The h Bar

Slereah

6:14 AM

Are we sure that spacetime has a metric structure anyway

Conformal structure still preserves lengths and mass ratios

More Anonymous

6:27 AM

@Slereah yes

Shiki Ryougi

6:41 AM

So next problem I'm studying is about an Ellis wormhole, with metric given by $ds^2=-dt^2+dr^2+(r^2+b^2)(d\theta^2 + \mathrm{sin}^2 \theta d\phi^2)$

Suppose we have a test particle falling freely and radially through the wormhole starting at position $r=r_1$. I have to calculate how much time it takes (given an initial speed $U=\frac{dr}{d\tau}$) for the particle to reach $r=-r_1$, in its own clock.

I have found the geodesic equations, the ones that I believe are useful for the problem are $\ddot{r}=0$ and $\ddot{t}=0$

Since the particle falls radially, I've set $d\theta=d\phi=0$

Next thing is to calculate the proper time $\Delta \tau = \int \sqrt[]{(\frac{dt}{d\tau})^2-(\frac{dr}{d\tau})^2}d\tau$

Using the geodesic equations we have $\frac{dr}{d\tau}=U$ and $\frac{dt}{d\tau}=a$. Therefore we get $$\Delta\tau = \int \sqrt[]{a^2-U^2}d\tau$$

(For some constant a which I do not know how to specify)

I feel like I've done something very wrong here, I'm not sure why

I should probably use $dt$ instead of $d\tau$

(be right back, doing some calculations)

Ok so I found that $\frac{dr}{dt}=\frac{U}{a}$

So the proper time would be $$\Delta \tau = \int \sqrt[]{1-(\frac{U}{a})^2}dt$$

Or better, since we want to find for given radius: $$\Delta\tau = \int \sqrt[]{(\frac{a}{U})^2-1}dr$$

Integrating from $-r_1$ to $r_1$ we get $\Delta \tau = 2r_1\sqrt[]{(\frac{a}{U})^2 -1 }$

I think it should be correct now, but the contant $a$ really bothers me

Yeah I don't really understand what it is

John Rennie

8:59 AM

@ShikiRyougi Why $dt/d\tau = a$? Isn't $a$ just $1$?

Shiki Ryougi

9:34 AM

Uhm, from the geodesic equation $\ddot{r}=0 \implies \dot{r}=c$, for some constant c. How do we know that $c=1$? I mean, why should time and proper time be equal?

I'm really missing something here

From some point on I started treating them as equal

But I have no idea why this is the case, I just know that it works with this method :p

Emilio Pisanty

10:53 AM

yo

@ACuriousMind

0

Q: What would stationary light look like?

I mean, have we ever witnessed what we understand and call as light at rest, and what would it look like at rest

I don't think that's a duplicate, and I've got an answer at ~85% completion

let me post it and convince you that it's not a dupe ;-)

ACuriousMind

hm, not a dupe because this question does specify "in vacuum"? I could see that

Emilio Pisanty

Slow light

Slow light is the propagation of an optical pulse or other modulation of an optical carrier at a very low group velocity. Slow light occurs when a propagating pulse is substantially slowed by the interaction with the medium in which the propagation takes place. Group velocities below c were known to be possible as far back as 1880, but could not be realized in a useful manner until 1991, when Stephen Harris and collaborators demonstrated electromagnetically induced transparency in trapped strontium atoms. Reduction of the speed of light by a factor of 165 was reported in 1995. In 1998, Danish...

in essence

ACuriousMind

yeah, alright, that's not a dupe

Emilio Pisanty

@ACuriousMind 👍🏼👍🏼👍🏼

thx

answer posted

Shiki Ryougi

11:53 AM

Why do metrics generally don't have cross terms like $\mathrm{d}x\mathrm{d}y$? Even more weirdly, I came across 2d deSitter space, in which after I transform the cartesian coordinates $(t,x,y)$ to some coordinates $(\tau,\phi)$ I found that in order to find the induced metric I had to assume that $\mathrm{d}\tau\mathrm{d}\phi + \mathrm{d}\phi\mathrm{d}\tau=0$, which seems an "out of nowhere" statement.

Slereah

@ShikiRyougi They do, but also those metrics are hard

So they are less commonly studied

Shiki Ryougi

But is there a physical reason as to why we choose such metrics other than their level of difficulty? What would a term with $dxdt$ mean?

ACuriousMind

it's perfectly possible to choose coordinates such that you get cross terms

e.g. Eddington-Finkelstein

Shiki Ryougi

12:13 PM

Pages 324-325 of Carroll GR:

In order to find 8.6, it seems like I should assume that cross terms shouldn't be present

Because they don't cancel out

Brb I will show you an example

ACuriousMind

you shouldn't need any "assumptions" there

Shiki Ryougi

12:37 PM

In 2d de Sitter you would have $u=\mathrm{sinh}t$, $x=\mathrm{cosh}t\mathrm{cos}\chi$, $y=\mathrm{cosh}t\mathrm{sin}\chi$. If you plug the transformations in the metric $ds^2=−\mathrm{d}u^2+\mathrm{d}x^2+\mathrm{d}y^2$ you will have a term

$$\mathrm{sinh}t\mathrm{sin}\chi\mathrm{cosh}t\mathrm{cos}\chi (\mathrm{d}t\mathrm{d}\chi+\mathrm{d}\chi\mathrm{d}t)$$

How am I supposed to get rid of this term?

Oof that was hard to write

ACuriousMind

you sure that isn't actually a - in between the two cross terms?

remember that the hyperbolic functions don't have a sign change in their derivatives while the trigonometric ones do

Shiki Ryougi

Oh no, I'm too tired. I'll correct my calculation

because I still think I would have a problem

be right back

Shiki Ryougi

12:55 PM

Indeed there is, but that would imply that the one forms commute. Why is that?

ACuriousMind

they aren't forms

when you write $\mathrm{d}t\mathrm{d}\chi$ what you really mean is something like $\frac{1}{2}(\mathrm{d}t\otimes\mathrm{d}\chi + \mathrm{d}\chi\otimes\mathrm{d}t)$

Shiki Ryougi

huh? symmetrized tensor product? I thought $dxdy$ meant $dx \otimes dy$

ACuriousMind

hmm

okay, let me express that differently:

Formally, there's a difference between the symmetric algebra and symmetric tensors (see Wiki), although they are isomorphic. The metric is a symmetric tensor, but when we use notation like $g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ (and so also stuff like $\mathrm{d}t\mathrm{d}\chi$, without tensor signs), we're talking in terms of the symmetric algebra. The symmetric algebra is commutative

here's a question that discusses various other ways to think about why the differentials in this notation commute

Shiki Ryougi

1:16 PM

That seams like a beautiful sidetrack to pursue, thank you

also turns out I had done another mistake and the terms cancel without the need for thinking about the commutation etc

they cancel each other perfectly :p

More Anonymous

1:45 PM

@Feynman_00 how are your lectures going? I'm sure your already wayyyy ahead of the class

More Anonymous

2:10 PM

Help please?

Consider the Schwarzschild metric:

$ ds^2 = -(1 -\frac{r_s}{r}) c^2 dt^2 + (1-\frac{r_s}{r})^{-1} dr^2 + r^2 g_{\Omega}$

Now, if we consider $2$ geodesics separated by a curve of constant t.

$ ds^2 = (1-\frac{r_s}{r})^{-1} dr^2 + r^2 g_{\Omega}$

Let us assume they are radially separated:

$ ds = (1-\frac{r_s}{r})^{-1/2}dr$

Thus if we have $2$ curves $\gamma_1$ and $\gamma_2$ such $\gamma(\tau,c_0) = \gamma_1$ and $\gamma(\tau,c_1) = \gamma_2$.

$ \partial \gamma^\mu/\partial \tau = \partial s^\mu/\partial r $

I think I've done something wrong

Feynman_00

4:53 PM

The ceremony is finally over. Dang, I was the last one :P

@MoreAnonymous Actually I know no QFT and just know some fundamentals of GR but know I can accelerate

aman

how do i teach myself GR. Any book that starts from scratch? I have four years of uni education in physics so I think I fulfill the pre-requisites.

Slereah

5:18 PM

you can try Carroll's book

It is a gentle introduction

rb3652

Hi

si-LV-er_and_b-LA-ck

Hi

rb3652

I have a question that may fall into thermodynamics. Can I ask here?

si-LV-er_and_b-LA-ck

Askaway

rb3652

OK

I don't need a full on explanation, but if someone could point me to the right terminology or resources to look up more information on this topic, that would be great.

Basically, I'm looking at a (program) file that claims to derive the "average stress tensor". This is in the context of molecular dynamics.

I'm not quite sure what the average stress tensor is or how it's related to MD.

DanielSank

5:45 PM

Hi, everybody.

si-LV-er_and_b-LA-ck

Hello

DanielSank

So Zeilinger won the Nobel prize for Bell violations.

si-LV-er_and_b-LA-ck

Yeah, I saw that.

DanielSank

I am pleased to say that the first author of the loophole free violation sits 20 meters from me :-D

Very cool.

si-LV-er_and_b-LA-ck

nice

offer them a virtual cocktail from the h-bar :P

DanielSank

5:54 PM

eheh

si-LV-er_and_b-LA-ck

;-)

🍸🍹🍻

Relativisticcucumber

6:37 PM

carroll is gentle? XD @Slereah

Slereah

I'm afraid it only gets worse after Carroll

Relativisticcucumber

also i wanted to ask - there is a quotation from a qft book i am reading that says "the condition that $\bar{x}(t)$ is a stationary point of the action means that $\delta S$, the variation of the action which is proportional to the first power of $\delta x$ vanishes." what is a stationary point of action? i only know about the stationary action principle as stating that action is unchanging but not what it means for there to be a stationary point of action

haha well i am eager @Slereah i have been starting carroll recently. for my GR class we used guidry which i found quite gentle in comparison

ACuriousMind

6:59 PM

@Relativisticcucumber I'm not sure where you see the difference here. The first sentence gives the definition of a stationary point, which is that $\delta S$ vanishes, which is exactly the same as your "the action is unchanging", just written with $\delta S$ for the change of the action

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