I did this long-term review on my normal UROP desktop computer with the 64-bit edition of the OS. Follow the jump to see how it fared. Also do note that there are more days logged because I intend to use it for about 60-80 full hours of work, which is the equivalent of 7-10 full days in the summer, though now I am working on a part-time basis as classes have started.
Read more »Monday, 8 April 2013
Sunday, 31 March 2013
There were two posts that got a few comments each, so I will repost most of those.
Commenter Juan Carlos García Ramírez had this to say: "I still prefer xfce :D so linux mint 14 (Nadia) xfce for me".
Commenter Mechatotoro had this bit of support: "@Prashanth, Thank you for your time with Pardus and your review. Good luck going back to school! @Mega, I just came from your blog and recommended you to read this review. I guess you are too fast :-)"
Thanks to all those who commented on those posts. I am back on campus now, and the semester isn't about to wait for me to settle down, so it will be back in full swing any minute now. This means that my post frequency will once again decrease through the rest of the semester. Anyway, if you like what I write, please continue subscribing and commenting!
Review: Linux Mint MATE 201303
Reader Gary Newell said, "This is a good in depth review. I am currently using the Mint 14 Cinnamon release (Ubuntu base). If you have a powerful enough PC then Cinnamon is the best desktop as far as I can tell. I prefer the Consort desktop used by SolusOS to the MATE desktop and if I have an older PC I actually overall prefer to use XFCE and so tend to run Xubuntu."Commenter Juan Carlos García Ramírez had this to say: "I still prefer xfce :D so linux mint 14 (Nadia) xfce for me".
Review: Pardus 2013 KDE
Reader Megatotoro shared, "I tested Pardus 2013 as well. In my case, I could test the repositories using Synaptic and could download the localization file for my language. I did notice that even so, some programs were still in Turkish (VLC, Synaptic). I have the Release Candidate installed on my laptop and it is greatly stable."Commenter Mechatotoro had this bit of support: "@Prashanth, Thank you for your time with Pardus and your review. Good luck going back to school! @Mega, I just came from your blog and recommended you to read this review. I guess you are too fast :-)"
Thanks to all those who commented on those posts. I am back on campus now, and the semester isn't about to wait for me to settle down, so it will be back in full swing any minute now. This means that my post frequency will once again decrease through the rest of the semester. Anyway, if you like what I write, please continue subscribing and commenting!
Friday, 29 March 2013
My spring break is coming to an end (I only have 1.5 more days), so I figured it might be nice to do another review while I still can. Today I'm reviewing Pardus 2013.
Pardus is a distribution developed at least in part by the Turkish military. It used to not be based on any other distribution and used its unique PISI package management system, which featured delta upgrades (meaning that only the differences between package versions would be applied for upgrades, greatly reducing their size). Since then, though, the organization largely responsible for the development of Pardus went through some troubles. One result was the forking of Pardus into PISI Linux to further develop the original alpha release of Pardus 2013. The other result was the rebasing of Pardus on Debian, abandoning PISI in that regard. Now Pardus 2013 is a distribution based on Debian 7 "Wheezy" that uses either KDE 4.8 or GNOME 3 (whatever version is packaged in the latest version of Debian, though I'm not sure what that is).
I reviewed Pardus on a live USB made with MultiSystem. Follow the jump to see what it's like.
Read more » |
| Main Screen + KDE Kickoff Menu |
I reviewed Pardus on a live USB made with MultiSystem. Follow the jump to see what it's like.
Wednesday, 27 March 2013
As an update to a previous post about my adventures in QED-land for 8.06, I emailed my recitation leader about whether my intuition about the meaning of the Fourier components of the electromagnetic potential solving the wave equation (and being quantized to the ladder operators) was correct. He said it basically is correct, although there are a few things that, while I kept in mind at that time, I still need to keep in mind throughout. The first is that the canonical quantization procedure uses the potential $\vec{A}$ as the coordinate-like quantity and finds the conjugate momentum to this field to be proportional to the electric field $\vec{E}$, with the magnetic field nowhere to be found directly in the Hamiltonian. The second is that there is a different harmonic oscillator for each mode, and the number eigenstates do not represent the energy of a given photon but instead represent the number of photons present with an energy corresponding to that mode. Hence, while coherent states do indeed represent points in the phase space of $(\vec{A}, \vec{E})$, the main point is that the photon number can fluctuate, and while classical behavior is recovered for large numbers $n$ of photons as the fluctuations of the number are $\sqrt{n}$ by Poisson statistics, the interesting physics happens for low $n$ eigenstates or superpositions thereof in which $a$ and $a^{\dagger}$ play the same role as in the usual quantum harmonic oscillator. Furthermore, the third issue is that only a particular mode $\vec{k}$ and position $\vec{x}$ can be considered, because the electromagnetic potential has a value for each of those quantities, so unless those are held constant, the picture of phase space $(\vec{A}, \vec{E})$ becomes infinite-dimensional. Related to this, the fourth and fifth issues are, respectively, that $\vec{A}$ is used as the field and $\vec{E}$ as its conjugate momentum rather than using $\vec{E}$ and $\vec{B}$ because the latter two fields are coupled to each other by the Maxwell equations so they form an overcomplete set of degrees of freedom (or something like that), whereas using $\vec{A}$ as the field and finding its conjugate momentum in conjunction with a particular gauge choice (usually the Coulomb gauge $\nabla \cdot \vec{A} = 0$) yields the correct number of degrees of freedom. These explanations seem convincing enough to me, so I will leave those there for the moment.
Another major issue that I brought up with him for which he didn't give me a complete answer was the issue that the conjugate momentum to $\vec{A}$ was being found through \[ \Pi_j = \frac{\partial \mathcal{L}}{\partial (\partial_t A_j)} \] given the Lagrangian density $\mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 - \vec{B}^2 \right)$ and the field relations $\vec{E} = -\frac{1}{c}\partial_t \vec{A}$ & $\vec{B} = \nabla \times \vec{A}$. This didn't seem manifestly Lorentz-covariant to me, because in the class 8.033 — Relativity, I had learned that the conjugate momentum to the electromagnetic potential $A^{\mu}$ in the above Lagrangian density would be the 2-index tensor \[ \Pi^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu})} .\] This would make a difference in finding the Hamiltonian density \[ \mathcal{H} = \sum_{\mu} \Pi^{\mu} \partial_t A_{\mu} - \mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right). \] I thought that the Hamiltonian density would need to be a Lorentz-invariant scalar just like the Lagrangian density. As it turns out, this is not the case, because the Hamiltonian density represents the energy which explicitly picks out the temporal direction as special, so time derivatives are OK in finding the momentum conjugate to the potential; because the Lagrangian and Hamiltonian densities looks so similar, it looks like both could be Lorentz-invariant scalar functions, but deceptively, only the former is so. At this point, I figured that because the Hamiltonian and (not field conjugate, but physical) momentum looked so similar, they could arise from the same covariant vector. However, there is no "natural" 1-index vector with which to multiply the Lagrangian density to get some sort of covariant vector generalization of the Hamiltonian density, though there is a 2-index tensor, and that is the metric. I figured here that the Hamiltonian and momentum for the electromagnetic field could be related to the stress-energy tensor, which gives the energy and momentum densities and fluxes. After a while of searching online for answers, I was quite pleased to find my intuition to be essentially spot-on: indeed the conjugate momentum should be a tensor as given above, the Legendre transformation can then be done in a covariant manner, and it does in fact turn out that the result is just the stress-energy tensor \[ T^{\mu \nu} = \sum_{\mu, \xi} \Pi^{\mu \xi} \partial^{\nu} A_{\xi} - \mathcal{L}\eta^{\mu \nu} \] (UPDATE: the index positions have been corrected) for the electromagnetic field. Indeed, the time-time component is exactly the energy/Hamiltonian density $\mathcal{H} = T_{(0, 0)}$, and the Hamiltonian $H = \sum_{\vec{k}} \hbar\omega(\vec{k}) \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k})) = \int T_{(0, 0)} d^3 x$. As it turns out, the momentum $\vec{p} = \sum_{\vec{k}} \hbar\vec{k} \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k}))$ doesn't look similar just by coincidence: $p_j = \int T_{(0, j)} d^3 x$. The only remaining point of confusion is that it seems like the Hamiltonian and momentum should together form a Lorentz-covariant vector $p_{\mu} = (H, p_j)$, yet if the stress-energy tensor respects Lorentz-covariance, then integrating over the volume element $d^3 x$ won't respect transformations in a Lorentz-covariant manner. I guess because the individual components of the stress-energy tensor transform under a Lorentz boost and the volume element does as well, then maybe the vector $p_{\mu}$ as given above will respect Lorentz-covariance. (UPDATE: another issue I was having but forgot to write before clicking "Publish" was the fact that only the $T_{(0, \nu)}$ components are being considered. I wonder if there is some natural 1-index Lorentz-convariant vector $b_{\nu}$ to contract with $T_{\mu \nu}$ so that the result is a 1-index vector which in a given frame has a temporal component given by the Hamiltonian density and spatial components given by the momentum density.) Overall, I think it is interesting that this particular hang-up was over a point in classical field theory and special relativity and had nothing to do with the quantization of the fields; in any case, I think I have gotten over the major hang-ups about this and can proceed reading through what I need to read for the 8.06 paper.
Another major issue that I brought up with him for which he didn't give me a complete answer was the issue that the conjugate momentum to $\vec{A}$ was being found through \[ \Pi_j = \frac{\partial \mathcal{L}}{\partial (\partial_t A_j)} \] given the Lagrangian density $\mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 - \vec{B}^2 \right)$ and the field relations $\vec{E} = -\frac{1}{c}\partial_t \vec{A}$ & $\vec{B} = \nabla \times \vec{A}$. This didn't seem manifestly Lorentz-covariant to me, because in the class 8.033 — Relativity, I had learned that the conjugate momentum to the electromagnetic potential $A^{\mu}$ in the above Lagrangian density would be the 2-index tensor \[ \Pi^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu})} .\] This would make a difference in finding the Hamiltonian density \[ \mathcal{H} = \sum_{\mu} \Pi^{\mu} \partial_t A_{\mu} - \mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right). \] I thought that the Hamiltonian density would need to be a Lorentz-invariant scalar just like the Lagrangian density. As it turns out, this is not the case, because the Hamiltonian density represents the energy which explicitly picks out the temporal direction as special, so time derivatives are OK in finding the momentum conjugate to the potential; because the Lagrangian and Hamiltonian densities looks so similar, it looks like both could be Lorentz-invariant scalar functions, but deceptively, only the former is so. At this point, I figured that because the Hamiltonian and (not field conjugate, but physical) momentum looked so similar, they could arise from the same covariant vector. However, there is no "natural" 1-index vector with which to multiply the Lagrangian density to get some sort of covariant vector generalization of the Hamiltonian density, though there is a 2-index tensor, and that is the metric. I figured here that the Hamiltonian and momentum for the electromagnetic field could be related to the stress-energy tensor, which gives the energy and momentum densities and fluxes. After a while of searching online for answers, I was quite pleased to find my intuition to be essentially spot-on: indeed the conjugate momentum should be a tensor as given above, the Legendre transformation can then be done in a covariant manner, and it does in fact turn out that the result is just the stress-energy tensor \[ T^{\mu \nu} = \sum_{\mu, \xi} \Pi^{\mu \xi} \partial^{\nu} A_{\xi} - \mathcal{L}\eta^{\mu \nu} \] (UPDATE: the index positions have been corrected) for the electromagnetic field. Indeed, the time-time component is exactly the energy/Hamiltonian density $\mathcal{H} = T_{(0, 0)}$, and the Hamiltonian $H = \sum_{\vec{k}} \hbar\omega(\vec{k}) \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k})) = \int T_{(0, 0)} d^3 x$. As it turns out, the momentum $\vec{p} = \sum_{\vec{k}} \hbar\vec{k} \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k}))$ doesn't look similar just by coincidence: $p_j = \int T_{(0, j)} d^3 x$. The only remaining point of confusion is that it seems like the Hamiltonian and momentum should together form a Lorentz-covariant vector $p_{\mu} = (H, p_j)$, yet if the stress-energy tensor respects Lorentz-covariance, then integrating over the volume element $d^3 x$ won't respect transformations in a Lorentz-covariant manner. I guess because the individual components of the stress-energy tensor transform under a Lorentz boost and the volume element does as well, then maybe the vector $p_{\mu}$ as given above will respect Lorentz-covariance. (UPDATE: another issue I was having but forgot to write before clicking "Publish" was the fact that only the $T_{(0, \nu)}$ components are being considered. I wonder if there is some natural 1-index Lorentz-convariant vector $b_{\nu}$ to contract with $T_{\mu \nu}$ so that the result is a 1-index vector which in a given frame has a temporal component given by the Hamiltonian density and spatial components given by the momentum density.) Overall, I think it is interesting that this particular hang-up was over a point in classical field theory and special relativity and had nothing to do with the quantization of the fields; in any case, I think I have gotten over the major hang-ups about this and can proceed reading through what I need to read for the 8.06 paper.
Tuesday, 26 March 2013
There were two things that I would like to post here today. The first is something I have been mulling over for a while. The second is something that I thought about more recently.
Time evolution in nonrelativistic quantum mechanics occurs according to the [time-dependent] Schrödinger equation \[ H|\Psi\rangle = i\hbar \frac{\partial}{\partial t} |\Psi\rangle .\] While this at first may seem intractable, the trick is that typically the Hamiltonian is not time-dependent, so a candidate solution could be $|\Psi\rangle = \phi(t)|E\rangle$. Plugging this back in yields time evolution that occurs through the phase $\phi(t) = e^{-\frac{iEt}{\hbar}}$ applied to energy eigenstates that solve \[ H|E\rangle = E \cdot |E\rangle \] and this equation is often called the "time-independent Schrödinger equation". When I was taking 8.04 — Quantum Physics I, I agreed with my professor who called this a misnomer, in that the Schrödinger equation is supposed to only describe time evolution, so what is being called "time-independent" is more properly just an energy eigenvalue equation. That said, I was thinking that the "time-independent Schrödinger equation" is really just like a Fourier transform of the Schrödinger equation from time to frequency (related to energy by $E = \hbar\omega$), so the former could be an OK nomenclature because it is just a change of basis. However, there are two things to note: the Schrödinger equation is basis-independent, whereas the "time-independent Schrödinger equation" is expressed only in the basis of energy eigenstates, and time is not an observable quantity (i.e. Hermitian operator) but is a parameter, so the change of basis/Fourier transform argument doesn't work in quite the same way that it does for position versus momentum. Hence, I've come to the conclusion that it is better to call the "time-independent Schrödinger equation" as the energy eigenvalue equation.
Switching gears, I was thinking about how the Biot-Savart law is derived. My AP Physics C teacher told me that the Ampère law is derived from the Biot-Savart law. However, this is patently not true, because the Biot-Savart law only works for charges moving at a constant velocity, whereas the Ampère law is true for magnetic fields created by any currents or any changing electric fields. In 8.022 — Physics II, I did see a derivation of the Biot-Savart law from the Ampère law, showing that the latter is indeed more fundamental than the former, but it involved the magnetic potential and a lot more work. I wanted to see if that derivation still made sense to me, but then I realized that because magnetism essentially springs from the combination of electricity and special relativity and because the Biot-Savart law relies on the approximation of the charges moving at a constant velocity, it should be possible to derive the Biot-Savart law from the Coulomb law and special relativity. Indeed, it is possible. Consider a charge $q$ whose electric field is \[ \vec{E} = \frac{q}{r^2} \vec{e}_r \] in its rest frame. Note that the Coulomb law is exact in the rest frame of a charge. Now consider a frame moving with respect to the charge at a velocity $-\vec{v}$, so that observers in the frame see the charge move at a velocity $\vec{v}$. Considering only the component of the magnetic field perpendicular to the relative motion, noting that there is no magnetic field in the rest frame of the charge yields, and considering the low-speed limit (which is the range of validity of the Biot-Savart law) $\left|\frac{\vec{v}}{c}\right| \ll 1$ so that $\gamma \approx 1$ yields $\vec{B} \approx -\frac{\vec{v}}{c} \times \vec{E}$. Plugging in $-\vec{v}$ (the specified velocity of the new frame relative to the charge) for $\vec{v}$ (the general expression for the relative velocity) and plugging in the Coulomb expression for $\vec{E}$ yields the Biot-Savart law \[ \vec{B} = \frac{q\vec{v} \times \vec{e}_r}{cr^2}. \] One thing to be emphasized again is that the Coulomb law is exact in the rest frame of the charge, while the Biot-Savart law is always an approximation because a moving charge will have an electric field that deviates from the Coulomb expression; the fact that the Biot-Savart law is a low-speed inertial approximation is why I feel comfortable doing the derivation this way.
Time evolution in nonrelativistic quantum mechanics occurs according to the [time-dependent] Schrödinger equation \[ H|\Psi\rangle = i\hbar \frac{\partial}{\partial t} |\Psi\rangle .\] While this at first may seem intractable, the trick is that typically the Hamiltonian is not time-dependent, so a candidate solution could be $|\Psi\rangle = \phi(t)|E\rangle$. Plugging this back in yields time evolution that occurs through the phase $\phi(t) = e^{-\frac{iEt}{\hbar}}$ applied to energy eigenstates that solve \[ H|E\rangle = E \cdot |E\rangle \] and this equation is often called the "time-independent Schrödinger equation". When I was taking 8.04 — Quantum Physics I, I agreed with my professor who called this a misnomer, in that the Schrödinger equation is supposed to only describe time evolution, so what is being called "time-independent" is more properly just an energy eigenvalue equation. That said, I was thinking that the "time-independent Schrödinger equation" is really just like a Fourier transform of the Schrödinger equation from time to frequency (related to energy by $E = \hbar\omega$), so the former could be an OK nomenclature because it is just a change of basis. However, there are two things to note: the Schrödinger equation is basis-independent, whereas the "time-independent Schrödinger equation" is expressed only in the basis of energy eigenstates, and time is not an observable quantity (i.e. Hermitian operator) but is a parameter, so the change of basis/Fourier transform argument doesn't work in quite the same way that it does for position versus momentum. Hence, I've come to the conclusion that it is better to call the "time-independent Schrödinger equation" as the energy eigenvalue equation.
Switching gears, I was thinking about how the Biot-Savart law is derived. My AP Physics C teacher told me that the Ampère law is derived from the Biot-Savart law. However, this is patently not true, because the Biot-Savart law only works for charges moving at a constant velocity, whereas the Ampère law is true for magnetic fields created by any currents or any changing electric fields. In 8.022 — Physics II, I did see a derivation of the Biot-Savart law from the Ampère law, showing that the latter is indeed more fundamental than the former, but it involved the magnetic potential and a lot more work. I wanted to see if that derivation still made sense to me, but then I realized that because magnetism essentially springs from the combination of electricity and special relativity and because the Biot-Savart law relies on the approximation of the charges moving at a constant velocity, it should be possible to derive the Biot-Savart law from the Coulomb law and special relativity. Indeed, it is possible. Consider a charge $q$ whose electric field is \[ \vec{E} = \frac{q}{r^2} \vec{e}_r \] in its rest frame. Note that the Coulomb law is exact in the rest frame of a charge. Now consider a frame moving with respect to the charge at a velocity $-\vec{v}$, so that observers in the frame see the charge move at a velocity $\vec{v}$. Considering only the component of the magnetic field perpendicular to the relative motion, noting that there is no magnetic field in the rest frame of the charge yields, and considering the low-speed limit (which is the range of validity of the Biot-Savart law) $\left|\frac{\vec{v}}{c}\right| \ll 1$ so that $\gamma \approx 1$ yields $\vec{B} \approx -\frac{\vec{v}}{c} \times \vec{E}$. Plugging in $-\vec{v}$ (the specified velocity of the new frame relative to the charge) for $\vec{v}$ (the general expression for the relative velocity) and plugging in the Coulomb expression for $\vec{E}$ yields the Biot-Savart law \[ \vec{B} = \frac{q\vec{v} \times \vec{e}_r}{cr^2}. \] One thing to be emphasized again is that the Coulomb law is exact in the rest frame of the charge, while the Biot-Savart law is always an approximation because a moving charge will have an electric field that deviates from the Coulomb expression; the fact that the Biot-Savart law is a low-speed inertial approximation is why I feel comfortable doing the derivation this way.
Monday, 25 March 2013
For those of you who have been waiting for a review, I think I may have said before that my writing would shift more to science-y stuff and away from distribution reviews. However, that does not mean that reviews will stop entirely. I'm on spring break now and have a little more time to do these reviews, so today I am reviewing Linux Mint MATE 201303, which came out earlier this week.
This is the version of Linux Mint based on Debian rather than Ubuntu. It uses a variant of a rolling-release model, in that while existing users can get the latest and greatest software simply by applying updates as usual, the updates come in large bundles (I almost want to say they are like the Microsoft Windows Service Packs, except that they work) rather than individual package files. This means that the most common packages used on a Debian-based Linux Mint system are tested so that they can be guaranteed to work not only individually but also together, so that the problem of an individual update breaking other dependencies becomes moot. Around the time of releasing a new update pack, a new ISO file snapshot of the distribution is released, as was the case this time around.
I reviewed the [32-bit] MATE edition using a live USB made with MultiSystem; I wanted to review the Cinnamon edition too, but it refused to boot, so I will leave my assessment of it at that. I also did an installation of this (which regular readers know is rare), so you will have to follow the jump to see what this is like.
Read more » |
| Main Screen + Linux Mint Menu |
I reviewed the [32-bit] MATE edition using a live USB made with MultiSystem; I wanted to review the Cinnamon edition too, but it refused to boot, so I will leave my assessment of it at that. I also did an installation of this (which regular readers know is rare), so you will have to follow the jump to see what this is like.
Thursday, 21 March 2013
In a post from a few days ago, I briefly mentioned the notion of imaginary time with regard to angular momentum. I'd like to go into that a little further in this post.
In 3 spatial dimensions, the flat (Euclidean) metric is $\eta_{ij} = \delta_{ij}$, which is quite convenient, as lengths are given by $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ which is just the usual Pythagorean theorem. When a temporal dimension is added, as in special relativity, the coordinates are now $x^{\mu} = (ct, x_{j})$, and the Euclidean metric becomes the Minkowski metric $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$ so that $\eta_{tt} = -1$, $\eta_{(t, j)} = 0$, and $\eta_{ij} = \delta_{ij}$. This means that spacetime intervals become $(\Delta s)^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$, which is the normal Pythagorean theorem only if $\Delta t = 0$. In general, time coordinate differences contribute negatively to the spacetime interval. In addition, Lorentz transformations are given by a hyperbolic rotation by a [hyperbolic] angle $\alpha$ equal to the rapidity given by $\frac{v}{c} = \tanh(\alpha)$. This doesn't look quite the same as normal Euclidean geometry. However, a transformation to imaginary time, called a Wick rotation, can be done by setting $\tau = it$, so $x^{\mu} = (ic\tau, x_{j})$, $\eta_{\mu \nu} = \delta_{\mu \nu}$, $(\Delta s)^2 = (c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ as in the usual Pythagorean theorem, and the Lorentz transformation is given by a real rotation by an angle $\theta = i\alpha$ (though I may have gotten some of these signs wrong so forgive me) where $\alpha$ is now imaginary. Now, the connection to the component $L_{(0, j)}$ of the angular momentum tensor should be more clear.
I first encountered this in the class 8.033 — Relativity, where I was able to explore this curiosity on a problem set. That question and the accompanying discussion seemed to say that while this is a cool thing to try doing once, it isn't really useful, especially because it does not hold true in general relativity with more general metrics $g_{\mu \nu} \neq \eta_{\mu \nu}$ except in very special cases. However, as it turns out, imaginary time does play a role in quantum mechanics, even without the help of relativity.
Schrödinger time evolution occurs through the unitary transformation $u = e^{-\frac{itH}{\hbar}}$ satisfying $uu^{\dagger} = u^{\dagger} u = 1$. This means that the probability that an initial state $|\psi\rangle$ ends after time $t$ in the same state is given by the amplitude (whose square is the probability [density]) $\mathfrak{p}(t) = \langle\psi|e^{-\frac{itH}{\hbar}}|\psi\rangle$. Meanwhile, assuming the states $|\psi\rangle$ form a complete and orthonormal basis (though I don't know if this assumption is truly necessary), the partition function $Z = \mathrm{trace}\left(e^{-\frac{H}{k_B T}}\right)$, which can be expanded in the basis $|\psi\rangle$ as $Z = \sum_{\psi} \langle\psi|e^{-\frac{H}{k_B T}}|\psi\rangle$. This, however, is just as well rewritten as $Z = \sum_{\psi} \mathfrak{p}\left(t = -\frac{i\hbar}{k_B T}\right)$. Hence, quantum and statistical mechanical information can be gotten from the same amplitudes using the substitution $t = -\frac{i\hbar}{k_B T}$, which essentially calls temperature a reciprocal imaginary time. This is not really meant to show anything more deep or profound about the connection between time and temperature; it is really more of a trick stemming from the fact that the same Hamiltonian can be used to solve problem in quantum mechanics or equilibrium statistical mechanics.
As an aside, it turns out that temperature, even when measured in an absolute scale, can be negative. There are plenty of papers of this online, but suffice it to say that this comes from a more general statistical definition of temperature. Rather than defining it (as it commonly is) as the average kinetic energy of particles, it is better to define it as a measure of the probability distribution that a particle will have a given energy. Usually, particles tend to be in lower energy states more than in higher energy states, and as a consequence, the temperature is positive. However, it is possible (and has been done repeatedly) under certain circumstances to cleverly force the system in a way that causes particles to be in higher energy states with higher probability than in lower energy states, and this is exactly the negative temperature. More formally, $\frac{1}{T} = \frac{\partial S}{\partial E}$ where $E$ is the energy and $S$ is the entropy of the system, which is a measure of how many different states the system can possibly have for a given energy. For positive temperature, if two objects of different temperatures are brought into contact, energy will flow from the hotter one to the colder, cooling the former and heating the latter until equal temperatures are achieved. For negative temperature, though, if an object with negative temperature is brought in contact with an object that has positive temperature, each object tends to increase its own entropy. Like most normal objects, the latter does this by absorbing energy, but by the definition of temperature, the former does this by releasing energy, meaning the former will spontaneously heat the latter. Hence, negative temperature is hotter than positive temperature; this is a quirk of the definition of reciprocal temperature, so really what is happening is that absolute zero on the positive side is still the coldest possible temperature, absolute zero on the negative side is now the hottest temperature, and $\pm \infty$ is in the middle.
This was really just me writing down stuff that I had been thinking about a couple of months ago. I hope this helps someone, and I also await the day when TV newscasters say "complex time brought to you by..." instead of "time and temperature brought to you by...".
In 3 spatial dimensions, the flat (Euclidean) metric is $\eta_{ij} = \delta_{ij}$, which is quite convenient, as lengths are given by $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ which is just the usual Pythagorean theorem. When a temporal dimension is added, as in special relativity, the coordinates are now $x^{\mu} = (ct, x_{j})$, and the Euclidean metric becomes the Minkowski metric $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$ so that $\eta_{tt} = -1$, $\eta_{(t, j)} = 0$, and $\eta_{ij} = \delta_{ij}$. This means that spacetime intervals become $(\Delta s)^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$, which is the normal Pythagorean theorem only if $\Delta t = 0$. In general, time coordinate differences contribute negatively to the spacetime interval. In addition, Lorentz transformations are given by a hyperbolic rotation by a [hyperbolic] angle $\alpha$ equal to the rapidity given by $\frac{v}{c} = \tanh(\alpha)$. This doesn't look quite the same as normal Euclidean geometry. However, a transformation to imaginary time, called a Wick rotation, can be done by setting $\tau = it$, so $x^{\mu} = (ic\tau, x_{j})$, $\eta_{\mu \nu} = \delta_{\mu \nu}$, $(\Delta s)^2 = (c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ as in the usual Pythagorean theorem, and the Lorentz transformation is given by a real rotation by an angle $\theta = i\alpha$ (though I may have gotten some of these signs wrong so forgive me) where $\alpha$ is now imaginary. Now, the connection to the component $L_{(0, j)}$ of the angular momentum tensor should be more clear.
I first encountered this in the class 8.033 — Relativity, where I was able to explore this curiosity on a problem set. That question and the accompanying discussion seemed to say that while this is a cool thing to try doing once, it isn't really useful, especially because it does not hold true in general relativity with more general metrics $g_{\mu \nu} \neq \eta_{\mu \nu}$ except in very special cases. However, as it turns out, imaginary time does play a role in quantum mechanics, even without the help of relativity.
Schrödinger time evolution occurs through the unitary transformation $u = e^{-\frac{itH}{\hbar}}$ satisfying $uu^{\dagger} = u^{\dagger} u = 1$. This means that the probability that an initial state $|\psi\rangle$ ends after time $t$ in the same state is given by the amplitude (whose square is the probability [density]) $\mathfrak{p}(t) = \langle\psi|e^{-\frac{itH}{\hbar}}|\psi\rangle$. Meanwhile, assuming the states $|\psi\rangle$ form a complete and orthonormal basis (though I don't know if this assumption is truly necessary), the partition function $Z = \mathrm{trace}\left(e^{-\frac{H}{k_B T}}\right)$, which can be expanded in the basis $|\psi\rangle$ as $Z = \sum_{\psi} \langle\psi|e^{-\frac{H}{k_B T}}|\psi\rangle$. This, however, is just as well rewritten as $Z = \sum_{\psi} \mathfrak{p}\left(t = -\frac{i\hbar}{k_B T}\right)$. Hence, quantum and statistical mechanical information can be gotten from the same amplitudes using the substitution $t = -\frac{i\hbar}{k_B T}$, which essentially calls temperature a reciprocal imaginary time. This is not really meant to show anything more deep or profound about the connection between time and temperature; it is really more of a trick stemming from the fact that the same Hamiltonian can be used to solve problem in quantum mechanics or equilibrium statistical mechanics.
As an aside, it turns out that temperature, even when measured in an absolute scale, can be negative. There are plenty of papers of this online, but suffice it to say that this comes from a more general statistical definition of temperature. Rather than defining it (as it commonly is) as the average kinetic energy of particles, it is better to define it as a measure of the probability distribution that a particle will have a given energy. Usually, particles tend to be in lower energy states more than in higher energy states, and as a consequence, the temperature is positive. However, it is possible (and has been done repeatedly) under certain circumstances to cleverly force the system in a way that causes particles to be in higher energy states with higher probability than in lower energy states, and this is exactly the negative temperature. More formally, $\frac{1}{T} = \frac{\partial S}{\partial E}$ where $E$ is the energy and $S$ is the entropy of the system, which is a measure of how many different states the system can possibly have for a given energy. For positive temperature, if two objects of different temperatures are brought into contact, energy will flow from the hotter one to the colder, cooling the former and heating the latter until equal temperatures are achieved. For negative temperature, though, if an object with negative temperature is brought in contact with an object that has positive temperature, each object tends to increase its own entropy. Like most normal objects, the latter does this by absorbing energy, but by the definition of temperature, the former does this by releasing energy, meaning the former will spontaneously heat the latter. Hence, negative temperature is hotter than positive temperature; this is a quirk of the definition of reciprocal temperature, so really what is happening is that absolute zero on the positive side is still the coldest possible temperature, absolute zero on the negative side is now the hottest temperature, and $\pm \infty$ is in the middle.
This was really just me writing down stuff that I had been thinking about a couple of months ago. I hope this helps someone, and I also await the day when TV newscasters say "complex time brought to you by..." instead of "time and temperature brought to you by...".
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