Are Christoffel symbols measurable?


by waterfall
Tags: christoffel, measurable, symbols
twofish-quant
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#109
Feb20-12, 10:01 PM
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Quote Quote by PAllen View Post
All measurements, even over time, constitute a finite amount of information.
I worry about statements like this. I don't have too much problem with the statement that "all measurements that physicists are used to making have characteristic X." But one of the points that I've been making is that there is a big difference between "measurements that physicists are used to making" and "all measurements."

You could argue that "all measurements can be reduced to all measurements that physicists are used to making." That's an extremely strong philosophical statement, and one that I'm inclined to consider to be false. There's one famous counter example called the "marriage meter." There is no set of physical measurements that you could make on me or my wife that could tell you whether or not we are married. By doing some sort of brain scan, you could establish that we *think* we are married. Since there is no such thing as a "marriage meter" then this means that things like marriage/divorce rates aren't physically measurable and you can put these things in vector spaces.

I run into this sort of thing all the time at work. Two of the big, big questions write now is "how do you measure liquidity?" and "how do you measure risk?" Which quickly gets you into some philosophical questions "what is liquidity?" and "what is risk?" The relevance of this to the current discussion is that it seems that whatever liquidity and risk are, they somehow involve rather complicated vector spaces and the same sort of math that you find in GR. (Correlation matrices from hell.)

In no sense I know of, can any collection of measurements be considered to constitute a vector field.
Color. Color requires three components to be specified, but color is independent of those components. You can specific color in terms of RGB, or CMYK or pantone or color temperature, but color is independent of those measurement. Because you need multiple components to specific color, and the existence of color is *independent* of those components, its a vector field, and more than a collection of measurements.

Stock portfolios have similar issues. There are multiple equivalent ways of representing the dynamics of stock portfolio, but the dynamics exists independent of those representations.

Vector spaces and the math associated with it comes in very handy when you have an "underlying reality" that's independent of the measurements. Relativity is one such example, but it's not the only one.

Any vector field you might associate with measurements is an abstraction. Thus, in my view no vector field is directly observable.
But you could argue that scalars are an abstraction. I mean when I measure light intensity, it goes into a meter that goes through my eyes into my brain where who knows what happens. The problem with saying that no vector field is directly observable is that you end up with a very restrictive definition of "observe" under which it's not clear that anything is observable.

There might be a physics reason to do this. In QM, to observe means to "collapse the wavefunction."

Thus any expression you give to a vector field has a significant contribution due to arbitrary convention.
True, but vectors are useful especially in cases where there is a "reality" that is independent of arbitrary convention. GR is one such use case but there are others.
twofish-quant
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#110
Feb20-12, 10:09 PM
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The other thing to be careful here is "proof by lack of imagination." In order to prove non-existence, you have to show that something really bad happens if something did exist. I can think of a lot of bad things that would happen if you had physical measurements that were none Lorenz invariant, or if quantum observations didn't reduce to a single number.

However, asserting that something is impossible because one can't think of counterexamples is a bad way of showing that something is impossible.

This is particularly true because vector spaces are really useful, and can represent things that are pure fantasy (i.e. any first person shooter video game has vector space representations of all sorts of imaginary things).
atyy
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#111
Feb20-12, 10:18 PM
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Quote Quote by twofish-quant View Post
I worry about statements like this. I don't have too much problem with the statement that "all measurements that physicists are used to making have characteristic X."
Proof for the modern age: it's stored on digital media.
Ben Niehoff
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#112
Feb21-12, 12:59 AM
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Quote Quote by twofish-quant View Post
The other thing to be careful here is "proof by lack of imagination." In order to prove non-existence, you have to show that something really bad happens if something did exist. I can think of a lot of bad things that would happen if you had physical measurements that were none Lorenz invariant, or if quantum observations didn't reduce to a single number.

However, asserting that something is impossible because one can't think of counterexamples is a bad way of showing that something is impossible.
Throughout this thread, we have been discussing the fact that all measured quantities must be scalars with respect to coordinate transformations of spacetime, a context to which none of your examples has been relevant.

However, this goes beyond merely spacetime scalars. Any measurement of a vector quantity (in the vector space sense, not in the computer science sense) must come by choosing a basis and projecting out components, thereby measuring a set of scalars. If you think about it for a moment, you will see that this statement is a trivial tautology. What I'm really saying here is that every measurement is a comparison.

After all, that is what we really mean, isn't it? When we say something is "2 meters", all we're saying is it's twice as long as a certain metal bar in France. (Or, in modern SI, it's twice the distance light travels during so many oscillations of Cesium 133.)

Color. Color requires three components to be specified, but color is independent of those components. You can specific color in terms of RGB, or CMYK or pantone or color temperature, but color is independent of those measurement. Because you need multiple components to specific color, and the existence of color is *independent* of those components, its a vector field, and more than a collection of measurements.
RGB, HSB, CMYK, and temperature are all coordinates on color space. They emphatically do not obey the linear transformation law and axioms of a vector basis. In fact, I'm not convinced color space is a vector space at all, if we mean the color space that is relevant to "perceived color". (Of course, we can define color spaces that are vector spaces, but these might have nothing to do with actual color perception).

You could probably geometrize the idea of color space if you like, and make it a manifold, possibly with notions of parallel transport. Who knows, maybe there's a useful way to model some psycho-physical process using a color space bundle over spacetime.

At any rate, "color" is measured by first comparing incoming light against certain bands of frequency; i.e., taking an inner product in frequency space against certain basis vectors in order to form a collection of scalars. These scalars can then be used as a coordinate system on color space. This method works for RGB, CMYK, and color temperature type coordinates. Mapping physical observables to HSB coordinates is more complicated, and will require projecting the incoming spectrum onto several bands and doing some analysis with the results.

Since frequency space is infinite-dimensional, there is no real reason for color space to be 3-dimensional; the ultimate reason is that we have 3 kinds of color receptors in our eyes (which project the spectrum onto 3 bands). Other animals have 2-dimensional, or sometimes 4-dimensional color spaces.
Ben Niehoff
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#113
Feb21-12, 01:25 AM
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Quote Quote by twofish-quant View Post
But we can stretch this into some absurd conclusions.

I take the a precinct-by-precinct map of the United States containing the election results of the Republican primary in 1980. The candidate votes form a vector and it's s perfectly good vector field. I can also form a vector field containing things like the price of real estate of different types of houses, the probability of default, divorce rates, crime statistics, etc. etc.

All of those are perfectly good vector fields.
No they are not. You can't just take a collection of data and call it a vector. This isn't computer science, it's geometry. Vector spaces have to obey certain axioms; you have to show how those axioms are physically reasonable if you want to say some physical quantity is modeled by a vector space.

Just thought of something ridiculous. Restaurant and movie reviews. I go on yelp.com or rottentomatoes.com. Restaurant rates form a vector (i.e. atmosphere, decor, service, etc.)
Nope, that's a collection of data, not a vector.

You can do movie reviews the same way (quality of plot, amount of action, quality of print, etc.)
Same problem.

So you are saying that it is *physically impossible* for me to measure restaurant atmosphere and service at the same time???!!!!
Nobody said you couldn't measure more than one scalar at a time. This isn't quantum mechanics.

In fact, I think I specifically mentioned that a collection of scalars can be used to construct a tensor quantity in the basis used to measure all the scalars. So, e.g., I can use a set of (scalar) measurements to deduce all 6 components of E and B in whatever basis I choose. The point is what constitutes a measurement.
twofish-quant
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#114
Feb22-12, 07:48 AM
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Quote Quote by Ben Niehoff View Post
Vector spaces have to obey certain axioms; you have to show how those axioms are physically reasonable if you want to say some physical quantity is modeled by a vector space.
And what axioms are violated by having election results represented by a vector? It seems to me to be a perfectly good vector space. All you have to do is to define addition and multiplication operations and you are done.

Axioms are axioms. I don't understand how "physically reasonable axioms" is a grammatical statement. Is "addition" physically reasonable? Now once you've defined addition and multiplication, you can then use them to make statements that are physically true or false. But the fact that you can make a statement that "the scalar multiplication of "red" by 2 gets us outside the set of physically valid colors" means that the axioms are defined.

And it's also a *useful* vector space. Once you've defined the operations, then you can define a "norm" which then describes how "close" two elections are. You can then do matrix transformations from one set of coordinates to another.

I define a C++ class RGB color, I define the operations of 2*Color and Color A + Color B. Once I've defined those operations, it's a vector space. I can even start do to tensor algebra.
twofish-quant
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#115
Feb22-12, 08:00 AM
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Quote Quote by Ben Niehoff View Post
What I'm really saying here is that every measurement is a comparison.
Which is an extremely strong statement that I'm not sure is true once you go outside the range of physical measurements that physicists are used to. Just to give a trivial example, how do you measure "bank liquidity"? If you state "physical measurement" then I don't object.

RGB, HSB, CMYK, and temperature are all coordinates on color space. They emphatically do not obey the linear transformation law and axioms of a vector basis.
All you need for a vector space to exist is for the transformations to be defined, and I can clearly define a set of vector addition and scalar multiplication operations for RGB numbers. Now whether I get something *physically* meaningful if I perform those operations is another issue.

RGB numbers are physically bound within a range, but if I measure x, y, z coordinates of the earth, there are some values which are invalid.

Also there *isn't* much of a difference between the mathematical concept of a vector space and the computer science one. All mathematics requires is that you have a defined addition and multiplication that has eight axioms. Once you have a collection of data for which you do that, then you have a vector space.
JDoolin
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#116
Feb22-12, 09:18 AM
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I'm not sure if this makes any difference to your argument, but in this "election-result-space" do you mean that the space is the discrete set of electoral districts, or do you mean the space is a continuous geographical/political map of the region?

Also, the "election-results" are vector-like, in the sense that they are multi-valued arrays, but they are non-vector-like in the sense that they do not have any direction. i.e. they don't point from one district toward another.
IsometricPion
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#117
Feb23-12, 01:21 AM
P: 177
Quote Quote by twofish-quant View Post
Which is an extremely strong statement that I'm not sure is true once you go outside the range of physical measurements that physicists are used to. Just to give a trivial example, how do you measure "bank liquidity"?
Suppose one has a collection of numbers arranged into categories. If these numbers are to represent quantitative data, then for each category there must be a unit reference against which all data in that category are compared (if two numbers do not share the same unit reference they cannot be immediately compared numerically). Therefore, in order to obtain any data representable as numbers or collections of categories thereof one must compare the quantity(ies) of interest to a reference value (or set thereof in the case of multiple categories). Thus, once a tuple of numbers and corresponding units are given the values of the quantities measured have been specified and any other description of the same values must yield the same (tuple of) numbers when converted into those units (and arranged by category into the same order in the tuple).

The measurement of a tuple of values either entails many different comparisions to unit references, and thus many measurements, or the the comparison of a smaller number of measurements to tuples of unit references. In any case, each comparison must yield precisely one number. It is the combination of this number and the associated unit tuple that represents the measured value. No matter what mappings are done on the collection of numbers, if the appropriate inverse mappings are done such that the necessary unit tuples map back to themselves the measured values must be represented by the same numbers (this places restrictions on what is considered a valid mapping and/or the types of valid measurements (in the context of this thread I prefer to think of it as the former rather than the latter)).

It is in this sense that all measurable quantities are collections of single numbers obtained from single measurements.

If the units involve space or time references, then they must pick out a set of vectors in space-time. Any such set of vectors can be used to construct a tensor of appropriate rank such that after any transformation, the evaluation of the resulting tensor on the images of the aforementioned set of vectors yields the same number as obtained originally (since one requires covarient tensors and vectors to transform in such a way that this is true).

Sorry if the vagueness (or triviality) of the above was excessive, I was going for an abstract approach but may have overreached the bounds of my knowledge and/or conventional nomenclature.
twofish-quant
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#118
Feb23-12, 02:27 AM
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Quote Quote by IsometricPion View Post
Therefore, in order to obtain any data representable as numbers or collections of categories thereof one must compare the quantity(ies) of interest to a reference value (or set thereof in the case of multiple categories).
No. I just thought of a counter example. Twenty questions.

I'm located in a spot on the earth. By asking me yes-no questions, you can figure out my latitude and longitude. Am I on land? Yes. Do I see taxicabs? Yes. Are they green? No. Are they yellow? No. Do I see water? Yes.

With each question, you can eliminate parts of the vector space. The fact that I see tax cabs and they are not yellow, means that I'm not in Manhattan. Now if you can ask enough questions, you can figure out my location and convert to GPS coordinates.

Note that you've figured out my GPS coordinates without actually measuring my latitude and longitude or doing any reference comparisons at all. You can show that no reference comparisons were done, because you can play this game without knowing anything about latitude and longitude at all, and it's the same game that you can play with things that are *not* vector spaces (i.e. words in a dictionary).

One other way of thinking about it is that you can specify points in a vector space as the interaction of subsets of that vector space, which allows you to specify a point in that space without reference to basis vectors at all.
twofish-quant
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#119
Feb23-12, 02:42 AM
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Quote Quote by JDoolin View Post
I'm not sure if this makes any difference to your argument, but in this "election-result-space" do you mean that the space is the discrete set of electoral districts, or do you mean the space is a continuous geographical/political map of the region?
I suppose it depends on the data. It's pretty easy to fit anything into a vector space. One thing about vector spaces is that scalars don't have to be real and neither do vectors. Election *results* certainly form a vector space (since you can add and scale vote totals). It's not obvious to me how to represent discrete electoral districts in a vector space, but that's just due to my lack of imagination.

Also, the "election-results" are vector-like, in the sense that they are multi-valued arrays, but they are non-vector-like in the sense that they do not have any direction. i.e. they don't point from one district toward another.
Again this depends on the structure of the data, but my point is that if you consider anything other than relativistic vectors to be "bags of unconnected data" you lose the structural information about the data. If you treat everything as "bags of data" you lose type information, which is a bad thing.

One thing that got me started thinking along these lines is the fact that you can call functions in C++ "covariant" and "contravariant". So what does tensor calculus have to do with C++. Well, that got me into the world of category theory.......
DaleSpam
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#120
Feb23-12, 09:25 AM
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Quote Quote by twofish-quant View Post
Election *results* certainly form a vector space (since you can add and scale vote totals).
They certainly do not. One of the requirements of a vector space is that there must be an operation where multiplication of a vector by a real number* leads to another vector. If you multiply an arbitrary election result by any negative number or by any irrational number you will get negative or fractional votes, neither of which are members of the space of possible election results.

You cannot scale election results by arbitrary real numbers, nor even by arbitrary integers.

*Vectors can be generalized to multiplication over other fields besides the real numbers, but the conclusion remains. There is an additive identity element, but no additive inverse in the space of election results.
IsometricPion
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#121
Feb23-12, 10:41 AM
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Quote Quote by twofish-quant View Post
Note that you've figured out my GPS coordinates without actually measuring my latitude and longitude or doing any reference comparisons at all. You can show that no reference comparisons were done, because you can play this game without knowing anything about latitude and longitude at all, and it's the same game that you can play with things that are *not* vector spaces (i.e. words in a dictionary).
I don't think any of my assertions would be altered if I replace "unit reference" with something like "way of mapping a number to a measured value".
JDoolin
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#122
Feb23-12, 04:38 PM
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Quote Quote by DaleSpam View Post
They certainly do not. One of the requirements of a vector space is that there must be an operation where multiplication of a vector by a real number* leads to another vector. If you multiply an arbitrary election result by any negative number or by any irrational number you will get negative or fractional votes, neither of which are members of the space of possible election results.

You cannot scale election results by arbitrary real numbers, nor even by arbitrary integers.

*Vectors can be generalized to multiplication over other fields besides the real numbers, but the conclusion remains. There is an additive identity element, but no additive inverse in the space of election results.
So, hypothetically, what if they changed the election rules so instead of voting yes/no, each voter would turn an analog dial to determine how much they liked each candidate, yielding some real number between zero and one?
DaleSpam
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#123
Feb23-12, 04:58 PM
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Quote Quote by JDoolin View Post
So, hypothetically, what if they changed the election rules so instead of voting yes/no, each voter would turn an analog dial to determine how much they liked each candidate, yielding some real number between zero and one?
It still wouldn't be a vector space because no candidate can have a negative result so the axioms of a vector space are not satisfied. I.e. if A is a non-null election result then there is no election result B such that A+B=0 where 0 is the null election vector.
JDoolin
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#124
Feb23-12, 05:24 PM
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Quote Quote by DaleSpam View Post
It still wouldn't be a vector space because no candidate can have a negative result so the axioms of a vector space are not satisfied. I.e. if A is a non-null election result then there is no election result B such that A+B=0 where 0 is the null election vector.
Hmmmm. And that motivates my next question, what if the dials are allowed to turn anywhere from -1 to 1?
DaleSpam
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#125
Feb23-12, 05:35 PM
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Quote Quote by JDoolin View Post
Hmmmm. And that motivates my next question, what if the dials are allowed to turn anywhere from -1 to 1?
That would overcome my previous objection. However, there are other problems, for instance you can always take an election result A representing 100% turnout with everyone voting the maximum allowed for one candidate. Then A+A would not be a valid election result since it would represent 200% turnout.
JDoolin
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#126
Feb23-12, 09:32 PM
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Quote Quote by twofish-quant View Post
No. I just thought of a counter example. Twenty questions.

I'm located in a spot on the earth. By asking me yes-no questions, you can figure out my latitude and longitude. Am I on land? Yes. Do I see taxicabs? Yes. Are they green? No. Are they yellow? No. Do I see water? Yes.

With each question, you can eliminate parts of the vector space. The fact that I see tax cabs and they are not yellow, means that I'm not in Manhattan. Now if you can ask enough questions, you can figure out my location and convert to GPS coordinates.

Note that you've figured out my GPS coordinates without actually measuring my latitude and longitude or doing any reference comparisons at all. You can show that no reference comparisons were done, because you can play this game without knowing anything about latitude and longitude at all, and it's the same game that you can play with things that are *not* vector spaces (i.e. words in a dictionary).

One other way of thinking about it is that you can specify points in a vector space as the interaction of subsets of that vector space, which allows you to specify a point in that space without reference to basis vectors at all.
In your twenty questions, aren't you likely to eventually ask something about a specific object? Not just the color of the taxi-cabs in the region in general, but you need to ask about a specific street-corner, or a specific building?

Once you pick a specific landmark, you now have a reference comparison--it's not latitude or longitude, but it is a reference comparison.


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