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amanuel2
amanuel2
01:49
Does anyone here understand why he set the Velocity of Center Mass = 0 here? He keeps setting the Velocity of center mass , and acceleration of center mass(on other questions) to zero which i dont comprehend why?
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PM 2Ring
PM 2Ring
02:11
@amanuel2 Initially, the astronaut & 2 wrenches are at rest, relative to each other. It makes the algebra simpler if we can work in a frame where the momenta sum to zero.
dmckee
dmckee
02:29
@amanuel2 Two questions. (A) What is the velocity of the CoM before anything happens? (B) What, if any, external forces act on the system that might change $v_{cm}$?
amanuel2
amanuel2
Yeah i understood thanks a lot!
I kinda need help understanding this question tho (#3)
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I set it up as MaV1 + MwV2 / (Ma+Mw) = 0
Ma being mass of astro, and Mw being mass of wrench. Is it the right setup?
@dmckee @PM2Ring
Sir Cumference
Sir Cumference
This spec rel?
amanuel2
amanuel2
Im starting to think this is conservation of momentum problem for #3-5? If so were suppose to learn that next lecture.
Sir Cumference
Sir Cumference
Throw a Lorentz transform at it
PM 2Ring
PM 2Ring
03:00
@amanuel2 Yes, this is a conservation of momentum question. The initial momentum is zero, and since there are no external forces, after she throws the 1st wrench the sum of her momentum plus the momentum of the thrown wrench is zero, and the centre of mass is still at the origin.
PranshuKhandal
PranshuKhandal
03:27
plz look at my edited question physics.stackexchange.com/questions/466776/…
now it is more, understandable
all upvotes are accepted
user351417
03:48
@PranshuKhandal Please don't periodically advertise your questions here and ask for upvotes.
user351417
Edits bump the question on the front page, so you can be sure that people see it.
Secret
Secret
0
Q: Physics breaking down - is it logically consistent?

Jakub KoniecznyI was just reading a sci-fi novel where physics "breaks down". While of course fiction is fiction and I don't expect this to happen in real life, when I tired to contemplate the concept I find that I cannot even imagine what it would mean for physics to break down. Is my imagination too limited o...

soft-question foundations
I guess when an observable phenomenon cannot be explained by any deductive reasoning, then it is a sign that physics breaks down
This is actually a pretty strict condition, since logic is basically deductive systems such that each rule is given semantic meaning
Sir Cumference
Sir Cumference
@Secret So magic?
Secret
Secret
Probably, but I think sufficiently complicated social dynamics mixed with a few dose of serendipity can also work
complex systems have enough phase transitions to give very surprising phenomenon that cannot be predicted beforehand
because those emergent phenomenon is inherently collective in nature, thus cannot be easily captured by knowing the theory that describes its parts or subsets of them
Sir Cumference
Sir Cumference
Eh just wait until the day computing power can explain all everyday phenomena
Maybe in a few centuries
Secret
Secret
03:58
Well, it will be interesting if that is true, cause we have no reason that the universe can be fully explained by deductive reasoning (we just have hints that maths work so well), so crossing fingers
Secret
Secret
04:10
@DanielSank It's not just behave like i, it IS i in matrix representation
yesterday, by DanielSank
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ p \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \nabla H(x,p) \, .$$
Rewriting $\mathbf{p} = \binom{x}{p}$, $\mathbf{i}= \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$, we have:
$$\frac{d \mathbf{p}}{dt} = \mathbf{i} \nabla H(\mathbf{p})$$
Compare this with Schroedinger equation:
PranshuKhandal
PranshuKhandal
@Chair sorry, for that, but I have been waiting too long for some answer.. so i thought of putting it here..
Secret
Secret
$$\frac{d \psi}{dt} = -\frac{i}{\hbar} \hat{H} \psi$$
Thus $\psi$, the wavefunction, is the quantum version of phase space
As for why in the classical version we have only grad(H) while in the quantum version we have Lap(H), I have no idea
Meanwhile, I think the two equations reads:
Hamilton's equation: The velocity of a point in phase space is given by the gradient of its total energy (taking account of its symplectic structure by i)
Schroedinger equation: The rate of evolution of a wave function is given by its total energy in units of $\hbar$
Phase-space formulation
The phase-space formulation of quantum mechanics places the position and momentum variables on equal footing, in phase space. In contrast, the Schrödinger picture uses the position or momentum representations (see also position and momentum space). The two key features of the phase-space formulation are that the quantum state is described by a quasiprobability distribution (instead of a wave function, state vector, or density matrix) and operator multiplication is replaced by a star product. The theory was fully developed by Hilbrand Groenewold in 1946 in his PhD thesis, and independently by Joe...
not exactly identical however
Also typo: Wavefunction does not really have an energy, it is the quantum state that has a spectrum of energy eigenvalues
 
1 hour later…
DanielSank
DanielSank
05:40
@Secret You're confusing two different things. $\nabla^2$ only shows up when writing the Schrodinger equation in a particular basis, namely the position basis.
Schrodinger's equation is still just $i\hbar (d/dt)\psi = H \psi$.
I want to post a question on main that says:
DanielSank
DanielSank
05:58
Since Hamilton's equation of motion in classical physics is $$\frac{d}{dt} \begin{pmatrix} x \\ p \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \nabla H(x,p) \, ,$$ why does everyone make a big deal about Schrodinger's equation, which is $$\frac{d}{dt} \begin{pmatrix} \text{Re}\Psi \\ \text{Im}\Psi \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \hat H \begin{pmatrix} \text{Re}\Psi \\ \text{Im}\Psi \end{pmatrix} \, ?$$
Oh by the way, the Hamiltonian is a stupid quantity. We should always work with $H / \hbar$, which has dimensions of frequency.
 
3 hours later…
Slereah
Slereah
08:52
Most people don't learn Hamiltonian mechanics
Also it's made of Hilbert rays which is entirely different :V
Secret
Secret
09:48
@DanielSank I think you should post that question. I don't recall many looked at the two Hamilton equations together in this matrix form before, which really highlight the similarities between them (even though technically speaking the schroedinger equation is based on quantising Hamiltonian mechanics)
and yes you are correct about the $\nabla^2$ thing. I got too used to the position basis
 
2 hours later…
bolbteppa
bolbteppa
11:24
$i \frac{d}{dt} (x,p) = \hat{\nabla} H$ is really different to $i \frac{d}{dt} \psi = \hat{H} \psi$
Mithrandir24601
Mithrandir24601
12:10
@DanielSank Or be a theorist about it and set $\hbar = 1$, essentially doing the same thing :P
I don't know what the fuss about the Schrödinger equation is either but then again why QM is taught in the way/order it's taught is something I find a mystery
Slereah
Slereah
As with a lot of physics, it is usually taught in historical order
which is a choice
dunno if it's the best but such it is
ACuriousMind
ACuriousMind
@DanielSank The big deal is not the equation itself, but the meaning of the variables. The form of the equation itself just says "the Hamiltonian is the generator of time translation", but surely you'll agree that classical position and momentum evolving in time are a rather different notion than the wavefunction of QM evolving in time.
If you want to make the similarity really obvious, just write the evolution equations for the observables. The classical equation is literally Heisenberg's evolution equation with the Poisson bracket instead of the commutator, no pesky additional $\nabla$ or what not
The big deal many introductory quantum texts make about the Schrödinger equation is due to the fact that their target audience are usually people who are not expected to be trained in classical Hamiltonian mechanics.
Emilio Pisanty
Emilio Pisanty
12:46
@DanielSank sorry if it came across as snark.
the point is that the gradient is a covariant vector
Slereah
Slereah
@EmilioPisanty gradients aren't vectors :V
Emilio Pisanty
Emilio Pisanty
so if $\begin{pmatrix}x\\p \end{pmatrix}$ transforms to $M\begin{pmatrix}x\\p \end{pmatrix}$, then $\nabla H$ transforms to $(M^{-1})^T \nabla H$
(or something like that)
so you end up requiring that $MJM^T = J$, or whatever the precise requirement for symplecticity looks like
@Slereah I know they're not. That's why I said they're something different =P
 
1 hour later…
Slereah
Slereah
13:55
"It is remarkable that Schwarzschild derived this solution while fighting in World War I (literally, in the trenches: in fact, he even got ill there and died shortly after the end of WWI)."
Can you stop with the mortars I'm trying to think!
MartianCactus
MartianCactus
14:33
When will dyson sphere be built
a guess?
JMac
JMac
No time remotely soon, as far as things seem. Just the amount of material required for an undertaking like that would be exceptional. It doesn't even seem like we're remotely near the advancement required to take advantage of such a project, let alone organize one.
I'd be honestly skeptical of humans ever reaching that point. It's cool to think about, but so much would have to change that trying to estimate it would be pointless currently
MartianCactus
MartianCactus
14:58
oh
well lets see :)
PM 2Ring
PM 2Ring
@MartianCactus Where do you propose obtaining the material to build a Dyson sphere?
Avnish Kabaj
Avnish Kabaj
Mars
PM 2Ring
PM 2Ring
15:22
Yeah, ok. That could work.
MartianCactus
MartianCactus
Mercury
is the best candidate
close to sun, and we could get robots to work there
theres a whole kurzgesagt video on it
PM 2Ring
PM 2Ring
I guess if you have the technology to dismantle planets the heat near Mercury is only a minor bother.
And Mercury (probably) has a huge iron-nickel core, which is rather nice.
vzn
vzn
15:42
(lol) talk about raping the planet(s)... re dyson sphere, solar energy is a simplified version right? which is advancing. what about orbiting solar energy harvesting? maybe not as far away. kurzgesagt also has a video on a space elevator, its very hard but expect that to be built decades earlier, and if it doesnt show up, maybe no hope for a dyson sphere... o_O
Slereah
Slereah
man the first paper on the conformal group of Minkowski space is from 1908
en.wikisource.org/wiki/…
You can tell it's old because he mentions LORD KELVIN
DanielSank
DanielSank
@EmilioPisanty It didn't come across as ssnark. I do appreciate the reminder that the gradient is covariant.
DanielSank
DanielSank
16:04
Hah, got on the HNQ list for a physics oriented post in World Building.
Emilio Pisanty
Emilio Pisanty
16:49
@DanielSank =P
I got on HNQ for saying that a thesis is a publication on academia.
(well, it was already on HNQ. but the answer did rise to second place.)
@DanielSank =)
is ssnark what ssnakes do?
Emilio Pisanty
Emilio Pisanty
17:00
BTW @DanielSank Do you know where I can go to wash off my karma? I just wrote a rather negative (though well-deserved, and as thorough and impartial as I could make it) referee report. And I'd rather it not come back to bite me on my next go-round as an author o.o
John Rennie
John Rennie
@EmilioPisanty oh no, you'll be reincarnated as an employee of Elsevier!
DanielSank
DanielSank
17:48
@EmilioPisanty Did you already submit the report?
@EmilioPisanty Yes, exactly.
Sir Cumference
Sir Cumference
18:02
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Emilio Pisanty
Emilio Pisanty
18:42
@DanielSank yes
For what it's worth, I got this a few hours after sending it in
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DanielSank
DanielSank
19:41
@EmilioPisanty Very good.
bolbteppa
bolbteppa
@Slereah man what's going on in that conformal paper
Slereah
Slereah
20:10
@bolbteppa it is a bit unorthodox
bolbteppa
bolbteppa
20:22
Trying to make sense of the intro, why he does what he does with the $l = x - i y$ stuff and the $l \lambda + m \mu + n \nu = 0$ thing
Slereah
Slereah
it is pretty confusing
It seems to be some reference to some existing method in optics
I'm guessing that was popular back then
bolbteppa
bolbteppa
21:05
$0 = x^2 + y^2 + z^2 + w^2 - x^2 - y^2 - z^2 - w^2 = (x + i y)(x - iy) + (z + i w)(z - i w) - (x^2 + y^2 + z^2 + w^2) = l \lambda + m \mu + n \nu$
 
2 hours later…
bolbteppa
bolbteppa
23:05
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