Discussion on tensor dimensions

[Orodruin]

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

Therefore it doesn't always make sense to talk about what units a tensor has.

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.

 

[bcrowell]

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.

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.)

 

[Orodruin]

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.

 

[bcrowell]

In the same way, the dual vector coordinate basis $dx^\mu$ would naturally have the same dimensions as the coordinates themselves.

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

 

[Orodruin]

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

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.

 

[bcrowell]

If the Ricci tensor has dimension of distance^2, would it not follow that T has dimension N(distance^2)?

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

 

[bcrowell]

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.

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 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 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 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.

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."

 

[stedwards]

From a few posts back,

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.

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, $d\tau^2 = d\tau^2[T^2]$?

 

[Orodruin]

If you want to assign units to VV, without having established any coordinate system, then I think the only sane thing to do is to say it's unitless, [V]=1[V]=1. Then with [∂r]=1/L[\partial_r]=1/L and [∂θ]=1[\partial_\theta]=1, we end up with [Vr]=L[V^r]=L and [Vθ]=1[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 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.

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 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.

They're not related, and really Orodruin and I have just hijacked the thread by having a side discussion about #2.

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

 

[bcrowell]

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

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.

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.

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

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.

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 $V=V^\mu \partial_\mu$.

 

[Orodruin]

I've never heard of anyone defining a four-velocity vector as having units of 1/L1/L. Is this something you're serious about and have thought through carefully?

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.

 

[bcrowell]

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.

 

[Orodruin]

If you do plain old SR in a plain old Minkowski coordinate system, its components are all unitless.

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 $L^2$.

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.

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 realize that $V^\mu$ for different $\mu$ may have different dimensions, depending on the dimensions of the coordinates.

 

[bcrowell]

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.

 

[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 $\hbar = 1$):

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

 

[bcrowell]

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.

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.

In Cartesian coordinates, we typically choose the displacements $dx^\mu$ to have units of length.

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.

 

[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:

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 $V=V^\mu \partial_\mu$.

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?

 

[bcrowell]

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.

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.

Looking at your #10:

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 $V=V^\mu \partial_\mu$.

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?

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.

 

[Ben Niehoff]

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.

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.

 

[Orodruin]

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.

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 $[\partial_\mu] = 1/[x^\mu]$.

 

[Ben Niehoff]

Bcrowell, you keep editing your posts after I've responded to them, which is making this conversation frustrating to follow. Correct typos, please, but don't add whole paragraphs.

 

[bcrowell]

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.

It's hard for me to believe that you mean this seriously. The vector's components are the things that we can measure directly. For example, a cop's radar gun measures the radial component of a car's velocity. In SI units, that component has units of m/s. In relativistic units, with c=1, that component is dimensionless. If you've concluded that it has units of meters, then you've reached an absurd conclusion.

 

[bcrowell]

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.

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 $[\partial_\mu] = 1/[x^\mu]$.

OK, I'm willing to buy this. However, this doesn't affect the validity of the point made in my first sentence, which is demonstrated in #17-19.

 

[Ben Niehoff]

It's hard for me to believe that you mean this seriously. The vector's components are the things that we can measure directly. For example, a cop's radar gun measures the radial component of a car's velocity. In SI units, that component has units of m/s. In relativistic units, with c=1, that component is dimensionless. If you've concluded that it has units of meters, then you've reached an absurd conclusion.

What? No, you can't measure the components directly. You can only measure invariant objects. The components are merely covariant (or contravariant, as the case may be).

To measure something you need a measuring device. In this case, the cop's radar gun is a device that projects velocity vectors onto a standard known velocity vector and compares them. Since both vectors have the same units (of velocity), this is perfectly cromulent. (Note that projection operators are always dimensionless.)

 

[Orodruin]

What? No, you can't measure the components directly. You can only measure invariant objects. The components are merely covariant (or contravariant, as the case may be).

This I agree with completely. The value of the components (and their dimensions!) are going to be completely dependent on the set of basis vectors you have chosen. You may have chosen an orthonormal basis where the components have the same dimensions as the vector itself, but this is not the only possible choice. I can just as well select to use a basis where the vector $\vec e_1$ is dimensional of magnitude 1 m in the direction of Paris. The corresponding vector component of the velocity vector will then have dimensions 1/T, telling me how many meters closer to paris I would get per time unit.

The only important thing is that any invariant you compute will obtain the correct dimensions.

 

[bcrowell]

What? No, you can't measure the components directly. You can only measure invariant objects. The components are merely covariant (or contravariant, as the case may be).

To measure something you need a measuring device. In this case, the cop's radar gun is a device that projects velocity vectors onto a standard known velocity vector and compares them. Since both vectors have the same units (of velocity), this is perfectly cromulent. (Note that projection operators are always dimensionless.)

The radar gun doesn't project velocity vectors onto a standard velocity vector. Velocity vectors are timelike. The radar gun projects velocity vectors onto a spacelike vector defined by the axis of the gun (i.e., a vector of simultaneity between two points on the gun, with simultaneity being defined in the gun's rest frame). If, as in your analysis, the gun were projecting a velocity vector $u$ onto a velocity vector $v$, then the only physically relevant choice for $v$ would be the velocity vector of the gun and the cop; but this would clearly not be a correct description of the operation of the gun, since then, e.g., the orientation of the gun would be irrelevant.

In general, I think there are good reasons why nobody actually uses the approach you seem to be suggesting, e.g., I don't believe that any physicist anywhere, including you, is in the habit of looking at an equation like $v'^\mu=v^\mu+\Delta v^\mu$ for velocity vectors and analyzing its units as "distance=distance+distance."

Suppose that we have a vector $v$ whose units are $L^\alpha$, and throughout this discussion let's assume a Cartesian basis. (In the example we were discussing, of a unitless velocity vector, we have $\alpha=0$.) Then according to your analysis, the contravariant components $v^\mu$, expressed in a Cartesian basis, have dimensions of $L^{\alpha+1}$, and if we apply the same reasoning to the covariant components we find that the $v_\mu$ have dimensions of $L^{\alpha-1}$. Examining an expression like $v^a=g^{ab}v_b$, we infer that $g^{ab}$ has units of $L^2$, while $g_{ab}$ would have units of $L^{-2}$. Continuing your approach in this way, we find that if a tensor has dimensions $L^\alpha$, then its components, when the tensor is expressed with type $(m,n)$, have units $L^{\alpha+m-n}$. Since the metric's components, when expressed with type $(2,0)$, have dimensions $L^2$, it follows that the metric itself is unitless (which would seem to contradict your #15).

I think it should be clear now what's going on. In your approach, we simply have a bunch of extra book-keeping to do. In index-free form, everything has the units we expect, but when we convert to concrete index notation, we have to multiply all the units by a factor of $L^{m-n}$. That's a lot of pointless extra work, which is presumably one of the reasons why nobody actually does it the way you suggest. By omitting all the factors of $L^{m-n}$, we get the standard system.

Another unfortunate feature of your approach is that you want a displacement vector $\Delta x$ to have units of $L$, which implies that its contravariant components $\Delta x^\mu$ have units of area, $L^2$. But since you insist that $dx^\mu$ have units of $L$, that means that you're going to assign different units to $\Delta x^\mu$ than to $dx^\mu$. Your approach also ends up assigning units of $L^2$ to $g^{\mu\nu}$ in a Cartesian basis, which would contradict every book and paper ever written on relativity, where these components are written as unitless values $\pm1$.

 

[Ben Niehoff]

The radar gun doesn't project velocity vectors onto a standard velocity vector. Velocity vectors are timelike. The radar gun projects velocity vectors onto a spacelike vector defined by the axis of the gun (i.e., a vector of simultaneity between two points on the gun, with simultaneity being defined in the gun's rest frame). If, as in your analysis, the gun were projecting a velocity vector $u$ onto a velocity vector $v$, then the only physically relevant choice for $v$ would be the velocity vector of the gun and the cop; but this would clearly not be a correct description of the operation of the gun, since then, e.g., the orientation of the gun would be irrelevant.

To be honest, radar measurement of velocities is more complex than mere projection, as it is a non-local measurement. But if the cop is sufficiently close to his target, we can approximate the measurement as local. In this case, the gun fires a beam of light in a given direction, and it is the velocity vector of this beam of light onto which we project the target's velocity. (Yes I know this is not how a radar gun works mechanically; I'm talking about how it works operationally).

This is easier to discuss if you want to measure something like proper length along some direction. That is a vector with dimensions of length. To measure an object, you project its length vector onto a "ruler", which is also a vector with dimensions of length. One obtains the ratio between the object's projected length and the ruler's length (a dimensionless number!), which is the actual physical quantity. It expresses the projected length of the object, in units of the ruler's length.

In general, I think there are good reasons why nobody actually uses the approach you seem to be suggesting, e.g., I don't believe that any physicist anywhere, including you, is in the habit of looking at an equation like $v'^\mu=v^\mu+\Delta v^\mu$ for velocity vectors and analyzing its units as "distance=distance+distance."

I'm not sure presuming to know how I (or "physicists anywhere") prefer to think about things is a worthwhile course of action. Ultimately, all units are a matter of convention. What I object to is you telling me my convention is inconsistent just because it is different from yours.

Suppose that we have a vector $v$ whose units are $L^\alpha$, and throughout this discussion let's assume a Cartesian basis. (In the example we were discussing, of a unitless velocity vector, we have $\alpha=0$.) Then according to your analysis, the contravariant components $v^\mu$, expressed in a Cartesian basis, have dimensions of $L^{\alpha+1}$

Here you have a logical error. Strictly speaking, you cannot think of vectors as objects that can have either contravariant components or covariant components according to your whim. Vectors and covectors belong to different spaces, and the isomorphism between those spaces is given strictly by the metric. Velocities are naturally vectors (i.e., tangents to curves!), not covectors, and you cannot presume to say what the units of $v^\flat = g(v, \cdot)$ are until you have specified the units of the metric tensor $g$.

I think it should be clear now what's going on. In your approach, we simply have a bunch of extra book-keeping to do.

Not really. It's quite natural and consistent. I don't typically have to use units, since my focus is more mathematical, but it is always possible to do calculations and then put in units consistently at the end. I find it easy to think my way.

In index-free form, everything has the units we expect, but when we convert to concrete index notation

I am using index-free notation. That's why I've been talking about vectors, tensors, etc.; the actual, invariant geometric objects, rather than their components. I've only written out components in order to explain how to consistently assign units to everything.

By contrast, abstract-index notation is never "index-free". :P

 

[bcrowell]

Here you have a logical error. Strictly speaking, you cannot think of vectors as objects that can have either contravariant components or covariant components according to your whim. Vectors and covectors belong to different spaces, and the isomorphism between those spaces is given strictly by the metric. Velocities are naturally vectors (i.e., tangents to curves!), not covectors, and you cannot presume to say what the units of $v^\flat = g(v, \cdot)$ are until you have specified the units of the metric tensor $g$.

I was carrying through the logical implications of your proposed system, so if you think there's a lack of clarity as to the units of the metric, then it's up to you to clarify. In #26 I worked out the implications of your system for the units of the metric, as well as the units of its contravariant and covariant components. If you disagree with those conclusions, then it's up to you either to clarify how your proposed system works, or to point out some error in the logic by which I worked out its implications. So no, I don't agree with your quoted material claiming that there was a logical error on my part.

In addition to the lack of consistency with the literature that I've pointed out, your approach seems to have an additional problem with a lack of self-consistency. You seem to want the following three things: (1) to assign nontrivial units to tensors, not just to their components; (2) to have the units of a product equal the product of the units of the factors; (3) to apply the same system to index-free expressions, as in the quoted material above. This clearly isn't going to work. Suppose we write down the following two index-free equations, $v^2=g(v,v)$ and $p^2=g(p,p)$. By assumptions 1 and 3, we can assign units to each side of each of these equations. Also by assumption 1, these units can be nontrivial, and therefore I'm free to say the the units in the two equations differ from one another. By assumption 2, the units of $g(v,v)$, which has only one factor, are simply those of the metric, and similarly for $g(p,p)$. Therefore $g(v,v)$ has different units from $g(p,p)$. But this contradicts assumption 1, which says that the metric tensor can be assigned definite units.

By the way, I've been using awkward phraseology throughout this thread to refer to the notation in which the $\partial_\mu$ are a basis for the space of vectors, and theirs duals are the covector basis. I looked through my notes and realized that this notation is due to Cartan, so for convenience we could refer to it as Cartan notation. (And it looks like its vintage may be quite a bit earlier than I believed, possibly as early as 1900; see http://hsm.stackexchange.com/questi...-his-notation-for-basis-vectors-and-covectors . Probably 1960 is more like the time at which physicists started using it.)

So here is my summary of where I think we stand right now:

In #14 I proposed a system for assigning units to both tensors and their components. This system is, as far as I can tell, consistent with how physicists think about units in practice, and it's also compatible with the last century's worth of physics literature, in which the components of the metric are always written as unitless numbers in a Cartesian coordinate system. This system is not compatible with the Cartan notation or with index-free notation.

Orodruin and Ben Niehoff also seem to want to build a system that assigns units to both tensors and their components, and you both have tried to make one that is compatible with the Cartan notation, and possibly with index-free notation as well. To the extent that you've made clear statements about what system you have in mind, its logical consequences seem to be that it's incompatible with the physics literature. Attempts to encompass index-free notation also seem to lead to a lack of self-consistency.

I don't think it's at all surprising that it doesn't work when you try to set up a system of dimensional analysis that works for so many different and incompatible systems of notation at once.

 

[Orodruin]

Attempts to encompass index-free notation also seem to lead to a lack of self-consistency.

I do not agree with this because I do not agree with g(v,v) having different units from g(p,p) being a problem. The assignment of dimensions to the metric tensor is based on the difference in dimensions between the object a linear operator acts on and the object it spits out. In that sense, it is perfectly compatible with g(v,v) differing in units by mass squared from g(p,p). The dimension of g(x,y) is [g][x][y], not [g].

 

[Ben Niehoff]

I do not agree with this because I do not agree with g(v,v) having different units from g(p,p) being a problem. The assignment of dimensions to the metric tensor is based on the difference in dimensions between the object a linear operator acts on and the object it spits out. In that sense, it is perfectly compatible with g(v,v) differing in units by mass squared from g(p,p). The dimension of g(x,y) is [g][x][y], not [g].

Yes. Units multiply under contractions.

However, I can see in this case why one might want to choose $g$ to be dimensionless. But if we choose $g$ to be dimensionless, then

$$\int_\gamma \sqrt{g_{\mu\nu} dx^\mu dx^\nu}$$
is no longer a length, unless we stick a dimensionful constant out front. So I guess you need to choose your poison.

 

[Ben Niehoff]

Let me add a bit more. Let's continue with my convention that the metric tensor $g$ has units of [L]^2, and that the coordinates $x^\mu$ have units of [L]. Then one has that $dx^\mu$ has units of [L], and $\partial/\partial x^\mu$ has units of [L]^(-1). Since

$$g = g_{\mu \nu} dx^\mu dx^\nu,$$
it follows that the components $g_{\mu\nu}$ are dimensionless. Notice that the exterior derivative operator $d$ is dimensionless (as $dx^\mu$ is just the exterior derivative of the function $x^\mu : M \to \mathbb{R}^n$).

If we write the metric in orthonormal frames

$$g = \delta_{ab} \, e^a \otimes e^b,$$
then it follows that the frames $e^a$ have dimension [L] (the Kronecker delta is always dimensionless). The connection 1-forms defined by

$$d e + \omega \wedge e = 0$$
are then dimensionless, and the curvature 2-form

$$\Omega \equiv d \omega +\omega \wedge \omega$$
is also dimensionless. Recall that the curvature 2-form is valued in a $\mathfrak{gl}(n)$ bundle over M. Choosing the coordinate basis for this bundle, we have that the components of the curvature 2-form are simply (the components of) the Riemann tensor:

$$\Omega^\mu{}_\nu \equiv \frac12 R^\mu{}_{\nu \rho \sigma} \, dx^\rho \wedge dx^\sigma.$$
Because the $dx^\mu$ have units [L], it follows that the components $R^\mu{}_{\nu \rho \sigma}$ have units [L]^(-2). This ought to look familiar.

I am happy to call this "Cartan notation", because these are the methods I typically use. But the consistency of unit conventions is an entirely separate issue from notation. Different notations are merely different ways to say the same thing. Given an abstract-index expression where the constituents are tensors, then I should be free to interpret it as a concrete-index expression in a basis of my choosing (including non-coordinate bases).

The difference is that an abstract-index tensor is defined as an object with indices that transform in a specified way, whereas the index-free definition of a tensor is a map between vector spaces which is function-linear in its arguments. These definitions are equivalent.Now, Bcrowell, let me ask you this. In your convention, you have chosen that the metric tensor to be unitless, and you have chosen that the coordinates, and hence the basis 1-forms $dx^\mu$, are unitless. It therefore follows, as far as I can tell, that the partial derivative operators $\partial / \partial x^\mu$ are unitless. In your convention, therefore, what are the units of the Riemann tensor? I think you might be surprised.

 

[bcrowell]

Here are a couple of other people's discussions of this sort of thing:

Dicke, "Mach's principle and invariance under transformation of units," Phys Rev 125 (1962) 2163 -- https://indico.kfki.hu/event/232/material/0/1.pdf

Terry Tao, https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/

Dicke has a nice discussion of how one can give differing descriptions based on one's definitions and assumptions, after which he goes ahead and picks the convention he wants. As in his treatment, consider this equation

$ds^2 = g_{ab}dx^a dx^b ,$

where nothing that follows depends on whether one considers these to be abstract or concrete indices. [EDIT: Changed my mind about this. It really needs to be abstract index notation. In concrete index notation, $dx^a$ would be an infinitesimal change in a coordinate, so we couldn't attribute units to it generically.] Generalizing his treatment, let's make no other assumption than that when something is written in multiplicative notation, the units multiply. Let the units of the factors be $[ds]=L^\sigma$, $[g_{ab}]=L^{2 \gamma}$, and $[dx^a]=[dx^b]=L^\xi$, where $L$ stands for units of length. We then have

$\sigma=\gamma+\xi .$

Some possible choices of these constants are:

$(1,1,0)$ - Dicke

$(0,\mp 1,\pm 1)$ - Tao

$(1,0,1)$ - what I've been advocating in this thread

Raising and lowering indices to form a tensor of rank $(r,s)$ makes units vary in proportion to $L^{\gamma(s-r)}$. In general, you can assign a physical quantity units $L^u$ that are a product of two factors, a "kinematical" or purely geometrical factor $L^k$, where $k=\gamma(s-r)$, and a dynamical factor $L^d\ldots$, which can depend on what kind of quantity it is, and where the ... indicates that if your system of units has more than just one base unit, those can be in there as well. Dicke uses units with $\hbar=c=1$, for example, so there's only one base unit, and mass has units of inverse length and $d_\text{mass}=-1$.

Although the units check out as long as the u exponents add up on both sides of the equation, it's also necessary for both the k's and the d's to add up, independently. To check the k's it suffices, in non-Cartan index notation, to check that the indices are written grammatically. (This fails when you mix in Cartan notation, e.g., $v=v^\mu\partial_\mu$ is notated as if it has (r,s)=(0,0), but it's actually a notation for a (1,0) vector.)

 

[bcrowell]

I do not agree with this because I do not agree with g(v,v) having different units from g(p,p) being a problem. The assignment of dimensions to the metric tensor is based on the difference in dimensions between the object a linear operator acts on and the object it spits out. In that sense, it is perfectly compatible with g(v,v) differing in units by mass squared from g(p,p). The dimension of g(x,y) is [g][x][y], not [g].

I think the issue here is simply that we have a customary, convenient procedure for checking units, and it's based on the idea that the units of a product are the product of the units of the factors. In index-free notation, there are a lot of cases where things get notated non-multiplicatively.

Let me add a bit more. Let's continue with my convention that [...] the coordinates $x^\mu$ have units of [L]. [...] the components $g_{\mu\nu}$ are dimensionless

So if I'm understanding you correctly, then in the notation I described in #32, you're saying you want $(\sigma,\gamma,\xi)=(1,0,1)$, which is the same convention I've been advocating.

In your convention, you have chosen that the metric tensor to be unitless, and you have chosen that the coordinates, and hence the basis 1-forms $dx^\mu$, are unitless.

No, I think you must have misunderstood me. I never said that I thought coordinates should be unitless. Of course coordinates can have any arbitrary mixture of units that we like, as in spherical coordinates.

 

[bcrowell]

I am happy to call this "Cartan notation", because these are the methods I typically use. But the consistency of unit conventions is an entirely separate issue from notation. Different notations are merely different ways to say the same thing. Given an abstract-index expression where the constituents are tensors, then I should be free to interpret it as a concrete-index expression in a basis of my choosing (including non-coordinate bases).

Of course all these notations are isomorphic ways of expressing the same ideas. However, (a) they may clash if you try to mix them, and (b) they may be inconvenient for certain purposes. There can be unpleasant clashes, for example, when you mix index and Cartan notation. Index-free notation can be inconvenient if you want to check units, since it uses a lot of non-multiplicative notation.

 

[bcrowell]

A velocity vector ought to have units of velocity. [...] the components $V^\mu$ have units of length. What exactly do you think is wrong about that?

[...] the coordinates $x^\mu$ have units of [L]. Then one has that $dx^\mu$ has units of [L][...] the components $g_{\mu\nu}$ are dimensionless.

I don't think these statements are logically consistent. We have

$ds^2=g_{\mu\nu}dx^\mu dx^\nu,$

and based on your statements above, the units on the right-hand side (in Cartesian coordinates) are $(L^0)(L^1)(L^1)$, so $[ds]=L$. Then since

$V^\mu = \frac{dx^\mu}{ds},$

and you say that $[dx^\mu]=L$, it follows that $V^\mu$ is unitless.