## September 30, 2006

### Dimensional Analysis and Coordinate Systems

#### Posted by John Baez

We had a nice conversation on dimensional analysis. Here are some things I learned.

As Terry Tao pointed out, systems of units are like coordinates. They’re both systems of arbitrary choices that let us describe reality using numbers. In both cases, we’re ultimately interested in learning things that don’t depend on these arbitrary choices. But to do that, we need to study how things change depending on these choices! We seek invariant quantities. But to get them, we must study covariant quantities.

Systems of units and coordinate transformations both cause a lot of debates in physics, but these debates tend to be separate. We should try to unify them. A system of units gives a coordinate system on some space of quantities we’re trying to measure. So, if we understand coordinate systems thoroughly, we should understand systems of units.

As Terry and Urs Schreiber noted, we can define a coordinate system to be an isomorphism

$f:X\stackrel{\sim }{\to }Y$

where $X$ is an object we’re trying to study and $Y$ is a “known” object. For example, $X$ could be a 4-dimensional spacetime and $Y$ could be ${ℝ}^{4}$. $X$ is like the landscape; $Y$ is like the map. Ideally, each point on the landscape corresponds to a unique point on the map.

If we pick different coordinates

$f\prime :X\stackrel{\sim }{\to }Y$

we can think of these as related to the original ones in either of two ways: by an “active” coordinate transformation, or by a “passive” one. This distinction causes a lot of grief and confusion, so let’s clarify it.

An active coordinate transformation is an isomorphism

$g:X\stackrel{\sim }{\to }X.$

It actually moves points around in the landscape. For example, if $X$ were the Earth, $g$ could describe how the Earth rotates. Given an active coordinate transformation $g:X\stackrel{\sim }{\to }X$ and a coordinate system $f:X\stackrel{\sim }{\to }Y$, we get a new coordinate system

$f\prime =fg.$

A passive coordinate transformation is an isomorphism

$h:Y\stackrel{\sim }{\to }Y.$

It moves points around on our map. For example, if $Y$ were a globe depicting the Earth, $h$ could describe how the globe rotates. Given a passive coordinate transformation $h:Y\stackrel{\sim }{\to }Y$ and a coordinate system $f:X\stackrel{\sim }{\to }Y$, we get a new coordinate system

$f\prime =hf.$

The confusion begins when someone has two coordinate systems $f,f\prime$ and they ask Tweedledum and Tweedledee how these coordinate systems are related.

Tweedledum says “by an active coordinate transformation, of course!” and writes

$f\prime =fg.$

Tweedledee says “by a passive coordinate transformation, of course!” and writes

$f\prime =hf.$

Either one could be right, but Tweedledum insists that we talk about

$g={f}^{-1}f\prime$

while Tweedledee insists that we talk about

$h=f\prime {f}^{-1}.$

This allows them to fight endlessly.

For example, when Daylight Savings Time starts, Tweedledum insists that the Sun is rising one hour later, while Tweedledee insists that we’ve just set our watch one hour forwards.

Or, when they’re riding in a train, Tweedledum insists that the scenery is rushing past him, while Tweedledee insists that he’s zipping through the scenery.

Or, when they’re doing quantum mechanics, Tweedledum insists on the Schrödinger picture, where the state of the universe is evolving in time. Tweedledee insists on the Heisenberg picture, where the state remains fixed and it’s observables that change with time.

There’s really no reason to get confused about this stuff - but when you have two equally good conventions for doing something, it’s easy to get stuck in endless arguments about which one to use.

(This is obvious when you have two factions involved, as Swift noted in his parable about the Big-Endians and Little-Endians: the Big-Endians insisted on cracking the big end of the hardboiled egg, while the Little-Endians insisted on cracking the little end. In fact this is not the best example, since the two alternatives are not symmetrical: isn’t it obviously easier to crack the big end? A better example is the dispute between Big-Endians and Little-Endians in computer science.

But, having two equally good choices can be a problem even if there’s just one of you, as Buridan’s Ass noted. This perfectly rational ass got stuck between two equidistant, equally good bales of hay and starved to death. Personally my problem with equally good choices is not making them but remembering which one I made. For example, I can never decide whether to denote the composite of two morphisms $f:X\to Y$ and $g:Y\to Z$ as $fg$ or $gf$. This adds an extra layer of confusion to the above dispute between Tweedledum and Tweedledee - but in case you’re wondering, I’m using the second convention here.)

Anyway, when it comes to systems of units, we are using active coordinate transformations when we imagine light going twice as fast, and passive coordinate transformations when we imagine using rulers half as long.

Everything I’ve said so far applies whenever $X$ and $Y$ are objects in any category. As Terry noted, there’s a lot of specially nice stuff that happens when this category is the groupoid of torsors for some group $G$. Then the group of active coordinate transformations is isomorphic to $G$, though not canonically so unless $G$ is abelian - and similarly for the group of passive coordinate transformations. But, since I’ve discussed torsors and physics elsewhere, I won’t go in that direction here.

Instead, I want to point out that so far our category might as well be a groupoid: I’ve only been talking about isomorphisms. My “coordinate systems”, “active coordinate transformations” and “passive coordinate transformations” were all isomorphisms.

But, sometimes coordinate systems aren’t isomorphisms.

For example, we often study an object $X$ using coordinate charts

$f:Y\to X$

where $Y$ is some “known” object. For example, $X$ could be an $n$-dimensional manifold and $Y$ could be ${ℝ}^{n}$. There are two new features here.

First, $f$ is pointing the opposite way: our “known” object is getting mapped into the object we’re trying to study. I believe this is not such a big deal, since we can switch the direction of arrows in a category using a purely formal trick, the “opposite category”. The opposite category of a category of “spaces” is often some sort of category of “rings of coordinate functions”, so when we map ${ℝ}^{n}$ into a manifold, we are turning functions on the manifold into functions on ${ℝ}^{n}$.

Second, $f$ is not an isomorphism anymore. I believe this more fundamental. Now we are considering situations where the map is not isomorphic to the landscape! That’s important, because most maps are indeed incomplete in some way. The unavoidability of this phenomenon may have first become clear when navigators sought to devise planar maps of the spherical Earth: since the plane and the sphere are not isomorphic, no such map can be perfect.

But, the importance of this phenomenon to physics became clear upon moving from special relativity to general relativity. Not only is the Earth not flat: the Universe is also not flat!

If we allow coordinate charts that are not isomorphisms, we should probably allows coordinate transformations that aren’t isomorphisms. What are these like?

Here I will consider passive coordinate transformations. In our new setup, with the arrows reversed, these are morphisms

$g:Y\prime \to Y$

and they act on any coordinate chart

$f:Y\to X$

to give a new coordinate chart

$f\prime :Y\prime \to X$

defined by

$f\prime =fg.$

We see this happening whenever we’re using a map and then we “zoom in” to a smaller map.

We also use them when we switch from a system of units to a simplified system of units. For example, the SI system has seven units, which measure length, mass, time, current, temperature, amount of substance, and luminous intensity. So, any physical quantity gives a point in ${ℝ}^{7}$ - the “dimensions” of this quantity in the SI system. But, we may choose to work with fewer units. For example, we may decide not to treat “amount of substance” as a dimensionful quantity. SI measures amount of substance in moles, but we can say a mole is just $6.0221415\left(10\right)×{10}^{23}$ - a dimensionless number, Avogadro’s number. Our new system of units assigns to each physical quantity a point in ${ℝ}^{6}$, and we have a “change of units”

$g:{ℝ}^{7}\to {ℝ}^{6}$

which is not an isomorphism. The ultimate extreme is to work in a system where all physical quantities are treated as dimensionless, so any physical quantity has dimensions living in ${ℝ}^{0}$. This is actually very popular in fundamental theoretical physics.

Anyway: we’ve seen that in any category, we can study an object $X$ by mapping “known” objects into it. The ultimate way to do this is to consider all possible morphisms

$f:Y\to X$

for all possible objects $Y$. This gives a very thorough description of $X$.

For example, we can study an $n$-dimensional manifold $X$ by looking at all possible coordinate charts

$f:Y\to X$

in this manifold, where $Y={ℝ}^{n}$. People like to use this trick; it’s called a maximal atlas. But, we can go even further and let $Y$ range over all possible manifolds!

Suppose we’re trying to understand an object $X$ in some category $C$. If we look at all possible “coordinate charts” on this object:

$f:Y\to X$

and how they are related by “passive coordinate transformations” of the form

$g:Y\prime \to Y,$

we get a gadget called a presheaf on $C$. This is just a functor

$F:{C}^{\mathrm{op}}\to \mathrm{Set}.$

In the case at hand, $F$ assigns to any object $Y$ the set of all “coordinate charts”$f:Y\to X$. And, it assigns to any “passive coordinate transformation” $g:Y\prime \to Y$ the function which eats a coordinate chart $f:Y\to X$ and spits out the transformed coordinate chart $fg:Y\prime \to X$.

If you’re wondering exactly what I mean now by the quoted phrases “coordinate chart” and “passive coordinate transformation”, it’s simple. In both cases, I mean simply morphism! We’ve left the warmup case where coordinate systems and coordinate transformations had to be isomorphisms: now we’re letting them be noninvertible.

But here’s the cool part:

The Yoneda Lemma says that everything we might want to know about our object $X$ can be recovered from the presheaf $F$ defined above! For a pretty good explanation of this lemma, try this. It takes a bit of category theory to understand the technical details.

But I don’t want to get into that here. I just want to say what this famous result says about coordinate systems:

We know an object completely if we know all coordinate charts on that object, together with how these coordinates transform.

Posted at September 30, 2006 1:45 AM UTC

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### Re: Dimensional Analysis and Coordinate Systems

I was going to post this comment in the previous discussion about dimensional analysis but comments finished there. I will try and tie it into the latest discussion towards the end.

When I teach about waves, the physics textbooks talk about frequency and the unit “hertz” as being “cycles per second” (or similar). However the textbooks, of course, correctly state that the unit Hz is equivalent to ${s}^{-1}$ and not actually “cycles per second”. Similarly, of course, wavelength is just measured in meters which is consistent with Hz simply being ${s}^{-1}$ so that, using dimensional analysis :-), wavelength multiplied by frequency gives us the correct units for a wave’s speed.

However, we could have Hz be “cycles per second” and wavelength as “meters per cycle” and speed would still work out just fine. This all sounds good and consistent to me… and more sensible… in the sense that it would be easier to justify when first taught.

Also, for example, isn’t saying a wavelength is “1m” like saying my speed is “1km”? We actually say 1km/h (or similar). So shouldn’t my wavelength be 1 m/cycle where it is explicit that we have to have knowledge of what a cycle is just as we have to have knowledge of what unit of time we are using?

The name “wavelength” is even misleading. It would be easier to have “meters per cycle”. After all we do not say speed is measured in “timelength” and assume a standard amount of time, but we say “kilometres per hour” (or similar). Yes, a “wavelength” has a canonical choice of “cycle” but we could equally (stealing John’s devil’s advocate role from the last dimensional analysis discussion) have chosen a definition that is equivalent to 2 waves or half a wave for our “cycle”.

Would it be sensible to instigate a campaign to have the “cycle” recognised as an SI base unit?! As, surely it is not a derived unit given its definition is “the distance between two consecutive points in phase” (or something similar). How would you express this in terms of the base units?

Linking in to the current main topic… it would seem like the current situation for SI base units is ${\Re }^{8}\to {\Re }^{7}$ [how do I get a blackboard R in itex?] though I am not sure it is clear to me how this trick has been done for a “cycle” in the same way as using Avogadro’s number for moles.

So maybe there is a difference here and thinking of possible morphisms is part of how the difference can be illustrated? Not sure I quite see this.

On the BIPM website there is a section that talks about dimensionless quantities that are ratios such as refractive index (easy to understand), but it also talks about counting quantities such as a number of molecules (less easy to understand – isn’t the mole one of these?). Perhaps “cycle” comes under this category as there are some in this category (such as the radian) that are given names. But in the case of the cycle we could change the definition (unlike counting the number of molecules for example – although could we talk about half molecules?) as mentioned above so does this really fit in this category and therefore how does it differ from other base units?

I had better stop here and ask… “So what do you all think?” Have I just missed something really obvious?

Posted by: Theo on September 30, 2006 11:04 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Theo said

When I teach about waves, the physics textbooks talk about frequency and the unit “hertz” as being “cycles per second” (or similar). However the textbooks, of course, correctly state that the unit Hz is equivalent to ${s}^{-1}$ and not actually “cycles per second”. Similarly, of course, wavelength is just measured in meters which is consistent with Hz simply being ${s}^{-1}$ so that, using dimensional analysis :-), wavelength multiplied by frequency gives us the correct units for a wave’s speed.

and

Would it be sensible to instigate a campaign to have the “cycle” recognised as an SI base unit?!

But suppose we wanted to measure sneezes per second? Or suppose an industrial packer of apples needed to measure apples per second?

Suppose we wanted to calculate the rate that air was expelled from a person’s nose during a sneezing fit. Then we could define the “sneezegust” as the volume of air expelled in a sneeze, in ${\mathrm{cm}}^{3}$, measure the frequency in ${s}^{-1}$, and multiply them together to get a figure in ${\mathrm{cm}}^{3}{s}^{-1}$. This is exactly analogous to calculating the wavespeed. Likewise we can use the mass of an average apple in kg and the frequency of apples in ${s}^{-1}$ to get the rate at which mass flows into the apple barrel in $\mathrm{kg}{s}^{-1}$.

But would we really want entries like the following in the SI standard?

“Sneezes. The SI quantity of sneezes is the Sneeze (Sn). The Sneeze is defined as the quantity of sneezing in 1 sneeze.

Apples. The SI quantity of apples is the Apple (Ap). Apple is defined as the quantity of apples in 1 apple.”

Of course, you are absolutely right that it is important to remember that the “wavelength” and “frequency” of a wave are based on one cycle of the wave. (Rather than, e.g. 1 radian of rotation of the phase, also a quite natural unit.)

The lesson as I see it is that, when doing calculations on measurements of a physical situation, it is essential to remember which physical quantity corresponds to which measurement, as well as what units the measurement is in. In a moment of confusion about what one is doing, or while teaching, it is often good to be more explicit than usual, and introduce extra units to make it clear what is going on. These needn’t be made into global standards, though, unless there is danger of confusion between different groups of professionals who are communicating with one another.

See also my parable of the cheese and the apples, where we have two different but equivalent units of weight, “pounds of cheese” and “pounds of apples”. Taking away some apples and substituting an equal weight of cheese has no effect for some purposes, e.g. estimating how much effort it will take to lug the picnic hamper around. In this case, splitting pounds into two different units is a needless excess. But for some purposes (e.g. whether people will enjoy eating the picnic food) it makes a difference, so people make the distinction. If the proportions of foods in the picnic were known from context (in the way that the use of “cycles” is known from context in the case of waves) then only the total quantity would be worth mentioning, and again we could dispense with the extra units.

Posted by: Tim Silverman on September 30, 2006 6:19 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

This is quite related to what John said about moles: the arbitrariness of what one considers to be the fundamental dimensions of a physical theory.

Under some circumstances, its quite useful to think of cycles and radians as two different units for the dimension rotational angle. Then wavelength has dimensions of [L]/[A], which may be expressed in (say) metres per cycle or miles per radian, while frequency has dimensions of [A]/[T], which may be expressed in (say) cycles per second or radians per hour.

Dimensional analysis then tells us that we can get a speed (dimensions of [L]/[T]), independent of our units of angle, by multiplying these. And we can see whether we need to convert units of angle as we do this: to get a speed in metres per second or miles per hour, we just multiply the corresponding wavelength and frequency; but to get a speed in metres per hour or miles per second, we’ll have to convert between cycles and radians (1 cycle equals 2π radians) first.

On the other hand, suppose you don’t want to think about rotational angle as a fundamental dimension. In that case, the difference between cycles and radians can still come up! But now, people think about two different quantities: wavelength versus reciprocal wave number, and similarly frequency versus angular frequency. The conversion factor of 2π still shows up, now in the formulæ k = 2π/λ (relating wave number k to wavelength λ) and ω = 2πν (relating angular frequency ω to frequency ν).

Ironically, since the SI gives official recognition to the radian as the unit of angle (albeit considered a derived unit), this suggests that Theo’s idea of making the cycle an SI unit has essentially been accepted, except that they chose the radian instead! This further suggests that there’s something slightly incorrect about using Hertz in the cycle-per-second sense to measure frequency; it would be more proper to use Hertz in the radian-per-second sense (which is regarded as equivalent to the official definition of the Hertz as an inverse second) to measure angular frequency. But in fact, people don’t pay any attention to that and happily speak of (say) 60-Hz alternating current, thinking that they’re using SI correctly. And they are using it correctly, of course —but only if you reject the dimension [A] of rotational angle and instead consider ν = ω/2π to be of dimension 1/[T].

Posted by: Toby Bartels on October 6, 2006 12:55 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Theo wrote:

Linking in to the current main topic… it would seem like the current situation for SI base units is ${\Re }^{8}\to {\Re }^{7}$ [how do I get a blackboard R in itex?]

You want to use $\mathbb{R}$ to get an $ℝ$.

You’re using $\Re$, which has always been the TeX symbol for $\Re$, meaning “real part”.

I urge anyone who has questions and complaints about the mechanics of posting to put them here.

Posted by: John Baez on September 30, 2006 8:48 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Apologies. Will ask questions where requested. It was just a spur of the moment thing that I should of looked up as I have used it before but couldn’t remember off of the top of my head. I was being lazy!

Posted by: Theo on September 30, 2006 10:55 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

I disagree (strongly) with the two points you highlight in the discussion of dimensional analysis. The points were,
- Dimensionless constants can depend on our choice of units - for example, the constant C=c/(m/s)=299,792,4
- Dimensionful constants often don’t depend on our choice of units - for example, the constant c=Cm/s=299,792,458m/s.

The problem is caused by an abuse of vocabulary. Let me try to clarify.

First, numbers are just numbers. They do not have units or dimension, they are just numbers.

Some aspects of the world are naturally associated with specific numbers – things that can be counted (the number of electrons on a carbon atom) or ratios of similar things (the ratio of Earth’s diameter to the height of the Washington Monument).

What if you want to associate a number with something that isn’t countable and isn’t a ratio, like the hight of the Washington Monument? Well, you can’t do it; it is impossible. What number would you choose? 6? 42? 10.786? 2.5x10^-35? The very idea is absurd.

What you can do is find the ratio of the thing you want to measure to some similar reference thing. For example, the ratio of the height of the Washington Monument to the length of a platinum rod that I keep in my basement is 169.29. If I want to tell somebody how tall the Washington Monument is, I have to give them the number (169.29) and I have to hand them my platinum rod (or a copy). This is very important: the number means nothing unless they have the physical rod in hand. Fortunately, there are people in the business of making and selling copies of this rod, often marked to show the location of various ratios, like all of the hundredths. They are big sellers.

Unfortunately, there is another rod somewhere that is shorter and people sell many copies of it as well (marked in 12ths, for some reason). Of course the ratio of the height of the Washington Monument to the shorter rod is a bigger number, 555 5/12. The rods have been given names (the meter and the foot) and when the numbers are shared, the standard rod is specified, e.g. 169.29 m and 555 5/12 ft. The meter and the foot are called standards of length.

Now for the vocabulary. If I can measure the ratio of some quantity to the length of one of these rods, then that quantity is said to have “dimensions of length” (not units of length). The ratio is a number and the specification that denotes which standard to use is called the “units.” We can treat the units as representing the length of the physical rods, so 169.29 m is 169.29 times the length of the meter rod, 1m. That is 169.29m = (169.29)(1m).

The height of the Washington monument does not depend on the standard used. However, the number assigned to the height of the Washington monument does depend on the choice of standard. If the standard of length is changed by a factor of x, then the number assigned to the height must be changed by a factor of 1/x.

As you know, we need several different units for measuring different things. Some of the units are distinct, like the second and the kilogram, others are derived, like the cubic meter or the meter per second. The derived standard of speed is the speed of something that travels 1 meter in 1 second.

Now we can correct your points.

The first one should say,
The number associated with a dimensional quantity always depends on our choice of units - for example, the number C, which is the ratio of the speed of light to the standard speed of 1m/s, C=c/(m/s)=299,792,458 depends on the choice of standard meter and second.

The second one should say,
The quantities themselves (whether they are dimensional or not) never depend on the choice of units - for example the speed of light, c = 299792458 m/s, is the same speed no matter what units are used as the standard to describe it.

The third point, which you did not make, is this,
The number associated with a dimensionless quantity does not depend on the choice of units.

Does the fact that the number C (which is dimensionless) changes when we change units violate the third point. No, because we are changing the dimensionless quantity C, not just the number associated with it. Consider a similar quantity H which is the ratio of the height of the Washington Monument to the length of the meter rod in my basement (H=169.29). H has no dimensions. If I decided to find this ratio by measuring the height of the Washington Monument in feet and the length of my meter rod in feet, then I would get the same number, H=169.29. So the quantity H defined as the ratio of the height of the Washington Monument to the length of my meter rod is independent of the choice of units. However, if I decide to measure a different quantity, for example the ratio of the height of the Washington monument to the length of my shoe, then I will get a different number representing this different quantity, H’=555 5/12. H’ is also dimensionless and does not depend on the units I use to measure the height and length. H and H’ are, however, different numbers because they represent different dimensionless quantities. So when you say C changes you are actually changing the dimensionless quantity that you measure, not just changing the same measurement to a different unit system.

You can avoid that mistake in the calculation by never allowing units to float on their own in an expression. The unit “m” is not a quantity, but “1m” is. So now we write

c = 299792458 m/s

The number 299792458 is the ratio of the speed or light to my speed standard, a toy car that travels 1m in 1s.
Since the speed of light is 299792458 times the speed of my speed standard, we can write.

c = (299792458)(1m/s)

Now let’s change to miles per hour. The ratio between the speed of light and my toy car will remain unchanged by this, so we keep the dimensionless 299792458 as it is. The speed of the toy car also remains unchanged, 1m/s, but I want to write that speed in different units. Measuring the ratio of the speed of the toy car to a new speed standard, a pet rabbit that runs one mile per hour, I find 1m/s = 2.237 mile/hr. Now I write c as

c = (299792458)(2.237miles/hr)

Notice that the dimensionless quantity, 229792458, did not change when I changed units, in agreement with point 2 and the number associated with it, 299792458, also did not change, in agreement with point 3. The quantity with dimensions of speed, 1m/s, did not change (also in agreement with point 2), but the number associated with it did, consistent with point 1. Finally, I can do the multiplication to find a new dimensionless quantity, the ratio of the speed of light to the speed of my rabbit, and use that number to represent the speed.

c = 670635729miles/hr

The point is that the 299792458 and 670635729 are not different numbers associated with the same dimensionless quantity, they are different numbers associated with different dimensionless quantities.

Sorry this post became so long. This is confusing and one has to be very careful with the language in order to be precise, something we do not do in our everyday use of units.

Posted by: Gavin Polhemus on September 30, 2006 2:42 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Experimentally, the choice of units is a technological issue. E.g., if I can compare a given mass to the standard kilogram to one part in 100 million, but only count atoms to one part in a million, then reducing SI units from 7 to 6 is a practical disaster, no matter what theoretical paradise ensues.

(FYI, trying to comment on this blog is extremely trying.)

Posted by: Arun on September 30, 2006 4:03 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Anyone who has trouble posting should read the FAQ. If that doesn’t help, post a question there and we will answer it.

Good point about technological limitations, Arun.

Posted by: John Baez on September 30, 2006 9:00 PM | Permalink | Reply to this
Weblog: The n-Category Café
Excerpt: Let's amass a collection of wisdom on gnarly issues in physics - starting with dimensional analysis.
Tracked: September 30, 2006 4:17 PM

### Re: Dimensional Analysis and Coordinate Systems

Good, this time I am not late, and it is the weekend…

So, there are two points about dimensionful constants I still don’t understand. In the following I refer to the constant itself, e.g the speed of light c, and not the number C expressing it’s value in certain units. We can agree the later does not appear in any laws of physics, so I am less interested in it.

So, my point before was that dimensionful constants parametrize physically equivalent theories- if you have two universes (or the same one at different times) where some dimensionful constants are different (but keeping all the dimensionless quantities fixed, lest we get confused), you get identical physical phenomena- though some measurements will be different, they’ll be related by precisely the same laws. This is very different from those dimensionless parameters (such as \alpha) which parametrize the space of inequivalent theories: a universe where \alpha =1 has (even qualitatively) different laws of physics from our own, a universe where the speed of light is twice as fast is physically indistinguishable. This is a fairly precise sense in which only dimensionless parameters can be “fundamental”.

The other point I’m confused about is that I think observers having access to measurements in only one universe will not be able even to define operationally what they mean by change in dimensionful constants (relative to other universe, or to the past). I think any such operational definition will end up making reference to dimensionless quantities only.

Posted by: Moshe on September 30, 2006 4:27 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

The other point I’m confused about is that I think observers having access to measurements in only one universe will not be able even to define operationally what they mean by change in dimensionful constants (relative to other universe, or to the past). I think any such operational definition will end up making reference to dimensionless quantities only.

I think this abstract discussion is going round in circles (John may feel otherwise). So let’s look at some examples, shall we?

${M}_{\text{pl}}$ is, I’m sure we can agree, a dimensionful constant, which sets a fundamental mass scale for (our) universe.

Would physics look different if ${M}_{\text{pl}}$ were different?

Sure it would. Among other things, the physics we know and love depends rather crucially on the ratio ${\Lambda }_{\text{QCD}}/{M}_{\text{pl}}$. And changing ${M}_{\text{pl}}$, leaving ${\Lambda }_{\text{QCD}}$ fixed, would certainly change the physics that we see. To keep the physics invariant, we’d also have to change ${\Lambda }_{\text{QCD}}$ … and the Electroweak scale, and …

How about John’s example, the speed of light? That sets a fundamental scale for velocities. Would a universe, with a different value for $c$, look different to its inhabitants?

For about 15 seconds, you might be tempted to think, “No, it would look the same to them.” But then you’ll realize that there are other quantities with dimensions of velocity. For instance,

$v=\sqrt{2{E}_{0}/{m}_{e}}$

where ${E}_{0}$ is the ground-state energy of hydrogen and ${m}_{e}$ is the electron mass. If we “just” changed $c$, then the dimensionless ratio $v/c$ would change, and physics would look different.

But this dimensionless ratio is, up to a trivial numerical factor, the fine structure constant, $\alpha$. So changing $c$, while holding $v$ fixed, is the same as changing $\alpha$.

Indeed, as Aaron pointed out, the only things we ever measure are dimensionless ratios of this sort. It’s quite meaningless to talk about changing the values of dimensionful constant, except to the extent that doing so alters the values of dimensionless ratios of the sort discussed above.

Now, you could ask whether some dimensionless ratios are more “fundamental” than others (some physically-important length-scale measured in units of the Bohr radius versus the same length, measured in units of the distance between two scratches in a bar of platinum).

That’s a matter of the choice of coordinates on your space of physical theories (as John would have it). Some coordinate choices are better than others …

[By the way, since the comments here are rather indistinguishable in form from the comments on John’s previous post, I wonder a bit at the wisdom of closing comments on that one and moving the discussion over here. Now we end up referring back to stuff that was said there or (more likely ) just end up repeating that conversation over here.]

Posted by: Jacques Distler on September 30, 2006 6:16 PM | Permalink | PGP Sig | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Yes Jacques, this is going around in circles, and becoming an exercise in pure semantics. I think you and me and Aaron all seem to agree that there exists an economical description of any physical law that makes no reference to dimensionful quantities, perhaps one should stick to that one.

Posted by: Moshe on September 30, 2006 6:40 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Indeed, as Aaron pointed out [here, u.s.], the only things we ever measure are dimensionless ratios of this sort.

Since Aaron and myself apparently couldn’t quite agree on - probably not the content, but in any case it seems the formulation - of this statement, maybe it is worthwhile to agree that

“measuring a dimensionless ratio $a/b$

is precisely the same as

“measuring $a$ in units of $b$”.

Posted by: urs on October 1, 2006 1:03 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

I agree with Urs that it’s a bit misleading to say “one never measures a dimensionful quantity”.

Fred pulls out his meter stick, slaps it down on the table three times and says “This table is 3 meters long. There! I just measured a dimensionful quantity, the length $L$ of this table. I got $L=3m$.

Ted says “No, no - you’re completely wrong. You just measured the dimensionless ratio $L/m$ and got 3.”

It’s hard to believe that Ted is “right” and Fred is “wrong” here. Even if we prefer Ted’s way of talking, it’s hard to argue that Fred is making a terrible mistake.

It’s much easier to believe that Ted and Fred have two slightly different, but equivalent, ways of talking.

So, instead of saying “You never measure dimensionful quantities,” I think it’s safer to say “You can do all of physics in terms of dimensionless quantities.” The former claims we’re unable to do something most people think they can do. The latter claims we’re able to take any equation involving dimensionful quantities, divide both sides by the same thing, and get an equation where both sides are dimensionless. The latter seems a lot less controversial.

Posted by: John Baez on October 2, 2006 3:38 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Even if we prefer Ted’s way of talking, […]

Fred’s way of talking is more explicit. Ted might assume things he does not make explicit.

A dimensionless ratio is only useful as long as we remember what was divided by what. This “what” is what Fred makes explicit.

If Ted wants to tell an experimentalist to tune his machine to a certain energy, he somehow has to convey to that experimentalist a torsor element. This cannot be done just using numbers.

Either Ted can assume the experimentalist implicitly knows what Ted did to compute his dimensionless ratios (for instance that he divided by electron volt), or else Ted will have to tell the poor guy.

Posted by: urs on October 2, 2006 5:43 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

John said a couple of things:

So, instead of saying “You never measure dimensionful quantities,” I think it’s safer to say “You can do all of physics in terms of dimensionless quantities.” The former claims we’re unable to do something most people think they can do. The latter claims we’re able to take any equation involving dimensionful quantities, divide both sides by the same thing, and get an equation where both sides are dimensionless. The latter seems a lot less controversial.

It is, as my recent experience at Wikipedia has been, a lot less controversial, but the latter is no more true (whatever the hell “truth” is) than the former, is it? I certainly cannot think of a measuring device, from a ruler to an ammeter to whatever, that is not comparing the measured quantity to a given standard that is defined by virtue of the construction of the measuring device. Is not the net measurement always a dimensionless number and then, given knowledge of the specifications of the measuring device, we humans attach a dimensionful meaning to that number?

The other thing you said, John, was

I agree that being dimensionless is not a sufficient condition for α to be “fundamental”. It’s fundamental because it says something about the strength of the electromagnetic force - at least if we pick units where c=−e=ℏ=1. If we pick units where c=4πϵ0=ℏ=1, it says something about the charge of the electron. Either way, it’s saying something about electromagnetism. There are other ways to slice the pie (see above), but these are stupider.

I really agree with this, too, but want to point out (as best as I can, being a neanderthal electrical engineer) that the effect is the same. Given a “constellation” of charged objects (all with a fixed integer multiple of the elementary charge on them) the force between the different objects will increase in proportion to e2 in the latter case or in proportion to (4πϵ0)-1 in the former case. and the difference between the two is “academic”. I prefer the version that α is proportional to e2, but that is inconsequential to reality. In both cases, the electrostatic force, relative to the other forces (including gravity) is proportional to α, no? α, or something proportional to it (I like 4πα, because I like “rationalized” Planck units with c = ℏ = 4πG = ϵ0 = 1 and the elementary charge is e = (4πα)1/2 = 0.30282212), is the salient quantity, no?

Posted by: robert bristow-johnson on October 4, 2006 12:30 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

John wrote:

So, instead of saying “You never measure dimensionful quantities,” I think it’s safer to say “You can do all of physics in terms of dimensionless quantities.” The former claims we’re unable to do something most people think they can do. The latter claims we’re able to take any equation involving dimensionful quantities, divide both sides by the same thing, and get an equation where both sides are dimensionless. The latter seems a lot less controversial.

robert bristow-johnson wrote:

It is, as my recent experience at Wikipedia has been, a lot less controversial, but the latter is no more true (whatever the hell “truth” is) than the former, is it?

Perhaps I should be a little less subtle about what I was trying to say.

The statement “You never measure dimensionful quantities” seems fairly obnoxious to me. It sounds like a statement of fact. But when you examine it, it’s really just an attempt to make us talk a certain funny new way. The only person I know who really sticks to this way of talking is Ted. And I really don’t like Ted.

For example: just yesterday, Ted’s wife said, “That diet is really working! I measured my weight the other day - turns out I weigh 140 pounds!”

And do you know what he replied? Instead of congratulating her, he said “No! You didn’t measure your weight - you can only measure dimensionless quantities! You really measured the ratio of your weight and one pound, and got the dimensionless number 140.”

It’s not clear that this clarifies anything. Ted is really just trying to push a funny way of talking on everyone. Frankly, I think Ted is headed for a divorce.

Posted by: john baez on October 4, 2006 4:16 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

I certainly cannot think of a measuring device […] that is not comparing the measured quantity to a given standard

Exactly. That’s what we all keep saying.

A measuring device measures a torsor element $e$. It does so by telling you the group element $g$ which you need to apply to some fixed torsor element ${e}_{0}$ such that

(1)$e=g\cdot {e}_{0}\phantom{\rule{thinmathspace}{0ex}}.$

${e}_{0}$ is what you called “a given standard”. $g$ is what you read off of the device’s display.

But the point is that $g$ alone does not tell you anything about the measurement. Only the combination $e=g\cdot {e}_{0}$ does.

If you happen to use another measuring device, one which uses the standard $e{\prime }_{0}$ instead of ${e}_{0}$ and you measure the same thing, you should still get $e$ as a result, but now the new device will in general display another group element $g\prime$, such that

(2)$e=g\prime \cdot e{\prime }_{0}\phantom{\rule{thinmathspace}{0ex}}.$

Notice that there will be some group element ${g}_{0}$ relating the two “standards” to each other

(3)$e{\prime }_{0}={g}_{0}\cdot {e}_{0}\phantom{\rule{thinmathspace}{0ex}}.$

It follows that the two group elements that are displayed by the two measuring devices are related by

(4)$g=g\prime \cdot {g}_{0}\phantom{\rule{thinmathspace}{0ex}}.$

To get an example, you might want to think throughout of the group $G$ that $g,g\prime$ and ${g}_{0}$ are elements of as the multiplicative group of the real numbers except zero, and of the torsor $E$ that $e,e\prime$ and so on are elements of as the torsor of units of length.

Posted by: urs on October 4, 2006 9:15 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Urs wrote in small part:

the multiplicative group of the real numbers except zero

Actually, the most directly relevant group is the multiplicative group of positive real numbers. If you want to make your unit of length 0.3048 metres, then you may be a backwards provincial, but you’re still a perfectly reasonable one. But if you want to make your unit of length −0.3048 metres, then you’re just weird. (This weirdness can be accomodated by using the larger group, but it’s still weird in a mathematically precise way.)

Posted by: Toby Bartels on October 6, 2006 2:12 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Actually, the most directly relevant group is the multiplicative group of positive real numbers.

In my example # I took the group to be what the display of our measuring device takes values in - which makes me want to include negative numbers.

So I blurred the distinction between “lengths” and “units of lengths”.

This is a slight oversimplification of the matter - since from this point of view I would also want to include the display value 0.

Posted by: urs on October 6, 2006 3:18 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Toby wrote:

If you want to make your unit of length 0.3048 metres, then you may be a backwards provincial, but you’re still a perfectly reasonable one. But if you want to make your unit of length −0.3048 metres, then you’re just weird.

No - you’re just a truly backwards provincial.

Don’t forget, America is the land whose homespun hero Benjamin Franklin is responsible for making the electron charge be called -e.

Posted by: John Baez on October 6, 2006 7:11 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

It seems to me that it really is an arbitrary choice —one way no better than the other— whether electrons or positrons have “positive” charge; while the length of the platinum rod in Paris is obviously a positive quantity. Does this mean that the group consists of the invertible real numbers in one case but the positive real numbers in another?

Posted by: Toby Bartels on October 13, 2006 10:59 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Toby wrote:

Does this mean that the group consists of the invertible real numbers in one case but the positive real numbers in another?

Maybe so! But it’s a bit subtle: position coordinates can equally well be positive or negative, while distances really want to be positive - or at least, positive ones are really different from negative ones.

Maybe this wouldn’t be true in a universe with an equal number of space and time dimensions… but maybe that’s why we don’t live in such a universe, despite what some believers in (1+1)-dimensional conformal field theory seem to think.

Posted by: John Baez on October 20, 2006 10:23 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

robert bristow-johnson wrote in small part:

I like “rationalized” Planck units with c = ℏ = 4πG = ϵ0 = 1

I like “rationalized” Planck units too, except I use 8πG = 1! That’s the trouble with Planck units.

Posted by: Toby Bartels on October 6, 2006 1:16 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Moshe wrote to Jacques Distler:

I think you and me and Aaron all seem to agree that there exists an economical description of any physical law that makes no reference to dimensionful quantities […]

I agree with this too. I just thought it was too uncontroversial to be worth restating! This “economical description” is the one I normally use.

As for whether the discussion is “going round in circles”, opinions will differ. People who work solely with dimensionless quantities will soon lose interest in a conversation about dimensional analysis - just as a winetasting party gets dull mighty fast when you’re on the wagon. The other people aren’t repeating themselves much yet; when they do I’ll delete their entries.

But, I would indeed appreciate some comments on the new blog entry I just wrote! I’m trying to connect Terry Tao’s remarks on torsors and Urs Schreiber’s remarks on models to the Yoneda Lemma.

Posted by: John Baez on September 30, 2006 7:19 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Thanks John, that makes perfect sense. The wine looks delicious from here, but maybe I should stick to coffee this early in the morning…

Posted by: Moshe on September 30, 2006 7:31 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Jacques Distler wrote:

Now, you could ask whether some dimensionless ratios are more “fundamental” than others (some physically-important length-scale measured in units of the Bohr radius versus the same length, measured in units of the distance between two scratches in a bar of platinum).

That’s a matter of the choice of coordinates on your space of physical theories (as John would have it). Some coordinate choices are better than others …

I agree with everything you say here. As a point of interest: in my entry I talked about a few classic examples of coordinates:

• coordinates on the surface of the Earth
• coordinates on spacetime
• coordinates on a space of dimensions

but not coordinates on a space of theories. That’s another good example, not to be confused with coordinates on a space of dimensions.

People working with the renormalization group are familiar with coordinates on a space of theories. It’s possible some people have never thought about “coordinates on a space of dimensions”, so let me illustrate what that means.

Say we use a system where physical quantities have dimensions

${L}^{a}{T}^{b}{M}^{c}$

where $L$ is length, $T$ is time and $M$ mass. Then any dimensioned physical quantity defines a point $\left(a,b,c\right)\in {ℝ}^{3}$ - its dimensions. So, ${ℝ}^{3}$ is our space of dimensions. Any system of three independent units like “meter, second, kilogram” or “horsepower, fortnight, troy ounce” determines coordinates on this space.

A space of dimensions is an abelian group, and then its group algebra, e.g. $ℝ\left[{ℝ}^{3}\right]$ is the algebra of dimensionful quantities, a typical element being something like

$3ML/T+{M}^{2}{L}^{1.5}+1/137$

Physicists like to avoid adding quantities with different dimensions, which amounts to restricting attention to homogeneous elements of this ${ℝ}^{3}$-graded algebra. But, there’s no mathematical reason we have to do that.

The math gets a little more fun when we consider maps between spaces of dimensions. For example, Theo wrote:

Would it be sensible to instigate a campaign to have the “cycle” recognised as an SI base unit?!

Sensible in what sense of “sensible”? You won’t see me leading protest marches on behalf of that cause. As Arun points out, we have funny units like “moles” due to technological limitations - we’re not so good at counting atoms. There’s no reason like this to use “cycles”. If we start counting cycles as a unit, next Tim Silverman will want to count apples! Indeed, it’s already a well-known violation of dimensional analysis to “compare apples and oranges”.

But, your system of units is logically possible, so let’s think about it.

As certain overly clever people love to point out, there are at least two options. We can take the “cycle” as a new unit but decree it to be dimensionless, so it’s just another name for the dimensionless constant $2\pi$. Or, we can invent a new dimension called “cyclicity”, say, and let the cycle have dimensions of cyclicity. Let me focus on the latter option.

Let’s say the SI system has ${ℝ}^{7}$ as its space of dimensions, with coordinate axes called “length, mass, time, electrical current, amount of substance, temperature, and luminous intensity”. Then your new system has ${ℝ}^{8}$ as its space of dimensions, with a new coordinate axis called “cyclicity”. How do we convert from your system to the old one? First, we pick a group homomorphism

$f:{ℝ}^{8}\to {ℝ}^{7}$

which projects down to the first 7 coordinates. This is our way of making “cyclicity” become dimensionless.

Any group homomorphism gives a homomorphism of group algebras in a standard way, so we get a homomorphism

${f}_{*}:ℝ\left[{ℝ}^{8}\right]\to ℝ\left[{ℝ}^{7}\right]$

which lets us convert physical quantities in your system to quantities in the SI system.

However, this “standard” homomorphism - the one mathematicians all know about - is not the right one to use, because it maps “1 cycle” to the number 1. Instead we want a different one, which maps “1 cycle” to the number $2\pi$.

With this slight wrinkle we see that changing from a system with more units to one with fewer can still be seen as a “coordinate transformation” between spaces of units, but of the noninvertible sort that my entry emphasizes.

Posted by: John Baez on September 30, 2006 8:02 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

I can’t say I was too serious about the campaign for “cycle” as a base unit though I think I could easily jump on to the bandwagon promoting “sneezegust” to go on some standards list… what a great word!

Thanks for the other examples too Tim. I think the examples illustrate what I was trying to highlight (though I do not think I was doing it very well). Although, I would personally find it useful to write down apples/s and kg/apple if I was working with such a fruity problem; especially if I was trying to describe the problem to someone else. I guess it would be useful to drop the apples when doing many problems about apples. But no, I wouldn’t promote apples as an SI base unit :-)

However, the missing “apples” and “cycles” units are illustrations of how even those who work using units may often be “missing units”. So there is not so much difference between those working with/without units. Further this links in to Arun’s comment that the choice of base units can be seen to be a technological issue. How is it, mathematically/theoretically justifiable to have a special number to count atoms but not have a unit that highlights the concept of a cycle? I don’t think it is (happy for someone to explain otherwise); it is more of a practical issue. Which leads, sort of, into John’s main question (I hope!?) asking what are the fundamental mathematical/theoretical differences between different types of quantities, dimensionful or not?

And thanks John for picking up on the bit of my comment that I was intending should lead to relevance to this thread. If we did have (my clunky?) system with the current 7 base units + “cycle” how do we get back to (the slick?) system of just the 7 base units? So we have a homomorphism ${ℝ}^{8}\to {ℝ}^{7}$ so, for example, $m/\mathrm{cycle}↦m$ and $m↦m$. There is also the associated homomorphism of group algebras such that, for example, $7m/\mathrm{cycle}↦7m$ and $7m↦7m$.

[By the way, I am flummoxed (sorry I am being slow, clarification welcome) by John saying that the homomorphism he describes is not what we want and that “1 cycle” gets mapped to $2\pi$.]

So, let me get this straight, when doing such a transformation we reduce the complexity of our system of units (so it becomes more economical – people must be trying to save ink/pixels in writing the units ;-) ) but introduce possible ambiguities such as $6{s}^{-1}$; possibly meaning six apples per second or six bananas per second or 6 cycles per second (though in a practical sense we are generally not going to get particularly confused)?

However, when we have a statement of a wavelength such as 6m, meaning 6m/cycle, we are still encoding the information that it is 6m/cycle (if only in a description of the situation that we pass from one person looking at the problem to another, i.e. by saying it is a wavelength). Here we have just used mathematical legerdemain to achieve economy yet not achieved anything theoretically “fundamental”?

So we could also really confuse things, get rid of the “m” as well (via a homomorphism reducing the dimension of our space of units yet again) and just be left with the number 6 describing our wavelength. So this number 6 is a number without units yet it is dimensionful (it is still a torsor? # ) ). However, if we try to construct an isomorphism between a model # containing this quantity and another, without the units, we are missing the data (units) to specify what the 6 maps to and we would need information regarding the definition of the cycle as well as the meter? However, if we also had a 1.4 that represented a refractive index in this model we would have no problem mapping this into the alternative model as it is not a dimensionful quantity?

Hopefully this illustrates any misunderstandings I might have in regards to the current discussion and someone might do me the honour of correcting them!

Posted by: Theo on October 1, 2006 8:26 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Physicists like to avoid adding quantities with different dimensions…[but] there’s no mathematical reason we have to do that.

I came across an example where you might want to add different dimensions recently. In geometric measure theory in n-dimensions you can associate ‘r-volumes’ to members of a class of body (eg. finite unions of convex polytopes) where r ranges from 0 to n (and the 0-volume is the Euler characteristic). The r-volume can be thought of as having dimension Lr. So there’s a kind of total volume, V, which is the sum of the differently dimensioned r-volumes. This notion has a degree of coherence to it, for example V(A×B)=V(A)V(B) where × is the usual cartesian product. The r-volumes of different dimensions sort themselves out correctly in the product. (I hope this is correct, my complete knowledge of geometric measure theory can be summed up in a few lines.)

Posted by: Dan Piponi on October 2, 2006 7:31 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

John wrote in part:

Physicists like to avoid adding quantities with different dimensions, which amounts to restricting attention to homogeneous elements of this ℝ3-graded algebra.  But, there’s no mathematical reason we have to do that.

Conversely, the homogeneous elements do have special mathematical properties, so the physicists’ penchant for them, while not mathematically necessary, is still mathematically justifiable.

However, the possibility of using nonhomogeneous elements explains how one can mix coordinate transformations like x′ = x + sin x (which mathematicians insist on allowing, at least for values of x far from integer multiples of π/2) with dimensionful coordinates like Schwarzschild’s coordinates for his black-hole solution (at least with the dimensions that physicists prefer to think of their having). Not a pleasant thing to do (who would want to apply that transformation to those coordinates?), but mathematically tractable.

Posted by: Toby Bartels on October 6, 2006 1:50 PM | Permalink | Reply to this

### Isomorphisms of Models

I think we all essentially agree now on

- the general notion of coordinates as certain (iso)morphisms

- the specific notion of isomorphisms between physical space(time) and our models thereof, leading to the concept of units of length

John has given a detailed discussion of units of length from this point of view, above and here.

What we still need to discuss are

- what in fact do we regard as an isomorphism of physical models # (going beyond mere isomorphisms related to physical space)

- in order to get a handle on non-length-like quantities.

As a simple example for this issue, I offered the class of FRW-models #.

In particular, we need to talk about how $\frac{1}{\hslash }\phantom{\rule{thinmathspace}{0ex}},$ to be thought of as arc length per action, establishes an identification of $ℝ$ with ${S}^{1}$.

I propose we solve the following

Exercise. A physical model for the uncharged spinless quantum particle propagating on a Riemannian (say) spacetime

(1)$\left(X,g\right)$

is a triple

(2)$\left(A,H,\Delta \right)$

consisting of a Hilbert space $H$ (of square integrable complex functions on $\left(X,g\right)$), an algebra $A$ (of smooth functions on $X$) and an operator $\Delta$ which is proportional to the Laplace operator on $\left(X,g\right)$.

1) describe the category of all such triples

2) identify the isomorphisms

3) identify the parameters characterizing the isomorphism classes of such models

4) identify which aspect of the model $\left(H,A,\Delta \right)$ (and in particular of our “coordinatizations” of it) gives rise to

$\phantom{\rule{thickmathspace}{0ex}}$ - a unit of length on $\left(X,g\right)$

$\phantom{\rule{thickmathspace}{0ex}}$ - the mass $m$ of the particle described by the model

$\phantom{\rule{thickmathspace}{0ex}}$ - the value of Planck’s constant $\hslash$ in this model.

Posted by: urs on October 1, 2006 1:35 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

Urs wrote:

I propose we solve the following Exercise

Sounds fun! You say an object of your category is:

a triple

(1)$\left(A,H,\Delta \right)$

consisting of a Hilbert space $H$ (of square integrable complex functions on $\left(X,g\right)$), an algebra $A$ (of smooth functions on $X$) and an operator $\Delta$ which is proportional to the Laplace operator on $\left(X,g\right)$.

Let’s make this a tiny bit more precise.

One nice category would consist of triples $\left(A,H,\Delta \right)$ where $A$ is a commutative C*-algebra of operators on a Hilbert space $H$ and $\Delta$ is a nonnegative self-adjoint operator on $H$.

I claim there’s an obvious notion of isomorphism of such triples: a C*-algebra isomorphism

$F:A\to A\prime$

and a unitary operator

$U:H\to H\prime$

such that

$Ua{U}^{-1}=F\left(a\right)$

for all $a\in A$, and

$U\Delta {U}^{-1}=\Delta \prime .$

This may not be the ultimate correct notion, but let’s use it for now and get a category that way - in fact a groupoid. Let’s call this groupoid $Y$.

In the case of a spinless particle of mass $m$ on a Riemannian manifold $\left(M,g\right)$, $A$ would be the algebra of bounded continuous functions on $M$, $H$ would be ${L}^{2}\left(M\right)$, and $\Delta$ would be ${\hslash }^{2}/2m$ times the (positive) Laplacian on $M$.

So, there’s another groupoid lurking around, where the objects are triples $\left(M,g,m\right)$ and the isomorphisms are isometric diffeomorphisms

$f:M\to M\prime$

together with equations

$m=m\prime .$

Let’s call this groupoid $X$.

The construction I just mentioned gives a functor

$F:X\to Y$

However - and surely this is part of Urs’ point - there’s another groupoid $X\prime$ that’s a bit like $X$, but more flexible. Here isomorphisms can rescale the metric while simultaneously rescaling the mass in such a way that ${\hslash }^{2}/2m$ times the Laplacian remains unchanged! We get a functor

$F\prime :X\prime \to Y$

and this is “better” than $F$ in a very precise sense: $F$ factors through it! In other words, we have an “inclusion” functor

$i:X\to X\prime$

such that

$F=F\prime i.$

Is this at all what you were thinking about, Urs?

Posted by: John Baez on October 2, 2006 4:10 AM | Permalink | Reply to this

### Re: Isomorphisms of Models

John wrote:

Urs wrote:

I propose we solve the following Exercise

Sounds fun! […]

I should maybe emphasize that I am not trying to tease the n-Café readers with an exercise which I think I have solved already. While I believe I know some aspects of the answer, I certainly haven’t taken the time to work out the full answer. Instead, I thought it would be fun to do so here in the Café together.

Let’s make this a tiny bit more precise.

Thanks, yes. As you say, maybe later on we will want to be more sophisticated about morphisms. But for the moment unitary transformations are certainly sufficient to get a feeling for what is going on.

In the case of a spinless particle of mass $m$ on a Riemannian manifold $\left(M,g\right)$, $A$ would be the algebra of bounded continuous functions on $M$, $H$ would be ${L}^{2}\left(M\right)$, […]

I have nothing to add to this part. But I will offer a different perspective on

[…] and $\Delta$ would be ${\hslash }^{2}/2m$ times the (positive) Laplacian on $M$.

This is the part where units and dimensions enter the game. My main intention with this little exercise was to examine how exactly units enter our description when we have an abstract quatum mechanical model of our world.

I shall argue now that it is not correct to say that $\Delta$ is ${\hslash }^{2}/2m$ times the Laplacian - but that instead $\hslash /m$ appears as an isomorphism which identifies the real $\Delta$ with our standard model for it.

I’ll denote the Laplacian on $M$ by ${\Delta }_{0}$. This is certainly an operator defined on our Hilbert space $H={L}^{2}\left(M\right)$.

But what is $\frac{{\hslash }^{2}}{2m}{\Delta }_{0}$?

Is this still an operator on $H$? No. Given any operator on $H$, I can multiply it with a real number and obtain another operator. But what does it mean to multiply an operator by an element not of $ℝ$, but of an $ℝ$-torsor? This is not a well defined operation.

(Or so I think. Everybody is kindly invited to diagree with me. But let’s beware of identifying $\frac{{\hslash }^{2}}{2m}$ with any dimensionless ratio obtained by comparing it to some other torsor element.)

Another way to see that we need to be careful is to spell out explicitly why we (driven by our standard QM textbooks) want to multiply ${\Delta }_{0}$ by $\frac{{\hslash }^{2}}{2m}$.

The reason is that our standard QM textbook makes us want to associate to

(1)${\Delta }_{0}=-{g}^{\mathrm{ij}}\frac{\partial }{\partial {x}^{i}}\frac{\partial }{\partial {x}^{j}}+\cdots$

the dimension one over length squared.

We tend to want to multiply hence by ${\hslash }^{2}/2m$ in order to get a quantity whose dimension is energy, such that when we further multiply by $t/\hslash$, we finally get something dimensionless.

But wait a moment. Where did all these dimensions come in suddenly? Didn’t we just talk very abstractly about a category of triples $\left(A,H,\Delta \right)$? All we need to interpret these triples is some functional analysis. Every mathematician - even of the sort who never have heard of dimensionful quantities, much less of Planck’s constant - could handle our category of triples - without ever using dimensional analysis.

I guess it is clear now what I am going to argue: I will now argue that all this $\frac{{\hslash }^{2}}{2m}$-business is a way to describe not the objects in our category of triples - but certain isomorphisms of triples.

In case you are worried, let me reassure you that I am not going to argue that there is no fundamental scale whatsoever (defined for instance by Planck’s unit, or the particle’s mass) in the game.

But instead of talking about what I am not going to argue, here is what I actually do want to say:

What do we really mean when we associate dimension $1/{L}^{2}$ to ${\Delta }_{0}$? Well, we had solved that part before.

In order to put a unit of length on our abstract Riemannian manifold $\left(X,G\right)$ of dimension $d$, we choose a monomorphism

(2)$e:{ℝ}^{d}\to TX$

such that it restricts to an isomorphism on each tangent space

(3)$e\left(x\right):{ℝ}^{d}\to {T}_{x}X\phantom{\rule{thinmathspace}{0ex}}.$

In other words, we choose $d$ linearly independent sections

(4)${e}_{1},{e}_{2},\cdots {e}_{d}$

of the tangent bundle of $X$ (which are to be thought of as the images of the canonical unit vectors in ${ℝ}^{d}$).

As usual, in terms of these the metric $g$ of the Riemannian manifold $\left(X,g\right)$ shall be denoted by $\eta$.

Using this basis, we can now write our Laplace operator ${\Delta }_{0}$ as

(5)${\Delta }_{0}=-\sum _{a,b=1}^{d}{\eta }^{\mathrm{ab}}{e}_{a}{e}_{b}+\cdots \phantom{\rule{thinmathspace}{0ex}}.$

Since $e$ is our choice of units on $TX$, this expression now is manifestly unit dependent. A different choice of $e$ will lead to a different choice of the coefficient ${\eta }^{\mathrm{ab}}$.

So we think of each $e$ as “being of dimension $1/L$”. Then ${\Delta }_{0}$ is “of dimension $1/{L}^{2}$”.

That’s space. Next we need to understand time (in our context of non-relativistic quantum mechanics).

In order to see how that works properly, it is best to consider, for a moment, something slightly more general than just our triples $\left(A,H,\Delta \right)$.

Namely, these triples are really already chosen isomorphism classes of an even more abstract concept, which is really at the heart of QM.

Namely, what we are really interested in are functors from 1-dimensional Riemannian manifolds to Hilbert spaces (which are equipped with a module structure for a ${C}^{*}$-algebra $A$).

For simplicity, let’s think of a fixed 1-dimensional manifold $W$ - called the worldline of our quantum particle.

The true nature of our Hamiltonian $\Delta$ is really something like a $\mathrm{End}\left(H={L}^{2}\left(M\right)\right)$-valued connection on $W$. Namely, given any such connection, we obtain a “propagator”, namely a functor from 1-dimensional cobordisms to Hilbert spaces by evaluating the parallel transport of that connection.

So in other words, the Hamiltonian comes from a morphism

(6)$TW\to \mathrm{End}\left(H\right)\phantom{\rule{thinmathspace}{0ex}}.$

We are however here for the moment only interested in a special case of all possible such morphisms, namely those which correspond to particles in background spaces which are in a “geometric phase”. Namely in those Hamiltonians which take values in multiples of the Laplace operator ${\Delta }_{0}$ of some $\left(X,g\right)$.

Since all these operators look like

(7)${\Delta }_{0}=-{g}^{\mathrm{ij}}{\partial }_{i}{\partial }_{j}+\cdots$

we are really only looking at Hamiltonians that come from morphisms

(8)$TW\to {T}^{\otimes 2}X\phantom{\rule{thinmathspace}{0ex}}.$

Let’s consider some such morphism. Call it $\omega$.

We are not content with just calling it $\omega$, however. We tend to want to know what $\omega$ “looks like”. By this we tend to mean how $\omega$ looks like in our chosen basis of ${T}^{\otimes 2}X$ and ${T}^{*}W$.

We already have a basis of ${T}^{\otimes 2}X$, induced from our choice $e$ above.

What we still need is a basis of ${T}^{*}W$. Usually we choose this basis by first choosing a “nonlinear isomorphism”

(9)$t:W\to ℝ\phantom{\rule{thinmathspace}{0ex}},$

known as a time coordinate on $W$. Usually, we pick the section of ${T}^{*}W$ given by

(10)$dt\phantom{\rule{thinmathspace}{0ex}}.$

With these choices made, our $\mathrm{End}\left(H\right)$-valued connection 1-form will now look like

(11)$c\phantom{\rule{thickmathspace}{0ex}}dt\otimes {\eta }^{ab}{e}_{a}{e}_{b}+\cdots \phantom{\rule{thinmathspace}{0ex}}.$

All the freedom (in this first term) is, after we have chosen bases, encoded in the number $c$.

We usually say that $dt$ is of dimension $T$. Recalling that ${e}_{a}$ is of dimension $1/L$, we find that $c$ must transform like something of dimension

(12)$\left[c\right]=\frac{{L}^{2}}{T}$

such that the connection 1-form itself is independent of how we describe it in a given basis - which certainly it is.

What is $c$?

Well, let me just give $c$ a different name. I’ll now call it

(13)$c=\frac{i}{2}\frac{\hslash }{m}\phantom{\rule{thinmathspace}{0ex}}.$

At this point, this is just a choice of notation. We found that $c$ has dimension ${L}^{2}/T$, hence also

(14)$\left[\hslash /m\right]=\frac{{L}^{2}}{T}\phantom{\rule{thinmathspace}{0ex}}.$

Actually, I smuggled in one assumption: $c$ could in principle depend on $t$. If so, we’d get time dependent quantum mechanics. Here we certainly first want to understand time-independent Hamiltonians.

With all these conventions in place, our connection 1-form reads now

(15)$i\frac{\hslash }{2m}\left(dt\right){\eta }^{ab}{e}_{a}{e}_{b}\phantom{\rule{thinmathspace}{0ex}}.$

We find the desired propagator by taking the path-ordered exponential

(16)$\begin{array}{rl}U\left({t}_{0},{t}_{1}\right)& =P\mathrm{exp}\left({\int }_{{t}_{0}}^{{t}_{1}}i\frac{\hslash }{2m}{\eta }^{\mathrm{ab}}{e}_{a}{e}_{b}\phantom{\rule{thickmathspace}{0ex}}dt+\cdots \right)\\ & =\mathrm{exp}\left(i\left({t}_{1}-{t}_{0}\right)\frac{\hslash }{2m}{\eta }^{\mathrm{ab}}{e}_{a}{e}_{b}+\cdots \right)\\ & =\mathrm{exp}\left(i\frac{1}{\hslash }\left({t}_{1}-{t}_{0}\right)\frac{{\hslash }^{2}}{2m}{\eta }^{\mathrm{ab}}{e}_{a}{e}_{b}+\cdots \right)\\ & \stackrel{X={ℝ}^{d}}{=}\mathrm{exp}\left(i\frac{1}{\hslash }\left({t}_{1}-{t}_{0}\right)\frac{{\hslash }^{2}}{2m}{\eta }^{\mathrm{ab}}{\partial }_{a}{\partial }_{b}\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

Here the last line gives the result for a particle in flat space, which is the form most easily compared to QM textbook notation.

This entire detour through propagation functors on worldlines was necessary to clearly see where the dimension of time enters the game. In terms of our triples, the above reads simply

(17)$U\left({t}_{0},{t}_{1}\right)=\mathrm{exp}\left(i\left({t}_{1}-{t}_{0}\right)\Delta \right)\phantom{\rule{thinmathspace}{0ex}}.$

The exponent is just an operator on $H$. It cannot be anything else. With our assumptions, there is a 1-parameter familily of such operators (namley all multiples of the Laplace operator on $\left(X,g\right)$).

All the dimensionful constants, like $\hslash$ and $m$, enter when we express this operator in terms of certain chosen isomorphisms.

And they all combine to describe for us a single number - which governs the ismorphism classes of our triples, namely the number which in our chosen basis is given by the quantity $c$, above, which encodes by which multiple our $\Delta$ differs from ${\Delta }_{0}$.

Posted by: urs on October 2, 2006 5:14 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

My comment above was too long. Nobody who has anything else to do will find the time to read it.

I should summarize what I said:

I considered the propagator $U\left({t}_{0},{t}_{1}\right)=\mathrm{exp}\left(i\left({t}_{1}-{t}_{0}\right)\Delta \right)$ of our particle.

I argued that $\Delta$ itself cannot carry units, but that units enter as soon as we express $\Delta$ with respect to a couple of chosen bases.

I argued that the dimensionful constant which is the “component of $\Delta$ in the bases defined by our units of length and time” is essentially

(1)$\hslash /m$

times the component of the metric on $X$ in the given basis.

In other words, $\hslash /m$ here is the fundamental dimensionful constant which tells us - with respect to our chosen basis - which multiple of the standard Laplace operator our Hamiltonian is.

Posted by: urs on October 4, 2006 1:12 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

Nothing of substance to add at the moment, but I’d very much like to see what category theory can do for a problem such as this, or this one. It would be very exciting for the philosophy of science if good things could be achieved.

Posted by: David Corfield on October 4, 2006 6:28 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

It would be very exciting for the philosophy of science if good things could be achieved.

There are a couple of straightforward things one could do next - if one were not heavily absorbed by other tasks.

By simply replacing Riemannian target space $\left(X,g\right)$ by pseudo-Riemannian spacetime, and replacing our former time variable by an affine worldline parameter, the discussion of the single spinless particle above immediately generalizes to the relativistic case - where our particle is then known as the Klein-Gordon particle.

Without much ado, we get the fundamental dimensionful quantity

(1)$c$

this way - the speed of light - simply by measuring the slope of a null vector at any point in terms of our chosen basis ${e}_{a}$ for the spacelike and the timelike tangent space.

In a given basis, our Klein-Gordon operator reads

(2)$\Delta ={K}^{2}{\eta }^{\mathrm{ab}}{e}_{a}{e}_{b}+\cdots$

as before. Hence $K$ transforms opposite to the basis $\left\{{e}_{a}\right\}$ and hence is said to have unit of length

(3)$\left[K\right]=L\phantom{\rule{thinmathspace}{0ex}}.$

Simply renaming this, we write

(4)$K=\frac{\hslash \prime }{m}\phantom{\rule{thinmathspace}{0ex}}.$

By taking the nonrelativistic limit to obtain the situation we had considered before, we find the our symbols $\hslash \prime$ and $\hslash$ are related by

(5)$\hslash \prime =\hslash /c\phantom{\rule{thinmathspace}{0ex}},$

hence the component of $\Delta$ in our basis ${e}_{a}$ is

(6)${\left(\frac{\hslash }{mc}\right)}^{2}{\eta }^{\mathrm{ab}}\phantom{\rule{thinmathspace}{0ex}}.$

So we find that the system is governed by a fundamental dimensionful constant whose dimension we call “length” - the Compton wavelength $2\pi \frac{\hslash }{\mathrm{mc}}$. We explicitly find this length as a sphere of vectors in each tangent space by multiplying the numerical value of $K$ with our chosen unit of length, given by the basis $\left\{{e}_{a}\right\}$.

Moreover, it is also straightforward to take the particle to be electrically charged, simply by replacing the Laplace operator in the previous case by the “covariant” Laplace operator with respect to some line bundle with connection.

Where before we had a 1-parameter family of operators (all multiples of the ordinary Laplace operator) we will now get a 2-parameter family, where the second parameter measures with which weight the connection modifies the original Laplace operator.

Where the first parameter gave us $\hslash /m$, the second parameter will give us the charge

(7)$e$

of our particle - in a completely analogous fashion, I think.

So now we have three fundamental dimensionful constants

(8)$\left\{\frac{\hslash }{m},\phantom{\rule{thinmathspace}{0ex}}e,\phantom{\rule{thinmathspace}{0ex}},c\right\}\phantom{\rule{thinmathspace}{0ex}}.$

The third doesn’t really depend on which isomorphism class of triples $\left(A,H,\Delta \right)$ we are looking it. It just tells us something about our choice of sections of the tangent bundle of the pseudo-Riemannian background spacetime.

The first two quantities, however, characterize (with respect to our chosen bases) isomorphism classes of triples $\left(A,H,\Delta \right)$, namely unitarily inequivalent choices of $\Delta$.

If we now also considered interactions, we’d be able to address the fine structure constant from this point of view.

Posted by: urs on October 4, 2006 8:42 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

David Corfield:

It would be very exciting for the philosophy of science if good things could be achieved.

I’m biased, but I think we’ve already achieved a lot in this thread. Needless to say, we won’t bother publishing it in a journal - the philosophers of science will just have to read this blog.

Posted by: John Baez on October 6, 2006 7:45 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

Urs wrote in small part:

Given any operator on H, I can multiply it with a real number and obtain another operator. But what does it mean to multiply an operator by an element not of ℝ, but of an ℝ-torsor? This is not a well defined operation.

It’s well defined, but the result is not an operator on H.

Given a field K, a linear K-space V, and a K-torsor T, there is linear K-space T ⊗ V, the space of T-twisted vectors. If you multiply an element of T by an element of V, then the result is an element of T ⊗ V. If V is the space of operators on some linear K-space H, then an element of T ⊗ V may act on an element of H to produce an element of T ⊗ H.

For H the Hilbert space of quantum states, then T ⊗ H is the T-twisted Hilbert space of T-twisted quantum states. T ⊗ H is a T-twisted Hilbert space, rather than simply a Hilbert space, because the inner product is valued in T ⊗ T ⊗ K, rather than in K itself.

Posted by: Toby Bartels on October 6, 2006 2:29 PM | Permalink | Reply to this

### Re: Isomorphisms of Models

It’s well defined, but the result is not an operator on $H$.

Agreed. You can define it, but not while staying within operators on $H$ (which we have to, in the application under discussion).

Posted by: urs on October 6, 2006 3:08 PM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

Urs wrote:

- the mass m of the particle described by the model

I’m confused. Do you mean a unit of mass? It looks like the mass of the particle would be a free parameter.

Christine
PS- Thanks to John Baez for annoying me back in the other dimensional analysis post. I feel happy when people annoy me like that! :) Thanks for the thought-provoking posts.

Posted by: Christine on October 2, 2006 3:25 AM | Permalink | Reply to this

### Re: Dimensional Analysis and Coordinate Systems

It looks like the mass of the particle would be a free parameter.

Yes, that’s right. The question is: given just abstract data of a model of physics, like a Hilbert space, a self-adjoint operator on it, and some algebra represented on it, how do you read off the mass of the particle whose propagation is described by this data? Or, if you find you cannot read it off, what else do you have to do such that you can?

Posted by: urs on October 2, 2006 5:24 PM | Permalink | Reply to this

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