The h Bar

Mew

09:01

@Kaumudi yes I have encountered dingo

my parents had a pet dingo

whatwhatwhat

@DanielSank when I'm important enough to be able to postpone talks without a drop in my grade/losing my chance at a diploma, then I'll take your advice :)

DanielSank

@whatwhatwhat Ah.

I see.

user228700

@DanielSank I wonder if u've put me on ignore .___. I pinged u awhile ago.

Mew

@DanielSank am i on ignore?

HALOOOOOOOOOOOO

@Kaumudi what pets do u have?

user228700

^ If u're actually on ignore, he won't be able to see any of ur messages so this "HALOOOOOOOOOO" is futile.

user228700

09:06

None.

Mew

oh

I think Daniel has us both on ignore

user228700

Never had any. I caught two fish from my pond one morning when I was about 10 but they mysteriously died by afternoon :-/

Mew

yeah they need water to survive

John Rennie

@whatwhatwhat there are lots of simple things you can do with a metric e.g. calculate time dilation.

user228700

@Mew He wouldn't put me to on ignore ;_; I offered to be his tour guide and everything!

Mew

09:08

must have come on too strong

user228700

@Mew Put them in a jar and everything, actually.

whatwhatwhat

@JohnRennie yea this is hitting me pretty hard. It's 3:08 AM where I am now and my presentation is at 10AM. Sleep? Pfbt.

user228700

@Mew Lol.

Mew

@whatwhatwhat what is your project on?

whatwhatwhat

deriving thorne's wormhole solution to the EFE

Mew

09:10

that sounds complciated

why didn't you pick something simple like entropy

but yeah i'd go to sleep bro

DanielSank

@Kaumudi Nope. Not ignoring you.

whatwhatwhat

If I sleep, I won't finish this on time.

DanielSank

You pinged me but I could not tell why.

Mew

what about me?

koolman

DanielSank

09:12

By Jove! I think I'm losing a significant amount of my SNR because of loss between the HEMT and the first room temperature amplifier!

user228700

@DanielSank I shared a video about street food.

DanielSank

@Mew eh?

Mew

yay

DanielSank

@Kaumudi U did?

I want street food!

koolman

In above question how they have written first equation . @JohnRennie

user228700

09:13

;_;

DanielSank

Delicious street food?

user228700

Yes!

whatwhatwhat

At this point I'm just trying to make a "skeleton" proof with minimal detail and explain to the audience that there was too much information for an undergrad student to understand.

Mew

@Kaumudi sweet balls?

user228700

JR watched it and said even he enjoyed the video!

user228700

09:13

@Mew :-P Don't think so.

John Rennie

4 hours ago, by Kaumudi

Indian Street Food Tour in Mumbai, India | Street Food in India BEST Curry

►

user228700

^ U are a savior :-D (:-P)

Mew

wtf

ur making me hungry

John Rennie

Though I'm not sure about the lamb brain curry. I had lambs brain en croute in Paris some years ago and I recall it had little in the way of texture or flavour.

DanielSank

@Kaumudi in Berkeley there is a restaurant called House of Curries.

Mew

09:15

if u eat brain u get smarter tho

DanielSank

They have an enormous stove top. Maybe eighteen burners. Next to the stove, there is a tower of pans.

koolman

Can anyone help in the above question

user228700

Whoa, I see.

DanielSank

You order something, a pan is filled, and it goes on the fire.

Mew

@koolman please direct your questionn to physics.quandaexchange.com

DanielSank

09:16

Then you get your curry. I go with friends and we order maybe six curries... and fresh bread.

user228700

It goes on fire. Wut? O.o

Mew

sorry i mean physics.qandaexchange.com

DanielSank

Everything is hot and fresh and delicious.

@Kaumudi The pan goes over the fire. Like normal.

user228700

Ohh :-P

DanielSank

I just mean that everything is cooked hot and fresh.

Mew

09:16

time to get food

laterz all

user228700

Sounds great :-)

DanielSank

It is.

user228700

@Mew Bye!

koolman

@Mew can we ask question on that like this

0

A: intensity at same point

The intensity of the wave is proportional to the square of the amplitude. In the center amplitude is doubled (A+A=2A), so intensity is quadrupled.

user228700

@DanielSank If u have the time, watch that vid and lemme know what u think.

DanielSank

09:17

Chana cooked in various Indian styles is one of my favorite foods ever.

I cook it in tomatoes and onion, with lots of tumeric.

user228700

Ah, I see. I'm a paneer fanatic :-)

DanielSank

And then I add the cilantro.

user228700

Coriander= Cilantro? Fancy.

whatwhatwhat

imgur.com/a/d9CW4 @JohnRennie what does the author mean in this step? We changed coordinate systems to simplify the components of the Riemann tensor and then contract the Riemann tensor to calculate the Ricci tensor and the scalar curvature R. What does that last part mean about the contracting?

user228700

Paneer is the savior of all Indian vegetarians, like me :-) It's so good <3

DanielSank

09:20

user228700

@JohnRennie I cannot comment for lack of knowledge.

John Rennie

Remember that in the Einstein notation repeated upper and lower indices are summed over. So $$R^\alpha{}_{\mu\alpha\nu} = \sum_{\alpha=0}^3 R^\alpha{}_{\mu\alpha\nu}$$

DanielSank

That is a photo of my poor attempt at Indian food.

user228700

@DanielSank Ey! That looks great!

user228700

I dunno of many people who have chana with rice tho. We have it with s'thing called poori :-)

John Rennie

09:22

This eliminates two of the indicies and the end result is to reduce the fourth rank Riemann tensor to a second rank tensor $R_{\mu\nu}$ called the Ricci tensor.

DanielSank

Yeah... I like rice though.

user228700

Also, beautiful tablecloth or whatever that is :-)

DanielSank

Thanks!

I like that tablecloth very much.

It is not Indian rice. It's Japanese rice.

John Rennie

The Ricci tensor is one of the objects that appears in Einstein's field equation.

user228700

Rice is staple food. It's lunch- every single day. The next time my mom makes a lot of fancy stuff for lunch/dinner, dyou want me to show u?

DanielSank

09:24

Hmmmm, the food in the video is very greasy so far...

@Kaumudi Yes. One thousand times, yes.

Please.

user228700

Yeah, street food tends to be pretty greasy :-/

DanielSank

@Kaumudi, a set of photos of such a thing would be a joy of my life not soon forgotten.

whatwhatwhat

@JohnRennie am I correct in thinking that the Riemann curvature tensor can be thought of as a large matrix and the Ricci tensor is a subset of that large matrix? Same for the scalar curvature R?

John Rennie

The Ricci tensor can be contracted to a scalar called the Ricci scalar using: $$R = g^{\mu\nu}R_{\mu\nu}$$ where $g^{\mu\nu}$ is the inverse of the metric tensor matrix. Again we're using the summation on repeated indicies.

Balarka Sen

@DanielS That's good stuff you're eating there.

user228700

09:25

@DanielSank There have been so many occasions these past couple of months! If I'd known, I'd have started taking pictures awhile ago! Damn.

DanielSank

@Kaumudi It's never too late.

@BalarkaSen Why, thank you!

user228700

Yep. Will do. I'll show u when I've enough pictures :-)

DanielSank

@BalarkaSen Are you familiar with my world famous chili?

John Rennie

@whatwhatwhat The Riemann tensor is a 4D matrix i.e. a 4x4x4x4 matrix. The Ricci tensor is not a subset of the Riemann tensor. Instead it's a contraction of it. In effect we squeeze the Riemann tensor down to 4x4 by discarding some of the data in it.

user228700

^ You've shared that about a thousand times till now :-)

DanielSank

09:27

@Kaumudi Yes, I am not shy about sharing something I think is truly good.

That chili recipe is certainly the highest quality creation of my life.

user228700

Wow :-)

Balarka Sen

@DanielSank No, but that looks fantabulous!

DanielSank

Higher than any physics I have done.

user228700

Oh my God :-O Even more than that book u've been working on?!

John Rennie

@whatwhatwhat have a quick scan through this paper

DanielSank

09:28

@BalarkaSen It is rather good. I shared it with an Indian friend who liked it enough to ask for the recipe. She prepared it for her husband and reported that he asked for her to prepare it again in the future.

In my humble opinion, when an Indian person asks for a repeat of a food, that indicates an ultimate level of success.

@Kaumudi I think so.

user228700

Man, I'm going to have to make it one day.

Amey Shukla

Hey guys, are there any good resources online for learning physics A to Z?

DanielSank

@Kaumudi Also, I only bring it up again if someone new hasn't seen it.

Amey Shukla

Like, I foudn this one staff.science.uu.nl/~gadda001/goodtheorist

btw, some are very tough. Ex. the optics section

DanielSank

@AmeyShukla Umm... I would say "no" only because learning physics A through Z is basically impossible :P

Amey Shukla

09:30

Guhh.

like what u learn in universities..

I knew someone was gonna say that.

I mean

user228700

@DanielSank: Unfortunately, almost none of the ingredients are available here :-/

user228700

Eg:

Balarka Sen

It certainly looks like that dish has Indian connotations. I admit I am rather fond of Indian foods (and that's one of the very few Indian things I am fond of).

Amey Shukla

That looks like chole :o

on rice..

Yep, definitly chole..

Balarka Sen

@DanielSank So, you're a food-theorist, eh?

John Rennie

09:32

@Kaumudi are you on Facebook? If so I have a link to a photo album of mine that might amuse (or possible appal) you

user228700

@JohnRennie Unfortunately, no, I'm not .__. Is there any other way I can see it? (Is it food food food? :-P)

DanielSank

@AmeyShukla Well... look... I've gone through university and graduate school and even now I cannot possibly tell you where to go learn everything I learned.

Amey Shukla

Oh

So like no university putting up notes?

DanielSank

Much of what I learned was cobbled together from a variety of (bad) books and required hours upon hours of studying to piece it all together.

@AmeyShukla There are lots of notes online!

Try MIT Open Courseware.

2

But do not expect to learn everything in one place!

John Rennie

@Kaumudi you can try this link though I suspect you need to be a Facebook member to see it.

Amey Shukla

09:34

Actually, havard one's are good but most of em are locked

DanielSank

@BalarkaSen More of a food experimentalist, in fact.

Amey Shukla

google.co.in/imgres?imgurl=http://www.tarladalal.com/members/…

You only eat this in mumbai

Rest will upset your stomach

Balarka Sen

Ya, that'd be a better choice of words. Good to hear; I like people with good food tastes.

DanielSank

@BalarkaSen I have a small set of recipes that I've worked hard on.

Balarka Sen

(Though I am not one myself - I just eat what's interesting to eat.)

DanielSank

09:37

Another is my chocolate cake. It requires 12 eggs and 1 kg chocolate.

Balarka Sen

yep I saw the chocolate cake here before I think

I like spicy things over sweets

DanielSank

@BalarkaSen A friend of mine recently prepared smoked beef ribs with Szechuan spices.

My goodness beyond the boundaries of the solar system that was so good.

user228700

@JohnRennie: Using my mom's account :-P (Don't tell her!) I'll get back to u once I've looked through...

Balarka Sen

haha, that's awesome

DanielSank

@BalarkaSen Still, in the end, my favorite foods are all vegetarian.

Fresh pizza, my chili, chana cooked in the Indian style...

Those are my favorites.

Many vegetarian foods should be eaten with yogurt.

user228700

09:42

@DanielSank: Man, I cannot wait for u to visit India sometime. You will positively die of joy :-)

Balarka Sen

I met with an american dude on a conference about a year ago in India and it turned out he didn't know what sweet yogurt was (that was in the menu)

DanielSank

@Kaumudi I know. I dearly want to visit and eat everything.

user228700

@JohnRennie: Is ur computer-table ur dinner table? :-P

John Rennie

Has anyone got plans to see Arrival?

Balarka Sen

Or rather, he didn't recognize it as yogurt when he was served with it.

@JohnRonnie I have heard that it's good

John Rennie

09:44

@Kaumudi I have a dinner table, but I like eating dinner at my desk so I can surf the web at the same time.

@BalarkaSen I've read the short story it's based on and it's very good. It's based on the block universe idea.

DanielSank

Thank you @Kaumudi for the video. I enjoyed it.

Balarka Sen

Huh, interesting.

DanielSank

I offer this other video about food in China. Many of the dishes are not vegetarian, but you may enjoy the scenery and the unique cooking equipment.

whatwhatwhat

@JohnRennie so I just talked with a classmate and he informed me of something that the professor told him in private: focus mostly on the interesting parts of the theory and leave out most of the derivation. That is, know the math, but don't show it as much. I also learned that he is a quantum theorist, so GR is on the other side of the world for him :P I hope...

Balarka Sen

@DanielSank Have you ever tried stuff (especially rice) with Dal's? I am not sure if an analogue of them is available outside of India.

DanielSank

09:50

@BalarkaSen Heh. My mother's family is from Lebanon. What you call "dal" I call "lentils", and eat them all the time.

One of my favorite dishes is m'jadara.

Balarka Sen

Ah.

user228700

@JohnRennie: And this whole time. You're almost as much of a foodie as DanielSank!

John Rennie

@Kaumudi Although I wouldn't describe any of my food as high cuisine :-)

I tend to like basic but tasty food.

DanielSank

@JohnRennie *@&! "high cuisine"!

@JohnRennie Amen!

user228700

:-) Man, u eat a lot of dessert!

John Rennie

09:52

@DanielSank you're speaking to a man who eats meat pie sandwiches. That is meat pies put in a burger bun and eaten like a burger.

DanielSank

^ delicious

John Rennie

@Kaumudi I eat a lot of everything :-)

Actually my weight is down this morning (65.6kg) so I need to eat a bit more.

DanielSank

@JohnRennie Jeez, you're light!

John Rennie

@DanielSank they are :-)

user228700

Haha :-D Man, I wish I could eat dessert all the time and still be thin.

DanielSank

09:53

I'm at 81.6 kg...

user228700

I eat like, one pastry a month :'-(

Balarka Sen

That sounds as if your weight was significantly higher the previous morning, J-R.

John Rennie

@DanielSank I'm a skinny runt. Everyone I meet says their first impression of me is that I'm skinny. Though I'm 1.78m tall so my BMI is in the normal range.

DanielSank

1.82m for me

No idea if my weight is healthy. Probably I'm a bit over the good range :(

user228700

Dammit, I'm hungry!

Balarka Sen

09:55

Chug some water

John Rennie

I'm unconvinced the BMI is that useful a measure. Can you run (run not jog) to the nearest restaurant without dying of exhaustion? If so you're fit and you can eat whatever you want when you get there :-)

user228700

^ Water is my one true love (for when lack of food screws me over). Oh, and sleep too.

John Rennie

@Kaumudi there aren't many calories in water ...

user228700

:-P I meant that when there's no food to eat, I drink loads of water but then that causes the levels of salt in my blood to drop or whatever.

user228700

So then I feel dizzy :-/

DanielSank

09:58

@JohnRennie Yeah I'm in reasonable shape.

Not great, but reasonable. I used to be an athlete and I think that somehow continues to keep me relatively healthy...

@Kaumudi Me too. Let's go get something...

John Rennie

@Kaumudi when I lived in Sudan we (European expats) used to take salt pills to keep our salt levels up.

user228700

@DanielSank Isn't it like, the middle of the night over there?

DanielSank

@Kaumudi Yes, perfect time for a snack :P

Actually I'll go to sleep instead.

user228700

@JohnRennie Oh, wow. Didn't know about their existence!

DanielSank

ciao

user228700

10:00

Night :-)

user228700

@JohnRennie: U're lucky u get to eat so much and still be described as "skinny".

user228700

And u told me u don't exercise that much!

John Rennie

When did I say I don't exercise much?

I don't go to the gym, but I cycle into town almost every day.

user228700

Oh, but u did!

user228700

I can't go back and look, it's been too long but OK, cycling is an excellent way to stay fit.

John Rennie

10:06

Oct 20 at 14:54, by Kaumudi

@JohnRennie: So, do you like cycling?

user228700

:-P OK.

user228700

October was a long time ago! :-o Anyhoo, gots to go out and run some errands. And it's about to rain >.<

user228700

Bye!

Electrodynamist

10:32

Hello! Anyone there?

This is just a spontaneous and arbitrary idea that came to me. Why don't we use a non-trivial metric to define a dual space for kets?

What would be the consequences if I do something like that?

whatwhatwhat

@JohnRennie Could you help me understand this part? What do they mean here by "take an equatorial slice"? imgur.com/a/zWar1

For ref: cmp.caltech.edu/refael/league/thorne-morris.pdf page 402

Electrodynamist

Would it all be still consistent but just complicated or there will be inconsistencies?

John Rennie

@whatwhatwhat The metric is given polar coordinates $r$, $\theta$, $phi$. If you're considering the trajectory of some object through the wormhole then in general all three variables will change with time.

Mew

@koolman of course you can

koolman

@Mew question of that format would be answered and would not be closed

John Rennie

10:37

However the metric is spherically symmetric, so we can choose our points $\theta=0$ and $\theta=\pi$ as we please. And it's convenient to choose them so the trajectory of the object lies in the plane $\theta=\pi/2$.

Mew

@koolman I don't know if they will be answered but they won't be closed

John Rennie

If we do that then only $r$ and $\phi$ change with time, and that makes the calculations of the trajectory simpler.

koolman

@Mew please do something as answers on that are answered very late

John Rennie

The spherical symmetry guarantees that the trajectory of any freely falling object will always lie in a plane. If you're intersted, the reason for this is that the spherical symmetry means angular momentum is conserved.

Mew

@koolman I can't control the pace at which people provide answeres tho

whatwhatwhat

10:40

@JohnRennie wait but this section is about the method of embedding

Electrodynamist

Would anyone like to help me understand this? Why don't we use a non-trivial metric to define a dual space for kets? Sorry for repeating the question, but it got surprisingly unnoticed!

whatwhatwhat

"We wish to construct, in 3-D Euclidean space, a 2-D surface with the same geometry as this slice, i.e., we wish to visualize this slice as removed from spacetime and embedded in Euclidean space."

John Rennie

@whatwhatwhat what page, and whereabouts on the page?

whatwhatwhat

@JohnRennie pg. 402, top right part of the page

koolman

@Mew I mean promote the site

Mew

10:43

@koolman you can if you want

koolman

@Mew Invite the active members

Mew

i'm not an expert in advertising

John Rennie

@whatwhatwhat OK. The spatial part of the metric is three dimensional ($r$, $\theta$ and $\phi$) but we can take a 2D slice by setting $\theta=\pi/2$. Since the metric is spherically symmetric we can do this without losing any important information.

The reason for doing this is that the 2D slice can have its curvature illustrated by embedding it in a 3D space.

koolman

@AmeyShukla of what level you want

whatwhatwhat

Ohhh...and it's because one of our assumptions was that the metric tensor was spherically symmetric!

facepalm.jpg

koolman

10:49

@Mew posted

Mew

ty

@koolman it appears you haven't posted the full question

are there 4 slits?

koolman

@Mew that is the full question

Mew

oh

whatwhatwhat

@JohnRennie so Figure 2 is NOT the 2-D slice then? (on pg. 403, top left part of the page)

koolman

@Mew Same as

0

A: intensity at same point

The intensity of the wave is proportional to the square of the amplitude. In the center amplitude is doubled (A+A=2A), so intensity is quadrupled.

Mew

10:52

oh i get it

John Rennie

@whatwhatwhat When you embed your 2d slice of constant $\theta$ in a 3D space you get something like figure 1(a). Figure 2 shows a vertical slice through this embedding i.e. it shows the profile as you pass through the wormhole.

Swapnil Das

@JohnRennie Why do u wanna stop your brain from astrophying?

John Rennie

@SwapnilDas That's an odd question? Why would anyone want their brain to atrophy?

Swapnil Das

It refers to astrophysics no? @JohnRennie

Oh god!

I'm so sorry. I read 'astrophying'

whatwhatwhat

@JohnRennie but doesn't Figure 2 seem a bit odd? Why is it lumpy along the z-axis? If I imagine a vertical slice of Figure 1a, I imagine a nice clean symmetrical tunnel-like wormhole.

In my imagination, there should be no lumps along the inner sides (next to the z-axis)

John Rennie

11:02

@whatwhatwhat Figure 2 shows a general wormhole i.e. the distribution of the exotic matter that keeps it open may not be smooth. So it shows a wormhole with a metric similar in general form to the Morris-Thorne metric but different in the details.

Mew

Hello @SwapnilDas

Swapnil Das

Hi @Mew !

whatwhatwhat

@JohnRennie ohhh ok. I was thinking that Figure 2 is part of his derivation. So question: can we call the shape in Figure 2 "spherically symmetric"?

I'm thinking no for the simple reason that we are not dealing with a sphere. It is more cylindrical, so I should use the phrase cylindrical symmetry, no?

John Rennie

Any metric of the form $$ds^2 = -a(\mathbf x)dt^2 + b(\mathbf x)dx^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2)$$ is spherically symmetric, where $a$ and $b$ can be arbitrary functions of $t$, $r$, $\theta$ and $\phi$.

So you have freedom to vary the geometry by choosing whatever functions you want for $a$ and $b$ while preserving the spherical symmetry.

whatwhatwhat

@JohnRennie ohhh ok I see.

John Rennie

11:10

I think the article means that figure 2 shows a geometry of this form. The bumpy curve is meant to show $a$ and $b$ can be anything.

whatwhatwhat

Btw, your help is not going unappreciated! My droopy eyes and the pain in my lower back thank you!!

:D

John Rennie

11:25

@SirCumference where did you get to with contracting spacetime?

user116211

@JohnRennie If I were a mod, I would have attacked it with \mathrms; alas ._.

3

John Rennie

@MAFIA36790 :-)

user116211

@JohnRennie That's a movie you won't want to miss.

John Rennie

@MAFIA36790 Have you seen it?

A friend went to see it when it was first screened in Chester and they said it was very good.

whatwhatwhat

@JohnRennie can you check my interpretation of this part? On the section called "The stress-energy tensor" on pg. 401, he explains how he derives his values for the stress-energy tensor. This is what I'm going to say: The EFE require that the stress energy tensor must be proportional to the Einstein tensor, which forces the stress-energy tensor to have the same algebraic structure as the Einstein tensor. Of the components that are nonzero,

each one must allow a static observer to make a measurement and return a physical interpretation.

user116211

11:44

@JohnRennie No; but I have highly positive reviews from the sources at Movies&TV SE chatroom I deem reliable.

John Rennie

@whatwhatwhat that seems OK. The same algebraic structure just means that if $G_{ab}=0$ for any values of $a$ and $b$ then $T_{ab}$ must also be zero. That's how he concludes which elements of $T$ can be non-zero.

For the meaning of the elements in $t$ see:

Stress–energy tensor

The stress–energy tensor (sometimes stress–energy–momentum tensor or energy–momentum tensor) is a tensor quantity in physics that describes the density and flux of energy and momentum in spacetime, generalizing the stress tensor of Newtonian physics. It is an attribute of matter, radiation, and non-gravitational force fields. The stress–energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity. == DefinitionEdit == The stress–energy tensor involves the use of superscripted...

The metric is time independent, so not of the elements of $T$ can be a function of time, and the spherical symmetry means none of the elements can be functions of $\theta$ and $\phi$. That's why the non-zero elements are functions only of $r$.

user116211

@0celo7 I miss thee :(

whatwhatwhat

12:02

@JohnRennie Ah ok. Good link, I think I should put in a section where I explain the stress-energy tensor as background info.

@JohnRennie am I correct on this next part? For the section on the absence of a horizon, pg. 403: For any static, asymptotically flat spacetime, the horizon is the region where $$g_{00} = -e^{2\Phi} \to 0$$. This means that the change in time (proper time) of an observer approaching the event horizon must go to zero. To remove any horizon from our wormhole solution, we must say that $Phi (r)$ is finite everywhere.

But I guess I don't fully see why there must be an absence of a horizon for a wormhole.

John Rennie

Suppose you're far from the wormhole where spacetime is flat, and there is an astronaut at some point $(t, r, \theta, \phi)$ near the wormhole. Suppose also that the astronaut is stationay in space. If they are stationary then $dr = d\theta = d\phi = 0$.

That means the metric simplifies so just $$ds^2 = -c^2e^{2\Phi}dt^2$$

Now what you need to know is that $ds^2 = -c^2d\tau^2$ where $\tau$ is the proper time, and the proper time is the elapsed time on a clock held by the astronaut. Sustituting this into the metric we get (after some rearranging):

user116211

@JohnRennie Aha!

John Rennie

$$ d\tau = e^\Phi dt $$

Or $$\frac{d\tau}{dt} = e^\Phi $$

That fraction $d\tau/dt$ is the time dilation i.e. the ration of the astronaut's time to your time, and at a horizon $d\tau/dt = 0$. And that can only happen as $\Phi \rightarrow -\infty$.

So what the article is saying is that to avoid a horizon we require that $\Phi > -\infty$ everywhere.

@MAFIA36790 I'm not doing it to annoy you :-) I'm trying to type fast and all those mathrms would really slow me down. It's OK in an answer where you can take the time to type the answer, but it's too slow in the chat.

user116211

12:17

@JohnRennie No offence; it's just a funny tone and nothing else; carry your work; no need to bother.

John Rennie

@whatwhatwhat ah, OK, I think I misunderstood. You're asking what the problem with a horizon is?

heather

hello everyone

John Rennie

If there's a horizon you can still pass through the wormhole but it's a one way trip i.e. there is no way to back the other way to where you started.

whatwhatwhat

@JohnRennie yes because the article is adamantly stating that there must be an absence of a horizon and I couldn't understand why

John Rennie

Geometries like this exist e.g. the Reissner-Nordstrom black hole. But it's a prett useless method of hyperspatial travel if you can only go one way.

Oh yes, and the infinite time dilation at the horizon means you take an infinite (external) time to pass through the wormhole so the outside universe will have ended before you get through :-)

Again, this presents operational difficulties :-)

So all in all it's probably good if our wormhole doesn't have a horizon.

John Rennie

12:38

@whatwhatwhat I have to go get lunch. Good luck with the presentation.

whatwhatwhat

@JohnRennie oh no!!!! Jk, you've been great. Enjoy!!

user116211

> Stack Exchange
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user116211

O.o

whatwhatwhat

Wait what

Mew

are there such thigns as wormholes

whatwhatwhat

13:08

we dont know

@JohnRennie if you happen to come back, I have a few more questions before my presentation.....let me know if you're there!!

Jim

13:29

@Mew ever been apple picking?

You'll see wormholes all the time

koolman

Hii @ZYX

Jim

@JohnRennie define "run" relative to "jog"

do I have to sprint the whole way?

What if the nearest restaurant to me is really far? Can't I still be fit if I can jog that without being exhausted?

Why am I taking your statement as the objective and decisive measure of fitness?

Secret

Caption: Checking associativity on the way go, without leaving the cayley table:

What associativity did on the cayley table is not obvious, but basically, it result in all entries located in the same row and column as the value of the product in question to be automatically associative.

As more entries were filled in, some of the triple product terms will be forced to be equal since the dark gray columns and rows overlap the previous ones established. This is where the control of the associativity is seen in limiting the possible choices of the elements

As an illustration, suppose I start with an empty cayley table of 4 elements a,b,c,d. Suppose I choose the entry bb. Now the value of bb can be equal to one of a,b,c,d. Regardless of what I choose, there's an algorithm that automatically multiply the column a,b,c,d on the left and row a,b,c,d on the right repectively to the term bb, and associativity will mean where I put the brackets will not matter.
If I choose bb=c, then the row and column containing c will be "grayed out, i.e. all possible left and right products of the term (bb) will be automatically associative, thus result in a total…

ACuriousMind

13:48

@JohnRennie You mean "venomous" :P

Secret

After that, I can pick the next element and decide what it has to be specified, and to make sure it does not contradict the previously established results

This is the reason why associativity is such a pain to check for large system, because what terms in the associativity table will possibly conflict with each other depends on the order of the elements specified

which manifest in the cayley table as how often the gray columns and rows overlap

ACuriousMind

14:08

@JohnRennie Can you explain to me how this is a duplicate? It's not particularly well-asked, but the two questions do not ask the same at all in my reading.

John Rennie

I thought the OP was asking about a 5D embedding. If you think it's not a duplicate I'll withdraw my close vote.

whatwhatwhat

@JohnRennie Could I ask you some more questions before my presentation??

John Rennie

I'm currently in an optician's waiting room and likely to be called any moment. But you're welcome to ask.

ACuriousMind

@JohnRennie I think OP is asking whether the background-independence of GR is a special feature of 4D or holds true in all dimensions. The question is badly written but imo not about an embedding

whatwhatwhat

@JohnRennie Ok. How does one apply the "standard equations" necessary to get the Cristoffel symbols and the components of the riemann Curvature Tensor? My interpretation is that you literally just plug in the metric, however, my classmate was doing a proof and he was applying them like operators.

For reference: imgur.com/a/pIYDg

ACuriousMind

14:21

@whatwhatwhat What do you mean "applying them like operators"?

whatwhatwhat

I thought I saw him doing something like $\Gamma^{\alpha}_{\alpha \beta} f$ where f was some function.

ACuriousMind

Your first formula gives the Christoffel symbols as a function of the metric. To compute the Christoffels for a given metric, you plug in that metric. I don't understand what alternative that there is.

whatwhatwhat

Ok that's what I was thinking too, but a memory of something my friend did with these symbols popped up and I didn't want to leave it unanswered.

ACuriousMind

@whatwhatwhat Well, you sure can write something like that but it's not going to help you compute the Christoffels.

whatwhatwhat

@JohnRennie I don't quite understand how to show my audience that I've solved for a wormhole solution.

What's the big finale? I think I found the shape function $b(r)$, but other than that it only talks about a constraint on $\Phi (r)$.

To my understanding, when I've solved for a solution to the EFE, I should be able to plug the solution back in and show that it works. But I don't see what I'm supposed to have at the end that can be plugged back in.

ACuriousMind

14:28

You will utterly bore your audience if you force them to watch you verify a solution to the EFE is indeed a solution

Better talk about the interesting features of the solution, i.e. where's the "wormhole", how is it different from the tamer vacuum solutions, etc.

whatwhatwhat

Ok. You saying that means my professor will think that, so I'm totally ok with not showing that.

@ACuriousMind but to your good point: I'd like to do that, but I don't even understand what the solution looks like. For reference, this is the paper I'm following. For me, the math is pretty intense: cmp.caltech.edu/refael/league/thorne-morris.pdf

ACuriousMind

As for what, in principle, you should be able to plug back in is just the metric. You usually solve the EFE for the metric given a stress-energy tensor (note that this is not unique, i.e. there can be several metrics compatible with a stress-energy)

Sanya

the problem with explicit calculations in a talk is that a) you need to be very good about guiding the audience along the main ideas of the calculation without letting them understand every step and b) you need to be extremely careful about the right speed and the right formatting (i.e., which steps need to appear together/reappear) - also in my experience, it is often not worth it ;)

whatwhatwhat

Well, it's a physics class about math needed by all physicists. He expected math heavy projects.

Sanya

yeah - but math heavy means explaining ideas and lines of thought I think - but in the end, I don't know your professor either, so don't worry too much about what I'm saying ;)

ACuriousMind

14:34

@whatwhatwhat To be frank, if you don't understand what your solution looks like then you are not suited to give a talk about it

So that should be your priority: Understand what it actually is that you are presenting here

Not the messy details, or the derivation, but the result. You can't confidently talk about a result when you have no idea what the result actually is

whatwhatwhat

This is true, which is why I asked for help :(

ACuriousMind

Well, I guess you should start by understanding why the terms in eq. (1) of your reference are called "shape" and "redshift" functions, and how adjusting them changes the properties of the wormhole

Sanya

@ACuriousMind and I wanted to ask you why I couldn't flag that harmonic oscillator question as homework ... xD

ACuriousMind

@Sanya Hmm? Did you get an error because I was faster? :D

Sanya

@ACuriousMind I only got three flag options

spam and ... I don't remember

and I was like "Oo what's happening"

ACuriousMind

14:44

Ah, yes, you got only the flag options for closed questions: spam, rude/abusive and "needs a mod"

Sanya

but it wasn't closed yet - so I was confused :D

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