# Discussion on tensor dimensions

1. Jun 26, 2015

### Orodruin

Staff Emeritus
<<Mentor note: This thread has been split from this thread due to going a bit off-topic.>>

I would actually disagree with this. Any tensor has well defined units, but its components may not have the same units as the tensor basis may consist of basis tensors with different units. For example, the stress tensor of classical mechanics always has units of pressure, but depending on the chosen basis may have components which do not. If I use unit vectors to describe a velocity, its components have units of length/time, but if I use basis vectors with units of length, the components would have units of 1/time, making the vector units length/time as before.

Unfortunately, physics texts often introduce tensors in such a way that the distinction between a tensor and its components in a given basis is not really clear.

Last edited: Jun 27, 2015
2. Jun 26, 2015

### bcrowell

Staff Emeritus
I don't think it matters whether one describes it my way or your way. To take the simplest possible example, infinitesimal displacements (but not finite ones) act like vectors. I would say that such a vector has no well-defined units (no vector ever does), but in a coordinate system where the coordinates all have units of distance, its components have units of distance. I guess you would say that that such a vector always has units of distance, but its components don't. Either description is self-consistent.

Another way of putting it is this. In the usual freshman physics picture, when we talk about the units of a quantity, we're talking about its transformation properties under certain types of scalings. But in GR, all vectors have the same transformation properties under any change of coordinates. (What it does make sense to discuss in GR is the weight of a tensor density.)

Last edited: Jun 26, 2015
3. Jun 26, 2015

### Orodruin

Staff Emeritus
I would say that the tangent vector coordinate basis has dimensions of 1/coordinate dimension. I think this is the most natural choice due to the fact that the coordinate basis is equivalent to the set of partial derivatives. In the same way, the dual vector coordinate basis $dx^\mu$ would naturally have the same dimensions as the coordinates themselves.

I would also say that infinitesimals do have dimensions, you cannot increase a distance by an infinitesimal increment of pressure. I think this way of assigning dimensions to infinitesimal quantities also makes it easier to keep track of dimensions. For example, if I want to write an infinitesimal increase in momentum, I would write this as $p\to p+dp$. Unless I assign dimensions of momentum to $dp$, this maks no sense dimensionally.

4. Jun 26, 2015

### bcrowell

Staff Emeritus
Sure, but the coordinates can have any units you like. They don't have to have units of distance.

5. Jun 26, 2015

### Orodruin

Staff Emeritus
Naturally, but a dual vector would be of the form $df = A_\mu dx^\mu$, where every term in the sum would have the same dimension, just as all terms in the coordinate expression for a tangent vector $V = V^\mu\partial_\mu$ will have the same dimensions. Since the bases $dx^\mu$ and $\partial_\mu$ may have different dimensions depending on the coordinates, the dimensions of the components must compensate for this.

6. Jun 27, 2015

### bcrowell

Staff Emeritus
The Ricci tensor has units of distance^-2, not distance^2.

7. Jun 27, 2015

### bcrowell

Staff Emeritus
So suppose I'm using coordinates $(t,r,\theta,\phi)$, and $V$ measures a small displacement in spacetime, so that I assume you would say it had units of distance. In your way of thinking about the units, what units would you ascribe to $V^r$, $\partial_r$, $V^\theta$, and $\partial_\theta$?

I don't see how this can make sense in the above example. It sounds like you're saying $[V]=L$, $[\partial_r]=1/L$, and $[\partial_\theta]=1$. If you then want the equation $V = V^\mu\partial_\mu$ to have units that make sense, then it seems like you need $[V^r]=L^2$ and $[V^\theta]=L$, which looks obviously absurd to me.

If you want to assign units to $V$, without having established any coordinate system, then I think the only sane thing to do is to say it's unitless, $[V]=1$. Then with $[\partial_r]=1/L$ and $[\partial_\theta]=1$, we end up with $[V^r]=L$ and $[V^\theta]=1$, which makes more sense. This also makes sense because when we describe the units of something, we're describing its transformation properties under scaling and how they differ from the transformation properties of other objects. Since all vectors have the same transformation properties under scaling, if you want to describe them as having units, they must all have the same units.

I would just take "$V$ has units of distance" to be shorthand for "if we pick coordinates that have units of distance, then the components of $V$ have units of distance." Likewise "$T$ has units of distance^-2" would be shorthand for "if we pick coordinates that have units of distance, then the components of $T$ have units of distance^-2."

I agree, with the proviso that "$dp$ has units of momentum" is shorthand for "if we pick coordinates that have units of distance, then the components of $dp$ have units of momentum."

Last edited: Jun 27, 2015
8. Jun 27, 2015

### stedwards

From a few posts back,
I like your ideas. Given a dual basis in $(t ,x)$ coordinates, having dimensions $dt = dt[T]$ and $dx=dx[L]$, then the dimensions of the metric and metric components follow.

$g[1]=g_{tt}[T^{-2}]dt^2[T^2]+g_{xx}[L^{-2}]dx^2[L^2]$.

How would you cajole the dimension assignments to obtain $T^2$ for the dimensions of the square interval, $d\tau^2 = d\tau^2[T^2]$?

9. Jun 27, 2015

### Orodruin

Staff Emeritus
I agree with this. Infinitesimal displacements in the manifold would be dimensionless (and the components thus having the same dimension as the coordinate). The metric tensor would have units of $L^2$ based on the displacements relating to an infinitesmal length.

I do not agree with this. You can divide and multiply by dimensional invariant quantities as you see fit. For example, the 4-velocity would have units of $1/L$ (using units where $L = T$ as it is space-time displacement per proper time. Consequently $g(V,V)$ would have dimensions $L^2/L^2 = 1$ as it should.

You are right, I am going to try some mentor magic. Closing the thread temporarily for moderation.

Last edited: Jun 28, 2015
10. Jun 27, 2015

### bcrowell

Staff Emeritus
Cool, thanks for moving this discussion to a separate thread.

You've quoted my sentence "Since all vectors...," twice, once saying you agree with it and once saying you disagree. I assume you didn't really mean to quote it at the tail end of the first quote, so you actually disagree with it.

You say you agree, but your third sentence contradicts my quoted material. The argument I was giving would apply equally to all tensors.

I've never heard of anyone defining a four-velocity vector as having units of $1/L$. Is this something you're serious about and have thought through carefully?

I suspect that you're running into problems here because you're not dealing successfully with a notational clash that has historical roots. In the early days of differential geometry, dx simply meant an infinitesimal change in a coordinate. Circa 1960, it became stylish to say that dx meant the dual of a derivative operator. It's perfectly possible to use infinitesimals in the old-school spirit and still do differential geometry in a rigorous way. In fact, I think there are at least two different ways that people have done that:

* using synthetic differential geometry: http://mathoverflow.net/questions/186851/synthetic-vs-classical-differential-geometry

* using non-standard analysis: Nowik and Katz, Differential geometry via infinitesimal displacements, http://arxiv.org/abs/1405.0984

IMO you're reaching conclusions that don't make sense because you're using the 1960s approach, which is not logically compatible with trying to assign units to the quantities in an equation such as $V=V^\mu \partial_\mu$.

11. Jun 27, 2015

### Orodruin

Staff Emeritus
Not overly, but it is the approach that makes sense to me if you want to assign a dimension to the vector (or tensor as a whole). It would follow directly from wanting to keep displacements with the same units as the coordinates (and thus $\dot x^\mu$, with the dot representing differentiation wrt a dimensionless curve parameter) transforming as the components of a vector). Wanting $g(\dot x,\dot x)$ to be a distance squared would set the dimension of the metric tensor.

The alternative would be to define displacements relative to the infinitesimal distance of displacement, i.e., $\dot x^\mu = dx^\mu/ds$ where $ds$ has units of length. This would make the vector components dimensionless and the vector itself to have units of $1/L$. To some extent I might agree that this makes more sense, you would have $df/ds = \dot x^\mu \partial_\mu f$ having dimensions of $[f]/L$, telling you how much $f$ changes per length. But I guess this would make the metric tensor have dimensions of $L^4$, which I am not really comfortable with.

Really, I am just thinking aloud at this point. I will go back to think about it a bit more.

12. Jun 27, 2015

### bcrowell

Staff Emeritus
There seem to me to be two reasons, each of them sufficient by itself, that if you want to assign units to the metric tensor (not just to its components), it has to be unitless:

1. If you do plain old SR in a plain old Minkowski coordinate system, its components are all unitless.
2. The metric is a universal piece of geometric apparatus, which is used to measure *all* vectors, not just one type of vector such as displacement vectors. Therefore it doesn't make sense to imagine that it would have units that are related in some way to the units we use for measuring displacements.

13. Jun 27, 2015

### Orodruin

Staff Emeritus
But this would still hold in a system where the metric tensor has units of $L^2$ due to the tensor basis (in coordinates with dimension $L$) having dimension $L^2$.

This is something I might agree on. However, the metric is also a notion of distance on the manifold from which the geometry is inherited, which also makes it viable for it to have dimensions of $L^2$. In this setting, tangent vectors and the corresponding dual vectors (related by contraction with the metric tensor) would differ in units by $L^2$. For example, the contravariant 4-velocity $V^\mu$ would have units of $1/L$ while the covariant 4-velocity $V_\mu$ would have units of #L#.

My thinking here is purely formal, derivatives generally lower the dimension of something by the dimension of the thing it is a derivative with respect to. Therefore, considered as a derivative, $V^\mu \partial_\mu$ would have dimensions of $[V^\mu]/[x^\mu]$.

In the end, I do not think it matters much as long as you are consistent and realise that $V^\mu$ for different $\mu$ may have different dimensions, depending on the dimensions of the coordinates.

14. Jun 29, 2015

### bcrowell

Staff Emeritus
Terry Tao has a blog post that touches on this: https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/ . See the paragraph beginning with "One can also assign dimensions to higher rank..." He identifies dimensional consistency with the grammar of indices in abstract index notation. His ideas are interesting, but not quite right IMO. Dimensional analysis is supposed to tell us whether a certain equation would stay valid if we changed the basic standards that define our system of units. A mismatch in indices often indicates that we will fail this requirement, because the transformation properties of the two sides are different, e.g., $A^{ij}=B^j$ becomes invalid under scaling of coordinates by $s$, which is a type of change of units. But such a mismatch doesn't always indicate that we fail the requirement, since we can make up ungrammatical equations that are nonsense as equations in abstract index notation, but not nonsense because of units, e.g., $\tau=dx^i$, where $\tau$ is a proper time and $dx$ is a displacement. We can also have things that are nonsense in terms of units, but are written grammatically, e.g., $v^i=a^i$, equating a velocity to an acceleration.

For another highbrow discussion of dimensional analysis, see https://www.math.ucdavis.edu/~hunter/m280_09/ch2.pdf . This presents a fancy group-theoretical perspective, but doesn't seem to get at the issues relating to tensors.

I would propose the following definition, which is just a more careful formulation of what I posted previously. Choose coordinates in a neighborhood of a point P such that the metric has the form $\operatorname{diag}(\pm 1,\ldots)$, with the ones being unitless. (This works in both Riemannian and semi-Riemannian spaces. It doesn't work in Galilean spacetime, where the metric is degenerate, but it does work in the Euclidean subspace of Galilean spacetime.) Then the units of a tensor at P (could be in the tangent space, cotangent space, ...) are the units of its components when expressed in these coordinates. This definition behaves properly under sums, products, and contractions. It also works regardless of whether you use geometrized or SI units. A tensorial equation can be dimensionally consistent by this definition, but still be ungrammatical in index notation, as with $\tau=dx^i$. By this definition the metric tensor is always unitless, which I think is the only sane possibility.

I think my definition works fine for tensor densities as well as pure tensors.

Suppose we make the following assumptions:
1. It's possible to assign units to tensors as a whole, and not just to their components.
2. The covariant form of the metric tensor is unitless.
3. Units stay the same under addition.
4. Units multiply under tensor products.
I think assumption 1 may be dubious, but if you want 1, you can't make a sensible system without assuming 2-4. Under these assumptions it follows that:
• The contravariant form of the metric is unitless.
• Raising and lowering indices doesn't change units.
• Contracting indices doesn't change units.

Last edited: Jun 29, 2015
15. Jun 30, 2015

### Ben Niehoff

Since the integral

$$\int \sqrt{g_{\mu\nu} dx^\mu dx^\nu}$$

gives a length, then the metric tensor $g_{\mu\nu} dx^\mu dx^\nu$ had better have units of length-squared. It is totally unambiguous to assign dimensions to tensors, because tensors are invariant objects. Note that $g_{\mu\nu} dx^\mu dx^\nu$ is a tensor, but $g_{\mu\nu}$ are the components of a tensor in a given basis, and its units will depend on the basis chosen (and my be different for different values of the indices!). Anyway, a tensor (not just its components!) has no free indices, so can be assigned units unambiguously.

In Cartesian coordinates, we typically choose the displacements $dx^\mu$ to have units of length. Then the metric tensor components $g_{\mu\nu}$ are unitless numbers. But in spherical coordinates, we typically choose $dr, dt$ to have units of length (having set $c=1$), while $d\theta, d\phi$ are unitless. In this case, $g_{tt},g_{rr}$ are unitless, whereas $g_{\theta \theta},g_{\phi \phi}$ have units of length-squared. This should be unsurprising, given that

$$g_{\theta \theta} = r^2, \qquad g_{\phi \phi} = r^2 \sin^2 \theta.$$

By contrast, in Cartesians, the partial derivatives $\partial/\partial x^i$ have units (length)^(-1), and again the inverse metric components $g^{\mu\nu}$ are unitless.

I should point out that this is the convention used for dimension counting in field theory, although typically one uses mass dimension rather than length dimension, so $dx^i$ gives you -1 and $\partial/\partial x^i$ gives you +1. Since the Lagrangian density $\mathcal{L}$ multiplies $d^4x$, it must have mass dimension +4 in order for the path integrand to be dimensionally-consistent (having set $\hbar = 1$):

$$e^{-i \int \mathcal{L} \, d^4x}$$

16. Jun 30, 2015

### bcrowell

Staff Emeritus
I think you're running into two different notational clashes here: (1) the one I referred to in #10, and (2) abstract index notation versus concrete index notation.

When you say that "$g_{\mu\nu}$ are the components of a tensor in a given basis," you're reasoning in concrete index notation (hence your Greek letters). All of my posts above have been using abstract index notation (hence my Latin indices). In abstract index notation, $g_{ij}$ is not a component, it's a tensor. When you say, "its units will depend on the basis chosen (and my be different for different values of the indices!)," that's fine for concrete index notation, but false for abstract index notation. In abstract index notation, there is no choice of basis, and indices don't take on values.

The metric tensor is unitless. Any other conclusion indicates a mistake in the reasoning leading up to it, since the metric is a generic piece of measurement apparatus that can be used on vectors that have any dimensions whatsoever.

Here you're running into the clash I referred to in #10. If you refer to the $dx^\mu$ as infinitesimal displacements, then you're using old-school notation (differential geometry before ca. 1960), which is fine, but then your statement that $g_{\mu\nu} dx^\mu dx^\nu$ is "the metric tensor" is incorrect. If you refer to $g_{\mu\nu} dx^\mu dx^\nu$ as the metric tensor, then you're using notation in which $dx^\mu$ is the dual of a derivative operator, not an infinitesimal displacement.

You can't use the post-1960 notation, in which $dx^\mu$ is the dual of a derivative operator, and reason about the units of expressions the way we're accustomed to doing, by inferring the units of a product from the products of the factors. In this notation, $\partial_\mu$ is a derivative operator, not a number that has units. Its dual $dx^\mu$, is also not a number that has units.

Last edited: Jun 30, 2015
17. Jun 30, 2015

### Ben Niehoff

The beauty of it is that an infinitesimal displacement is the dual of a derivative operator, which is why the notation is overloaded in the first place.

Looking at your #10:

Of course it's logically consistent to assign units! Not sure why you would think otherwise. I assign units to my formulas all the time to check their consistency.

A velocity vector ought to have units of velocity. Since we've set $c=1$, this means it must be dimensionless. In Cartesians, the coordinate basis $\partial_\mu$ has units of (length)^(-1), which means the components $V^\mu$ have units of length. What exactly do you think is wrong about that?

18. Jun 30, 2015

### bcrowell

Staff Emeritus
No, they're completely different concepts, as correctly suggested by your word "overloaded." (Your "is" contradicts your "overloaded.") I think what you mean is that you've been taught not to interpret dx according to its historical interpretation, as an infinitesimal, and instead to interpret it as something else. This was a reasonable attitude ca. 1960, because there was doubt at that point about whether the notion of infinitesimals could be defined rigorously. That was cleared up by non-standard analysis. One of the disadvantages of the 1960 redefinition was that it made it impossible to talk about the units of a quantity like dx -- the ability to do this was one of the original, cool design features of the Leibniz notation.

The $V$we were discussing was not a velocity vector, it's just some arbitrary contravariant vector. But anyway, let's say it is a velocity vector. What's wrong with your analysis is that you've concluded that in a Cartesian basis, the components of a velocity vector have units of length, which is false. You started from the assumption that one could assign units to the factors in this equation, and you ended up with a false conclusion. This is a proof by contradiction that your assumption was false. You can't assign units to the factors in this equation.

I'm not saying that it's impossible to do dimensional analysis on tensor equations. I'm saying it's incompatible with that particular style of writing tensors in a coordinate basis, with $\partial_\mu$ as a basis for covectors and the notation $dx^\mu$ for their duals being used as a basis for vectors.

Last edited: Jun 30, 2015
19. Jun 30, 2015

### Ben Niehoff

This statement is absurd. It is not "false", it is a matter of convention. It is a direct consequence of assigning units of length to the Cartesian coordinates.

All that is physically relevant is that the velocity vector have units of velocity. Not sure why you seem perturbed that its components, in some (dimensionful) basis, might have other units.

20. Jun 30, 2015

### Orodruin

Staff Emeritus
I do not agree with this. I would argue that any linear operator can be associated with a physical dimension, namely the dimension that differs between the dimensions of the quantity it is acting on and the dimensions of the result. This makes perfect sense in terms of the definitions of derivatives as limits, expressions of the type $(f(x+a)-f(x))/a$ has dimensions of [f]/[x].

It also means that it is perfectly fine to assign dimensions to dual vectors and you will find that $[dx^\mu] = [x^\mu]$ simply by the fact that $1 = [\delta^\mu_\nu] = [dx^\mu(\partial_\nu)] = [dx^\mu][\partial_\nu]$ and the argument we just applied for the derivatives, which leads to $[\partial_\mu] = 1/[x^\mu]$.

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