Minkowski Spacetime KVF Symmetries

[cianfa72]

TL;DR Summary

About linear combination of isometries (KVF) of a given spacetime

Hi,

reading Carrol chapter 5 (More Geometry), he claims that a maximal symmetric space such as Minkowski spacetime has got $4(4+1)/2 = 10$ indipendent Killing Vector Fields (KVFs). Indeed we can just count the isometries of such spacetime in terms of translations (4) and rotations (6).

By definition a KVF is a vector field $V$ such that the Lie derivative of the metric tensor $g$ along it vanishes: $$\mathcal L_V g = 0$$
I believe the Lie derivative operator is actually linear in the vector field $V$. If it is the case, said $\partial / \partial_t, \partial / \partial_x, \partial / \partial_y, \partial / \partial_x$ the 4 translation isometries of Minkowski spacetime, then any linear combination of them should result in another KVF (i.e. an isometry) for the given spacetime.

Does it make sense ? Thank you.

 

[Orodruin]

The Lie derivative is linear in the vector field only with constant coefficients.

Also, the number of symmetries counts only linearly independent symmetries, i.e., if you have x and y translations as symmetries, translations in the (1,1)-direction is not a new symmetry.

 

[vanhees71]

Let's find the Killing vectors of Minkowski space. Let $\xi^{\mu}$ be the Killing vector, and we work of course in Galilean coordinates, $x^{\nu}$. The metric components are $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ and thus the Christoffel symbols al vanish in these coordinates, and the Killing equation
$$(\mathcal{L}_{\xi} g)_{\mu \nu} =0$$
simplifies to
$$\partial_{\mu} \xi_{\nu}+\partial_{\nu} \xi_{\mu}=0.$$
To find the general solution the trick is to take one more partial derivative,
$$\partial_{\rho} \partial_{\mu} \xi_{\nu} + \partial_{\rho} \partial_{\nu} \xi_{\mu}=0. \qquad(1)$$
Now we write down the other two cyclic permutations of the indices:
$$\partial_{\mu} \partial_{\nu} \xi_{\rho} + \partial_{\mu} \partial_{\rho} \xi_{\nu}=0, \qquad (2)$$
$$\partial_{\nu} \partial_{\rho} \xi_{\mu} + \partial_{\nu} \partial_{\mu} \xi_{\rho}=0. \qquad(3)$$
Now add equations (2) and (3) and subtract equation (1), and you get
$$\partial_{\mu} \partial_{\nu} \xi_{\rho}=0.$$
So you get by integration
$$\partial_{\nu} \xi_{\rho}=\omega_{\rho \nu}=\text{const}.$$
With another integration
$$\xi_{\rho} = \omega_{\rho \nu} x^{\nu}+a_{\rho}, \quad a_{\rho}=\text{const}.$$
Now we plug this into the original Killing equation, leading to
[index confusion corrected in view of #4]
$$\partial_{\mu} \xi_{\nu} + \partial_{\mu} \xi_{\nu} = \omega_{\mu \nu} + \omega_{\nu \mu} \stackrel{!}{=}0,$$
i.e., $\omega_{\mu \nu}=-\omega_{\nu \mu}$. So $\omega_{\mu \nu}$ is antisymmetric and thus has 6 independent degrees of freedom. The $a_{\rho}$ are arbitrary making 4 more degrees of freedom, i.e., indeed you have 10 linearly independent Killing fields.

It's also clear that the corresponding infinitesimal transformations are the general Lorentz transformations, corresponding to the three $\omega_{0j}$ (boost generators) and the three $\omega_{jk}=\phi_l \epsilon_{ijk}$ (rotation generators), where $j,k,l \in \{1,2,3 \}$ and four translations $a_{\mu}$. The symmetry group is, as to be expected, the (proper orthochronous) Poincare group.

 

[Orodruin]

Now we plug this into the original Killing equation, leading to
∂μξρ+∂ρξμ=ωμν+ωρμ=!0,
i.e., ωμν=−ωνρ.

Just to mention the index mismatch and subsequent copy-paste error. Otherwise

 

[cianfa72]

Maybe I missed the point. The Killing equation for the metric tensor $g$ requires that the diffeomorphism $\phi$ associated to the vector field $V$ is such that the pullback of metric tensor $g$ is equals to itself, namely $$ \phi^{*}g = g$$
From an intuitively point of view, I'm stuck with the following point: take a generic (smooth) vector field $X$ on Minkowski spacetime. Then locally at each point it can be decomposed (through non-constant functions of spacetime as coefficients) as linear combination of 'coordinate' vector fields $\partial_t, \partial_x, \partial_y, \partial_z$.

Since each of them is actually a KVF of Minkowski tensor metric (i.e. an isometry) why the given (generic) smooth vector field is not ?

 

[Ibix]

Generally the vector field will be expressible as four functions $V^a$ times those four Killing fields. But if the functions are non-trivial functions of the coordinates then you'll get extra terms when you take your derivatives, and they won't cancel out.

Physically, we're saying that, for example, if we take some flat spacelike hypersurface and displace it slightly in the $t$ direction it "looks the same". But if you displace it a bit more in some places and a bit less in others then in general the resulting plane won't be flat.

 

[Orodruin]

Maybe I missed the point. The Killing equation for the metric tensor $g$ requires that the diffeomorphism $\phi$ associated to the vector field $V$ is such that the pullback of metric tensor $g$ is equals to itself, namely $$ \phi^{*}g = g$$
From an intuitively point of view, I'm stuck with the following point: take a generic (smooth) vector field $X$ on Minkowski spacetime. Then locally at each point it can be decomposed (through non-constant functions of spacetime as coefficients) as linear combination of 'coordinate' vector fields $\partial_t, \partial_x, \partial_y, \partial_z$.

Since each of them is actually a KVF of Minkowski tensor metric (i.e. an isometry) why the given (generic) smooth vector field is not ?

Because, as I mentioned in #2, the Lie derivative is not linear unless you restrict yourself to constant coefficients.

Edit: This is essentially why the Lie derivative does not satisfy the requirements of an affine connection.

 

[cianfa72]

Btw, assigning a metric structure on a smooth manifold (i.e. defining a Riemann or pseudo-Riemann manifold) does turn it in a metric space ?

In other words are the axioms of metric space satisfied for it ?

 

[Orodruin]

Btw, assigning a metric structure on a smooth manifold (i.e. defining a Riemann or pseudo-Riemann manifold) does turn it in a metric space ?

In other words are the axioms of metric space satisfied for it ?

This is a completely different question and should therefore probably gobin its own thread, but yes. Any Riemann manifold is a metric space with the distance function defined as the infimum of the length of paths between the points.

A pseudo-Riemannian manifold is not a metric space (for example, distance between distinct points may be zero etc)

 

[vanhees71]

The intuitive meaning of the Lie derivative of a vector field $V^{\mu}$ in direction of a vector $a^{\nu}$ answers the question, what change of $V^{\mu}$ an observer sees when he's moving from a point $P$ (coordinates $q^{\mu}$) to an infintesimally close point $P'$ in direction of $a^{\nu}$, i.e., to $\bar{x}^{\mu}=x^{\mu} + \delta a^{\mu}$ and taking his local coordinates with him. This refers to an infinitesimal change of coordinates,
$$x^{\prime \mu}=x^{\mu} -\delta a^{\mu}, \quad V^{\prime \mu}=V^{\mu}-\delta a^{\nu} \partial_{\nu} V^{\mu}.$$
So he uses as components of $V^{\mu}$ at $P'$ the components
$$V^{\prime \mu}(P')=(\delta_{\nu}^{\mu} -\delta \partial_{\nu} a^{\mu}) V^{\nu}(x+\delta a) = V^{\mu}(P) + \delta [a^{\nu} \partial_{\nu} V^{\mu}(x)-\delta \partial_{\nu} a^{\mu} V^{\nu}(x)]+\mathcal{O}(\delta^2).$$
This leads to the definition of the Lie derivative in direction of $a^{\mu}$:
$$\mathcal{L}_{a} V^{\mu} = \lim_{\delta \rightarrow 0} \frac{1}{\delta} [V^{\prime \mu}(P') - V^{\mu}(P)] = a^{\nu} \partial_{\nu} V^{\mu}-V^{\nu} \partial_{\nu} a^{\mu}.$$
In the same way you can derive the Lie derivative of a one form (or covariant vector components)
$$\mathcal{L}_a V_{\mu} = a^{\nu} \partial_{\nu} V_{\mu} + V_{\nu} \partial_{\mu} a^{\nu}.$$
You can also show that instead of the partial derivatives in the definition of the Lie derivatives you can write covariant derivatives, i.e., $\partial_{\mu} \rightarrow \nabla_{\mu}$. The terms with the Christoffel symbols cancel.

For higher-rank tensors you correspondingly apply the Lie-derivative rule "to each index". Particularly for the covariant components of the metric you get
$$\mathcal{L}_a g_{\mu \nu} = a^{\rho} \nabla_{\rho} g_{\mu \nu} + g_{\rho \nu} \nabla_{\mu} a^{\rho} + g_{\mu \rho} \nabla_{\nu} a^{\rho}.$$
Since $\nabla_{\rho} g_{\mu \nu}=0$ ("compatibility of the metric with the affine connection" of Riemann spaces) you get
$$\mathcal{L}_a g_{\mu \nu} = g_{\rho \nu} \nabla_{\mu} a^{\rho} + g_{\mu \rho} \nabla_{\nu} a^{\rho}.$$
Since further you can commute the index lowering with the metric and covariant derivatives, this can be written as
$$\mathcal{L}_a g_{\mu \nu} =\nabla_{\mu} a_{\nu} + \nabla_{\nu} a_{\mu}.$$
Now from the above intuitive meaning of the Lie derivative, if this experession vanishes, it means that the metric looks the same at point $P'$ as at point $P$ when moving from $P'$ to $P$ in direction of $a^{\mu}$, i.e., it provides a symmetry of the Riemannian space under consideration, and that's why such Killing vectors are generators of symmetry transformations. This defines a Killing vector $a$ as such a vector for which
$$\mathcal{L}_a g_{\mu \nu}=0 \; \Leftrightarrow \; \nabla_{\mu} a_{\nu} + \nabla_{\nu} a_{\mu}=0.$$

 

[cianfa72]

Now from the above intuitive meaning of the Lie derivative, if this experession vanishes, it means that the metric looks the same at point $P'$ as at point $P$ when moving from $P'$ to $P$ in direction of $a^{\mu}$, i.e., it provides a symmetry of the Riemannian space under consideration

So in case of Minkowski spacetime the metric looks the same when moving in whatever direction $a^{\mu}$ from $P$ (i.e. your expression vanishes for every $a^{\mu}$), I think.

 

[vanhees71]

Sure, that's translation invariance and a subgroup of the (proper orthochronous) Poincare group. In addition you have boosts and rotations (making up the other 6 independent one-parameter subgroups of this spacetime manifold). Minkowski space is of course a space of "maximum symmetry", i.e., it has 10 independent Killing vectors.

 

[cianfa72]

Sorry, maybe the point unclear to me is the following: when we claim the above Lie derivative vanishes for every $a^{\mu}$ the latter is actually a vector field so when evaluated in a neighborhood of point $P$ it results let me say in a 'continuous' set of vectors pointing in the same direction.

I believe what is confusing me is that Lie derivative actually involve a vector field and not just a vector defined at some point.

 

[vanhees71]

The dimension of a Lie group is defined as the dimension of any differentiable manifold, and the Poincare group is 10-dimensional.

Also it's not any vector field that is a Killing vector of Minkowski space but only the ones we have derived above from the Killing equation for the given Minkowski metric, i.e.,
$\xi_{\mu}(x)=\omega_{\mu \nu} x^{\nu}+a_{\mu},$
where $\omega_{\mu \nu}=-\omega_{\nu \mu}=\text{const}$ and $a_{\mu}=\text{const}$.

If you read this as a coordinate transformation, it's clear that this provides the "infinitesimal transformations". The finite transformations follow via exponentiation, as usual in Lie groups.

E.g., for the Lorentz subgroup you can choose the 6 matrices $\hat{K}^{\mu \nu}=-\hat{K}^{\mu \nu}$, where, e.g., $\hat{M}=\hat{K}^{01}$ is the matrix where $M_{01}=-M_{10}=1$ and all other matrix elements are 0. Then you can write any finite dimensional Lorentz transformation as
$$\hat{L}=\exp \left (\frac{1}{2} \omega_{\mu \nu} \hat{K}^{\mu \nu} \right).$$
You can easily verify that you get for $\omega_{01}=-\omega_{10}$ and all other $\omega_{\mu \nu}=0$ a Lorentz boost in $x^1$ direction.

For more details in somewhat different notation, see Sect. 1.7 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[PeterDonis]

Since each of them is actually a KVF of Minkowski tensor metric (i.e. an isometry) why the given (generic) smooth vector field is not ?

Because the coefficients of each of the coordinate basis vector fields, in general, will vary from point to point, and so the vector field $V$ will not be a linear combination of KVFs with constant coefficients.

 

[cianfa72]

Also it's not any vector field that is a Killing vector of Minkowski space but only the ones we have derived above from the Killing equation for the given Minkowski metric, i.e.,
$\xi_{\mu}(x)=\omega_{\mu \nu} x^{\nu}+a_{\mu},$
where $\omega_{\mu \nu}=-\omega_{\nu \mu}=\text{const}$ and $a_{\mu}=\text{const}$.

Ah ok, I take it as in a neighborhood of point $P$ any Killing vector field (KVF) of Minkowski spacetime has to be in the form quoted above (when expressed in the Minkowski standard coordinates).

 

[Orodruin]

The intuitive meaning of the Lie derivative of a vector field Vμ in direction of a vector aν answers the question, what change of Vμ an observer sees when he's moving from a point P (coordinates qμ) to an infintesimally close point P′ in direction of aν, i.e., to x¯μ=xμ+δaμ and taking his local coordinates with him.

To me that is not a very intuitive meaning as I prefer intuition not based on any coordinates for things that are really coordinate independent. Call me weird, but it kind of destroys the ideas for me.

The thing with the Lie derivative is that it refers to a change under a transformation of the entire space, comparing how the transformation leaves the object differentiated to the objects original value. It then becomes clear that it is the entire field $a$ that is relevant rather than its value at the particular point of differentiation as the flows of curves in the space (and consequently their tangents) depend on how the field generating the flow depends on position.

 

[cianfa72]

The thing with the Lie derivative is that it refers to a change under a transformation of the entire space, comparing how the transformation leaves the object differentiated to the objects original value. It then becomes clear that it is the entire field $a$ that is relevant rather than its value at the particular point of differentiation as the flows of curves in the space (and consequently their tangents) depend on how the field generating the flow depends on position.

Yes, this should be the point of view of Carrol's lecture notes in chapter 5 (More geometry).

 

[vanhees71]

To me that is not a very intuitive meaning as I prefer intuition not based on any coordinates for things that are really coordinate independent. Call me weird, but it kind of destroys the ideas for me.

The thing with the Lie derivative is that it refers to a change under a transformation of the entire space, comparing how the transformation leaves the object differentiated to the objects original value. It then becomes clear that it is the entire field $a$ that is relevant rather than its value at the particular point of differentiation as the flows of curves in the space (and consequently their tangents) depend on how the field generating the flow depends on position.

Well, I'm an old-fashioned Ricci-calculus guy. Of course as any covariant concept you can formulate everything in a coordinate- and basis-independent way.

 

[Orodruin]

Well, I'm an old-fashioned Ricci-calculus guy. Of course as any covariant concept you can formulate everything in a coordinate- and basis-independent way.

That’s fine of course. To each their own as long as it builds correct intuition.

 

[cianfa72]

$$(\mathcal{L}_{\xi} g)_{\mu \nu} =0$$

Let's start with Happy new year !

A doubt about the notation employed above. Inside the parentheses () there is a (0,2)-type tensor which is the Lie derivative of tensor metric $g$ along the vector field $\xi$. Then the two lower indices $\mu,\nu$ really 'apply' to this tensor enclosed in ().

On the other hand when we come across $\mathcal{L}_{\xi} g_{\mu \nu}$ is it actually a shorthand for $(\mathcal{L}_{\xi} g)_{\mu \nu}$ ?

 

[Orodruin]

Using parentheses is just clearer snd technically less probe to misunderstanding. The possible other interpretation of $\mathcal L_\xi g_{\mu\nu}$ would be the Lie derivative of the component $g_{\mu\nu}$, which is a scalar field and would therefore be equal to $\xi^\rho \partial_\rho g_{\mu\nu}$, which would probably be used instead.

I would probably just write $\mathcal L_\xi g = 0$ in this case anyway. (The (0,2) tensor is the (0,2) zero tensor - which is of course equivalent to the components being zero.)

 

[vanhees71]

Of course $\mathcal{L}_{\xi} g_{\mu \nu}$ are 2nd-rank tensor-field components. Written out it's, by definition,
$$\mathcal{L}_{\xi} g_{\mu \nu} = \nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu},$$
i.e. components of a 2nd rank symmetric tensor. You can also write $\mathcal{L}_{\xi} g$ as an invariant 2nd-rank tensor field.

 

[Orodruin]

Of course $\mathcal{L}_{\xi} g_{\mu \nu}$ are 2nd-rank tensor-field components. Written out it's, by definition,
$$\mathcal{L}_{\xi} g_{\mu \nu} = \nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu},$$
i.e. components of a 2nd rank symmetric tensor. You can also write $\mathcal{L}_{\xi} g$ as an invariant 2nd-rank tensor field.

Nitpicking, but I would not say that is by definition of the Lie derivative, which exists independent of the existence of a connection on the manifold. I would rather call it a consequence of introducing a metric and the corresponding Levi-Civita connection on a manifold.

 

[vanhees71]

Well, yes. I guess it's sufficient to assume an arbitrary metric-compatible connection to define the corresponding Lie derivative, which is of course dependent on the choice of the connection. To get the Levi-Civita connection you have to assume in addition that the space is a (pseudo-)Riemannian manifold, i.e., that the connection should be torsion free. That's what I tacitly assumed, because we are talking about standard GR, right?

 

[cianfa72]

I would rather call it a consequence of introducing a metric and the corresponding Levi-Civita connection on a manifold

You are really saying that the covariant derivative operator $\nabla_{\mu}$ employed here
$$\mathcal{L}_{\xi} g_{\mu \nu} = \nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu}$$ is the Levi-Civita connection corresponding of the metric tensor $g$ assigned on a manifold.

 

[Orodruin]

Well, yes. I guess it's sufficient to assume an arbitrary metric-compatible connection to define the corresponding Lie derivative, which is of course dependent on the choice of the connection. To get the Levi-Civita connection you have to assume in addition that the space is a (pseudo-)Riemannian manifold, i.e., that the connection should be torsion free. That's what I tacitly assumed, because we are talking about standard GR, right?

Sure, the only thing I reacted to was the ”by definition”.

 

[vanhees71]

You are really saying that the covariant derivative operator $\nabla_{\mu}$ employed here
$$\mathcal{L}_{\xi} g_{\mu \nu} = \nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu}$$ is the Levi-Civita connection corresponding of the metric tensor $g$ assigned on a manifold.

As discussed with @Orodruin of course I made the hidden assumption that we talk about general relativity, where by assumption the spacetime manifold is a pseudo-Riemannian, i.e., torsion-free one. By definition this implies that we use the then unique metric-compatible connection, which is the Levi-Civita connection.

 

[Orodruin]

I guess it's sufficient to assume an arbitrary metric-compatible connection to define the corresponding Lie derivative, which is of course dependent on the choice of the connection.

I reread this part. The Lie derivative is independent of the choice of connection and indeed does not need a connection to be defined at all. It is defined by pulling back the field being differentiated along the flow lines of the vector field $\xi$ (with the flow distance going to zero and divided by the flow distance as per typical differentiation)
$$
\mathcal L_\xi T = \lim_{s\to 0}[(\phi_{s\xi}^*T - T)/s].
$$
As the pullback defines a map between different points in the manifold, we are not in need of a connection in order to compute the Lie derivative. The question may then be ”why do we not use the Lie derivative as a connection?”, the answer to which is that it is not linear in $\xi$, which is one of the requirements for a connection.

An example of this is the Lie derivative of a vector, $\mathcal L_X Y = [X,Y]$ where the vector commutator does not depend on any connection.

 

[vanhees71]

That's the deeper reason, why within the Ricci calculus the Lie derivative of tensor fields can be expressed with partial derivatives and yet lead to tensor components. For torsion-free connections you can as well write covariant derivatives. For connections with torsion you get additional terms involving the torsion.

 

[cianfa72]

$$
\mathcal L_\xi T = \lim_{s\to 0}[(\phi_{s\xi}^*T - T)/s].
$$ As the pullback defines a map between different points in the manifold, we are not in need of a connection in order to compute the Lie derivative.

A doubt about the limit involved in the Lie derivative defintion. At first sight that limit makes sense when we pick a chart in the differentiable atals and do the calculation there. Does it imply a notion of limit from an invariant coordinate-free point of view ?

 

[vanhees71]

Sure, this is a definition completely independent of any choice of coordinates. Since all the manipulations used to define the limit are manifestly covariant, it maps the arbitrary tensor field $T$ to another tensor field of the same rank.

 

[cianfa72]

Sure, this is a definition completely independent of any choice of coordinates. Since all the manipulations used to define the limit are manifestly covariant

Ah ok, indeed the operations involved in the definition are part of tensor space structure that exist at each point of the manifold. Namely a difference between (0,2) tensors defined at the same point (i.e. elements of the same tensor space) and a scalar multiplication for $1/s$.