Data types in physics
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pioneer1 - 07 Apr 2008 21:44 GMT Hi,
I submitted this question to sci.math.research but it was rejected for being related to physics or computer science and not math perse.
I have been trying to classify data types in physics
http://www.densytics.com/wiki/index.php?title=Data_types_in_Physics
and I am confused about what is considered string (in the sense of computer language terminology) and what is considered number or quantity or magnitude in physics.
More specifically, looking at F in F=ma I see something like a pointer to an address, not the address itself. Or if ma is number in a spreadsheet cell, F is the name of that cell.
I know this is not how physicists see it. I am looking to find the correct mathematical terminology so that I can state the problem clearly. Do you know a field of physics or math that studies these things?
Thank you for the help.
Hans Aberg - 08 Apr 2008 18:13 GMT > I have been trying to classify data types in physics > [quoted text clipped - 7 lines] > to an address, not the address itself. Or if ma is number in a > spreadsheet cell, F is the name of that cell. I am not sure exactly what you want to do:
But you might try Haskell <http://haskell.org/>, using say Hugs <http://haskell.org/hugs>. One can write say: module Physics where
type Real = Double
-- Mass, acceleration, force data Measurable = M(Real) | A(Real) | F(Real) deriving (Eq, Show)
instance Num Measurable where (M m)*(A a) = F(m*a) (A a)*(M m) = F(a*m)
This represents some measurable quantities as SI-unit floating point numbers wrapped up as objects. Then in Hugs, for example: Physics> (M 1.5)*(A 9.8) F 14.7
In a more ambitious approach, one might introduce abstract physical units: type Real = Double
-- Mass, time, length data Unit = M | T | L deriving (Eq, Show)
data UnitDimension = Unit:^Integer | UnitDimension:*UnitDimension deriving (Eq)
instance Show UnitDimension where show (x:^k) = show x ++ "^" ++ show k show (x:*y) = show x ++ "*" ++ show y so acceleration can be represented by: Physics> (L:^1):*(T:^ -2) L^1*T^-2
In this approach, the exact data types are not as important as the functions used to define a user interface. For example, I might have defined more directly -- Mass, time, length data UnitDimension = (Integer, Integer, Integer) and if I define an operator (*) only its definition will change with the choice of data types.
> I know this is not how physicists see it. I am looking to find the > correct mathematical terminology so that I can state the problem > clearly. Do you know a field of physics or math that studies these > things? The stuff above does not give any chance of doing mathematics; for that, a theorem prover. But current theorem provers do not admit doing very advanced mathematics. An intermediate would be a symbolic algebra propgram, i.e., which does not treat formulas axiomatically. A theorem provers is an enhanced form of Prolog and constraint logic programs such as CLP(R). So you might search for those words, and "automated deduction", say in the Wikipedia.
Hans Aberg
Joseph Warner - 10 Apr 2008 21:58 GMT >> I have been trying to classify data types in physics >> [quoted text clipped - 16 lines] > <http://haskell.org/hugs>. One can write say: > module Physics where I not sure either of what the person wants. But he may want to do both vectors and dimension analysis at the same time.
MathCad allows that. A variable can be assigned a dimension whether that is a scaler, vector and an array element. Through the manipulation of the calculations the program keeps the dimensions correct and report it out when you want to see a quantity.
But of data types Physics as far as I know only uses real numbers with a certain number of digits or integers like or ladder operators. But all of the numbers except for some constants like pi have a dimension associated with it.
So the original poster may want to look at MathCad or similar programming languages.
pioneer1 - 12 Apr 2008 14:30 GMT > 0type Real = Double But you already defined Force as number. As I mentioned to John Forkosh I am trying to understand if F in F=ma is type string or type number? In programming there is no ambiguity. F is clearly defined as number. In physics, there is an ambiguity in the equality sign. It may mean F :: a (F is proportional to a) or it may mean F = ma (F is a name for ma, we may substitute F for ma).
Phillip Helbig---remove CLOTHES to reply - 12 Apr 2008 23:09 GMT In article <2ee95a6f-2ed3-4370-862d-b3e81e4354a7@w8g2000prd.googlegroups.com>, pioneer1 <1pioneer1@gmail.com> writes:
> > 0type Real = Double > > But you already defined Force as number. As I mentioned to John > Forkosh I am trying to understand if F in F=ma is type string or type > number? Number.
> In programming there is no ambiguity. F is clearly defined as > number. In physics, there is an ambiguity in the equality sign. It may > mean F :: a (F is proportional to a) or it may mean F = ma (F is a > name for ma, we may substitute F for ma). The idea of proportionality is closer. In fact, this is essentially the definition of intertial mass (the proportionality constant between force and acceleration). Depending on the context, one could substitute F for ma; after all, they are equal.
I think no-one yet really understands what your question is, though.
Hans Aberg - 12 Apr 2008 23:09 GMT > As I mentioned to John > Forkosh I am trying to understand if F in F=ma is type string or type > number? In programming there is no ambiguity. You question suggests you think that physics theories already have a natural computer language representation, but the best hope is to model some only aspects.
> F is clearly defined as > number. No: F is a physically measurable quantity. By fixing a unit of force u_F, disregarding the direction that force may have, F can be measured using a number r, as in F = r*u_F In theoretical physical theories, r is modeled as a real number (needing differential equations), and in practical measurements, an approximation thereof.
> In physics, there is an ambiguity in the equality sign. It may > mean F :: a (F is proportional to a) or it may mean F = ma (F is a > name for ma, we may substitute F for ma). The formally correct version of this equation is F = k m a where k is a dimension-less constant depending only on the units chosen for measuring the physical quantities force, mass and acceleration; by suitable choice of measurements units, it can be chosen to be equal to 1.
Hans
J. J. Lodder - 13 Apr 2008 19:49 GMT > The formally correct version of this equation is > F = k m a > where k is a dimension-less constant depending only on the units chosen > for measuring the physical quantities force, mass and acceleration; by > suitable choice of measurements units, it can be chosen to be equal to 1. I don't understand this remark. Are you implying that every physical formula should be thought of as containing an invisible extra constant, dimensionless and of value 1?
You can of course think so, but what good would that do?
Jan
Hans Aberg - 13 Apr 2008 23:59 GMT >> The formally correct version of this equation is >> F = k m a [quoted text clipped - 6 lines] > should be thought of as containing an invisible extra constant, > dimensionless and of value 1? This is how the formulas will be, if one does not settle for a specific set of measurement units.
> You can of course think so, > but what good would that do? In order to be formally correct, but the formulas quickly become unreadable. As illustration:
In theoretical physics, one may set c = 1, so all lengths can be measured in seconds. So how about buying 20 ns (nano seconds) of rope?
Hans
J. J. Lodder - 16 Apr 2008 15:44 GMT > >> The formally correct version of this equation is > >> F = k m a [quoted text clipped - 9 lines] > This is how the formulas will be, if one does not settle for a specific > set of measurement units. Dimensions and units can in principle be chosen independently of each other. The need for dimensioned constants arises when you insist on using a system of dimensions that is too rich for the system of equations describing the physics.
A trivial example is the maximal system of dimensions, in which every physical quantity has its own dimension. In consequence every formula must contain a dimensioned constant (possibly of value 1) to balance the dimensions. This looks like what you were talking about above.
A non-trivial case occurs in practice when people insist on using the MKSA system of dimensions with natural units, instead of the natural system of dimensions. (c=1, but not dimensionless) Then the c's must be kept in the formulae, which destroys some of the elegance of the natural units.
In such cases particle masses for example are given in GeV/c^2 The usual comprimise is to use natural units throughout and to do this only for final results.
> > You can of course think so, > > but what good would that do? [quoted text clipped - 4 lines] > In theoretical physics, one may set c = 1, so all lengths can be > measured in seconds. So how about buying 20 ns (nano seconds) of rope? That's of course just what we would do, if we could start afresh with our unit systems. Backward compatibility does have it's claims however. And in actual fact you are buying rope measured in nanoseconds, it is just that you (and everybody else) prefer to use some inconvenient numerical constants as well.
Jan
Hans Aberg - 18 Apr 2008 18:52 GMT > A trivial example is the maximal system of dimensions, > in which every physical quantity has its own dimension. That would be the most general dimension system, in the sens that the others arise by imposing some equivalence relations.
>> In theoretical physics, one may set c = 1, so all lengths can be >> measured in seconds. So how about buying 20 ns (nano seconds) of rope? > > That's of course just what we would do, > if we could start afresh with our unit systems. I think one use say the charge of the electron as basic unit.
> Backward compatibility does have it's claims however. That is one reason. Another is the ability to measure the physical constants. For example, mass, time and length can be measured to a higher accuracy than the gravitational constant, making it unsuitable as a conversion constant.
> And in actual fact you are buying rope measured in nanoseconds, > it is just that you (and everybody else) > prefer to use some inconvenient numerical constants as well. There is another factor: even though time and length can be treated as the same in a physical theory, they have in practical experience quite different characteristics. So giving them different dimensions is in line with that experience.
Hans Aberg
J. J. Lodder - 19 Apr 2008 23:50 GMT > > A trivial example is the maximal system of dimensions, > > in which every physical quantity has its own dimension. > > That would be the most general dimension system, in the sens that the > others arise by imposing some equivalence relations. All dimension systems are equally general. (all are finite dimensional algebras, mathematically speaking)
It just that some have more dimensions than others. Some are more trivial than others though. The maximal system of dimensions (every physical quantity its own dimension) is as trivial as the one-dimensional system, (every physical quantity dimensionless) for both can't be used for anything useful.
> >> In theoretical physics, one may set c = 1, so all lengths can be > >> measured in seconds. So how about buying 20 ns (nano seconds) of rope? [quoted text clipped - 3 lines] > > I think one use say the charge of the electron as basic unit. Planck did that, originally, in Planck units v1. The disadvantage is having to many 137-s popping up in less logical places.
> > Backward compatibility does have it's claims however. > > That is one reason. Another is the ability to measure the physical > constants. For example, mass, time and length can be measured to a > higher accuracy than the gravitational constant, making it unsuitable as > a conversion constant. That's no problem, in principle. The same situation occurred for the Ampere. The solution adopted was to have both an absolute ampere, and an international ampere. The absolute ampere/international ampere was effectively just another fundamental constant to be measured, and to be adjusted in the overall best fit.
And to pick a nit: length can't be measured at all, nowadays. All we can measure is time, with either a ruler or a stopwatch. The reason is the same: time and speed (of light) can be measured to greater accuracy than length, so the obvious solution is to define c, and not to measure length independently anymore.
> > And in actual fact you are buying rope measured in nanoseconds, > > it is just that you (and everybody else) [quoted text clipped - 4 lines] > different characteristics. So giving them different dimensions is in > line with that experience. It is just a choice that can be made freely, subject only to consistency. You are for example free to use natural units with c = 1, both with [c] = [I] and with [c] = [L][T]^{-1} as you find suitabl for some purpose.
Best,
Jan
Hans Aberg - 20 Apr 2008 22:47 GMT >>> A trivial example is the maximal system of dimensions, >>> in which every physical quantity has its own dimension. [quoted text clipped - 3 lines] > All dimension systems are equally general. > (all are finite dimensional algebras, mathematically speaking) If the dimensions are part of a mathematical theory then one looses information by equation dimensions, but if one has physical information on how physical units change, then it can be retrieved.
> It just that some have more dimensions than others. > Some are more trivial than others though. [quoted text clipped - 3 lines] > (every physical quantity dimensionless) > for both can't be used for anything useful. No. They are all equally non-trivial: in theoretical physics, one may set c = 1, G = 1, /h = 1, and so forth, without loosing physical generality.
> And to pick a nit: length can't be measured at all, nowadays. > All we can measure is time, with either a ruler or a stopwatch. This is just a definition of a length unit. It does not affect the physical measurement methods of length.
Hans Aberg
Paul Danaher - 21 Apr 2008 21:44 GMT ..
> And to pick a nit: length can't be measured at all, nowadays. > All we can measure is time, with either a ruler or a stopwatch. > The reason is the same: time and speed (of light) > can be measured to greater accuracy than length, > so the obvious solution is to define c, > and not to measure length independently anymore. I may have been mislead by sensationalism, but I recall a report a few weeks ago that the various international kilogram platinum-iridium masses have been showing an unexplained divergence when compared with each other. Two questions: has anybody come up with an explanation, and what's the state of debate on what to replace the cylinders with?
Boo - 23 May 2008 13:27 GMT > I may have been mislead by sensationalism, but I recall a report a few > weeks ago that the various international kilogram platinum-iridium > masses have been showing an unexplained divergence when compared with > each other. Out of interest can anyone tell me why they use platinum-iridium for reference masses ? I assume it has good oxidation characteristics but is the any other reason for the ue of this material ?
Also, do they keep them under a vacuum ?
 Signature Boo
J. J. Lodder - 23 May 2008 17:28 GMT > > I may have been mislead by sensationalism, but I recall a report a few > > weeks ago that the various international kilogram platinum-iridium [quoted text clipped - 4 lines] > masses ? I assume it has good oxidation characteristics but is the any other > reason for the ue of this material ? Resistance to wear.
> Also, do they keep them under a vacuum ? <http://www.bipm.org/en/scientific/mass/prototype.html> says it all,
Jan
Phillip Helbig---remove CLOTHES to reply - 21 Apr 2008 22:11 GMT In article <rvPOj.58991$097.58918@newsfe21.lga>, Paul Danaher <paul.danaher@watwinc.com> writes:
> I may have been mislead by sensationalism, but I recall a report a few > weeks ago that the various international kilogram platinum-iridium > masses have been showing an unexplained divergence when compared with > each other. I've heard of that as well. As far as I know, it is unexplained, but there are many hypotheses. The problem is, it is difficult to do experiments on the reference kilograms.
> Two questions: has anybody come up with an explanation, and what's the > state of debate on what to replace the cylinders with? There are a few different approaches to redefining the basic measurement references. In the case of the kilogram, it won't be replaced by a similar reference object, but will be defined by, say, a certain number of atoms of a certain isotope. Perhaps it will be tied to something else, say multiply the number of electrons corresponding to a coulomb by a certain number to get the number of electrons in a kilogram.
There are two tendencies, which sometimes compete: One is to find a (conceptually) simple measurement. The other is to make sure that if something else which is defined by a measurement becomes more exact in the future, that the other quantities don't shift too much.
A while back, there was an article in the Physikalische Blätter (now called Physik Journal---the member magazine of the German Physical Society) which reviewed the status of the various proposals.
At http://www.ptb.de/en/wegweiser/einheiten/si/kilogramm.html one can read:
The scientists of the workin group "Ion akkumulation" try to trace the unit of mass back to an atomic mass: Individual gold ions from a gold ion beam are accumulated on a collector. The problem (not yet completely solved) here is to collect an ion mass accessible to weighing.
Maybe you can find more information from the link above (which is from the German equivalent of the National Bureau of Standards).
Jonathan Thornburg [remove -animal to reply] - 22 Apr 2008 18:11 GMT > There are a few different approaches to redefining the basic measurement > references. In the case of the kilogram, it won't be replaced by a > similar reference object, but will be defined by, say, a certain number > of atoms of a certain isotope. Perhaps it will be tied to something > else, say multiply the number of electrons corresponding to a coulomb by > a certain number to get the number of electrons in a kilogram. Here are 3 references with more information about this:
There was a pretty good (actually I though _excellent_ for a general newspaper!) article about this in the Los Angeles Times last week: http://www.latimes.com/news/nationworld/nation/la-sci-kilogram17apr17,0,7998161, print.story
physorg.com also had a pretty good short piece in 2005: http://www.physorg.com/printnews.php?newsid=3178
There's also a nice Fermilab colloquium presentations by Richard Steiner (NIST) on "How Measuring the Planck Constant gets to an Electronic Kilogram Standard" on 1 Aug 2007: search for Steiner in http://www-ppd.fnal.gov/EPPOffice-w/colloq/colloq_06_07.html
 Signature -- "Jonathan Thornburg [remove -animal to reply]" <J.Thornburg@soton.ac-zebra.uk> School of Mathematics, U of Southampton, England "If English was good enough for Jesus, it's good enough for the schoolchildren of Texas." -- Texas governor James Ferguson, 1917, explaining why he vetoed a bill funding the teaching of foreign languages in Texas schools.
Paul Danaher - 23 Apr 2008 04:43 GMT <snip>
> Here are 3 references with more information about this: > [quoted text clipped - 4 lines] > physorg.com also had a pretty good short piece in 2005: > http://www.physorg.com/printnews.php?newsid=3178 Uh-oh - to quote: "Electrical power can be related to the Planck constant, defined as the ratio between the frequency of an electromagnetic particle such as a photon of light and its energy. This experimental method of defining the kilogram relies on selecting a fixed value for the Planck constant, which is currently determined experimentally based on the fixed value of the kilogram artifact."
The circularity is disturbing.
> There's also a nice Fermilab colloquium presentations by Richard Steiner > (NIST) on "How Measuring the Planck Constant gets to an Electronic Kilogram > Standard" on 1 Aug 2007: search for Steiner in > http://www-ppd.fnal.gov/EPPOffice-w/colloq/colloq_06_07.html Avogadro's number looks better here ...
Hans Aberg - 23 Apr 2008 22:22 GMT >> There was a pretty good (actually I though _excellent_ for a general >> newspaper!) article about this in the Los Angeles Times last week: >> >> http://www.latimes.com/news/nationworld/nation/la-sci-kilogram17apr17,0,7998161, print.story
>> physorg.com also had a pretty good short piece in 2005: >> http://www.physorg.com/printnews.php?newsid=3178
> Uh-oh - to quote: > "Electrical power can be related to the Planck constant, defined as the [quoted text clipped - 5 lines] > > The circularity is disturbing. There are no logical circularity problems here:
The current definition of the mass and some other physical units produce a value of the Planck constant. But the process can be reversed: fixing the Planck constant and the other constants in order to define the mass unit.
The problem is, using the current physical units, to get a value of the Planck constant sufficiently exact that its fixing does not alter the current mass unit too much. And there must be a measurement method that can define the new mass unit at least as well as the current mass unit.
It seems me to be an interesting method of defining mass, as the Planck constant is so fundamental in modern physics. But the main thing, for now, is to find a measurement method that can fix the mass unit at least as good as the current one, without reference to an artifact, but by a generally accessible and invariant reproducible physical phenomenon.
Hans Aberg
Phillip Helbig---remove CLOTHES to reply - 23 Apr 2008 22:24 GMT In article <5kpPj.58006$QC.54917@newsfe20.lga>, Paul Danaher <paul.danaher@watwinc.com> writes:
> Uh-oh - to quote: > "Electrical power can be related to the Planck constant, defined as [quoted text clipped - 5 lines] > > The circularity is disturbing. No. CURRENTLY the kilogram is fixed and Planck's constant is derived. This is one alternative. The other is to reverse the situation. In the past, the meter was fixed and the speed of light was measured. Now the situtation is reversed. Same idea.
> > There's also a nice Fermilab colloquium presentations by Richard Steiner > > (NIST) on "How Measuring the Planck Constant gets to an Electronic Kilogram > > Standard" on 1 Aug 2007: search for Steiner in > > http://www-ppd.fnal.gov/EPPOffice-w/colloq/colloq_06_07.html > > Avogadro's number looks better here ... In a sense, yes, it's more straightforward. On the other hand, the idea of fixing some constants of nature and deriving the standard measures from them, as the meter is derived from the speed of light, also has its attractions.
Paul Danaher - 22 Apr 2008 22:39 GMT > In article <rvPOj.58991$097.58918@newsfe21.lga>, Paul Danaher <snip>
> There are a few different approaches to redefining the basic measurement > references. In the case of the kilogram, it won't be replaced by a [quoted text clipped - 14 lines] > At http://www.ptb.de/en/wegweiser/einheiten/si/kilogramm.html one can > read: <snip>
Thank you! The direct English link to "Ion Akkumulation" seems to be broken, but the German works fine. http://www.ptb.de/de/org/1/12/124/_index.htm has a diagram (without error bars) which seems to show a drift, and the German text is quite definite in stating that changes in reference masses have been established. The English page http://www.ptb.de/en/org/1/12/124/_index.htm is much briefer, but there's a useful list of publications http://www.ptb.de/en/org/1/12/124/_index.htm.
However, these are technical (although interesting) and have no discussion of the underlying phenomenon (if any). The BIPM has a list of comparisons at http://kcdb.bipm.org/AppendixB/KCDB_ApB_result.asp?cmp_idy=432&cmp_cod=APMP.M.M- K1&page=1&search=1&cmp_cod_search=&met_idy=6&bra_idy=2= 0&epo_idy=0&cmt_idy=0&ett_idy_org=0&lab_idy=0 which include a report by the UK National Physical Laboratory in 2000. This concludes that the "drift" is within the experimental uncertainties involved in transporting and comparing the reference bodies.
So it seems the "problem" may be the result of experimental inaccuracy.
Phillip Helbig---remove CLOTHES to reply - 14 Apr 2008 22:35 GMT In article <1ifbvtl.4kohut1kw41ddN@de-ster.xs4all.nl>, nospam@de-ster.demon.nl (J. J. Lodder) writes:
> > The formally correct version of this equation is > > F = k m a [quoted text clipped - 9 lines] > You can of course think so, > but what good would that do? I see what he is driving at. In this case, the extra constant is dimensionless and has value 1. However, that is because m (the intertial mass) is DEFINED as the proportionality constant between F and a. However, let's assume that it has previously been defined in some other way, say as the gravitational mass m'. Wonder of wonders, it is STILL 1 and dimensionless, i.e. m = m'. This is a great mystery in Newtonian mechanics.
In other cases, say gravitational attraction, we have
F = G m1 m2 ----------- r²
In this case, the constant G is neither dimensionless nor does it have the value 1. In many cases, units are often used so that G is dimensionless and of value 1. Such definitions are often made for many such constants at once, and one sees notes like G = h = c = 1.
In other words, the quantity of interest (F in the examples above) is proportional to various things we can measure (like r, or m). In general, to make the proportionality an equality, we need a constant.
J. J. Lodder - 16 Apr 2008 15:44 GMT Phillip Helbig---remove CLOTHES to reply <helbig@astro.multiCLOTHESvax.de> wrote:
> In article <1ifbvtl.4kohut1kw41ddN@de-ster.xs4all.nl>, > nospam@de-ster.demon.nl (J. J. Lodder) writes: [quoted text clipped - 20 lines] > STILL 1 and dimensionless, i.e. m = m'. This is a great mystery in > Newtonian mechanics. Indeed, when discussing things like the Eotvos experiment one needs to write m_g and m_i explicitly. For such a system of equations MKSA dimensions won't do. It is of course quite possible to set up a matching system of dimensions, with M-i and m-G each having a dimension of their own.
However, those working at this level of sophistication are mature physisists who understand their stuff, not kiddies having to learn it.
So I don't think anyone has ever bothered to formally set up such a system of dimensions, for those to whom it matters can easily do without, without making too many mistakes.
> In other cases, say gravitational attraction, we have > [quoted text clipped - 10 lines] > proportional to various things we can measure (like r, or m). In > general, to make the proportionality an equality, we need a constant. See also my reply to Hans Ahlberg for the remark that the choices of c = 1 and [c] = [I] are independent. The value and the dimension of c can be chosen independently of each other.
More generally: units do not 'have' dimensions, despite the widespread misunderstanding to the contrary.
Best,
Jan
Hans Aberg - 18 Apr 2008 18:52 GMT >>> The formally correct version of this equation is >>> F = k m a >>> where k is a dimension-less constant depending only on the units chosen >>> for measuring the physical quantities force, mass and acceleration; by >>> suitable choice of measurements units, it can be chosen to be equal to 1. ..
> In this case, the extra constant is > dimensionless and has value 1. However, that is because m (the [quoted text clipped - 3 lines] > STILL 1 and dimensionless, i.e. m = m'. This is a great mystery in > Newtonian mechanics. There are several issues involved here. First, assume that mass is measured in pounds, acceleration using the Paris acceleration in vacuum, and F using the Paris archive spring - then there is good chance that k will not be 1, though dimensionless. - Decades ago, it was mentioned to me somebody trying to write out all these constants, but the formulas because unreadable.
Second, is k above always dimensionless?...
> In other cases, say gravitational attraction, we have > [quoted text clipped - 6 lines] > dimensionless and of value 1. Such definitions are often made for many > such constants at once, and one sees notes like G = h = c = 1. The reason that k is dimensionless is probably because the force F was discovered first via this equation, and defined using it. Suppose that the gravitational equation would have been discovered first - then G might have been made into a dimensionless constant. Doing some physical analysis, setting [F] = F, [mi] = M, [r] = L, [G] = 1, [F] = [(m1 m2)/r^2] = [m1][m2]/[r]^2 = M^2 L^-2 and setting [a] = A, [F] = [k m a] = [k][m][a] = [k] M A So [k] = M L^-2 A^-1 If the physical dimension of time is T, then A = L T^-2, so [k] = M T^2 L^-3
It is somewhat unusual, but I do not see a contradiction here.
Then there is the question of the equality of the inertial mass m and the gravitation mass m' (defined by these two equations). If the equations have constants in them, the formulation must be that for all measured masses m, m' one has m = k' m', once physical units of measurements have been chosen. This part is real physics, describing a relation of measurements from seemingly different equations in different physical context (because we have not been able to unify them as coming from a single model).
> In other words, the quantity of interest (F in the examples above) is > proportional to various things we can measure (like r, or m). In > general, to make the proportionality an equality, we need a constant. In order to make measurements of a physical unit, we need another physical unit as a reference. Then the number gotten from a physical measurement is relative that reference. And if the physical units are not selected carefully enough, there will be various constants showing up
J. J. Lodder - 19 Apr 2008 23:50 GMT > >>> The formally correct version of this equation is > >>> F = k m a [quoted text clipped - 19 lines] > > Second, is k above always dimensionless?... Once again: being or not being dimensionless is NOT a property of a physical quantity, or of a term in an equation. It is a choice you can freely make.
In the example above you can for example choose F to have a dimension of it's own. [F] Then the constant in F = kma must have the dimension [k] = [F][M]^{-1}[L]^{-1}[T]^2
Dimensional analysis has nothing to do with reality. It is a meta-analysis, of systems of equations that are used to describe reality. Dimensions can be assigned in any way we please, subject only to consistency.
Best,
Jan
Hans Aberg - 20 Apr 2008 22:47 GMT >>>>> F = k m a ..
>> Second, is k above always dimensionless?... > > Once again: being or not being dimensionless > is NOT a property of a physical quantity, > or of a term in an equation. Right, formally, these are different physical theories.
> It is a choice you can freely make. Or rather, a choice of physical theory - once it is fixed, the dimensions are fixed.
> In the example above you can for example > choose F to have a dimension of it's own. [F] > Then the constant in F = kma must have the dimension > [k] = [F][M]^{-1}[L]^{-1}[T]^2 Yes. I though this was clear from the context.
> Dimensional analysis has nothing to do with reality. > It is a meta-analysis, of systems of equations > that are used to describe reality. It is just an analysis of the chosen physical theory, which in its turn is a modeling of the physical reality, specifically physically measurable quantities.
> Dimensions can be assigned in any way we please, > subject only to consistency. Different physical theories yield different dimension systems, but one concern is the ability to measure the different quantities. For example, the gravitational constant cannot be measured as accurately as time, mas and length, and so cannot be used as a conversion constant.
Hans Aberg
pioneer1 - 28 Apr 2008 18:22 GMT On Apr 14, 5:35 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig--- remove CLOTHES to reply) wrote:
> In other words, the quantity of interest (F in the examples above) is > proportional to various things we can measure (like r, or m). In > general, to make the proportionality an equality, we need a constant. Do I understand correctly that you are saying that F itself is not a measurable quantity? We measure other quantities such as r (radius of orbit) and m (mass of satellite) and then infer the value of F. I am not sure we can measure these quantities observationally to obtain force. Force and radius are never together in formulas used to compute orbits because force is written down temporarily and always cancels. The same is true for m since that too cancels out of the formulas to compute orbits. We can only calculate force from given values not from observations. If r is given then, from F=GM/r2 we can compute F. But since that computed value of F refers to force in F=ma and since m cancels, our calculation amounts to computing acceleration. We then call acceleration force. This suggests to me that orbits are independent of force.
Phillip Helbig---remove CLOTHES to reply - 11 May 2008 20:01 GMT In article <65445ccb-a4a7-499e-a938-ce4a70875efd@a9g2000prl.googlegroups.com>, pioneer1 <1pioneer1@gmail.com> writes:
> On Apr 14, 5:35=A0pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig--- > remove CLOTHES to reply) wrote: [quoted text clipped - 5 lines] > Do I understand correctly that you are saying that F itself is not a > measurable quantity? Define "measurable quantity".
> since that computed value of F refers to force in F=ma and since m > cancels, our calculation amounts to computing acceleration. We then > call acceleration force. This suggests to me that orbits are > independent of force. What you might be thinking is that, since m cancels, orbits are independent of the mass of the orbiting object. This is a mystery in Newtonian physics, but a natural consequence of General Relativity.
J. J. Lodder - 15 May 2008 22:51 GMT Phillip Helbig---remove CLOTHES to reply <helbig@astro.multiCLOTHESvax.de> wrote:
> In article > <65445ccb-a4a7-499e-a938-ce4a70875efd@a9g2000prl.googlegroups.com>, [quoted text clipped - 19 lines] > What you might be thinking is that, since m cancels, orbits are > independent of the mass of the orbiting object. Only if inertial mass equals gravitational mass.
> This is a mystery in Newtonian physics, but a natural consequence of > General Relativity. The other way round. General relativity cannot possibly explain this, since it went into it at the start as a postulate,
Jan
pioneer1 - 28 May 2008 03:06 GMT On May 11, 3:01=A0pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig--- remove CLOTHES to reply) wrote:
> Define "measurable quantity". I'm not sure. A quick search did not reveal much online, except this paper
http://www.ihpst2005.leeds.ac.uk/papers/Lauginie.pdf
where they define measurable quantity as a "quantity for which an additive operation and the multiplication by a scalar (a real number) can be physically defined." I don't exactly understand this definition but the example they give about historical measurements of the speed of light in the section "An observable phenomenon" I think applies to the case of force as well. The speed of light measurements were delayed for centuries because people thought that light had infinite speed and that there was nothing to measure. Newton's force is defined as having infinite speed and similarly by that reasoning it could not be measured.
But what I had in mind in the case of orbits was that force is not a quantity that we can measure because it is eliminated and does not enter orbital formulas. I would think that if the term we are trying to measure does not enter the equations it cannot be measured within that problem. In this case, F does not enter orbital formulas so we cannot measure force in orbits. Does this make sense?
> This is a mystery in > Newtonian physics, but a natural consequence of General Relativity. Why is this a mystery? To me if a quantity cancels out of the equation there is no mystery: The problem described by that equation is independent of that term. Can you explain why you think there is a mystery?
John Forkosh - 08 Apr 2008 19:22 GMT > I have been trying to classify data types in physics > [quoted text clipped - 12 lines] > clearly. Do you know a field of physics or math that studies these > things? I believe you may be barking up the wrong "tree" (sorry). I'm not aware of a field of physics that studies these things, per se, but I believe dimensional analysis is what you're looking for. In other words, what you call data types is what physics calls units. Consider your own F=ma example, and consider the rhs, ma, as a lambda expression in a typed lambda calculus. Then "input" would be a variable of type mass and a variable of type acceleration (or would be other lambda expressions that reduce to these types). Seems pretty straightforward. And I'd hazard a guess it's already been discussed (or maybe just mentioned) somewheres, though I'm not aware of any references.
 Signature John Forkosh ( mailto: j@f.com where j=john and f=forkosh )
J. J. Lodder - 11 Apr 2008 01:06 GMT > > I have been trying to classify data types in physics > > [quoted text clipped - 18 lines] > looking for. In other words, what you call data types > is what physics calls units. Yes, provided that it is understood that units and dimensions inherently have nothing to do with physics. Units can in principle be chosen freely, subject only to consistency. Units are tools to describe reality, to map physical quantities into numbers and any -physical- result must be independent of the chosen description.
Calling a dimension a data type may be misleading, for it suggests that dimensions do have an inherent physical meaning. This is not the case. Systems of dimensions can be chosen freely and arbitrarily, again subject only to consistency,
Jan
John Forkosh - 11 Apr 2008 17:22 GMT >> > I have been trying to classify data types in physics >> > http://www.densytics.com/wiki/index.php?title=Data_types_in_Physics [quoted text clipped - 10 lines] >> looking for. In other words, what you call data types >> is what physics calls units.
> Yes, provided that it is understood that units and dimensions > inherently have nothing to do with physics. [quoted text clipped - 11 lines] > again subject only to consistency, > Jan Wandering a bit from the OP's question, but I'd think that units are more fundamental than numbers. Just because there are many different/equivalent "representations" (e.g., theoretical units), doesn't mean units aren't fundamental. On the other hand, there are no directly measurable numbers, per se, in nature (though there are dimensionless ratios). You can't measure "one", just one apple, one erg, etc.
Numbers, at least vis-a-vis physics, are a completely human construction (despite Kronecker's claim that "God invented the integers"). We find numbers very useful to describe measurements, but maybe that's just because we first found numbers useful to calibrate our measuring instruments in the first place. And then we find numbers useful to formulate theories correlating our already-numerical measurements. Now, I agree there are many holes in the above reasoning, and I have zero idea how physics might be done without numbers. But maybe that's just how human brains work. To that extent, we're all trapped in Whorf's "conceptual boat" (lots of google hits for that).
By looking for numeric-like data types that are physically fundamental, I think the OP was mistakenly elevating numeric-like entities to a foundational level they don't really occupy. For a recent discussion vaguely along these lines (though they aren't, as far as I can tell, actually trying to do away with numbers), try http://arxiv.org/abs/0803.0417/
 Signature John Forkosh ( mailto: j@f.com where j=john and f=forkosh )
J. J. Lodder - 12 Apr 2008 23:09 GMT > >> > I have been trying to classify data types in physics > >> > http://www.densytics.com/wiki/index.php?title=Data_types_in_Physics [quoted text clipped - 32 lines] > (e.g., theoretical units), doesn't mean units aren't > fundamental. In my use of words human constructs are by definition not fundamental. Only things that are dependent of the way in which they are described can be. I don't see how you can find levels of fundamentality here. The way I see it a unit system (with appropriate measurement procedures) is a mapping, for physical quantities into numbers.
> On the other hand, there are no directly measurable > numbers, per se, in nature (though there are dimensionless ratios). > You can't measure "one", just one apple, one erg, etc. But you can count, cows, or electrons for example. And no, he other way, with sufficient perversity dimensionless quantities are not exempt. You can for example measure angles either in radians (aka m/m) or in feet/mile. Angles need not even be dimensionless, for you can assign a different dimension to the feet and mile mentioned above. (discussed here long ago as 'pilot's units'.
> Numbers, at least vis-a-vis physics, are a completely human > construction (despite Kronecker's claim that "God invented the > integers"). That's a matter for philosophers, hence off-topic here.
> We find numbers very useful to describe measurements, > but maybe that's just because we first found numbers useful to calibrate > our measuring instruments in the first place. And then we find numbers > useful to formulate theories correlating our already-numerical measurements. > Now, I agree there are many holes in the above reasoning, > and I have zero idea how physics might be done without numbers. It seems fairly obvious that you can't.
> But maybe that's just how human brains work. To that extent, we're all > trapped in Whorf's "conceptual boat" (lots of google hits for that). Describing reality at sufficient sophistication, to get for example the electron g-factor correct to ten decimal places or so, will require setting up a very elaborate mathematical apparatus, no matter how your brain works. At it won't give the right answer without being equivalent.
> By looking for numeric-like data types that are physically fundamental, > I think the OP was mistakenly elevating numeric-like entities to a > foundational level they don't really occupy. Agreed,
Jan
pioneer1 - 12 Apr 2008 14:30 GMT On Apr 8, 2:22=A0pm, John Forkosh <j...@please.see.sig.for.email.com> wrote:
>. . . what you call data types > is what physics calls units. I am not sure because I am trying to understand if force in F=ma is type string or type number. Only type number has units and dimensions. The distinction between string and number is called types too, correct?
Phillip Helbig---remove CLOTHES to reply - 12 Apr 2008 23:09 GMT In article <afeec54e-fa8c-4303-b02e-e2ed7cc4f1c5@a5g2000prg.googlegroups.com>, pioneer1 <1pioneer1@gmail.com> writes:
> On Apr 8, 2:22pm, John Forkosh <j...@please.see.sig.for.email.com> > wrote: [quoted text clipped - 4 lines] > I am not sure because I am trying to understand if force in F=ma is > type string or type number. Only type number has units and dimensions. It is type number, not string.
> The distinction between string and number is called types too, correct? Yes, in programming languages, there are various data types, some of which correspond to numbers and some of which correspond to strings.
John Forkosh - 12 Apr 2008 23:09 GMT >>. . . what you call data types >> is what physics calls units. > > I am not sure because I am trying to understand if force in F=ma is > type string or type number. Only type number has units and dimensions. > The distinction between string and number is called types too, correct? I tried to answer, as best I can, what I think you've been asking (both here and in your parallel thread) in the comp.theory ng. So I'll reproduce that answer (modulo a few typos) below...
> If I understand correctly, then in physics symbols are loaded and > they can represent either a string or a number and a physicist will > know how to interpret the symbol correctly depending on the context. > Is this correct? Well, yes, that's kind of correct and I think it answers the question you were originally asking. Context is important, but it's a little more complicated than I think you're suggesting. Consider your F=ma and alsp consider the electrostatic force F=qE on a charge q in an electric field E. That's two different contexts for F, in the following way. F=ma is kind of "definitional". If you see a mass m with an acceleration a, then you infer that there must be some force F=ma somewhere, somehow, of some sort, acting on m. But you have no clue what F is or where it comes from. In this sense F might be called a "label" (like you said in a previous comp.theory post) in the F=ma context. F=qE is a "quantity" (like you said in that same post). If you put a charge q in an electric field E, then that charge q will experience an electrostatic force F=qE. ( Now, if that charge q also has mass m, then you'll see an acceleration given by F=ma=qE so a=(q/m)E. )
Note that the physical units of F (force) are the same regardless of "context" (in the above sense). But your "data type" might not be invariant. That makes it difficult for me to see what purpose it might serve.
 Signature John Forkosh ( mailto: j@f.com where j=john and f=forkosh )
sr - 18 Apr 2008 18:52 GMT > On Apr 8, 2:22=A0pm, John Forkosh <j...@please.see.sig.for.email.com> > wrote:
>>. . . what you call data types >> is what physics calls units.
> I am not sure because I am trying to understand if force in F=ma is > type string or type number. Only type number has units and dimensions. > The distinction between string and number is called types too, correct? > [...] OK, here's an attempted answer to the original question from someone who is a C++ expert, but also knows a reasonable amount about maths and physics... ----------------------------------- By "string", I assume you mean "array of char" or "array of byte". Yes, this is of different "type" than (say) int, float, etc.
Force, mass, acceleration are all distinct types. In C++, one might represent them like this (sketch only) :
class Force; // Forward declaration.
enum Units { SI, NATURAL, ..... };
class Acceleration { float my_val; Units my_units; public: // constructor(s), etc, ..... };
class Mass { float my_val; Units my_units; public: // constructor(s), etc, .....
Force operator*(Acceleration const & a const; { return Force(....construct from <my_val> times <a.my_val>, and correct for respective units....); } };
class Force { ... similar.... };
Then one can write stuff like Mass m(....); Acceleration a(....); Force f = m * a; // Initialization.
or
if (f == m*a) { ..... } else { ..... }
One can of course make all this far more sophisticated. --------------------------
The important thing is to maintain a clear distinction between the abstract mathematical type (e.g., real number, sequence of octet, Lie algebra generator, Group element, vector space, dual vector space, etc, ....) and the respective concrete representations in terms of programming structures. The abstract algebras/relationships in which the abstract types may (or may not) participate in can then be expressed in your program via C++ types and operator overloading, etc.
Hope that helps.
David Golden - 26 Apr 2008 03:43 GMT > pioneer1 <1pioneer1@gmail.com> wrote: >> I know this is not how physicists see it. I am looking to find the [quoted text clipped - 7 lines] > looking for. In other words, what you call data types > is what physics calls units. FWIW, a far-from-exhaustive grab-bag:
*** "Frink" [1] and "Fortress" [2] are examples of science ior engineering oriented programming language projects where dimensions and units are intended to be supported as built in dynamic (frink) or static (fortress) language features.
The Fortress language specification has quite a lot of formalism - the "1.0 Beta" specification (at [2]) chapters 18, 29 and 35 are a good attempt to explore rigorous support for statically checked physical dimensions and units in a practical programming language (re "1.0 Beta": note that the "1.0" specification (temporarily) dropped dimensions and units, the particular topic of interest here!)
In the 1.0 Beta spec, dimensions are described as "type-like constructs" and units as "constructs that modify types and values" (page 155), though those statements should really be read in context of the spec.
*** The "Qi" [3] programming language has a powerful/meta type system using sequent notation. In Qi, the programmer should be able to construct types that allow static checking of dimensions and units, as (nearly) shown by [3] (the specifics of [3]'s system may not be entirely satisfactory compared to [2]'s, but it's still a nice example)
*** "Q" [4] is an example of a programming language using an equational paradigm. Not _directly_ to do with physical dimensions and units in programming, but reading its documentation should help a lot when thinking about what the symbol "=" can actually mean in different contexts in computing, since in most programming languages it's quite the false friend for mathematicians...
[1] http://futureboy.us/frinkdocs/#HowFrinkIsDifferent
[2] http://projectfortress.sun.com/Projects/Community/wiki/FortressDocumentation
[3] http://www.lambdassociates.org/studies/study11.htm [4] http://q-lang.sourceforge.net/examples/
J. J. Lodder - 11 Apr 2008 01:06 GMT > I submitted this question to sci.math.research but it was rejected for > being related to physics or computer science and not math perse. [quoted text clipped - 6 lines] > computer language terminology) and what is considered number or > quantity or magnitude in physics. There are no data types in physics, just numbers. For convenience one usually chooses the reals, but (given that everyting in physics has finite precision) the rationals or even the integers would do as well. One can group numbers into more involved structures (complex numbers, vectors, tensors, etc) but that is in principle a matter of convenience and economy of notation only.
> More specifically, looking at F in F=ma I see something like a pointer > to an address, not the address itself. Or if ma is number in a > spreadsheet cell, F is the name of that cell. Just a multiplication here, (if desired a vector multiplication) whatever else you want to call it,
Jan
Bossavit - 12 Apr 2008 14:30 GMT J.J. Lodder:
>There are no data types in physics, just numbers. This is a bit too radical, I think. The opposite stand was argued in, e.g.,
H. Whitney: "The mathematics of physical quantities, Part I: Mathematical models for measurement", Am. Math. Monthly, 75, 2 (1968), pp. 115-38, H. Whitney: "The mathematics of physical quantities, Part II, Quantity structures and dimensional analysis", Am. Math. Monthly, 75, 3 (1968), pp. 227-56
that the OP might find useful, although Whitney did not invoke "data types" in the modern sense.
Also relevant,
M. Parkinson: "An Axiomatic Approach to Dimensions in Physics", Am. J. Phys., 32, 3 (1964), pp. 200-5.
J.F. Price: "Dimensional analysis of models and data sets", Am. J. Phys., 71, 5 (2003), pp. 437-47.
AB
============================================================================== Alain Bossavit LGEP, CNRS,91192 Gif sur Yvette CEDEX, France http://natrium.em.tut.fi/~bossavit/
pioneer1 - 22 Apr 2008 22:39 GMT On Apr 12, 9:30am, Bossavit <bossa...@lgep.supelec.removethis.fr> wrote:
Thanks for the references. Unfortunately, except for the first one I couldn't find them online. Do you know if they exist online?
> H. Whitney: "The mathematics of physical quantities, Part I: > Mathematical models for measurement", Am. Math. Monthly, 75, 2 (1968), [quoted text clipped - 19 lines] > Alain Bossavit > LGEP, CNRS,91192 Gif sur Yvette CEDEX, Francehttp://natrium.em.tut.fi/~bossavit/ Bossavit - 27 Apr 2008 04:57 GMT Pioneer1:
>Thanks for the references. Unfortunately, except for the first one I >couldn't find them online. Do you know if they exist online? The American Journal of Physics is indeed online, the whole collection. But one must be a subscriber to the paper version to get electronic access (with a small additional fee).
Another reference, germane to some turns the discussion has taken after your first query:
J.M. Levy-Leblond: "On the Conceptual Nature of the Physical Constants", Rivista del Nuovo Cimento, 7, 2 (1977), pp. 187-214.
Kyle - 18 Apr 2008 18:52 GMT Equations contain only values. To display the equation itself you would display it as a string: "F=ma". Programmatically these would all be floating point values.
>Hi, > [quoted text clipped - 19 lines] > >Thank you for the help.
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