Different authors seem to attach different levels of importance to keeping track of the exact tensor valences of various physical quantities. In the strict-Catholic-school-nun camp, we have Burke 1980, which emphasizes that you don't always have a metric available, so it may not always be possible to raise and lower indices at will. Burke makes firm pronouncements, e.g., that force is a covector (I recapped his argument here). At the permissive end of the spectrum, Rindler 1997 has a disclaimer early on in the book that he doesn't want to worry about distinguishing upper and lower indices, and won't do so until some later point in the book. Sometimes it feels a little strained to try to maintain such distinctions, especially in relativity, which we don't even know how to formulate without a nondegenerate metric. E.g., Burke argues that momentum is really a covector, because you can get it by differentiation of the Lagrangian with respect to $\dot{x}$. But then a perfectly natural index-gymnastics expression like $p^i=m v^i$ becomes something wrong and naughty.

I find this particularly confusing when it comes to higher-rank tensors and questions about which form of a tensor is the one that corresponds to actual measurements. Measurements with rulers measure $\Delta x^i$, not $\Delta x_i$, which is essentially a definition that breaks the otherwise perfect symmetry between vectors and covectors. But for me, at least, it gets a lot more muddled when we're talking about something like the stress-energy tensor. For example, in this question, I was working through a calculation in Brown 2012 in which he essentially writes down $T^\mu_\nu=\operatorname{diag}(\rho,P,0,0)$ for the stress-energy tensor of a rope hanging in a Schwarzschild spacetime. It's not obvious to me that this corresponds better with measurements than writing down the same r.h.s. but with $T^{\mu\nu}$ or $T_{\mu\nu}$ on the left. Misner 1973 has a nice little summary of this sort of thing on p. 131, with, e.g., a rule stating that $T^\mu_\nu v_\mu v^\nu$ is to be interpreted as the density of mass-energy seen by an observer with four-velocity $v$. Most, but not all, of their rules are, like this one, expressed as scalars. This is very attractive, because we have identities such as $a^ib_i=a_ib^i$, which means that it makes absolutely no difference whether we discuss an object like $a^i$ or its dual $a_i$, and we never have to discuss which form of a tensor matches up with measurements, because our measurements are scalars.

Is this approach, of reducing every measurement to a scalar, universally workable in GR? Is it universally desirable? Is it valid philosophically to say that all measurements are ultimately measurements of scalars?

A few examples:

Some quantities like mass are defined as scalars, so we're good.

Mass-energy is $p^i v_i=p_i v^i$, where $v$ is the velocity vector of the observer.

In the case of a Killing vector, there is no way to reduce it to a scalar, but a Killing vector isn't really something you can measure directly, so maybe that's OK...?

Relations like $\nabla_i T^{ij}=0$ and $\nabla_i \xi_j+\nabla_j \xi_i=0$ could be reduced to scalars, e.g., $v_j\nabla_i T^{ij}=0$, but there is no real need to do so, because we're saying a tensor is zero, and a zero tensor is zero regardless of how you raise or lower its indices.

I would be particularly interested in answers that spelled out how one should reason about examples like the hanging rope. In the treatment on p. 131 of Misner, they give $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$ for a perfect fluid; this is not scalar-ized, and in fact appears to contradict Brown's use of $T^\mu_\nu$.


After chewing this over with Cristi Stoica and Trimok, I think I've understood the issue about $T^\mu_\nu$ versus $T_{\mu\nu}$ better. Contrary to what I said above, the expression $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$ (with $-+++$ signature) for a perfect fluid is really scalar-ized, in the sense that $\rho$ and $P$ have no tensor indices, so they're notated as scalars. This makes sense, because $\rho$ and $P$ are defined by reference to a particular frame of reference, the rest frame of the fluid. This is exactly analogous to the way in which we define the scalar proper time by reference to the rest frame of a clock.

Now suppose we have coordinates in which the metric is diagonal, e.g., the Schwarzschild metric written in Schwarzschild coordinates. Let $g_{\mu\nu}=\operatorname{diag}(-A^2,B^2,\ldots)$. Furthermore, let's assume that we have some perfect fluid whose rest frame corresponds to zero coordinate velocity in these coordinates. The velocity vector of this rest frame is $v^\mu=(A^{-1},0,0,0)$, or, lowering an index, $v_\mu=(-A,0,0,0)$. Simply plugging in to the expression for $T$, we have $T_{00}=A^2\rho$, $T_{11}=B^2P$, $T^0_0=-\rho$, $T^1_1=P$. So this is the justification for Brown's use of the mixed-index form of $T$ -- it simply happens to be the one in which the factors of $A$ and $B$ don't appear. But that doesn't mean that the mixed-index form is the "real" one. What a static observer actually measures is the scalars $\rho$ and $P$. Similarly, it's not really $\Delta x^0$ or $\Delta x_0$ that an observer measures on a clock, it's the scalar $\Delta s$. When we say that coordinate differences correspond to upper-index vectors $\Delta x^i$, we're actually making a much more complicated statement that refers not to a simple measurement with a single device but rather to some much more extensive setup involving surveying, gyroscopes, and synchronization of clocks.

I think one of the pitfalls here is that I wasn't keeping in mind the fact that in an equation like $T_{\mu\nu}=\operatorname{diag}(\ldots)$, the right-hand side circumvents the rules of index gymnastics. This makes it preferable to work with equations like $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$, where both sides are valid index-gymnastics notation, and we can raise and lower indices at will without worrying about invalidating the equation.

Finally, in the example above, suppose that we have an asymptotically flat spacetime, and suppose that at large distances we have $A^2\rightarrow 1$ and $B^2\rightarrow 1$. Then $|T^0_0/T_{00}|=A^{-2}>1$ corresponds to a gravitational redshift factor seen by an observer at infinity.

Related: Type/Valence of the stress tensor


Brown, "Tensile Strength and the Mining of Black Holes," http://arxiv.org/abs/1207.3342

Burke, Spacetime, Geometry, Cosmology, 1980

Misner, Thorne, and Wheeler, Gravitation, 1973

Rindler, Essential Relativity, 1997

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    $\begingroup$ Even when you're measuring $\Delta x^i$ you're actually measuring it's projection on your measuring apparatus, which is actually an inner product and hence a scalar. $\endgroup$
    – Siva
    Commented Sep 24, 2013 at 0:25
  • $\begingroup$ @Siva: That seems reasonable. If you look at the methods MTW uses for scalar-izing things, they use both inner products with the observer's four-velocity and inner products with unit vectors $\hat{n}$. E.g., they interpret $T_{\mu\nu}v^\mu n^\nu$ as the component of the four-momentum density along $\hat{n}$, in the frame of an observer defined by $v$. I guess a ruler measurement would be scalar-ized as $\Delta x^i n_i=\Delta x_i n^i$. $\endgroup$
    – user4552
    Commented Sep 24, 2013 at 0:44
  • $\begingroup$ Possibly relevant: physics.stackexchange.com/q/64988/353 asking about all measurement reducing to position and time. There's a not-famous-enough quote about all measurements in physics ultimately being readings of the position of a needle (i.e. a voltmeter)- was it de Broglie? $\endgroup$
    – DarenW
    Commented Sep 24, 2013 at 3:14
  • $\begingroup$ Excellent update, it makes perfect sense! $\endgroup$ Commented Sep 28, 2013 at 5:00

2 Answers 2


1. Let $p\in M$ be a point in the manifold $M$. A tensor of type $(r,s)$ at $p$ is an element of the tensor product between $r$ copies of the tangent space at $p$ and $s$ copies of the cotangent space at $p$. To evaluate the tensor, you plug in the vectors from the frame and the coframe. For instance, if the frame is $(e_i)$, and $(e^i)$ is its dual, the component $T_{i_1,...,i_r,j_1,\ldots\,j_s}$ is evaluated by $$T_{i_1,...,i_r,j_1,\ldots\,j_s}=T(e_{i_1},\ldots,e_{i_r},e^{j_1},\ldots,e^{j_s}),$$ which is indeed a scalar. So, as a general rule, to find the components of a tensor, find some scalars (which depend on the frame, because the components depend on the frame).

begin update

It has been commented that $T_{i_1,...,i_r,j_1,\ldots\,j_s}$ can't be a scalar, because it depends on the frame. It is true that when you change $(e_i)$, $T_{i_1,...,i_r,j_1,\ldots\,j_s}$ changes, and I already said this above. But there is no contradiction between that, and the fact that $T_{i_1,...,i_r,j_1,\ldots\,j_s}$ is a scalar. Consider, for simplicity, that $T$ is a covector. If we contract it with a vector $v$, we get a scalar $T(v)=T_jv^j$. If we express now $T$ and $v$ in another frame, we get the same scalar $T(v)=T'_j v'^j$, where $T'_j$ and $v'^j$ are the new components. Everybody agrees that when we contract a covector with a vector we get a scalar. But, recall that $(e_i)$ are also vectors. Why, if we replace the vector $v$ with the vector $e_i$, the result would not be a scalar? In this case we get $T(e_i)=T_je_i{}^j$. If we express now in another frame both $T$ and $e_i$, we get the same value $T(e_i)=T'_je_i'{}^j$. So $T(e_i)$ is a scalar, which depends on the vector fields $(e_i)$. It is the same scalar when we express both $T$ and $(e_i)$ in another frame.

end update

2. There are two main reasons why we can do index gymnastics. Firstly, because $g_{ij}$ is nondegenerate, its reciprocal (inverse) $g^{ij}$ is uniquely defined, and they both give isomorphisms between the tangent and cotangent bundles. With these isomorphisms, we can raise and lower indices. Secondly, when covariant derivative defined by the Levi-Civita connection is involved, $\nabla g=0$. Because of this, when applying the Lebniz rule to tensors that are contracted with the metric, the term containing $\nabla g$ vanishes, hence index raising and lowering commute with covariant derivatives.

3. One topic discussed in the question is whether it is always justified to identify different tensors by using index lowering or index raising isomorphisms. Apparently, because index raising and lowering operations are isomorphisms, it is. But I will explain below that this led to long standing problems in relativity.

The metric is dynamical, it evolves in time, and its evolution is determined by Einstein's field equation. Einstein's equation shows how curvature is connected with the stress-energy of matter (In "matter" I include also bosonic fields. Anything that comes with a stress-energy tensor.). Now, there is no reason why this evolution would not lead to degenerate metrics. And, as it is known by Penrose's and Hawking's singularity theorems, singularities are obtained, in very generic situations.

The singularities obtained are usually seen as a good reason to give up General Relativity, and replace it with something more radical. In most of the times the "more radical" approaches are also plagued with singularities, or when they are not, it is because they change the dynamics (the Einstein equations).

Singularities are inevitable, but they come in many flavors. In some cases, the metric has singular components, for instance, for the Schwarzschild metric, $g_{tt}=1-\frac {2m}{r}$ is singular at $r=0$. In other cases, the metric is degenerate, that is, $\det g=0$. In this case, $g_{ij}$ doesn't determine an isomorphism between the tangent and cotangent spaces, and we can't define $g^{ij}$. Hence, we cannot raise indices. We can lower them, but in this case we lose information. Since $g^{ij}$ is part of defining the geometric objects needed in semi-Riemannian geometry (hence in General Relativity), such as the Levi-Civita connection, and the Riemann, Ricci, and scalar curvatures. Hence, Einstein's equation doesn't make sense.

The solution is simple, but it turned out difficult to implement. Find a way to reconstruct all semi-Riemannian geometry, without making use of $g^{ij}$. Surprisingly, this is possible, and it was done in arXiv:1105.0201. A special class of singularities, named semi-regular, turned our to have smooth Riemann curvature $R_{ijkl}$. Note that the $R^i{}_{jkl}$ version remains in general singular when the metric is degenerate, because it is obtained from $R_{sjkl}$ by contracting with $g^{is}$. The usual approach defined first $R^i{}_{jkl}$, but if $g^{ij}$ is singular, you can only define $R_{ijkl}$. Other geometric objects were defined in that reference. More concrete examples were built in arXiv:1105.3404.

This method allows the description of singularities in terms of nonsingular geometric invariants. For instance, it worked for finding a nonsingular description of the FLRW singularity (arXiv:1112.4508, arXiv:1203.1819), for finding more Big-Bang solutions which admit nonsingular descriptions, and even satisfy Penrose's Weyl curvature hypothesis (arXiv:1203.3382).

But what we do when components of the metric $g_{ij}$ are singular? For instance, the Schwarzschild metric, $g_{tt}=1-\frac {2m}{r}$ is singular at $r=0$. The component $g_{rr}=(1-\frac {2m}{r})^{-1}$ is also singular for $r=2m$, on the event horizon, but this singularity is due to the coordinates, as seen by using Eddington-Finkelstein or Kruskal-Szekeres coordinates. These coordinates reveal that the metric is regular on the event horizon, but it appeared singular because the Schwarzschild coordinates are singular at $r=2m$. Couldn't it be that the singularity $r=0$ is also due to the coordinates? Well, this singularity is genuine, since the invariant $R_{ijkl}R^{ijkl}\to\infty$. But maybe it is a well-behaved singularity. In arXiv:1111.4837 it was shown that there are coordinates making the Schwarzschild metric analytic at the singularity $r=0$. Hence, in these new coordinates, $g_{ij}$ is degenerate, and also smooth. Moreover, this singularity is semi-regular, and we can apply the geometry developed in arXiv:1105.0201. Nonsingular coordinates were found also for the charged singularities (arXiv:1111.4332) and the rotating ones (arXiv:1111.7082).

Bottom line: one should not identify the lower and upper indices, because the resulting tensors are different. It is true that so long as the metric is nondegenerate, we can identify them, but if the metric becomes degenerate, then huge problems arise by this identification.

Update. I will comment the main example, that of the stress-energy tensor in Brown's paper.


In the treatment on p. 131 of Misner, they give $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$ for a perfect fluid; this is not scalar-ized, and in fact appears to contradict Brown's use of $T^\mu_\nu$.

In MTW p131, they are talking about a perfect fluid. Because of this, the pressure is the same in all directions. In Brown, the stress-energy for a spherically symmetric matter distribution (which is not always a fluid) is $T^\mu_\nu=\operatorname{diag}(-\rho,p_r,p_\theta,p_\phi)$. Note that Brown considers that there are no off-diagonal components (no shear stress, because of spherical symmetry, and no momentum density, because this black hole is static). Only energy density and pressure.

Later, Brown takes the stress-energy to be $T^\mu_\nu=\operatorname{diag}(-\rho,P,0,0)$, under the assumption that there is only radial tension. This assumption is made to model a thread (actually, a thread for each radial direction), and this is why it differs from MTS, where it is considered a fluid.


It's not obvious to me that this corresponds better with measurements than writing down the same r.h.s. but with $T^{\mu\nu}$ or $T_{\mu\nu}$ on the left.

In this case, $T^\mu_\nu=\operatorname{diag}(-\rho,P,0,0)$ is expressed in an orthogonal frame (Brown, eq. 3). But the frame is not orthonormal, since $\chi^2$ and $f^2$ are not equal to $1$. So, switching to $T^{\mu\nu}$ or $T_{\mu\nu}$ on the left changes the values. In general, when people write $T^{\mu\nu}=\operatorname{diag}(\rho,P,P,P)$ or stuff like this, they consider it in an orthonormal frame, and in this case, $T^\mu_\nu=\operatorname{diag}(-\rho,P,P,P)$.


Misner 1973 has a nice little summary of this sort of thing on p. 131, with, e.g., a rule stating that $T^\mu_\nu v_\mu v^\nu$ is to be interpreted as the density of mass-energy seen by an observer with four-velocity $v$.

This doesn't help, because it refers to $T_{00}$ only. If we consider the frame of the observer, say $(e_0,e_1,e_2,e_3)$, then $v=e_0=(1,0,0,0)$, hence, $T^\mu_\nu v_\mu v^\nu=T_{00}$. To find the other components of $T_{\mu\nu}$, one should contract with other elements of the frame.

  • 1
    $\begingroup$ +1 for the interesting informations. However, in $1$, the quantity $T_{i_1,...,i_r,j_1,\ldots\,j_s}=T(e_{i_1},\ldots,e_{i_r},e^{j_1},\ldots,e^{j_s}),$ is not a scalar (in the sense of frame invariant). It is only a real value (by opposition to $T$) $\endgroup$
    – Trimok
    Commented Sep 24, 2013 at 9:51
  • 1
    $\begingroup$ @Trimok: I said that it is scalar, and that it "depends on the frame", but there is no contradiction. Say $T$ is a covector. Contract it with a vector $v$; you will get a scalar $T(v)=T_jv^j$. Express now $T$ and $v$ in another frame. You get the same scalar $T(v)=T'_j v'^j$, where $T'_j$ and $v'^j$ are the new components. Now, $(e_i)$ are vectors. Replace $v$ with $e_i$; you get $T(e_i)=T_je_i{}^j$. Express in another frame both $T$ and $e_i$, you get the same value $T(e_i)=T'_je_i'{}^j$. So $T(e_i)$ is a scalar, which depends on the vector fields $(e_i)$, and there is no contradiction. $\endgroup$ Commented Sep 24, 2013 at 10:40
  • 1
    $\begingroup$ @Ben Crowell: I updated my answer with a discussion of the stress-energy tensor from Brown's paper. $\endgroup$ Commented Sep 24, 2013 at 20:52
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    $\begingroup$ @Trimok: I called "frame" a collection of $n$ vector fields $(e_i)$, which at each point form a basis of the tangent space. I called "coframe" its dual, $(e^i)$, made of covectors (1-forms) so that $e^i(e_j)=\delta^i_j$. In 4D, the frame is also called tetrad, and the coframe inverse tetrad, as you mention. From the viewpoint of linear algebra, they are just bases. $\endgroup$ Commented Sep 25, 2013 at 5:20
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    $\begingroup$ @Trimok: In the original answer, I used $i,j$ for indices of any frame. In the part about $T_{\mu\nu}$ I used $\mu$ and $\nu$ just to keep the notations from the question, but represent the same thing. $e_i{}^j$ simply represent the components of the vector $e_i$ in another frame. I considered both frames as general as can be. But in the tetrad formalism, in general one of the frames is orthonormal, and the other is a coordinate frame $(\partial_\mu)$. In general, people denote the coordinate indices differently than those from orthonormal vector frames. I did not use this in my answer. $\endgroup$ Commented Sep 25, 2013 at 6:15

I think we don't measure scalars.

If we think about a local measurement of the stress-energy tensor, it should always corresponds to a covariant quantity $T_{ij}$. If a local measurement of energy/momentum is possible, it should always corresponds to a covariant quantity $p_i$. This comes from the definition of the momentum, if we think to Lagrangian and actions, in classical or quantum mechanics.

A scalar quantity, build from the stress-energy tensor and contravariant vectors may correspond to a measured quantity, but it is not a "measured scalar". Moreover, the information living in scalar quantities is interesting, but very partial. Yes, $T_{ij}u^iu^j$ is the energy density seen by the observer with 4-velocity $u^i$. But, I don't think you can obtain , for instance, the momentum flux seen by this observer, in a scalar expression. In the same way, the scalar expression $p_iu^i$ is the energy of the particle of energy/momentum $p$ seen by the observer of 4-velocity $u^i$, but I think that you cannot obtain a scalar expression for the momentum of the particle, seen by the same observer.

Quantities like $T_i^j$ are interesting, but in considering local stress-energy tensor quantities, seen by a distant observer. For instance, in a Schwarzschild metrics, $T_0^0$ and $T$ are local stress-energy tensor, but seen by the observer at infinity, that is red-shifted. This is interesting when we calculate the total mass (with a killing vector $\xi^a = \delta_0^a$) with expressions like :

$M = -2\int d^3x \sqrt{g}[T_0^0 - \frac {T}{2}]$ (Padmanabhan 6.208 p 287)

  • 3
    $\begingroup$ But the point is that to measure a vector quantity, you have to take four measurments of different scalar quantities, because you fix a frame for your measurements, and then measure relative to the frame. $\endgroup$ Commented Sep 24, 2013 at 16:30
  • $\begingroup$ Interesting answer, thanks. But, I don't think you can obtain , for instance, the momentum flux seen by this observer, in a scalar expression. The way MTW deals with this is to make expressions like $T_{ij}u^in^j$, where $\hat{n}$ is a unit vector. $\endgroup$
    – user4552
    Commented Sep 24, 2013 at 19:04
  • $\begingroup$ Quantities like Tji are interesting, but in considering local stress-energy tensor quantities, seen by a distant observer. Interesting...this might explain why Brown works with that form. Is there some general justification for this interpretation, in any asymptotically flat spacetime? Doesn't it depend on the assumption that the coordinates are chosen such that the metric approaches $\operatorname{diag}(1,-1,-1,-1)$ far away? $\endgroup$
    – user4552
    Commented Sep 24, 2013 at 19:05
  • $\begingroup$ @BenCrowell : The whole formula $(6.208$ p $287)$ in Padmanabhan, is that we have a global conserved quantity $ I=-2\int d^3 \Sigma_m \xi^n [T_n^m - \frac{1}{2}\delta_n^m T]$, where $\xi^n$ is a Killing vector. In the case of a Killing vector representing time-invariance, the conserved quantity is the global mass/energy. In the Schwarzschild case, we have $\xi^a = \delta_0^a$. It is said in Padmanabhan that "this is true for any source which is confined to a finite region in space and the metric perturbations are small at large distances" $\endgroup$
    – Trimok
    Commented Sep 24, 2013 at 19:25
  • $\begingroup$ @Trimok: I haven't done the calculation, but I"m sure that that's just the ADM mass and ADM momentum... so? $\endgroup$ Commented Sep 24, 2013 at 20:55

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