Re: [boost] Re: Dimensional Analysis Interest?

From: Walter Landry (wlandry_at_[hidden])
Date: 2003-10-16 19:45:44

Deane Yang <deane_yang_at_[hidden]> wrote:
> Jan Langer wrote:
> > Jan Van Dijk wrote:
> > 
> >> Unfortunately the algebra (the transformation properties) is not 
> >> linear. Just consider conversion from the temperature-'base vector' K 
> >> (kelvin) to C (Celsius), C=K-273.15.
> >> In technical terms: obviously the K->C base transformation cannot be 
> >> described with a (metric) 00-tensor (or scalar) M via the linear 
> >> transformation relation C=MK.
> >>
> > when thinking about it, i cannot imagine a case where this is important. 
> > it makes imho no sense to add °C to °F or something similar. what is the 
> > result of 20°C plus 10°C? 30°C or (293.17+283.17)°K or just 30 Kelvin? 
> > it is difficult to calculate with units which cannot be used as 
> > differences. and if they can be used as differences they are linear. IMHO.
> > jan
> > 
> 
> These are excellent points (as well as others made in other posts that I 
> will comment on in another posting).
> 
> Temperature is an example of what I would call an "affine" dimension
> and not a "linear" one, which is what we've been focusing on so far.
> 
> When we think of grams, meters, or seconds, we are thinking of relative
> measurements, namely the difference between two absolute measurements.
> So there is a natural meaning for 0 and the measurement lives naturally
> in a vector space. So, as a measurement of the difference between two 
> temperatures, both °C (= °K) and °F make perfect sense as linear units.
> 
> If you want, however, to represent absolute temperatures in Kelvin,
> Centigrade, or Fahrenheit, I would say that you need to define the 
> notion of an affine dimension. An affine dimension corresponds to
> a 1-dimensional affine space, where there is no natural origin but
> given any two points in the space, there is a natural vector associated 
> with the two points (the difference between the two points).

Actually, I consider temperature to be equivalent to energy.
Boltzmann's constant and all that.

Regards,
Walter Landry
wlandry_at_[hidden]