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3:02 AM
Yo anyone around here good at classical Hamiltonian physics?
You can kinda express Hamilton's equations of motion like this:
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ p \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \nabla H(x,p) \, .$$
Is there a decent way to understand coordinate transformations in this representation?
(By the way, the incredible similarity between that equation and Schrodinger's equation is pretty cool. That matrix there behaves a lot like the complex unit $i$ in that it is a rotation by 90 degrees and has eigenvalues $\pm i$.)
3:41 AM
schroedingers eqn has strong parallel(s) to the wave eqn of classical physics (part of its inception) but it seems this connection is rarely pointed out/ emphasized/ seriously investigated anywhere...
2 hours later…
5:54 AM
Q: Connection between Schrödinger equation and heat equation

Kevin KwokIf we do the wick rotation such that τ = it, then Schrödinger equation, say of a free particle, does have the same form of heat equation. However, it is clear that it admits the wave solution so it is sensible to call it a wave equation. Whether we should treat it as a wave equation or a heat e...

3 hours later…
8:26 AM
@JohnRennie physics.stackexchange.com/questions/468349/… What Georgio said. Can't you add additional dupe targets? Gold badgers on SO can, but maybe they haven't extended that functionality to Physics.SE.
@PM2Ring fixed! :-)
That was quick! Thanks.
9:04 AM
This question prompted me to do a price check on osmium. I was surprised that it's relatively cheap compared to the rest of the platinum group, but that seems to be because it has a small & steady supply & demand, so it's not attractive to traders in precious metals.
I guess the toxicity of its oxide (& other tetravalent compounds) is also a disincentive. ;) Wikipedia says it sells for around $1000 USD per troy ounce, but other sites quote a starting price of $400.
These guys sell nice looking ingots of just about every metal you can think of, apart from the alkalis & radioactives. I think I'll pass on the osmium & iridium, but I suppose I could afford a 1 oz tungsten ingot. :) Sure, it's not quite as dense as osmium / iridium, but its density is still pretty impressive.
1 hour later…
10:25 AM
@DanielSank the term you want to look for is "complexification" of a symplectic manifold
also this:
In mathematics, a complex structure on a real vector space V is an automorphism of V that squares to the minus identity, −I. Such a structure on V allows one to define multiplication by complex scalars in a canonical fashion so as to regard V as a complex vector space. Every complex vector space can be equipped with a compatible complex structure, however, there is in general no canonical such structure. Complex structures have applications in representation theory as well as in complex geometry where they play an essential role in the definition of almost complex manifolds, by contrast to complex...
I thought Arnol'd had a more in-depth discussion, but there's only a brief mention in §41.E
1 hour later…
11:44 AM
I just don't get why someone would think that proving a mathematical thing like Pythagoras' theorem is something that physics can do. physics.stackexchange.com/questions/468504/… I guess it'd be reasonable in Euclid's day, or even Newton's, but certainly not since the development of non-Euclidean geometries.
2 hours later…
1:33 PM
Hi. Am I wrong about this solution?
this question is from tensor algebra
Hi, @Leyla. I hope you don't think my previous reply is rude, but it's much better if you write equations in MathJax. My old eyes can barely read the equations in that photo, especially on my phone. And MathJax is a lot easier to search than equations in images.
@PM2Ring Ohh sure, if it's the case then I can type them in mathjax
There are bookmarklets & extensions that can be used to render MathJax in chatrooms. Stack Exchange won't build it into chatrooms because they don't want to impose the overhead on chat users...
@LeylaAlkan in a word, yes
your formulation is incorrect
unless there's crucial bits of context that you've omitted, the tensor cannot be assumed to be symmetric
indeed if $t_{ijk]$ were indeed totally symmetric, then $t_{(ijk)}$ would be identically zero and there would be no need to consider it
you're correct as far as $$t_{[321]} + t_{(321)} = \frac{2}{3!} \left[ t_{321} + t_{213} + t_{132} \right] $$ goes, but that's as far as you can take the calculation
this is enough to ensure that $t_{321} \neq t_{[321]} + t_{(321)} $ for an arbitrary rank-three tensor
particularly because it is perfectly possible for there to exist a rank-three tensor $t$ and a reference frame $R$ such that the components of $t$ on $R$ are such that $t_{321}=1$ and the rest of its components vanish.
2:08 PM
> Write out $t_{(321)}$ and $t_{[321]}$ .
Show that $t_{321}\neq t_{(321)}+t_{[321]}$

My solution:

$t_{(321)}=\frac 1 {3!}(t_{321}+t_{312}+t_{231}+t_{213}+t_{132}+t_{123})$

$t_{[321]}=\frac 1 {3!}(t_{321}-t_{312}-t_{231}+t_{213}+t_{132}-t_{123})$

$t_{(321)}+t_{[321]}=\frac 1 {3!}2(t_{321}+t_{213}+t_{132})$

Since $(3,0)$ tensor $t_{ijk}$ is totally symmetric, so it's independent of ordering of indices.

So,$t_{(321)}+t_{[321]}=\frac 1 {3!}2(t_{321}+t_{321}+t_{321})=t_{321}$

This how I done it first.
For @PM2Ring
Oh okay great @EmilioPisanty
2:32 PM
@LeylaAlkan tensors are just vectors in a vector space. It's extremely important that you understand how these linear-independence and linearity arguments work, and that you get comfortable in producing them when they're needed.
i.e. the core take-home message you should be extracting from this is how the counter-example was generated and why it works.
1 hour later…
3:47 PM
@JohnRennie what do you mean by superposition?
4:08 PM
@Akash.B it's like position
only better
1 hour later…
5:33 PM
@DanielSank What exactly do you want to understand? Any canonical transformation is just going to leave that equation unchanged, right?
6:19 PM
Why are superstrings so hard
@EmilioPisanty Excellent
@ACuriousMind I suppose, but I'm trying to see it algebraically. In some sense I'm asking how to represent a canonical transformation in the notation used in my comment.
6:54 PM
@DanielSank in general?
it'll just be an arbitrary function
your notation won't be helped much
the cases where it gets interesting is if you want a linear transformation
in which case it's required to be symplectic
does that keyword get you closer to the core of your question?
Q: When is separating the total wavefunction into a space part and a spin part possible?

mithusengupta123The total wavefunction of an electron $\psi(\vec{r},s)$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s}$$ where $\phi(\vec{r})$ is the space part and $\zeta_{s,m_s}$ is the spin part of the total wavefunction $\psi(\vec{r},s)$. In my notation, $s=1/2, m_s=\pm 1/2$. Questio...

in other news, this random thing has been on HNQ for most of the day
colour me perplexed.
I mean, not that I don't appreciate the rep-cap hit
but still
I do look forward to the gibbous-moon question getting flooded with outsiders, though =P
is the weak isospin (quantum number) the so-called flavor (quantum number)?
7:28 PM
or does flavor (quantum number) also involve weak hypercharge (quantum number)?
7:56 PM
I don't know if there is such a rule that only particles with nonzero flavor would undergo weak interaction. I read from [Wikipedia-Weak isospin](https://en.wikipedia.org/wiki/Weak_isospin) that "Fermions with positive chirality ("right-handed" fermions) and anti-fermions with negative chirality ("left-handed" anti-fermions) have $T = T_3 = 0$ and form singlets that do not undergo weak interactions." and "... all the electroweak bosons have weak hypercharge $Y_ w = 0$
, so unlike gluons and the color force, the electroweak bosons are unaffected by the force they mediate."
but $W^+$ has weak isospin 1 and $W^-$ has weak isospin -1, not zero, so they should participate weak interaction.
so I am confused as to what quantum number determines whether a particle participates weak interaction.
@EmilioPisanty I guess I'm asking how to transform the gradient.
Suppose I pick new variables that are related to the previous ones through a linear transformation. I know what to do on the left hand side, but on the right I have to do something to the gradient.
8:16 PM
When two harmonic waves going opposite directions collide do the completely cancel each other out?
1 hour later…
9:26 PM
can we really assume spinors are more fundamental than vectors?
if a manifold by chance doesn't admit spin structures, can we still assume spinors are more fundamental than vectors?
but if a manifold doesn't admit spin structures, how do you discuss fermions?
1 hour later…
10:47 PM
@DanielSank that's what the chain rule is for, right?
11:09 PM
@EmilioPisanty Yeah yeah fine I get the point. "Do the damned calculation yourself."
11:44 PM
@CaptainBohemian If a manifold does not admit spinors, you don't discuss spinors.

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