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LPZ

 Discussion on question by moonblink: …

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LPZ
Dec 3, 2024 01:18
I would argue that $\hbar$ hides the $h$ rather than $h$. It's just that physicists do not want to write the $2\pi$ in the Fourier transforms $e^{ixp/\hbar}$ or $e^{-itE/\hbar}$.
 

 Discussion on question by lolloriffic…

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LPZ
Oct 27, 2024 04:23
I would suggest to reframe your question. Instead of asking why the speed if light is this value, rather why the human scale velocities $1m/s$ is so slow compared to the speed of light. It is like how Schrödinger answers why atoms are so small even though their size is fixed by physics (Bohr radius) by explaining why humans must be so large.
 

 Discussion on question by lolloriffic…

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LPZ
Oct 24, 2024 18:15
I would suggest to reframe your question. Instead of asking why the speed if light is this value, rather why the human scale velocities $1m/s$ is so slow compared to the speed of light. It is like how Schrödinger answers why atoms are so small even though their size is fixed by physics (Bohr radius) by explaining why humans must be so large.
 

 Discussion between Jagerber48 and LPZ

Imported from a comment discussion on physics.stackexchange.co...
LPZ
Jun 28, 2024 16:18
Perhaps you are mixing up the adjoint representation and this specific representation where the Lie bracket is implemented as commutators of matrices. Once you know the Lie structure, the underlying implementation is irrelevant, it is purely a question of $\mathfrak{so}(3)$
LPZ
Jun 28, 2024 16:15
I think my previous comment answers your question. To make things even more explicit, it’s best to write:$$[J_i,J_j]=\epsilon_{ijk} \\ (J_i)_{jk}=\epsilon_{ijk}$$ Compare that to the general formula:$$[e_i,e_j]=f_{ij}^ke_k\\ad(e_i)_j^k=f_{ij}^k$$
LPZ
Jun 28, 2024 16:10
I think my previous comment answers your question. To make things even more explicit, it’s best to write:$$[J_i,J_j]=\epsilon_{ijk}(J_i)_{jk}=\epsilon_{ijk}$$ Compare that to the general formula:$$[e_i,e_j]=f_{ij}^ke_k\\ad(e_i)_j^k=f_{ij}^k$$
LPZ
Jun 28, 2024 07:40
Disregard my previous message. Things are usually done in reverse. The relation that you want to prove is assumed to define $J$. In a general Lie algebra, the adjoint action is: $$ad(x)y = [x,y]$$. Given a basis $e_i$, you have the structure coefficients:$$[e_i,e_j] = f_{ij}^ke_k$$. This imposes the coefficients of the matrix:$$ad(e_i)_j^k = f_{ij}^k$$
LPZ
Jun 28, 2024 07:35
It comes from the definition of $J$. In a general Lie algebra, given a basis $e_i$, you have the structure coefficients:$$[e_i,e_j] = f_{ij}^ke_k$$ This is used to define the the adjoint action:$$ad(e_i)_j^k = f_{ij}^k$$ $f$ is the generalisation of $J$, and it is immediate that $$ad(x)y = [x,y]$$
LPZ
Jun 26, 2024 23:32
Ok I'll write up a proper answer when I find the time
LPZ
Jun 26, 2024 23:30
You can render mathjax in chat! math.ucla.edu/~robjohn/math/mathjax.html
LPZ
Jun 26, 2024 22:05
this is seen explicitly by the Lie bracket. Let $x,y\in\mathbb R^3$, then $$[x\cdot s,y\cdot s] = (x\times y)\cdot s$$ with $s=\sigma/2$, i.e. the Lie bracket structure of $\mathfrak{so}(3)$
LPZ
Jun 26, 2024 22:03
No, for the adjoint, it is only a matter of $\mathfrak{so}(3)$. The 3D space spanned by $\sigma_x,\sigma_y,\sigma_z$ is a representation of $\mathfrak{so}(3)$.
LPZ
Jun 26, 2024 21:59
This is where I feel the isomorphism between $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ is needed to convert the $\sigma$ that appears on the RHS into a $J$
LPZ
Jun 26, 2024 21:59
Like, I guess the LHS is $Ad(\exp(i(\theta/2)(\hat{n}\cdot \sigma))(\sigma)$. I see that this equals the LHS of (*).

So the formula Ad/ad/exp formula would imply $\exp(ad(i(\theta/2)(\hat{n}\cdot\sigma))(\sigma)$ is the RHS. But then the question is how to show that this equals $e^{\theta \hat{n}\cdot J} \sigma$
LPZ
Jun 26, 2024 21:58
In the ad/Ad formula it seems we would calculate $\exp(ad(i\theta/2 \hat{n}\cdot \sigma)$, but in the expression in the question we have $J$ appear on the RHS, not $\sigma$
LPZ
Jun 26, 2024 21:57
So we have $Ad \circ \exp = \exp \circ ad$. I'm trying to understand exactly how that applies in this case. We have $U^{\dagger}\sigma U = R\sigma$. The LHS can be written as $Ad_{U} \sigma$ and $U$ is $\exp(i(\theta/2)(\hat{n}\cdot \sigma))$ so that already covers the LHS. The RHS is tricker.
LPZ
Jun 26, 2024 21:56
Just copying your message with the dollars
LPZ
Jun 26, 2024 15:53
Well, there are already many answers to this on this site. You can check out a previous answer of mine physics.stackexchange.com/q/805505/333454, and in the linked answers of QMechanics as well. Is there something ore specific you are interested in?
LPZ
Jun 26, 2024 14:58
More formally, what only matters is the action of $SO(3)$ on the components of $\sigma$ which is precisely the adjoint representation of $\mathfrak{so}(3)$
LPZ
Jun 26, 2024 14:56
In your case, it is applied to $\mathfrak{so}(3)$. $\mathfrak{so}(2)$ just enters the game if you want to implement the representation
LPZ
Jun 26, 2024 14:54
Yes, the proof that I provided generalises and this is why the more abstract answers refer to
LPZ
Jun 26, 2024 14:52
But yes, it all boils down to the adjoint formula $Ad\circ\exp = \exp\circ ad$
LPZ
Jun 26, 2024 14:52
Similarly, you would not have a Pauli vector anymore, you would rather have a Pauli bivector.
LPZ
Jun 26, 2024 14:50
The $J_{ij}$ are the generators of rotations in the $ij$ plane. Using the conversion in 3D $J_{ij} = \epsilon_{ijk}J_k$ you can check the equivalence
LPZ
Jun 26, 2024 14:49
The pattern that generalises in arbitrary dimensions is that a rotation is decomposed in terms of elementary rotations along planes that are orthogonal to each other
LPZ
Jun 26, 2024 14:44
It's just that you need to be more careful with your book keeping since you cannot use the Levi-Civita symbol. But the relations are the same:$$[J_{ij},J_{kl}] = \delta_{jk}J_{li}-\delta_{ik}J_{lj}+\delta_{il}J_{kj}-\delta_{jl}J_{ki}$$
LPZ
Jun 26, 2024 14:44
Well, that's usually how you define $J$, not from writing out the matrices as you did. It comes from the adjoint representation, the components are the structure coefficients of the Lie algebra.
LPZ
Jun 26, 2024 14:44
There were also some errors, which didn't help for clarity... I've added some details and there shouldn't be any mistakes now
 

 Resistively and Capacitively Shunted …

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LPZ
May 22, 2024 19:20
Once again, the most obvious improvement is the SDE solver, you can try using smaller dt to see if it changes anything. The best way would be to use ready coded package with adaptative time steps. But once again, matching experimental data is actual research. I have no idea where your data comes from so the assumptions may not hold. Depending on it, you may not have the correct noise modeling or the entire model may not be appropriate.
LPZ
May 22, 2024 18:58
But before all of that, a pretty obvious starting point is to look at the SDE solver. How did you choose dt and did you try with other methods? (Euler is pretty basic after all)
LPZ
May 22, 2024 18:56
I should test the code to make sure. But a good place to start if you have different experimental data is to check whether the difference in $\Gamma$ is systematic or not (there could be just a prefactor missing or somthing). Furthermore, as the slides later explain, the white noise spectrum may not be realistic for experiments. To reproduce the 1/f spectrum, you’ll need to apply a low pass filter
LPZ
May 21, 2024 21:05
Actually, just realized it’s the opposite, your first curve is $\Gamma=.006$ since it is less smoothed out. Are you sure you used the correct formula for noise? For $\Gamma=.006$, $\sqrt{4\Gamma}\sim 1.5$ which is suspiciously close to your $.2$…
LPZ
May 21, 2024 19:32
Just to clarify, the first blue curve is for $\Gamma=.2$ while the second blue curve is for $\Gamma=.006$ and the red curves are the same experimental data?
 

 Discussion between LPZ and Luthier415Hz

Imported from a comment discussion on math.stackexchange.com/q...
LPZ
Apr 2, 2024 11:19
Thanks you too
LPZ
Apr 2, 2024 11:19
And I meant $x^2+x$ for the quadratic example, not $x^2$.
LPZ
Apr 2, 2024 11:18
Oh ok I get what you meant. Actually, the pattern is more to add a linear term. The idea is that generically, having $f'(p)=1$ is not stable, which is why you add linear terms to make it generic.
LPZ
Apr 2, 2024 11:16
Saddle-node bifurcation
In the mathematical area of bifurcation theory a saddle-node bifurcation, tangential bifurcation or fold bifurcation is a local bifurcation in which two fixed points (or equilibria) of a dynamical system collide and annihilate each other. The term 'saddle-node bifurcation' is most often used in reference to continuous dynamical systems. In discrete dynamical systems, the same bifurcation is often instead called a fold bifurcation. Another name is blue sky bifurcation in reference to the sudden creation of two fixed points.If the phase space is one-dimensional, one of the equilibrium points is...
LPZ
Apr 2, 2024 11:16
No, still the same, $x^2$ is not stable, you can have a saddle-node bifurcation by adding a linear term $x^2+\epsilon x$
LPZ
Apr 2, 2024 11:15
I'm only looking at the dynamic in $\mathbb R$. It's essentially like $x^2$, with the stable fixed point at $0$ and an unstable fixed point at the other real root of $x^4+x^2=x$
LPZ
Apr 2, 2024 11:12
actually your example $x^4+x^2$ is stable, perhaps you mean $x^4+x$? If so, then yes, a linear perturbation still does the trick
LPZ
Apr 2, 2024 11:10
it won't always work, but it helps for simple bifurcations like this one
LPZ
Apr 2, 2024 11:09
I don't get your question. I just perturbed $T_{-1}\to T_{-1}+\epsilon x$ which changes the slope by $\epsilon$ at the origin without changing the fixed point. The idea is actually pretty standard, it's a pitchfork bifurcation
LPZ
Apr 2, 2024 11:09
Intuitively, the structural instability is due to the behaviour at the fixed point $x=0$. The fixed point is critical, which is not stable by a $C^1$ perturbation by modifying the local slope. The simplest way to do that is to add a small linear term.
LPZ
Apr 2, 2024 11:09
consider instead $x^3+(1+\epsilon)x$
 

 Discussion between LPZ and MRule

Imported from a comment discussion on physics.stackexchange.co...
LPZ
Jun 5, 2023 15:00
Glad it was useful! I'll try to come up with examples of systems using the triharmonic equation if I can.
LPZ
Jun 5, 2023 14:01
Well, you can look at the literature for the triharmonic equation, but I don't know about physical applications on the top of my head. Perhaps the physically relevant quantity is not your original field $u$, but rather its derivatives?
LPZ
Jun 5, 2023 13:53
it's just to avoid cluttering the comment section
LPZ
Jun 5, 2023 13:52
I think they ar public
LPZ
Jun 5, 2023 13:52
Yes I understand what you are trying to do, and I have no problem with the objective. It's just the methodology that I am investigating, since you don't leave equations
LPZ
Jun 5, 2023 13:49
and when you say 1s,2p,3d, since you are in 2D, I guess that the first number is the number of the Bessel function zero, but you don't have orbital number, only the "magnetic" one. Ok, but if you only have five points, the fit better be perfect. Perhaps including a graphe of the fit would help