$\frac{d \theta}{dt} = \frac{q}{mc}-\frac{k}{mc}(\theta-T)$ theta, t - degrees and time q - energy absorbed per time k - energy/degrees x time T - degrees m - mass c - ?
So the first part is just like I said earlier, you need a reference temperature and you have one in your equation -- T. So $\overline{\theta} = \theta/T$ where the \overline{} means a nondimensional variable
So you plug in $\theta = \overline{\theta} T$ for $\theta$ and get:
Now the "trick" is how to determine what to use to non-dimensionalize time. As you get better at it, you can just "spot" it. But here, you can say $\tau = t/A$ where A is unknown
Now, you know the units need to work out. Your differential is non-dimensional and the $(\overline{\theta} - 1)$ is nondimensional, and both are multiplied by $T$ times something else
So you equate the coefficients as: $T A = T \frac{k}{m c}$
but I know how to pick new variables and name it like for example
m = theta/theta star n= t/ t star multiply through to get the new values of it sub them into the equation divide throughout and there should be some t* or theta* by itself pick the RIGHT MATCHING unit and then wham we can cancel
I'm trying to figure out c... subsituting with new variables is not my problem
but q - energy absorbed per time which is calories /time m - mass but it also said units of time... it's also grams... so I'm missing degrees here for q/mc
mhm, but I can see the k/mc(theta-T) as having degrees/time... it's q/mc which is tricky to see unless I know what c really is
I) My first question would be "why should classical systems obey the principle of least action ?" When we find out the propagator in quantum physics, we find the amplitude to be equal to the sum over all possible paths, weighted by the phase $e^{iS/\hbar}$. Then one arrives at the conclusion that...
so it does require those ... what is that called it was in M L T or something.. I saw a list and it showed the values of it... I was watching about the buckingham pi theorem which was similar, but the calculations are really lengthy
So the first part is just like I said earlier, you need a reference temperature and you have one in your equation -- T. So $\overline{\theta} = \theta/T$ where the \overline{} means a nondimensional variable
That whole section there
When you equate the coefficients, you get the factor that makes time non-dimensional and you can proceed to non-dimensionalize the equation. Without needing to know what the actual units are for c
Right. So hopefully what we found from this whole exercise -- non-dimensionalization is different from dimensional analysis; you don't need to know the units of everything to find non-dimensional parameters (but it helps to check if you are right); degrees is an ambiguous unit because it could be a length (rotation) or a temperature
@ACuriousMind : ahhh! What did you do? Restore please the editing that I did. Can you? It was beautifully organized, with numbering of the equations. How can one refer to non-numbered equations. Ohhhh !
@ACuriousMind : it was so nice, all the equations aligned, a pleasure !
@ACuriousMind Ohhh ! It was so nicely organized, all the equations aligned. Well, is it a way to align equations so as to begin on a same column? It's not convenient to insert $ \ \ \ $
