 

[vanhees71]

Minkowski spacetime is an affine pseudo-Euclidean (Lorentzian) space, i.e., a set called "points" and a vector space with a fundamental bilinear form of signature (1,3) (or equivalently (3,1)). For two points $A$ and $B$ there is always a unique vector $\overrightarrow{AB}$. The vectors can be identified with directed straight lines connecting two points. For three points $A$, $B$, and $C$ vector addition is defined as $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. Two vectors $\overrightarrow{AB}$ and $\overrightarrow{A'B'}$ are equal, if one can parallel shift the corresponding directed straight lines into each other.

The full symmetry group, connected smoothly to the identity is the proper orthochronous Poincare group, generated by translations (homogeneity of spacetime, no spacetime point is distinguished from any other), and the Lorentz group, which is again generated by the rotation-free boosts (building NOT a subgroup), describing the invariance of the special-relativistic laws of nature when switching from one inertial reference frame to another inertial reference frame, and rotations (building a subgroup), describing the isotropy of space as observed by any inertial observer.

 

[cianfa72]

The full symmetry group, connected smoothly to the identity is the proper orthochronous Poincare group, generated by translations (homogeneity of spacetime, no spacetime point is distinguished from any other), and the Lorentz group, which is again generated by the rotation-free boosts (building NOT a subgroup), describing the invariance of the special-relativistic laws of nature when switching from one inertial reference frame to another inertial reference frame, and rotations (building a subgroup), describing the isotropy of space as observed by any inertial observer.

ok, from the point of view of the representation of a Lorentz group element (i.e. a Lorentz transformation) by a (Lorentz) matrix we need to select a basis in the vector space part of the definition of $M$ as affine space (as said above in this thread actually an orthonormal basis), right ?

 

[vanhees71]

$$\newcommand{\bvec}[1]{\boldsymbol{#1}}$$
Sure, a Lorentzian (inertial) frame of reference can be defined by choosing an arbitrary point $O$ (origin) of Minkowski space and 4 Minkowski-orthonormal basis vectors, $\bvec{e}_{\mu}$ with $\bvec{e}_{\mu} \cdot \bvec{e}_{\nu}=\eta_{\mu \nu}$. Then under a Poincare transformation the spacetime coordinates transform as
$$x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu},$$
where ${\Lambda^{\mu}}_{\rho}$ is a Lorentz-transformation matrix with $\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}$ and $a^{\mu}$ arbitrary four-vector components.

[EDIT: corrected the typos $L \rightarrow \Lambda$ mentioned in #65.

 

[cianfa72]

$$\newcommand{\bvec}[1]{\boldsymbol{#1}}$$
Sure, a Lorentzian (inertial) frame of reference can be defined by choosing an arbitrary point $O$ (origin) of Minkowski space and 4 Minkowski-orthonormal basis vectors, $\bvec{e}_{\mu}$ with $\bvec{e}_{\mu} \cdot \bvec{e}_{\nu}=\eta_{\mu \nu}$. Then under a Poincare transformation the spacetime coordinates transform as
$$x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu},$$

A doubt on coordinates $x^{\mu}$ and $x^{\prime \mu}$. They are used for the components of vectors (in a given orthonormal basis $\{ \bvec{e}_{\mu} \}$) that live into the 'translation' vector space involved in the definition of $M$ as affine space.

where ${\Lambda^{\mu}}_{\rho}$ is a Lorentz-transformation matrix with $\eta_{\mu \nu} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma} = \eta_{\rho \sigma}$ and $a^{\mu}$ arbitrary four-vector components.

I think $\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}$ you meant.

Here as four-vector components $a^{\mu}$ do you mean the components of an arbitrary vector $\mathbf a$ that lives in the 'translation' vector space in the same basis $\{ \bvec{e}_{\mu} \}$ ?

In other words $x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu}$ actually is not a linear transformation of the 'translation' vector space into itself (since the term $a^{\mu}$), right ?

 

[vanhees71]

The full group of Minkowski space as an affine space includes space-time translations, i.e., you can shift the origin arbitrarily around without changing the physics, i.e., the physical laws look the same at any time and any place.

 

[cianfa72]

The full group of Minkowski space as an affine space includes space-time translations, i.e., you can shift the origin arbitrarily around without changing the physics, i.e., the physical laws look the same at any time and any place.

Surely, my point was to better understand the notation used for Lorentz transformations. To me it seems we are basically 'turning' the affine space into a vector space by selecting a given point in it. Hence the coordinates $x^{\mu}$ and $x^{\prime \mu}$ entering in $x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu}$ are really the coordinates for the 'translation' vector space part of the affine space definition (i.e. the operator $+$ inside it is actually a sum of two vectors in the 'translation' vector space and is not the sum of a vector and a point in the affine space we started with).

Thanks for your patience

 

[vanhees71]

$$\newcommand{\bvec}[1]{\boldsymbol{#1}}$$
Well, $x^{\prime \mu}$ are the spactime coordinates in an inertial reference frame $\Sigma'$, which is defined by the corresponding origin $O'$ and the tetrad $\bvec{e}_{\mu}'$ and $x^{\rho}$ are the spacetime coordinates in an inertial frame $\sigma$ with origin $O'$ and the tetrad $\bvec{e}_{\rho}$.

Let $P$ be an arbitrary point in Minkowski space. The relation of the coordinates is given as follows:
$$\overrightarrow{OP}=x^{\rho} \bvec{e}_{\rho}, \quad \overrightarrow{O'P}=x^{\prime \mu} \bvec{e}_{\mu}'.$$
Now
$$x^{\prime \mu} \bvec{e}_{\mu}'=\overrightarrow{O'P}=\overrightarrow{O'O}+\overrightarrow{OP}=\bvec{a}+x^{\rho} \bvec{e}_{\rho}=a^{\prime \mu} \bvec{e}_{\mu}' + \bvec{e}_{\mu}' {\Lambda^{\mu}}_{\rho} x^{\rho}.$$
From this you get
$$x^{\prime \mu}=a^{\prime \mu}+ {\Lambda^{\mu}}_{\rho} x^{\rho}.$$

 

[cianfa72]

ok thanks.

Sorry to come back to the original topic: Lorentz transformations are actually isometries of $M$ that leave fixed the (chosen) origin. Hence we can consider them as (linear) isometries on the 'translation' vector space (say it $V$) part of the definition of $M$ as affine space, I believe.

Now, if the group of Lorentz transformations was itself a vector space on $\mathbb {R}$, then we could 'apply' the canonical isomorphism as described in the aforementioned link in order to 'identify' a Lorentz transformation (i.e. its matrix representation into a given basis of $V$) with a (1,1) tensor element of $V^{*} \otimes V$.

I was thinking that the group of Lorentz transformations actually is not itself a vector space on $\mathbb {R}$ since the multiplication of a Lorentz transformation for a real number is not a Lorentz transformation anymore.

Maybe that is the reason why we cannot 'build' the canonical isomorphism as above.

 

[vanhees71]

The Lorentz group is a subgroup of the symmetry group of Minkwoski space as an affine space. The Lorentz transformations themselves don't form a vector space. As a Lie group their Lie algebra is a vector space though.