Lorentz invariance of the Minkowski metric

["Don't panic!"]

I understand that in order to preserve the inner product of two four vectors under a change of coordinates $x^{\mu}\rightarrow x^{\mu^{'}}=\Lambda^{\mu^{'}}_{\,\, \nu}x^{\nu}$ the Minkowski metric must transform as $\eta_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\, \mu^{'}}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\, \nu^{'}}$, but how does this imply that the metric tensor $\eta_{\mu\nu}$ is Lorentz invariant? I mean, doesn't an arbitrary (0,2) tensor transform $B_{\mu\nu}$ as $B_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\, \mu^{'}}\Lambda^{\beta}_{\,\, \nu^{'}}B_{\alpha\beta}$, but tensors aren't generally invariant under Lorentz transformations, so how is it obvious that the metric tensor is Lorentz invariant?

The reason I ask is that I've been reading up on the cosmological constant problem and it is mentioned that the vacuum energy density contribution to the energy momentum tensor must be of the form $T^{vac}_{\mu\nu}=-\rho^{vac}g_{\mu\nu}$, the reasoning being that the vacuum is Lorentz invariant and the only Lorentz invariant tensor is the Minkowski metric?!

 

[Nugatory]

but tensors aren't generally invariant under Lorentz transformations, so how is it obvious that the metric tensor is Lorentz invariant?

It's not obvious, as far as I know. You can calculate it out - write down the components of the metric tensor in your favorite coordinate system (Cartesian x, y, z, t are strongly recommended) and then apply the Lorentz transforms.

On the other hand, it would be somewhat amazing if it didn't come out properly - we chose Minkowski space instead of Euclidean because we wanted to preserve the invariance of the spacetime interval under Lorentz transforms. So maybe it is obvious after all.

 

[bcrowell]

but tensors aren't generally invariant under Lorentz transformations

Components of tensors certainly aren't invariant under Lorentz transformations. Many people do think of tensors as being invariant objects. This depends on philosophy and definitions.

 

[haushofer]

It does not, I'd say. Covariance is something different than an isometry; the latter 'involves primes' and is in general more constraining. In spec.rel with Minkowski spacetime they happen to be the same group, but eg in gen.rel. for the Schwarzschild solution they are not. Zee's book has a discusson on this, but i can't find the section right now.

Isometries lead via their corresponding Killing vectors to conserved charges for a given particle action, but covariance does not. In the context of sigmamodels one speaks of pseudosymmetries.

Admittedly, this can be very confusing. For me, at least :P

 

[haushofer]

Let me elaborate, also for myself, from the point of GR and field theories :P

In GR we have particle dynamics and spacetime dynamics.

Spacetime (gravitational) dynamics is governed by the D-dimensional Einstein-Hilbert action. This action, defined on the D-dimensional target space, has as fundamental field the metric g(x). Acting with a gct on the metric (let's forget about other possible fields in the target space) keeps the action invariant. We call this general coviarance. Hence there is a concerved charge in the target space: energy-momentum.

Particle dynamics is governed by the 1-dimensional point particle action. This action can be considered as defining a 1-dimensional field theory on the wordline of the particle. The fundamental fields on this worldline are the D embedding coordinates X(tau). This action is also general covariant if one transforms both the metric AND the embedding coordinates. This does not generate any conserved charges for the particle and hence is, from the worldline point of view, a pseudosymmetry. From the wordline point of view, the metric g(X) can be regarded as "couplinging" the fundamental fields X. If you want to find the conserved charges associated to the particle, you only must vary the fundamental fields X which will induce a variation in the metric (in a field theory you only vary the fields to get symmetries, not the couplings). This gives you the well-known Noether charges for the particle in a fixed background. Such a background "breaks", from the target space point of view, the gct's to a certain subset of them.

Now, without curvature GR reduces to SR, and the metric is given by the Minkowski-metric which is maximal symmetric. This means it has the maximum amount of Killing vectors (isometries) which generate the Poincare algebra. You can now also restrict the covariance group to the Poincaré group, but you don't have to. (e.g. the connection is a Lorentz tensor, but not a gct-tensor).

Maybe I'm making this overly complicated here, but this is how I understand it.

 

["Don't panic!"]

Components of tensors certainly aren't invariant under Lorentz transformations. Many people do think of tensors as being invariant objects. This depends on philosophy and definitions.

Sorry, the wording of my original post wasn't the best. Aren't all tensors Lorentz invariant (they surely must be since Lorentz transformations are simply changes of coordinate systems and tensors are by definition coordinate independent)?! What confuses me is that generally the components of tensors transform covariantly or contravariantly under a Lorentz transformation such that the tensor itself is left invariant, for example $B=B_{\mu^{'}\nu^{'}}dx^{\mu^{'}}\otimes dx^{\nu^{'}}=\Lambda^{\alpha}_{\,\,\mu^{'}}\Lambda^{\beta}_{\,\,\nu^{'}}B_{\alpha\beta}\Lambda^{\mu^{'}}_{\,\,\lambda}\Lambda^{\nu^{'}}_{\,\,\sigma}dx^{\lambda}\otimes dx^{\sigma}=\delta^{\alpha}_{\,\,\lambda}\delta^{\beta}_{\,\,\sigma}B_{\alpha\beta}dx^{\lambda}\otimes dx^{\sigma}=B_{\alpha\beta}dx^{\alpha}\otimes dx^{\beta}$, but from what I've read it is stated that the only Lorentz invariant tensors (by which I think they mean the components of the corresponding tensor) are the metric tensor and the Levi-Civita tensor (density). I can't seem to find anywhere that justifies this statement.

 

[haushofer]

No. E.g. Newton's laws are tensorial equations under the Galilei group, but most certainly not under the Lorentz group. And Maxwell's equations are not tensorial under the Galilei group, but under the Lorentz group. You can check this for yourself.

Covariance is not invariance. The components of the Minkowski metric and epsilon tensor (not density) are invariant under Lorentz transfo's, but covariant under gct's.

 

["Don't panic!"]

The components of the Minkowski metric and epsilon tensor (not density) are invariant under Lorentz transfo's,

How does one show that they are Lorentz invariant though? Shouldn't one end up with an equation of the form $\eta_{\mu '\nu '}=\eta_{\mu\nu}$?

 

[bcrowell]

from what I've read it is stated that the only Lorentz invariant tensors (by which I think they mean the components of the corresponding tensor) are the metric tensor and the Levi-Civita tensor (density). I can't seem to find anywhere that justifies this statement.

This sounds wrong to me. Counterexample #1: any scalar has a sole "component" that is invariant under Lorentz transformations. Counterexample #2: the tensor product $\mathbf{g}\otimes\mathbf{g}$.

How does one show that they are Lorentz invariant though?

Why not just do an explicit calculation?

 

["Don't panic!"]

Would it be in some sense correct to say that it is a two step procedure:
If we transform from one inertial frame to another, we require that the line element is Lorentz covariant, such that if it has the form $\eta '_{\mu\nu}x'^{\mu}x'^{\nu}=\eta_{\alpha\beta}x^{\alpha}x^{\beta}$. From this, in principle, $\eta '_{\mu\nu}$ does not have to equal $\eta_{\alpha\beta}$, however, we require that the line element actually be Lorentz invariant, i.e. if it has the form $\eta_{\mu\nu}x^{\mu}x^{\nu}=-(x^{0})^{2}+\mathbf{x}\cdot\mathbf{x}$ in one inertial frame, then in another inertial frame it will also have the form $-(x'^{0})^{2}+\mathbf{x}'\cdot\mathbf{x}'=\eta_{\alpha\beta}x'^{\alpha}x'^{\beta}$, such that $\eta_{\alpha\beta}x'^{\alpha}x'^{\beta}=\eta_{\mu\nu}x^{\mu}x^{\nu}$, and then the condition on the form of the Lorentz transformations follows from this?!

 

[bcrowell]

You need to start by deciding what you're taking as an assumption and what you want to be a derived fact.

First, we require that the line element is Lorentz covariant, such that if it has the form $-(x^{0})^{2}+\mathbf{x}\cdot\mathbf{x}=\eta_{\mu\nu}x^{\mu}x^{\nu}$ in one inertial frame, then it has this form in any other inertial frame, i.e. in another inertial frame it will have the form $-(x'^{0})^{2}+\mathbf{x}'\cdot\mathbf{x}'=\eta_{\alpha\beta}x'^{\alpha}x'^{\beta}$.

I don't think this is a standard way of defining the term "covariant." Maybe @haushofer could clarify, but I think s/he is simply using it to mean that the metric transforms as a covariant tensor, i.e., a tensor with two lower indices.

Furthermore, we require that the line element actually be Lorentz invariant, i.e. $\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\alpha\beta}x'^{\alpha}x'^{\beta}$, and then the condition on the form of the Lorentz transformations follows from this?!

If the metric is a tensor, then the line element is automatically invariant simply because it's constructed from tensors and has no indices.

 

[haushofer]

How does one show that they are Lorentz invariant though? Shouldn't one end up with an equation of the form $\eta_{\mu '\nu '}=\eta_{\mu\nu}$?

Do the calculation, as bcrowell suggests. E.g., what do you get when you lorentz-transform the epsilon-symbol? You should get something e involving the determinant of the lorentz transfo matrix.

 

[ChrisVer]

invariant means that the components don't change. covariant means that although they can change the equations they satisfy are the same.
Take for example the line element in a RF A:
$ds_A^2 = n^{(A)}_{ab} dx^a dx^b$
and do a Lorentz transformation to a RF B:
$ds_B^2 = n^{(B)}_{mn} du^m du^n$ (with $u= \Lambda x$ )
Now apply the invariance of the line element $ds^2_A= ds^2_B$ to obtain the relation between the $n^A$ and the $n^B$.
Now if your $n^A$ has the known form, what form will $n^B$ have?

 

[George Jones]

For me, the symbol $\eta$ implies a specific form for the matrix of components $\eta_{\mu \nu}$, so

$$\eta_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\, \mu^{'}}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\, \nu^{'}}$$

has different meaning than does, e.g., a transformation of the components of the electromagnetic field tensor

$$F_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\, \mu^{'}}F_{\alpha\beta}\Lambda^{\beta}_{\,\, \nu^{'}}.$$

 

["Don't panic!"]

Now if your nAn^A has the known form, what form will nBn^B have?

My confusion is mainly over this point, in principle why could you not have $ds_{A}^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$ and $ds_{B}^{2}=\tilde{\eta}_{\mu\nu}d\tilde{x}^{\mu}d\tilde{x}^{nu}$, such that for $ds_{A}^{2}=ds_{B}^{2}$ $$\tilde{\eta}_{\mu\nu}d\tilde{x}^{\mu}d\tilde{x}^{nu}=\tilde{\eta}_{\mu\nu}\Lambda^{\mu}_{\;\;\alpha}\Lambda^{\nu}_{\;\;\beta}dx^{\alpha}dx^{\beta}=\left(\Lambda^{\mu}_{\;\;\alpha}\tilde{\eta}_{\mu\nu}\Lambda^{\nu}_{\;\;\beta}\right)dx^{\alpha}dx^{\beta}=\eta_{\alpha\beta}dx^{\alpha}dx^{\beta}$$ Of course, this implies that $\Lambda^{\mu}_{\;\;\alpha}\tilde{\eta}_{\mu\nu}\Lambda^{\nu}_{\;\;\beta}=\eta_{\alpha\beta}$, and since we know the form of $[\eta_{\alpha\beta}]\equiv\eta =\text{diag}(-1,1,1,1)$ we end up with the following relations $$\Lambda^{\mu}_{\;\;0}\tilde{\eta}_{\mu\nu}\Lambda^{\nu}_{\;\;0}=-1 \\ \Lambda^{\mu}_{\;\;1}\tilde{\eta}_{\mu\nu}\Lambda^{\nu}_{\;\;1}=1\\ \Lambda^{\mu}_{\;\;2}\tilde{\eta}_{\mu\nu}\Lambda^{\nu}_{\;\;2}=1\\ \Lambda^{\mu}_{\;\;3}\tilde{\eta}_{\mu\nu}\Lambda^{\nu}_{\;\;3}=1$$ But I don't see how the components of the metric are invariant unless we assume also that $[\tilde{\eta}_{\mu\nu}]\equiv\tilde{\eta} =\text{diag}(-1,1,1,1)$?

Apologies if I'm being really stupid here, but I just can't seem to see the wood for the trees at the moment :-/

 

[ChrisVer]

this in general could work (but in the other way around; take tilde as the -+++ metric) and see what's the other metric
In other words this is the same as if you took your coordinates (t,x,y,z) and do a Lorentz transformation (t',x',y',z') and trying to form the line element expression. I am pretty sure you would end up to: $(dx^2 +d y^2 +d z^2) -d t^2 = (dx^{\prime 2}+dy^{\prime 2}+dz^{\prime 2})-dt^{\prime 2}$. Also I am pretty sure that it is enough to show that only for a Lorentz transf. along the t,x components.

 

[George Jones]

Different people understand things in different ways, and I cannot understand this solely using indices. Here is how I understand this, which might not work for others.

Roughly, Minkowski spacetime is a pair $\left( V, \bf{g}\right)$, where $V$ is a 4-dimensional vector space and $\bf{g}$ is a symmetric, non-degenerate bilinear form that maps pairs of 4-vectors to real numbers, i.e., $\bf{g}:V\times V\rightarrow \Bbb{R}$. $\left( V,\bf{g}\right)$ is such that an orthonormal basis $\left\{ \bf{e}_{0},\bf{e}_{1},\bf{e}_{2},\bf{e}_{3}\right\}$ with signature $\left( -,+,+,+\right)$ exists. The components of $\bf{g}$ with respect to an orthonormal basis $\left\{ \bf{e}_{\mu }\right\}$ are often denoted but by $\eta _{\mu \nu }:=.\bf{g}\left( \bf{e}_{\mu },\bf{e}_{\nu }\right)$. Note that there exist bases of $V$ that are not orthonormal (or even orthogonal). In this case, even for Minkowski spacetime, the components of $\bf{g}$ should be denoted by $g_{\mu \nu }$.

A Lorentz transformation is DEFINED to be a linear transformation $\Lambda :V\rightarrow V$ that preserves $\bf{g}$, i.e.,

$$\bf{g}\left( \bf{u},\bf{v}\right) =\bf{g}\left( \Lambda \bf{u},\Lambda \bf{v}\right)$$

for every $\bf{u}$ and $\bf{v}$ in $V$. The electromagnetic field tensor can also be viewed as a map $\bf{F}:V\times V\rightarrow \Bbb{R}$, but $\bf{F}$ does not satisfy the above property.

Now, in order to look at all of this in terms of coordinates, intro an orthonormal basis $\left\{ \bf{e}_{\mu }\right\}$. Define a new set of 4-vectors, $\left\{ \bf{e}_{\mu ^{\prime }}\right\}$ by

$$\bf{e}_{\mu ^{\prime }}=\Lambda \bf{e}_{\mu }.$$

Then,

$$\begin{align}
\bf{g}\left( \bf{e}_{\mu ^{\prime }},\bf{e}_{\nu ^{\prime }}\right) & =\bf{g}\left( \Lambda \bf{e}_{\mu },\Lambda \bf{e}_{\nu }\right)\\
& =\bf{g}\left( \bf{e}_{\mu },\bf{e}_{\nu }\right) ,
\end{align}$$

so also an orthonormal basis, which allows us to write (a bit jarringly)

$$\begin{align}
\bf{g}\left( \bf{e}_{\mu ^{\prime }},\bf{e}_{\nu ^{\prime }}\right) & =\bf{g}\left( \bf{e}_{\mu },\bf{e}_{\nu }\right)\\
g_{\mu ^{\prime }\nu ^{\prime }}& =g_{\mu \nu }\\
\eta _{\mu ^{\prime }\nu ^{\prime }}& =\eta _{\mu \nu }.
\end{align}$$

The last equality can only be written.because the primed and unprimed bases are orthonormal, and thus the metric tensor $\bf{g}$ has the same matrix of components with respect to both bases.

Each primed basis vector can be written as a linear combination of the unprimed orthonormal basis vectors $\left\{ \bf{e}_{\mu }\right\}$,

$$\begin{align}
\bf{e}_{\mu ^{\prime }}& =\Lambda \bf{e}_{\mu }\\
& =\Lambda^{\alpha }{} _{\mu ^{\prime }}\bf{e}_{\alpha },\\
\end{align}$$

so

$$\begin{align}
\bf{g}\left( \bf{e}_{\mu ^{\prime }},\bf{e}_{\nu ^{\prime }}\right) & =\bf{g}\left(\Lambda^{\alpha }{} _{\mu ^{\prime }}\bf{e}_{\alpha },\Lambda^{\beta }{} _{\nu ^{\prime }}\bf{e}_{\beta }\right)\\
& =\Lambda^{\alpha }{} _{\mu ^{\prime }} \Lambda^{\beta }{} _{\nu ^{\prime }} \bf{g}\left( \bf{e}_{\alpha },\bf{e}_{\beta }\right)\\
g_{\mu ^{\prime }\nu ^{\prime }}& =\Lambda^{\alpha }{} _{\mu ^{\prime }} \Lambda^{\beta }{} _{\nu ^{\prime }}g_{\alpha \beta }\\
\eta _{\mu ^{\prime }\nu ^{\prime }}& =\Lambda^{\alpha }{} _{\mu ^{\prime }} \Lambda^{\beta }{} _{\nu ^{\prime }}\eta _{\alpha \beta }
\end{align}$$

The second-last equation is true for any tensor $\bf{g}$ (not just the metric tensor) and any transformation $\Lambda$ (not just Lorentz transformations) on $V$, and any basis $\left\{ \bf{e}_{\mu }\right\}$ (not just an orthonormal basis). The last equation is true because $\bf{g}$ is the metric tensor, and because $\Lambda$ is a Lorentz transformation that, by definition, preserves $\bf{g}$, and because $\left\{ \bf{e}_{\mu }\right\}$ is an orthonormal basis. As shown above, this means the matrices of components of the metric tensor $\bf{g}$ with respect to the primed and unprimed bases both take the $\eta$ form.

 

["Don't panic!"]

The last equality can only be written.because the primed and unprimed bases are orthonormal, and thus the metric tensor g\bf{g} has the same matrix of components with respect to both bases.

Ah, so is this essentially the point - we want the inner product to be invariant, so as it has the form $\eta_{\mu\nu}u^{\mu}w^{\nu}$ only when the basis $\lbrace e_{\mu}\rbrace$ is orthonormal, then in order for this to remain invariant we must have that Lorentz transformations map orthonormal bases to orthonormal bases, such that $e_{\mu}\rightarrow e_{\mu^{\prime}}=\Lambda^{\mu}_{\;\;\mu^{\prime}}e_{\mu}$, and $\eta\left(e_{\mu^{\prime}},e_{\nu^{\prime}}\right)=\eta_{\mu\nu}$ (i.e. $\eta$ has the same components in the $\lbrace e_{\mu^{\prime}}\rbrace$ since it is also orthonormal).

I have found it confusing, since in a couple of sets of notes that I have read, they don't discuss the issue at all, apart from saying that we want the spacetime interval to be Lorentz invariant and going straight from this to the calculation $$x'^{\mu}x'_{\mu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}=\eta_{\mu\nu}\Lambda^{\mu}_{\;\;\alpha}x^{\alpha}\Lambda^{\mu}_{\;\;\beta}x^{\beta}=\Lambda^{\mu}_{\;\;\alpha}\eta_{\mu\nu}\Lambda^{\mu}_{\;\;\beta}x^{\alpha}x^{\beta}=\eta_{\alpha\beta}x^{\alpha}x^{\beta}=x^{\alpha}x_{\alpha}$$ and they don't at any point justify why in the first equality we can simply write $\eta_{\mu\nu}$ in the "primed" frame (in other words, they simply assume that the metric tensor has the same components in both frames).
In principle couldn't one have that $\eta$ had different coordinate components in some other frame, i.e. $\eta '_{\mu\nu}$, but that the interval was still invariant, i.e. $\eta '_{\mu\nu}x'^{\mu}x'^{\nu}=\eta_{\alpha\beta}x^{\alpha}x^{\beta}$, or is the point that the form of the interval should remain the same, since this is the only way that the speed of light, $c$ will be frame-independent?!

 

["Don't panic!"]

A Lorentz transformation is DEFINED to be a linear transformation Λ:V→VΛ:V→V\Lambda :V\rightarrow V that preserves gg\bf{g}

Would it also be correct to approach it from the opposite reasoning i.e. that we require the spacetime interval to be invariant, which is equivalent to the requirement that the coordinate components of the metric are invariant. We then look for a set of transformations that satisfy these requirements, which are exactly the Lorentz transformations?!

 

[samalkhaiat]

In principle couldn't one have that $\eta$ had different coordinate components in some other frame, i.e. $\eta '_{\mu\nu}$, but that the interval was still invariant, i.e. $\eta '_{\mu\nu}x'^{\mu}x'^{\nu}=\eta_{\alpha\beta}x^{\alpha}x^{\beta}$,

No, that would not be a Lorentz transformation. Lorentz transformation is defined by the invariance of metric components $\bar{g}_{\mu\nu} = g_{\mu\nu} = \eta_{\mu\nu}$.

or is the point that the form of the interval should remain the same, since this is the only way that the speed of light, $c$ will be frame-independent?!

No, because conformal transformation (which is non-linear) also preserves the light cone structure. The linearity of the map is essential.

Try to prove the following theorem:
Let $\Lambda : \mathbb{R}^{(1,3)} \to \mathbb{R}^{(1,3)}$ be a linear map on Minkowski space-time $M \cong (\mathbb{R}^{(3,1)},g)$. Then the following are equivalent

(1) $\Lambda$ preserves the inner product of $\mathbb{R}^{(3,1)}$, i.e., $g(\Lambda \mathbf{x}, \Lambda \mathbf{y}) = g(\mathbf{x}, \mathbf{y})$ for all $\mathbf{x}$ and $\mathbf{y}$ in $\mathbb{R}^{(1,3)}$.

(2) $\Lambda$ preserves the quadratic form of $\mathbb{R}^{(3,1)}$, i.e., $Q(\Lambda \mathbf{x}) = Q(\mathbf{x})$ for all $\mathbf{x}$ in $\mathbb{R}^{(3,1)}$ $\left( Q \ \mbox{is defined by} \ Q(\mathbf{x}) = g(\mathbf{x} , \mathbf{x}) = \mathbf{x} \cdot \mathbf{x} \right)$.

(3) $\Lambda$ carries any orthonormal basis, $\big \{ e_{\mu}\big \}$, for $\mathbb{R}^{(3,1)}$ onto another orthonormal basis, $\{ \bar{e}_{\mu}\} \equiv \big \{ \Lambda (e_{\mu}) \big \}$, for $\mathbb{R}^{(3,1)}$, i.e., $g(\Lambda (e_{\mu}) , \Lambda (e_{\nu}) ) = g(e_{\mu} , e_{\nu}) = \eta_{\mu\nu}$.

Hint: some parts of the proof can be established using George post.
This theorem tells you that Lorentz transformations can be defined by anyone of the above three statements.

 

[samalkhaiat]

we require the spacetime interval to be invariant, which is equivalent to the requirement that the coordinate components of the metric are invariant.

No, that is not correct. The space-time interval is invariant (scalar) even under general arbitrary coordinate transformations $\bar{x}^{\mu} = \bar{x}^{\mu} (x)$:
$$\bar{g}_{\mu \nu} d\bar{x}^{\mu} d\bar{x}^{\nu} = \frac{\partial x^{\rho}}{\partial \bar{x}^{\mu}} \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} g_{\rho \sigma} d\bar{x}^{\mu} d\bar{x}^{\nu} = g_{\rho \sigma} dx^{\rho} dx^{\sigma} .$$

We then look for a set of transformations that satisfy these requirements, which are exactly the Lorentz transformations?!

Again, that is wrong. To see why: Set $\bar{g}_{\mu \nu} = g_{\mu \nu} = \eta_{\mu \nu}$ in the transformation law of $g_{\mu \nu}$ under general coordinate transformation,
$$\bar{g}_{\mu \nu} = \frac{\partial x^{\rho}}{\partial \bar{x}^{\mu}} \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} g_{\rho \sigma} ,$$
then solve the resulting partial differential equations with the condition $\bar{x}^{\mu}(0) = 0$.
Hint: assume that the transformation $\bar{x}^{\mu} = \bar{x}^{\mu}(x)$ is smooth and thrice differentiable.

 

["Don't panic!"]

No, that is not correct. The space-time interval is invariant (scalar) even under general arbitrary coordinate transformations x¯μ=x¯μ(x)\bar{x}^{\mu} = \bar{x}^{\mu} (x):

g¯μνdx¯μdx¯ν=∂xρ∂x¯μ∂xσ∂x¯νgρσdx¯μdx¯ν=gρσdxρdxσ.​

This is what I find confusing though, as in general relativity the spacetime interval is invariant, but the components of the metric do change under a coordinate transformation (indeed, as you put in your post $g_{\mu\nu}\rightarrow\bar{g}_{\mu\nu}=\frac{\partial x^{\alpha}}{\partial\bar{x}^{\mu}}\frac{\partial x^{\beta}}{\partial\bar{x}^{\nu}}g_{\alpha\beta}$). What is the physical argument then why in special relativity we choose transformations that preserve the components of the metric? Why do we require that an orthonormal basis is mapped to an orthonormal basis?

I understand that given the metric tensor and two vectors $v$ and $w$, then $g:(v,w)\mapsto g(v,w)$, and then given a basis, $\lbrace e_{\mu}=\frac{\partial}{\partial x^{\mu}}\rbrace$, we have that $v=v^{\mu}e_{\mu}$ and $w=w^{\mu}e_{\mu}$, such that $g(v,w)=v^{\mu}w^{\nu}g(e_{\mu},e_{\nu})=v^{\mu}w^{\nu}g_{\mu\nu}$. Now, if the basis $\lbrace e_{\mu}\rbrace$ is orthonormal, then $g_{\mu\nu}=\eta_{\mu\nu}$. If we then consider another basis $\lbrace f_{\alpha}=\frac{\partial}{\partial \bar{x}^{\alpha}}\rbrace$, then we have that $v=\bar{v}^{\alpha}f_{\alpha}$ and $w=\bar{w}^{\alpha}f_{\alpha}$ such that $g(v,w)=\bar{v}^{\alpha}\bar{w}^{\beta}g(f_{\alpha},f_{\beta})=\bar{v}^{\alpha}\bar{w}^{\beta}\bar{g}_{\alpha\beta}$. If this basis is also orthonormal then we also have that $\bar{g}_{\alpha\beta}=\eta_{\alpha\beta}$, but the point is that $v^{\mu}w^{\nu}g_{\mu\nu}=\bar{v}^{\alpha}\bar{w}^{\beta}\bar{g}_{\alpha\beta}$, regardless of whether or not $g_{\mu\nu}=\eta_{\mu\nu}=\bar{g}_{\mu\nu}$, so why is this a requirement in Minkowski spacetime? Is the point that all inertial frames in special relativity are considered equivalent, and furthermore, a global coordinate system can be constructed for each which naturally comes with an orthonormal basis (the canonical basis). Accordingly, in a given inertial coordinate frame we have that $v^{\mu}w^{\nu}g_{\mu\nu}=v^{\mu}w^{\nu}\eta_{\mu\nu}$, and we require that in order for a coordinate transformation to map from one inertial frame to another, the metric should be preserved such that $v^{\mu}w^{\nu}\eta_{\mu\nu}=\bar{v}^{\alpha}\bar{w}^{\beta}\eta_{\alpha\beta}$. We then consider such transformations for which this is the case (which of course turn out to be the group of Lorentz transformations). Would this be correct at all?

 

[samalkhaiat]

What is the physical argument then why in special relativity we choose transformations that preserve the components of the metric?

Why is it that Lorentz group is the symmetry group of relativistic system? 1) experiments show that is the case, and 2) we believe that “admissible observers” are equivalent if and only if they are connected by the action of the symmetry group.

Why do we require that an orthonormal basis is mapped to an orthonormal basis?

An orthonormal basis we think of defining a frame of reference established by some “admissible (inertial) observer”.

Is the point that all inertial frames in special relativity are considered equivalent, and furthermore, a global coordinate system can be constructed for each which naturally comes with an orthonormal basis (the canonical basis). Accordingly, in a given inertial coordinate frame we have that $v^{\mu}w^{\nu}g_{\mu\nu}=v^{\mu}w^{\nu}\eta_{\mu\nu}$, and we require that in order for a coordinate transformation to map from one inertial frame to another, the metric should be preserved such that $v^{\mu}w^{\nu}\eta_{\mu\nu}=\bar{v}^{\alpha}\bar{w}^{\beta}\eta_{\alpha\beta}$. We then consider such transformations for which this is the case (which of course turn out to be the group of Lorentz transformations). Would this be correct at all?

Yes, that is more or less correct.

 

[samalkhaiat]

... the only Lorentz invariant tensors (by which I think they mean the components of the corresponding tensor) are the metric tensor and the Levi-Civita tensor (density).

Yes, absolutely correct.

I can't seem to find anywhere that justifies this statement.

Invariant tensors are treated in most group theory textbooks. And if you have studied the rules of combining angular momentum in QM , you probably know how to justify it. For example, if you couple 4 spin one states in $SO(3)$, i.e., $[3]\otimes [3] \otimes [3] \otimes [3]$, you get the singlet (spin zero) state $[1]$ appearing exactly 3 times. This shows that a rank-4 tensor $T^{ijkl}$ that is invariant under $SO(3)$ must have the form
$$T^{ijkl} = c_{1} \delta^{ij} \delta^{kl} + c_{2} \delta^{ik} \delta^{jl} + c_{3} \delta^{il} \delta^{jk} .$$
Because $\delta^{ij}$ and $\epsilon^{ijk}$ are the only independent invariant tensors in $SO(3)$, all other invariant tensors can be expressed in terms of the $\delta$ and the $\epsilon$. Similar story occurs in the Lorentz group $SO(1,3)$:
First, as we said many times before, the invariance of $\eta^{\mu\nu}$ is the defining property of the Lorentz group:
$$\eta^{\mu\nu} = \Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} \ \eta^{\rho\sigma} . \ \ \ \ \ (1)$$
This means that there exists no independent rank-2 invariant tensor, and the invariant part of a tensor $T^{\mu\nu}$, must be (1) symmetric, and (2) proportional to $\eta^{\mu\nu}$. You can see this by decomposing $T^{\mu\nu}$ (which is, $[4] \otimes [4]$, the tensor product of two 4-vector) into irreducible tensors:
$$[4] \otimes [4] = [1] \oplus [6] \oplus [9]$$
So, the invariant part of $T^{\mu\nu}$ must belong to the singlet $[1]$. Indeed, if you write
$$T^{\mu\nu} = \frac{1}{4} \eta^{\mu\nu} T + \frac{1}{2}T^{[\mu \nu]} + \frac{1}{2} \left( T^{(\mu\nu)} - \frac{1}{2} \eta^{\mu\nu} T \right) ,$$
you see that
$$\frac{1}{4} \eta^{\mu\nu} T \in [1] , \ \ \ T \equiv \mbox{Tr}(T) = \eta_{\rho \sigma}T^{\rho \sigma}.$$
As for the invariance of $\epsilon^{\mu\nu\rho\sigma}$, this follows from the fact that $\det | \Lambda | = 1$ in the Lorentz group $SO(1,3)$:
$$\begin{align*}<br /> \bar{\epsilon}^{\mu\nu\rho\sigma} &= \Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} \epsilon^{\alpha\beta\gamma\tau} \\<br /> &= \epsilon^{\mu\nu\rho\sigma} \det |\Lambda | \\<br /> &= \epsilon^{\mu\nu\rho\sigma} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)<br /> \end{align*}$$
From (1) and (2) it follows that there exists no independent higher rank invariant Lorentz tensors.
Okay, here is an exercise for you. Suppose that $T^{\mu\nu\rho\sigma}$ is invariant under the Lorentz group:
$$T^{\mu\nu\rho\sigma} = \Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} T^{\alpha\beta\gamma\tau} . \ \ \ \ \ (3)$$
Use (1), (3) and (3) to show that $T^{\mu\nu\rho\sigma}$ is of the form
$$T^{\mu\nu\rho\sigma} = c_{1} \eta^{\mu\nu} \eta^{\rho\sigma} + c_{2} \eta^{\mu\rho} \eta^{\nu\sigma} + c_{3} \eta^{\mu\sigma} \eta^{\nu\rho} + c_{4} \epsilon^{\mu\nu\rho\sigma} .$$

 

[&quot;Don&#039;t panic!&quot;]

An orthonormal basis we think of defining a frame of reference established by some “admissible (inertial) observer”.

So, is it simply taken that an orthonormal basis is canonical for an inertial observer (because there is a canonical global coordinate system, i.e. the Cartesian coordinate system) an as such this is taken to be what is meant by an inertial frame, hence transforming between inertial frames is equivalent to transforming between two orthonormal bases?!

Yes, that is more or less correct.

Would you able to improve on it at all (so that I'm not thinking about it incorrectly)? I like this way of thinking about it (it seems intuitive to me) as it starts from considering why the invariance of the spacetime interval between inertial observers leaves the metric components invariant and then leads onto the Lorentz transformations.

Also, aren't the components of the metric, $\eta_{\mu\nu}$ essentially Lorentz invariant by construction, since we require that Lorentz transformations are the set of transformations that map orthonormal bases to orthonormal bases (and $g_{\mu\nu}=\eta_{\mu\nu}$ is always true in an orthonormal basis)?!