Recent content by I_laff
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Undergrad What Determines the Maximum Number of Microstates at Equilibrium?
Ah of course. Thanks, that cleared things up.- I_laff
- Post #5
- Forum: Thermodynamics
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Undergrad What Determines the Maximum Number of Microstates at Equilibrium?
Apologies if the answer is obvious, but why does ## \Omega(E_1)\Omega(E_2) ## give the total number of states?- I_laff
- Post #3
- Forum: Thermodynamics
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Undergrad What Determines the Maximum Number of Microstates at Equilibrium?
## \Omega(E_1)## is the number of microstates accessible to a system when it has an energy ##E_1## and ##\Omega(E_2)## is the number of microstates accessible to the system when it has an energy ##E_2##. I understand that each microstate has equal probability of being occupied, but could...- I_laff
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- microstates statistical mechanics thermal thermal physics
- Replies: 4
- Forum: Thermodynamics
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Graduate Where does the equation of the velocity of a free vortex come from?
Thanks for the reply, my question better put is, where does the second equation, [SIZE=16px][FONT=Noto Sans]$$\vec{v}(\vec{r})=\frac{c}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$ come from. Is it simply that equation because the velocity field of that equation matches the description... -
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Graduate Where does the equation of the velocity of a free vortex come from?
I am aware of the equation of the velocity of a forced vortex, which is simply ## v = r\omega ##. However, the velocity for a free vortex is ## v = \frac{c}{r} ## (## c ## is the 'circulation constant', ## r ## is the radius, ## v ## is the velocity and ## \omega ## is the angular velocity). I... -
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Undergrad Boltzmann Distribution and microstate probabilities
For a canonical ensemble the probability of occupying a certain microstate varies depending on the energy, however I thought that every microstate has an equal chance of being occupied. So what part of the canonical ensemble have I misunderstood?- I_laff
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- Boltzmann Boltzmann distribution Distribution Probabilities Thermal physics
- Replies: 2
- Forum: Thermodynamics
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Undergrad What is a symmetric ODE / what does it mean when an ODE is symmetric?
How can an ODE be symmetric? How would you plot an ODE to show off this property? (i.e. what would be the axes?)- I_laff
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- Differential equations Mean Ode Symmetric
- Replies: 1
- Forum: Differential Equations
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Undergrad Particle in a box, boundary co-ordinate change
Actually would the wavefunction be ##\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})## instead?- I_laff
- Post #2
- Forum: Quantum Physics
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Undergrad Particle in a box, boundary co-ordinate change
If you have a particle in a 1-d box with a finite potential when ##0 < x < L ## and an infinite potential outside this region, then the normalised wavefunction used to describe said particle is ## \psi (x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})##. However, if you had say instead a finite...- I_laff
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- Boundary Box Change Particle Quantum Quantum mechanics
- Replies: 3
- Forum: Quantum Physics
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Undergrad Fourier's Trick and calculation of Cn
Ah, that's a nice way to think about it :ok:- I_laff
- Post #9
- Forum: Quantum Physics
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Undergrad Fourier's Trick and calculation of Cn
I understand the properties of the Kronecker delta, what I didn't understand was why ##c_m## came out. So, let me get this right. When applying the Kronecker delta to ##\sum_{n = 1}^{\infty} c_n##, you will get a coefficient value that both ##\psi_n## and ##\psi_m## share (due to the properties...- I_laff
- Post #7
- Forum: Quantum Physics
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Undergrad Fourier's Trick and calculation of Cn
It was from Griffiths, Introduction to quantum mechanics second edition chapter 2- I_laff
- Post #5
- Forum: Quantum Physics
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Undergrad Geometrical Optics - Light ray angles on a spherical mirror
I can't see how the textbook produces the following relationships between angles: $$ \theta = \phi + \alpha \qquad (1)$$ $$ 2\theta = \alpha + \alpha ' \qquad (2)$$ My thinking is that the exterior angle theorem for triangles was used to create expression ##(1)##, but I am unsure as to how... -
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Undergrad Fourier's Trick and calculation of Cn
I referred to it as a trick because that's what the material I was reading referred to it as. I'll take a look into Fourier transforms then.- I_laff
- Post #3
- Forum: Quantum Physics
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Undergrad Fourier's Trick and calculation of Cn
I understand that the solutions to the time-independent Schrödinger equation are complete, so a linear combination of the wavefunctions can describe any function (i.e. ##f(x) = \sum_{n = 1}^{\infty}c_n\psi_n(x) = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} c_n\sin\left(\frac{n\pi}{a}x\right)## for...- I_laff
- Thread
- Calculation Fourier Quantum Wavefunctions
- Replies: 8
- Forum: Quantum Physics