Recent content by hmparticle9

We can say $$f'(b) < f'(b+\epsilon) < f'(b+2\epsilon) < ... < f'(c)$$ for some ##b < c##.

I think I understand where you are coming from now. If we let ##a## tend to ##b## then since ##c## is in ##(b,a)## we must have ##f'(b) < f'(c)##. But is that enough?

I just realized that my proof is even more wrong. I need to show that ##f'(x) > f'(y)## for ##x > y##. Sorry guys. I am well off.

What difference would that make?

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem...

I am going through this course on collision detection: https://siggraphcontact.github.io/ In this link is a PDF called course notes. Scrolling down to section 1.3, called constraints. In this section it is said that we can write bilateral constraints as ##\phi(\mathbf{x}) = 0## and unilateral...

The transitions occur between 5 energy levels, from ##n=0## to ##n=4##. I only included transitions between neighbouring energy levels because #26. I should really include all possible transitions. Well if we limit transitions between shown energy levels from post #43, then from the 5 shown we...

"Can you use this diagram to make a list of the energies of the missing photons that appear as dips?" $$0, 2\frac{\hbar^2}{2I}, 6\frac{\hbar^2}{2I}, 12\frac{\hbar^2}{2I}$$ There are 4 transitions.

So I said, in words, "The red line shows a dip in the intensity of detected radiation. This means that the sample of CO has absorbed energy." And you said that was correct. Okay. So we want a mathematical expression for this energy. Surely this dip in energy has to equal the energy of the...

Are we interested in: $$\Delta \nu = \frac{\hbar^2}{I}?$$

##\hbar## is the same as you. Mass of carbon ##2 \times 10^{-26}## and mass of oxygen ##2.656 \times 10^{-26}## To make the dimensions work out I said: $$\Delta \nu \approx 4 \text{ cm}^{-1} = 12 \times 10^{10} \text{ s}^{-1}$$

Look at post #9. my expression for ##a## is correct

I get ##2.8 \times 10^{-10} \text{m}##

@TensorCalculus @kuruman As part of my maths degree I did vector calculus, classical mechanics, fluid dynamics. Maybe I should get Y and F. I have read Shankar's "Fundamentals of Physics I and II". I did all the problems. These books cover a wide array of physics.

I don't understand. What is wrong with my post #29? $$ a = \sqrt{\frac{\hbar}{4} \frac{m_1+m_2}{m_1m_2}}$$