The h Bar - 2015-10-21 (page 1 of 3)

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0celo7

00:04

@ACuriousMind :(

dmckee

00:32

@0celo7 I think someone got carried away. Not that the problem is conceptually difficult, but there is a lot of algebraic complexity for no real purpose.

David Z

Ugh, sorry I missed the chat session. Blame the great firewall.

1 hour later…

Stan Shunpike

01:42

Sup y'all

@0celo7 whatchu been listening to lately?

0celo7

Defqon.1 2015

Got an iTunes gift card for my Bday, time to expand the collection

Waiting for an album by Radical Redemption (raw hardstyle) and Bassleader 2015 (hard dance festival)

Terry Bollinger

02:24

@0celo7 Peace; we all have hot buttons, as I also just demonstrated. I hope everyone can get back to physics, and I respect and appreciate your clear desire to do so.

Hi @DavidZ, hope you are well. Goodnight all, it's late here...

0celo7

@ACuriousMind I'd appreciate a hint for the problem on the top of page 225 if you don't mind.

@ACuriousMind symplectic and complex = orthogonal I have

0celo7

03:00

@ACuriousMind Well, I somehow proved all 3 equalities there :)

@ACuriousMind I don't see how these intersections are unitary though...

@ACuriousMind Just ignore all of that, I get it

3 hours later…

Galois in the Field

06:02

What can I do with $[r^{-1},r^{-1}q_j]$?

Ahhh wait I think I have it

3 hours later…

Secret

08:46

Back to the future day is today
https://www.facebook.com/uniladmag/videos/1842522722437445/

Everything except time travel

"Cristy Elizabeth Caplan It didn't come true. It was made true.
Like · Reply · 4 · 11 hours ago"

This is why I like scifi
This is, aruguably why I study physics: To make the scifi real whenever possible

user685252

09:06

I would definitely argue that reason.

Nature is not fictitious.

Secret

Well, the fact is, nature has a lot more things better than scifi. For example, anyone think about metamaterials like 20 years ago?

Secret

09:40

(cont., philosophical) And this is why some said me studying physics is a conspiracy, or a lie, because my passion on physics is not as selfless as for chemistry

One could say I have an agenda on studying physics: and that agenda is $Re(scifi)\neq 0$

Emilio Pisanty

10:12

Any bibliography-minded folks around?

I'm trying to track down

Never mind :)

It seems there's two different Ann. Phys.

Secret

10:46

NbCeID 20
http://www.sciencedaily.com/releases/2015/10/151019122240.htm

New photonics in town, zero refractive index and constant waves

NbCeID 21
http://www.sciencedaily.com/releases/2015/10/151013103227.htm

glueglue->glueglue
glueglueglueglue
gleuglueglueglueglueglue...

Slereah

11:06

what

Secret

glueballs, that's what
and then metamaterials

Slereah

I wonder what's the decay of glueballs

Let's see

Meson pairs

Figures

Secret

always check the journal articles whenever you see a particle physics news item, because the media is often very bad at presenting it properly

Slereah

I basically don't read science journalism anymore

Secret

? that's strange, I thought they are often reliable enough

Slereah

11:32

Eh.

Have you ever read a newspaper article on a topic that you actually knew well and thought "This is exactly how things are"

Secret

11:43

O sorry, I thought you mean you don't read journal articles anymore

as in academic journals

0celo7

@Slereah My new avatar is pretty classy

Slereah

it is

Secret

For me, I still read science magazines to catch up on science outside my speciality, but will dig further into the journal articles whenever the context is oversimplified

Galois in the Field

What relations does one have for $[r^{-1},r^{-1}q_j]$?

I am trying to prove some identities in regards to the Hamiltonian of the non-relativistic hydrogen atom

(Just to learn how to manipulate the math)

Slereah

12:00

0?

Coordinates commute with themselves

$[f(x), g(x)] = 0$

Galois in the Field

I wasn't sure how to treat that, since $[r^2,q_j]=0$ surely, but then for $r$ I have a sqrt and then for $1/r$ I have 1/sqrt

Slereah

You can expand them as Taylor series, if you wish

Galois in the Field

Oh I haven't seen that before, what is that called?

Slereah

And $[x^n, x^m] = 0$

Galois in the Field

That's true

What is a good resource to read more about this exact topic?

Slereah

12:03

I mean at its core $f(\hat x) \psi = f(x) \psi$

And since c-numbers commute, functions of the position will always commute

Galois in the Field

I am a math student, I think this is physics notation so I am confused

Slereah

Well math-wise it's operators on a Hilbert space

Galois in the Field

So why does $[r^{-1},r^{-1}q_j]=0$ mathematically?

Secret

and commutators are a type of lie algebra

Galois in the Field

Commutators aren't a type of lie algebra haha, I know that much

Slereah

12:06

Well the easiest way to see it is to apply it to a Hilbert space vector

Wait

In that case I don't even think that's necessary

The commutator is linear

You can get $q_j$ out

Galois in the Field

I only have seen $r^2$ defined as $r^2=q_1^2+q_2^2+q_3^2$

Slereah

And an operator always commute with itself

Secret

@GaloisintheField really? that's how my quantum course taught us and I also worked with one of my peers and the commmutator satisfy the cyclic rule and other properties of lie algebra

Galois in the Field

That's not what I meant sorry, I just meant that you are talking about the Lie algebra of the associative algebra, which has the Lie bracket being the commutator

So commutators aren't a type of lie algebra per say, but they are a valid choice of bracket

Secret

I see...

Galois in the Field

12:10

Sorry to be pedantic

Slereah

Commutators are just defined as $[\hat A, \hat B] \psi = \hat A \hat B \psi - \hat B \hat A \psi $

Galois in the Field

Wait so the bracket does represent the standard commutator bracket

Slereah

Well I assume, I don't know what you're working on

Hydrogen atom is probably commutator, yes

Galois in the Field

I had the result $[p_j,r^n] = -inr^{n-2} q_j$

Would you expect that from the commutator?

Slereah

Sounds about right

Galois in the Field

12:13

Oh, great!

Slereah

(if you set $\hbar = 1$)

Galois in the Field

Thanks, I wasn't sure how that was formed, so I assumed it was some strange bracket

@Slereah Why does $p_jr^n - r^np_j = -inr^{n-2}q_j$ (if it isn't a pain to explain)

@Secret Sorry I just saw it like saying multiplication is a group, but without specifying the set its applied, it isn't really

Slereah

The easiest way to deal with it really is to put the operators in their usual form in QM

$\hat p_i = \partial_{x_i}$

Secret

ok

Slereah

Then you just have to compute $\partial_x (r^n \psi) - r^n \partial_x \psi$

Galois in the Field

12:21

$\partial_x = \frac{\partial}{\partial x}$?

ahhh don't worry, I will have to read more, otherwise I'll just keep asking dumb questions probably

I'll tell you when I work it out what I worked out

Slereah

@GaloisintheField yes

Secret

For reference, I tend to ask a lot of dumb questions, lol

Slereah

Basically there's two ways to deal with QM, mainly

Galois in the Field

@Secret Me too

Slereah

Either the Schrödinger way, where you just solve differential equations on a wavefunctions

or the Heisenberg way, in a more abstract manner, with operators on a Hilbert space

Both methods are equivalent

But Schrodinger is a bit easier to get into

Secret

12:26

I usually prefer matrix methods

because it is a lot more "chunky"

i.e. one step at a time

0celo7

What?

@Slereah who's the guy in my avatar again?

Slereah

One of the heir of the throne of the Byzantine empire

0celo7

Bow down, female dogs

skill patrol

da raiders bow to nobody

boy

0celo7

So that's why they suck

skill patrol

12:31

:P

0celo7

Dude the Skins are gonna die

D:

skill patrol

i saw

0celo7

12:49

@Slereah what football team do you support

2

Q: Why does the Palatini formalism of GR works?

We can get Einstein field equations of GR via two distinct methods: by taking the metric as the only degree of freedom, and imposing right away that the connection is the levi-civitta one. In that case the lagrangian will be a functional only of the metric; or we can follow the Palatini methods: ...

general-relativity lagrangian-formalism

@Slereah

I don't know the answer D:

I thought it was just a coincidence...maybe I should read that section in Straumann

Huh, he doesn't cover that formalism

I guess that's why I don't remember reading it

Danu

@foobarrr a response like that is a very good way to drive people away from helping with your question — David Z ♦ 1 hour ago

I wonder what was said there...

ACuriousMind

@0celo7 Y'know, when you think a question is good, you could edit it to make it more readable and correct spelling/grammar :P

Danu

@ACuriousMind People should do this more often.

0celo7

@ACuriousMind Didn't notice anything wrong with it.

ACuriousMind

@0celo7 Look at the title, for starters: "Why does the Palatini formalism works?"

"Levi-Civitta"

Danu

13:02

this sounds like someone destined for blogs and amazon self-published books. — casey Oct 16 at 0:14

0celo7

He's probably a non-native speaker.

Danu

Hmm... :P

0celo7

Lol

ACuriousMind

@0celo7 That is a reason not to hold that against OP, not a reason to not correct it.

skill patrol

:P

btw have you guys seen the new "highlights" option at the top of the chat page?

ACuriousMind

13:15

@skillpatrol Yeah. Not sure how it chooses those

Galois in the Field

I have now

skill patrol

by starred comments

ACuriousMind

@skillpatrol Nope, it gives me a lot of conversation snippets where there's no star in sight

skill patrol

really?

Galois in the Field

$[p_i^2,r^{-1}q_j]$, what about that @Slereah

skill patrol

13:17

the center comment should be starred :-/

Galois in the Field

What can I do with this?

ACuriousMind

@GaloisintheField Use $[ab,c] = a[b,c] + [a,c]b$.

Galois in the Field

Ahh yes of course, thank you

This is precisely why I am playing around with the identities xD, I don't see these things automatically yet

ACuriousMind

@skillpatrol No, doesn't look like it.

0celo7

@ACuriousMind The reason I didn't correct it is I didn't notice anything wrong with it.

St.Clair Bij

13:35

@DavidZ I have read your comment and the link in the helpcentre you sent me. I think I used a completely mainstream definition of time and have not asked to evaluate anything outside of the context of mainstream physics. I do not understand why the question is closed, is it possible for you to re-evaluate the question?

@David the question was: is time related to motion. (about what would happen with time if motion would stop)

skill patrol

link please

0celo7

Poor TA, doing Gram-Schmidt in 4 dimensions

Probably means we're gonna have to do that on a quiz or test. Fun.

St.Clair Bij

the link is: physics.stackexchange.com/questions/212617/…

ACuriousMind

@St.ClairBij Your question does not make sense. What does "Suppose all matter stopped motion" mean? How is this different from just stopping time, and how do you deduce that time is derived from motion and not motion from time?

0celo7

@ACuriousMind Brick walls stop motion.

I don't see the issue.

St.Clair Bij

14:04

@ACuriousMind Have you read the first two parts of the question though? It is exactly that it does not differ from stopping time, which is the point. If a particular aspect of motion does not differ from time, what does it mean when you say time is dependent upon the speed of light. Time as such is a more vague concept than motion.

ACuriousMind

@St.ClairBij Oh, you're mixing up different physical pictures here. "Time", as one particular dimension, exists only in classical non-relativistic mechanics. There time and motion are pretty much the same, and time does not "depend on the speed of light".

But in relativity, there is no unique time direction, and motion is described as worldlines in spacetime which each have their own proper time. If you extend those proper times to coordinate systems, you find that the directions of the proper times of different worldlines do not agree.

And time is not a "vague concept". It's a dimension of spacetime, and it gives robust, quantitative predictions.

St.Clair Bij

@ACuriousMind that's pretty interesting. (And like the answer I was looking for). But what do you mean by: there is no unique time direction? (is this empirically verifiable?) Or stated differently: what does no unique time direction mean?

ACuriousMind

@St.ClairBij (Special) relativity is invariant under the Lorentz transformations, which mix the former time and space directions. The phenomena of length contraction and time dilation are consequences of the fact that the time axes in two inertial coordinate systems related by a Lorentz transformation are not the same.

(And yes, time dilation is empirically tested)

St.Clair Bij

@ACuriousMind Is time dilation the same as a different direction in time? (Does dilation change direction)

ACuriousMind

@St.ClairBij Time dilation means that observers that move at different speeds will not agree on the time it takes one of them to get from one place to another (and neither will they agree on the distance traveled because of length contraction). You may interpret this as both of them using time-space directions that are rotated with respect to each other.

You should go and learn about special and general relativity in detail before you try to ask questions about time again, I think.

Secret

14:18

user image

St.Clair Bij

@ACuriousMind In terms of vectors: not only the magnitude but also the direction of time changes?

Secret

@ACuriousMind so like 4 velocities of each observer are rotations of each other?

ACuriousMind

@St.ClairBij The two observers between which time dilation occurs would not agree on the "direction" of time, no.

Danu

You got a nice bunch of bonus votes due to my bounty @ACuriousMind :D

ACuriousMind

@Danu That's the power of bounties :)

St.Clair Bij

14:20

@ACuriousMind (in terms of vectors: not only the magnitude but also the direction of the vector changes for the observers?)

Galois in the Field

Is there something I am missing, I have a discrepancy, but perhaps they are equivalent for reasons I can't see:

Should equal: $\frac{i}{\mu}r^{-3} p_mq_mq_j - \frac3\mu r^{-3}q_j - \frac{i}{\mu}r^{-1}p_j$

I obtained: $\frac{1}{2\mu}\left( p_mir^{-3}q_mq_j+ ir^{-3}q_mp_mq_j\right) $

(einstein summation)

ACuriousMind

@St.ClairBij Well, the total magnitude of the four-vector containing position and momentum doesn't actually change - what changes is the coordinate system its components are interpreted in.

An analogy: Let's say you look at a line and say it's 1m in the x-direction and 1m in the y-direction. Someone else stands at an 45° angle and says it's sqrt(2)m in the x-direction and 0m in the y-direction. Because you have different notions of "x-direction" and "y-direction", the other person is experiencing "x-dilation" and "y-contraction".

Galois in the Field

So the first term is equivalent other than a factor of $\frac12$ but the other two seem just plain wrong

ACuriousMind

@GaloisintheField Did you try moving the $q$ past the $p$ in your second summand?

Galois in the Field

They don't commute(in the multiplicative sense) do they?

St.Clair Bij

14:27

@ACuriousMind ok, the reason i asked about direction of time is also because this way of thinking leads people to think "time travel" would be possible. (manipulating the direction of time) This to me would be in direct contradiction to realistic physics. But thanks for the info A curious mind

ACuriousMind

@GaloisintheField Exactly. I meant like: $qp = [q,p] + pq$

Galois in the Field

@ACuriousMind Ahh that's clever, hopefully it remedies my problem

I'll give it a shott

Secret

If my memory serves, Lorentz transformation cannot flip any future directing timelike 4 vector to point in the past direction, so you cannot time travel via boosting to a different frame

ACuriousMind

@Secret The non-orthochronous transformations can flip the time direction.

Secret

but are those physical?

ACuriousMind

14:31

I don't know what that means.

Galois in the Field

@ACuriousMind It literally just flipped them(which is cool because the bracket went to $0$) but it didn't remedy the difference, I guess I worked it out wrong

Secret

I understood the orthochronous ones corresponds to an observer boosting to another frame by travelling at a different velocity wrt some observer in the rest frame, but how can an observer do a non orthochrnous transformation?

ACuriousMind

@Secret Ah. Yes, the time flip is not a boost transformation.

So, in usual Minkowski space, you cannot flip time just by accelerating.

Secret

how does non orthochrnous transformation went into play if an observer cannot perform something to do the transformation, is it only a geometric thing telling how the observers are related but does not correspond to a physical process that can carry it out?

ACuriousMind

@Secret THe Lorentz transformations are just the transformations preserving the Minkowski metric. They do not come with a relation to "observers" a priori.

Why "petrol", @skullpetrol?

skull petrol

14:39

Fuel for thought :-)

Galois in the Field

@ACuriousMind Oh that bracket didn't go to zero, oops $[q_i,q_k]=0$ and same for $p$, but that bracket goes to something else

Secret

I see. When do we need to care about the non orthochronous transformation when we investigate GR problems? I.e. when are they play important role and what kind of role they play?

ACuriousMind

@Secret Uh, you don't need to especially "care" about them, I think. But, they're there.

Secret

ok

Galois in the Field

Does $[r^{-1},q_j]=0$ and $[r^{-1},r^{-1}]q_j=0$, more to the point does $[r^n,q_j]=0$ and/or $[r^n,r^k]=0$?

St.Clair Bij

14:48

@ACuriousMind is it possible in Minkowski space to flip time at all? Or is there some other "space system" where such an operation would be allowed. (And does such a system have a physical analogue)

Galois in the Field

I'm still not entirely sure if I am working with the commutator bracket since I can't derive any of the results that come out of the bracket exclusively

if $r^n$ commutes in the multiplicative sense with $q_j$, and I am using the commutator bracket, my question above follows immediately

skull petrol

Do you want to reverse cause and effect? @St.ClairBij

ACuriousMind

@St.ClairBij Actual time travel is impossible in Minkowski space. There exist spacetimes in general relativity though where all sorts of weird things can happen. Those are hoped to be unphysical, but there is hitherto no a priori reason to exclude them from the allowed solutions of Einstein's equations.

@GaloisintheField Are you doing classical Hamiltonian mechanics or quantum mechanics, i.e. is that the Poisson bracket or the commutator of operators?

Secret

The Godel metric contains CTCs, where time can loop back for some worldlines. The light cones in this metric are basically twisted so they joined back to back

Galois in the Field

@ACuriousMind That's part of my confusion, I am in quantum mechanics

Secret

14:50

but there is no evidence for their existence

ACuriousMind

@GaloisintheField Okay, so you know $[q_i,p_j] = \delta_{ij}$, right?

Galois in the Field

@ACuriousMind $\times i$?

ACuriousMind

Ah, yes

Damn $\mathrm{i}$.

All commutators between operators corresponding to functions of $q$ and $p$ (like $r$) can be deduced from that commutation relation.

St.Clair Bij

@skullpetrol yeah lol... it seems to me some physicists have smoked too much weed. (And the definition of time as it stands now is the cause for it. (is my hypothesis).

Secret

The consensus for solutions in GR that allow CTCs are often the "you cannot change the past" type, thus it does not allow inconsistent <I forgot the term> like the grandfather paradox to happen

Galois in the Field

14:53

@ACuriousMind My confusion stemmed from $r^{-1} = \frac{1}{\sqrt{q_1^2+q_2^2+q_3^2}}$ and hence I couldn't use that commutation relation

ACuriousMind

@GaloisintheField Yeah, strictly speaking, that is not an admissible operator.

0celo7

@ACuriousMind you need to care about them if you're interested in CPT, IIRC

ACuriousMind

$r^{-1}$ is just not defined in the algebra of operators. You can define $r^2 = \sum_i q_i^2$.

Galois in the Field

I am trying to show some identity they have listed holds which has RHS:

$$[H,r^{-1}q_j]$$ Where $H$ is the hamiltonian of the non-relativistic hydrogen atom

ACuriousMind

@GaloisintheField As I said, that is, strictly speaking, nonsense. The operator $r^{-1}$ does not exist, or at least you can't simply write it down and claim it exists.

Since $r^2$ is Hermitian, one can define the square root $r$, but it is not guaranteed at all that $r$ is invertible, and since $0$ is an allowed position, I'm pretty sure it is not invertible.

Galois in the Field

14:58

Well that's unfortunate :(

ACuriousMind

Yeah, physicists are sometime pretty bad at this operator business :P

Galois in the Field

I spent an hour manipulating nonsense HAHA

I guess I learned methods so that's really what I wanted haha

ACuriousMind

If we're doing ill-defined things anyway, we could write $r^{-1}q_j = \partial_j r$ and compute $[H,r]$, then symbolically differentiate w.r.t $q_j$ to get the result.

Galois in the Field

Specifically I was trying to get:

$$\left[\frac{p^2}{2\mu} - Ze^2r^{-1}, r^{-1}q_j\right]=\frac{i}{\mu}r^{-3} p_mq_mq_j - \frac{3}{\mu}r^{-3} q_j - \frac{i}{\mu} r^{-1}p_j$$

@ACuriousMind I mean I can try that, but I haven't done this before(or seen it in the notes)

ACuriousMind

Hm, I'm trying to recall how we did this back in the day

Galois in the Field

15:03

@ACuriousMind Have you done this explicit problem(or something similar)

ACuriousMind

@GaloisintheField I'm pretty certain I've done stuff like that for the hydrogen atom.

@GaloisintheField Hm, we never treated this in the abstract operator formalism. If you just stay on wavefunctions $L^2(\mathbb{R}^3)$, you just solve differential equations and never have to worry about such commutators.

Galois in the Field

@ACuriousMind $L^2$ refers to $L-p$ space?

0celo7

@ACuriousMind whatever happened with your QCD code

Galois in the Field

@ACuriousMind Well thanks, I'll tell you if I get it(unfortunately I am an algebra student ahaha)

0celo7

@ACuriousMind yessss

Wave mechanics is the best!

ACuriousMind

15:10

@GaloisintheField Yeah. There's some subtleties because they're not all differentiable and so on but at least the $\Delta + r^{-1}$ is a good differential operator. The abstract $r^{-1}$ just doesn't exist.

@0celo7 I thought I'd told you that I didn't find the error?

Galois in the Field

It does say at one point in the notes: take positive definite square root, call it $r$, assume $r^{-1}$ exists

ACuriousMind

@0celo7 For actually solving stuff it is! I just don't like it if it is used as the conceptual starting point.

0celo7

@ACuriousMind D:

you broke QCD

nobel when?

ACuriousMind

@0celo7 Nah, alarge found that other simulation that also didn't see anything, and I found another. It seems the simulation just isn't as easy as people claim it is.

David Z

@St.ClairBij Nothing about the question has changed that would affect my decision. Very little about your question is mainstream.

ACuriousMind

15:13

@GaloisintheField Yeah, you can't assume that when $r$ has an eigenvalue of 0.

0celo7

@ACuriousMind stupid question: why not just invert $r$ on the space where it has nonzero eigenvalues

ACuriousMind

@0celo7 Well, you can do that. But then it's not defined on your whole space. Additionally, you get that issue that the eigenvectors of the position operators aren't actual memeber of the Hilbert space. It is not clear to me how the actual prescription to invert the operator has to look.

@GaloisintheField: Do your notes at least have an instruction what something like $[r^{-1},p_i]$ is and how to compute it?

Secret

Have finished reading the paper referred by Slereah but cannot get back to it yet, not before I finish the data analysis of some spectroscopy thing

Galois in the Field

@ACuriousMind Different notes from 2 years ago have a little thing on it

user image

The hint just has an arrow to $[L_j,H]=0$

Where $L_j$ is a basis element of an $o(3)$-module

'quantum hamiltonian has o(3)-symmetry'

ACuriousMind

I don't think one can determine $[r,p_i]$ from just $[r,q_i] = 0$.

Slereah

15:23

I wonder what spot on earth has the lowest gravity

I guess some mountain top near the equator

Galois in the Field

@Slereah Some area that has a giant void underneath it iirc

Slereah

neat

Galois in the Field

5

Q: Lowest gravity on Earth's surface?

I am trying to determine which on Earth's surface has the lowest gravity. Googling is not finding anything concrete. My natural inclination would be to think of Mount Chimborazo in Ecuador, being on the equator (centripetal force) and also being the furthest point from the Earth's center. However...

gravity specific-reference measurement geophysics planets

Slereah

$[r,p]$ can only be determined either by definition or by saying what the operators are

ACuriousMind

However, I think that $[r,p_i] = \mathrm{i}r^{-1}q_i$

Slereah

15:25

or by a quantization procedure

ACuriousMind

@Slereah They explicitly say to treat $r$ as independent from the coordinate operators.

Slereah

Like $\{r, p\} \rightarrow \frac {i}{\hbar} [r,p]$

ACuriousMind

So no quantization, although that's just what I did to get my guess.

Galois in the Field

I have to go now, I'll work it out tomorrow I imagine and let you know what I did haha

ACuriousMind

Ahhhh

Slereah

15:26

What definitions do we have?

Galois in the Field

Thanks for your thoughts!

ACuriousMind

I think I get it

Hm, no.

0celo7

15:44

@ACuriousMind Unfortunately, I still don't see why GL(n,C) is the group that preserves the complex structure.

And I'm not sure why GL(n,C)=GL(n,R)xGL(n,R)

ACuriousMind

15:58

@0celo7 Every complex matrix in $\mathrm{GL}(n,\mathbb{C})$ with entries $a_{ij}+\mathrm{i}b_{ij}$ corresponds to two matrices in $\mathrm{GL}(n,\mathbb{R})$ with entries $a_{ij}$ and $b_{ij}$.

As for why those are those are those who preserve the complex structure, just write the complex structure as $\left(\begin{matrix}0 & -E_n \\ E_n & 0\end{matrix}\right)$ and just examine which block matrices $\left(\begin{matrix} A & B \\ C & D\end{matrix}\right)$ with $A,B,C,D\in\mathrm{GL}(n,\mathbb{R})$ commute with it.

Slereah

Hm

You know

A simple way to check for tachyons in a non-free case would be like

The $\phi^4$ solution with Jacobian elliptic functions for tachyons

Or maybe breather solutions

I dunno

Although they don't really have compact support

Hm, how to do front velocity on non-compact supports

Maybe you can make them compact in some limit

Let's see

I think the solution was $\approx nd(z,1)$

0celo7

@ACuriousMind Oh, I knew that.

@ACuriousMind Well, I knew that too.

Thanks for confirmation.

@Slereah what is that?

@Slereah If you want something to research, do a study of causality in string theory.

@Slereah Try to show if the quantum Einstein equations have a well defined Cauchy problem or not.

Slereah

16:25

@0celo7 Ugh

@0celo7 That's a solution to the $\phi^4$ theory

$(\square + m^2) \phi = \phi^3$

0celo7

@Slereah What?

@Slereah no what does $nd(z,1)$ mean

Slereah

It's one of the Jacob elliptic function

It obeys the ODE $nd''(z,k) = (1 + k'^2) nd(z,k) - 2 k'^2 nd^3(z,k)$

obe

@0celo7 I've been reading SR in general frames and it's a lot different from the stuff I've learned before, is it useful though?

For example how he defines the minkowski space-time on page 25.

Slereah

I would say learning SR in general frames isn't terribly useful

It is best to learn GR and do it in GR

obe

Well the first few chapters of that book cover regular SR though with a lot of detail.

ACuriousMind

16:35

@0celo7 Why do you keep asking me about things you know, then? :P

Slereah

although

If you can, learn about Born coordinates

Because if you do a lot of SR, you will run into the usual jackass

"Hey look, FTL in rotating frames!"

Gotta learn about Born coordinates and tell him to shut up

0celo7

@ACuriousMind reasons

@obe Eh, it's more of a novelty book.

Some of the later chapters are interesting.

But in the beginning he uses way too much detail.

But if you need to calculate time dilation in an accelerating, rotating frame, that's the book you need.

Slereah

what book is it

Is it Callahan

Sounds a bit like Callahan

0celo7

no

@ACuriousMind Does the construction of Hamilton's equations on page 236 have any physical content?

He never actually identifies $H$ with any physical quantity, right?

obe

@Slereah This book.

Slereah

16:44

o

0celo7

The best part of the book is that he does the twin paradox with a $C^\infty$ trajectory.

Slereah

You can do that in GR :p

Easy as pie!

Just do a Fermi transport on the twin's trajectory

use Fermi coordinates

0celo7

Reference?

Slereah

Fermi coordinates are the coordinates for an observer along a curve

I guess Wald has it?

0celo7

I know what they are

Slereah

16:50

Probavbly

0celo7

I want a reference for someone doing the twin paradox in GR with a $C^\infty$ trajectory.

> 75 The proof of this theorem which is presented in the excellent book by Landau and Lifshitz (Mechanics, Pergamon, Oxford, 1960) is incorrect.

#rekt

Slereah

Let's see

arxiv.org/pdf/physics/0004024.pdf

this maybe?

I dunno

It's not overly difficult to write the twin paradox for arbitrary trajectories in GR

just gotta rewrite the coordinates so that the connection disappears along the path

0celo7

ACM keeps coming and going

Slereah

just like autoerotic asphyxiation

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