Recent content by Omega0

Hello, I would like to go step by step trough the process of the generation of a star. An AI suggests for a simple model an adiabetic case (an improvement of the isothermal case) with ##\gamma=5/3##: $$ \begin{align} T &= T_0 \left(\frac{p}{p_0}\right)^{\gamma-1} \\ P &= \rho...

Well, I have another feeling... what about the following mathematical problem: Given the following equation: $$\begin{align*} 6^x+6y &= 42 \\ x+y &= a \end{align*}$$ Find the solutions for ##a \in \mathbf{N}## so that ##x## and ##y## are natural numbers. This is algebra but has nothing to...

No, again, it is no misconception if you understand my setup. Nevertheless... I believe I got something wrong, indeed. You said, that in 2D a symmetric function is symmetric to the line ##x=y## which I didn't valued. Sorry, but that's the point, visually the contour lines of a non-linear...

About solving non-linear equation sets, I would say - and the solution space. Okay, thank you but I am not convinced that Lie groups are the right choice for non-linear problems?

This is not "to unnecessarily complicate things" but what my original post was about: A set of equations... Well, not sure how confused it is... you posted an example where the solution is neccessarily ##x=y## which is perfectly symmetric, right? See perhaps the definition, if it helps. Okay...

Yes, this is clear. I am quite sure that for a set of equations which is invariant under the exchange of all variables there is something special happening. Like we know from tensor symmetries etc. I would like to understand, what - and naively expected an equality of the variables themselves...

Thanks. Unfortunately, this example doesn't fit because in the case that you have only ##x+y=3## you have a function which is invariant under variable exchange but it is exactly the case I mentioned in my post: Now you have an unlimited number of solutions. This doesn't count. I am talking...

Yes, I know, but what is behind this? That is my question. Again: A linear set of equations which is invariant under exchange of the variables has an infinite number of solutions. As soon as we have a non-linearity, the solution space seems to collapse. I would like to know why and how. Thanks.

Okay, I should have said invariant under an exchange of the variables? This problem is not about lines or so, variable names are just dust in the wind.

Hello, please let me split me split my question into 3 blocks. The first: The problem and the solution. The second: The question. The third: Maybe weird thoughts about about a similar problem. The problem and the solution $$ \begin{align*} 6^x+6^y &= 42 \\ x+y &= 3 \end{align*} $$ This is...

Okay, let's say this is a reconstruction - and I try to ignore the problems which we already have with an "exact shape". Sure, you mean the different statistics for Bose or Fermi particles? Well, in this statistics there isn't any specific information any more, right? Statistics is a big...

Okay, I appear to be too bad in QM... you say it not practical to retrieve the information any more. If I take a sugar cube and eat it, how would you theoretically - if not practically - reconstruct it a day later?

We know that there is no law of conservation for the entropy. It is quite the contrary: If we have a closed system without exchange of heat the entropy cannot get less. It will reach the max. If we have not a closed system but a stream of entropy only into a system, the entropy will increase...

0 apples + 0 oranges = 0 fruit I know a bit dimensional analysis but also I know a bit about the empty set in mathematics. As you might have seen in my post, in the top, this was not about a philosophical discussion if 1 apple +2 oranges may in the end be n bananas. It was about the number 0...

"Sorry, professor, I am not to very convinced by your arguments. I have questions... I learned in math that I do just do things betweens numbers. I learned that there are mathematical objects like numbers or vectors. I learned the rules how to do mathematics between them. As an example, I can...