Recent content by Oliver321

1. I General solution of the hydrogen atom Schrödinger equation

Thank you! But why is it that ‚the velocity you are trying to define has nothing to do with any velocity that could has been ascribed to the particle prior the first position measurement‘? Is that only a problem arising due to measurement ‚inperfection‘? This would be unsatisfying.
2. I General solution of the hydrogen atom Schrödinger equation

Thank you, this is helping me a lot! But I don’t understand another thing: If I measure the energy of an electron, how do I know in wich state (e.g. a superposition of eigenstates) it was before? For example: Could it be that a single electron in hydrogen is in a superposition of 1s and 2s...
3. I General solution of the hydrogen atom Schrödinger equation

Hello everyone! I have two questions wich had bothered me for quite some time. I am sorry if they are rather trivial. The first is about the general solution of the hydrogen atom schrödinger-equation: We learned in our quantum mechanics class that the general solution of every quantum system is...
4. B Gravitational force between two masses

Thank you all, I think I do understand!
5. B Gravitational force between two masses

Thanks for the awnser. But why not? Isn’t the force on m1 the same as on m2 because of Newton’s third law? Thank you!
6. B Gravitational force between two masses

Thank you! But would this mean, that the following is true: So let’s say, that we have two masses m1, m2. Mass m1 is accelerated toward m2 with a magnitude of $$a1=G\frac{m2}{r^2}$$ And then also m2 would be accelerated by $$a2=G\frac{m1}{r^2}$$ So the acceleration of the approaching of both...
7. B Gravitational force between two masses

Hello everyone. Probably this question is trivial, but nevertheless I am confused about newtons law of motion: $$F=G\frac{m_1m_2}{r^2}$$ Now, some sources say, that F is the force between the two masses m1 and m2. Other sources say, that F is the force that m1 exhibits on m2. But isn’t this a...

Thank you!

Yes, that’s absolutely right. I have a mistake in my calculation above. So here is what I think: Let’s talk about only one sort particle: H=U+pV=ST-pV+N##\mu##+pV=ST+N##\mu## So if H is really ST+N##\mu## than d(H-ST-N##\mu##)=0 This is true because of Gibbs dulem...

I have also ignored it in the definition of H=U-pV. U has the chemical potential in it. If I make the calculation with the chemical potential from the beginning on I get: d(H-TS)=Vdp-SdT-N##\mu## wich is zero.

But Vdp-SdT=0 because of the Gibbs Dulem relation (I am ignoring the chemical Potential)