The h Bar

DanielSank

12:10 AM

I want my hats back.

Bernardo Meurer

@DanielSank Come get them

DanielSank

12:56 AM

@BernardoMeurer You don't has.

Bernardo Meurer

No, but I have flip flops

vzn

6:07 AM

you guys all like neil degrasse tyson a lot? re the many refs in here? check out this guy! tanmay bakshi, coder extraordinaire... kind a like an IT justin bieber? o_O :P ... heard about him on another SE chat room, hes got an acct on the site! etc

13 Year Old on a Mission - to Connect the Disconnected w/Ai | Tanmay Bakshi | TEDxCincinnati

►

dmckee

I would say that the poster of this question has worked much harder than average to make his question agree with the homework policy.

John Rennie

6:44 AM

@dmckee I think I originally voted to close that question then on reflection voted to reopen

dmckee

@JohnRennie I think it was clearly a homework question at first, but this user has been willing to take advice and pare it down to the conceptual core. He's a keeper.

Secret

That's a very interesting question layout. Had I not read those comments, I will never realise it is a homework question

Shing

8:01 AM

I am reading this, and thinking if we can create a box with potential that is greater than energy of a particle. Then we somehow manage to trap the particle into the box *without* affecting its energy, then what will happen?
https://physics.stackexchange.com/questions/107235/how-will-a-particle-with-energy-less-than-v-rm-min-behave

Secret

8:23 AM

Such a scenario means the potential basically act as a barrier to the particle, and the particle inside the barrier will be an exponentially decaying wavefunction

John Rennie

8:49 AM

@Shing a free particle by definition must have an energy greater than zero. So if we create a potential well with an energy less than zero we cannot trap the particle in the box unless there is some mechanism for transferring energy away from the particle.

As a concrete example, a free electron cannot be captured by a proton to form a hydrogen atom unless it radiates some of its energy away as a photon.

Shing

10:20 AM

@JohnRennie thanks for the elaborating. that makes sense

Slereah

11:15 AM

Thermodynamics question

My heater isn't working

What do I do

John Rennie

Put on a coat

Paint yourself silver

Go into a vacuum chamber and pump out all the air from it

Slereah

I'd die

John Rennie

But you'd die warm ...

CooperCape

11:29 AM

If you have $\int_{r_0}^r f_1(x)-f_2(x)$ you can split that up into $\int_{r_0}^r f_1(x)-\int_{r_0}^r f_2(x)$, right?

John Rennie

Yes

CooperCape

sweet

Slereah

integrals are linear yeah

CooperCape

time for fun with partial fractions then innit

ugh

Secret

@JohnRennie I thought shing mean this

Anonymous

11:37 AM

@CooperCape Try proving it :)

CooperCape

Maybe later got integrals to do...

Secret

0

Q: Does the " time " of reaching thermal equilibrium under the same physical conditions always equal regardless of the number of trials?

Does the " time " of reaching thermal equilibrium under the same physical conditions always equal regardless of the number of trials? eg) which one is more probable ?

and weird question lol

Shing

@Secret Yeah, I originally meant this, but I am not so sure if it is meaningful to say $V_{min}>0$, as I can't picture the actual setup for $V_{min}>0$ (say the jump in potential represents conducting plates)

Secret

11:53 AM

but the problem of that is the particle will be inside the potential barrier, thus its wavefunction will exponentially decay. I am not sure if tunneling inside an infinite wide barrier make sense though

Coto TheArcher

@JohnRennie I'm here

John Rennie

@CotoTheArcher Hi :-)

Coto TheArcher

Hello

CooperCape

Btw @CotoTheArcher I jsut posted an answer to your q. going a different (and imo less intelligent) way than John's :p

Coto TheArcher

Checking

John Rennie

11:57 AM

Suppose the Earth-Moon distance is $d$ and you're at some distance $x$ from the Earth. Then your distance to the Moon is $d-x$. OK so far?

Coto TheArcher

What is W @CooperCape ?

CooperCape

work done.

Coto TheArcher

`and you're at some distance $d$ from the Earth.`

You mean $x$?

Shing

oh man, calculating a potential with three dirac dalta potential's eigenenergy is killing me...

John Rennie

@CotoTheArcher oops, yes. Now corrected.

Coto TheArcher

11:59 AM

Ok so far @JohnRennie

John Rennie

Then the gravitational acceleration due to the Earth is $$ a_{Earth} = \frac{G M_{Earth}}{x^2} $$

And the gravitational acceleration due to the Moon is $$ a_{Moon} = \frac{G M_{Moon}}{(d-x)^2} $$

The balance point between them is the point where the two accelerations are equal (and opposite)

So we get:

Coto TheArcher

$$ \frac{G M_{Earth}}{x^2} $$ = $$ \frac{G M_{Moon}}{(d-x)^2} $$

John Rennie

Yes

Solve for $x$

Coto TheArcher

Ok

let me replace the other constants

John Rennie

@CooperCape I had flagged the question as homework ...

Coto TheArcher

12:03 PM

d = distance from center of earth to lunar?

CooperCape

I didn't really do much in terms of answering.

John Rennie

That's why I commented instead of answering

Coto TheArcher

It's not homework btw

CooperCape

But I will delete it if you want it was nice revision anyway.

John Rennie

@CotoTheArcher $d$ is the distance from the centre of the Earth to the centre of the Moon.

Coto TheArcher

12:04 PM

Ok

x = d/2 ???

Doesn't sound right, I'll recheck

John Rennie

@CotoTheArcher You would only get $x = d/2$ if $M_{Earth} = M_{Moon}$

Coto TheArcher

Ehm ok?

So what does x show

Being in the center of earth & moon would pull the payload towards the earth considering earth's mass > moon's mass

CooperCape

That's why $x$ is only $d/2$ if M_earth=M_moon

John Rennie

@CotoTheArcher I'm not sure what you are asking. $x$ is the distance from the centre of the Earth to the balance point. If the masses were equal obviously the balance point would be exactly halfway between them.

Coto TheArcher

Yeah but they're obviously not, so that's not the balance point. Taking in consideration the actual mass of the earth and the moon, I want to find the balance point

Shing

12:13 PM

damn, I thought I can solve it analytically, but it seems I have to solve the 3 dirac potentials with graph method.

John Rennie

CooperCape

How do you diagram so quick...

Coto TheArcher

Yes, I want to find x ^

John Rennie

OK, for convenience (to avoid the subscripts) I'm going to call the mass of the Earth $M$ and the mass of the Moon $m$

So our equation is: $$\frac{GM}{x^2} = \frac{Gm}{(d-x)^2} $$

OK so far?

Coto TheArcher

Ok

John Rennie

12:18 PM

I'm going to rearrange this to: $$ \frac{(d-x)^2}{x^2} = \frac{m}{M} $$

Still with me?

Secret

@Shing Ok I just did my back of the envelope calculation. Taking the usual finite barrier problem but changing the position of the barrier to $(-\infty,\infty)$ you will get the transmitted, reflected and incident wavefunctions to all blow up for one of the exp and go to zero for the other exp term, thus there is no wavefunction for the case of infinite barrier

Coto TheArcher

Yes

John Rennie

Then square root both sides to get $$ \frac{d-x}{x} = \sqrt{\frac{m}{M}} $$

Coto TheArcher

Ok

Shing

@Secret thanks for the help!

John Rennie

12:21 PM

The left side can be rewritten as: $$ \frac{d}{x} - \frac{x}{x} = \frac{d}{x} - 1 = \sqrt{\frac{m}{M}} $$

So we get: $$x = \frac{d}{1 + \sqrt{m/M}} $$

Coto TheArcher

So that's the balance point?

John Rennie

Yes.

Suppose the masses are equal, what does that equation give for $x$?

Coto TheArcher

And what would be the required velocity to reach that balance point?

If m = M then x = d/2

since sqrt(1)=1

John Rennie

@CotoTheArcher Which is exactly halfway. Just what you would expect.

Coto TheArcher

Yes but I'm interested to find the balance point which is given by the equation $$x = \frac{d}{1 + \sqrt{m/M}} $$

John Rennie

12:25 PM

That's a quick sanity test that the equation does make sense and we haven't made a mistake in the working.

Coto TheArcher

Yep

How can I find the minimum velocity required to reach x?

taking in consideration the altitude launch/distance from center of earth

John Rennie

Trajectories are time symmetric i.e. if you launch from the surface of the Earth with some velocity $v$ and get to $x$ with zero velocity that is the reverse of starting at $x$ with velocity zero and falling back to Earth's surface with velocity $v$.

The point of this is that it's easier to do the reverse calculation

Coto TheArcher

But with any tiny force towards earth from point X it would return to earth

Wouldn't the reverse calculation be 1/v?

approaching infinity as v approaches 0

John Rennie

Gravitational potential energy is given by $$ U = - \frac{GMm}{r} $$

For convenience we'll assume our object has unit mass, and again I'll use $M$ for the mass of the Earth and $m$ for the mass of the Moon.

CooperCape

Isn't mass irrelevant anyway because you take $KE=U$?

John Rennie

12:32 PM

So the potential energy at the balance point is: $$ U = -\frac{GM}{x} - \frac{Gm}{d-x} $$

OK so far?

Coto TheArcher

Sure

John Rennie

When we reach the surface of the Earth our dustance from the Earth's centre is $r$ (r = radius of Earth) and the distance from the Moon is $d - r$

Coto TheArcher

Ok

John Rennie

So the potential energy at the Earth's surface is: $$ U' = -\frac{GM}{r} - \frac{Gm}{d-r} $$

And the change in the potential energy is $U - U'$

Since total energy is conserved the change in the potential energy must be equal to the change in the kinetic energy, so this means we can calculate the kinetic energy of our object when it reaches the surfaceof the Earth.

Coto TheArcher

ok

John Rennie

12:37 PM

And KE = $\tfrac{1}{2}mv^2$, or just $\tfrac{1}{2} v^2$ since we assumed our object has unit mass.

Coto TheArcher

Ok

John Rennie

So we get $$ v = \sqrt{ 2(U - U')} $$ and that gives us the velocity you're trying to calculate.

Coto TheArcher

Replacing U and U'?

John Rennie

Yes

The full equation is going to be rather messy, but you can do it bit by bit.

First calculate $x$, then use $x$ to calculate $U$ and $U'$

I recommend using a spreadsheet to do calculations like this as you can lay out the calculation ina nice way

Coto TheArcher

I'm having problems replacing cause I think I do something wrong with addition/subtraction. Can you replace U and U' and give me the full equation?

and I'll see if it's any "simplificatable"

John Rennie

12:43 PM

I have to say I feel no great to write out the whole long equation ...

Coto TheArcher

Ok I'll write it and you just confirm if that's possible

$$ v = \sqrt{ 2(-\frac{GM}{x} - \frac{Gm}{d-x} + \frac{GM}{r} - \frac{Gm}{d-r})} $$

That's what I got

Btw I'm using latex2png.com to translate all these

John Rennie

This is what I get for $x$ - using Google Docs:

Do you agree?

Coto TheArcher

Yeah

John Rennie

Let me do U and U' ...

Coto TheArcher

Sure

John Rennie

12:51 PM

And taking the last step:

Coto TheArcher

So $$ v = \sqrt{ 2(6.13E+0.7)} $$

Sounds right?

John Rennie

Well the escape velocity from the Earth's surface is a shade under 11,200 m/sec so our answer certainly seems reasonable.

Coto TheArcher

Wait, what is the launch radius for v = 11.06km/s ?

escape velocity being 11.2km/s is at sea level

= 6,378 km from center of earth (radius)

John Rennie

That's launching from the surface of the Earth r = 6,371,000 m

Coto TheArcher

Can you give me the spreadsheet link?

John Rennie

12:58 PM

Remember we don't need escape velocity because we aren't trying to escape from the Earth. We're only trying to get to a distance $x = 346,018$ km.

Coto TheArcher

Yeah

But it's almost as much as escape velocity

John Rennie

@CotoTheArcher yes, because the the balance point is a long way from the Earth ...

Coto TheArcher

Okay

John Rennie

The spreadsheet is here

I think I've set the permissions properly ...

Coto TheArcher

I've copied the file to my drive

John Rennie

1:01 PM

I have to go now. I'll be around later if you want to discuss if further.

Coto TheArcher

You answered my question. Thank you very very much for your time! :)

John Rennie

You're welcome :-)

Coto TheArcher

Good day

Future Historian

1:19 PM

Hello.

So......turns out that I may have a bit of a rocket science problem.

More specifically, how long to decelerate from 2.8 light years to enter Saturn's orbit in a realistic scenario, then from there, head to Earth with nuclear pulse fusion engines.

So, how do I get that science figured out in realistic terms, when considering an object of 734 km in length slowing down to enter our Solar System at that velocity, and a 10 km long, 6 km in diameter monster heading to Earth from there?

Sorry if I disrupted anything. Just wanted to know, since WorldbuildingSE is quiet.....a bit too quiet.

:(

0celo7

1:50 PM

@BalarkaSen PUBG is actually an awful game

It gives me heart palpitations

Future Historian

@0celo7. Is that not the point?

Balarka Sen

@0celo7 Exactly, the working principle is that it's more of a surviving game than a shooting game

0celo7

@BalarkaSen you don’t know until you play it

Balarka Sen

True

0celo7

You spend 20 minutes gathering supplies only to be shot in the head by a kar 98

Balarka Sen

1:56 PM

lol

0celo7

Playing this game is literally insanity

Future Historian

That is kind of the point.

Survive for as long as possible.

It is not called a "Battle Royale" game for nothing.

0celo7

the blue circle of death is too fast

It’s too easy to get stuck if you don’t have a car

Poor design choice

Balarka Sen

That's actually a problem, yes

I think they're gathering data to find out what the optimum speed of the blue circle should be

Future Historian

^

NSA-style data gathering.

:P

laughs evilly

Balarka Sen

2:01 PM

PUBG needs a lot of development

0celo7

It needs to die

Waste of money

Balarka Sen

lol

rip

0celo7

The gameplay is really terrible

It should be on the frostbite engine or something

Right now it’s way too clunky for a competitive FPS

Balarka Sen

I agree

0celo7

You don’t even play

Future Historian

2:02 PM

@0celo7. It started out with an independent developer.

Not like they can afford the Frostbite engine yet.

Balarka Sen

I played it once actually :) I don't have it on my machine, but had a few tournament on a friend's machine

0celo7

Can your computer run it?

My rig runs it on ultra but basically melts in the process

Balarka Sen

No I don't have the requirements

Not a gamer here myself

Future Historian

Does anyone remember that one question I had about the kinetic rods?

Because I just realised: how deep can those things penetrate anyway?

ACuriousMind

I don't remember any kinetic rods and find the question without context vaguely unsettling :P

0celo7

2:07 PM

@FutureHistorian up to the cervix, usually.

Balarka Sen

@ACuriousMind There, settled

Future Historian

Oh.

Basically, @ACuriousMind. It was a question I posted a few months ago as to what one would visually see as a kinetic rod hits Earth's surface.

0celo7

Kinetic rod = penis, no?

Oh

Future Historian

And it was part of a fictional world I am working on where Earth comes under attack by an extraterrestrial force.

ACuriousMind

@FutureHistorian You expected someone here to remember a question you posted months ago off the top of their head?

Future Historian

2:08 PM

But since WorldbuildingSE is too quiet.............well......you get the point.

Er.....to a degree, yes.

:(

Balarka Sen

@0celo7 so crude

ACuriousMind

@FutureHistorian There's enough stuff in this chat that you can be lucky if people remember last week ;)

Loong

@ACuriousMind Who are you again?

2

Balarka Sen

loooool

ACuriousMind

:'(

Future Historian

2:11 PM

Oh.

Well, @ACuriousMind. It is a closed question, though.

However, I can point out that......well, the details for the rods are in there, and I may have not fully finished the designs neither now or at the time of posting.

ACuriousMind

Ah, you're talking about physics.stackexchange.com/q/371531/50583

0celo7

@BalarkaSen ugh so my talk is just going to be me rambling. I think if I try to figure out any more physics i might die

ACuriousMind

Rambling talks are the best talks

Balarka Sen

study h principles

h principles > physics

ACuriousMind

@BalarkaSen What an appropriate thing to say in the h bar :P

Balarka Sen

2:15 PM

hahah

0celo7

I have a general idea of what I want to cover and I know the material pretty well, so I think it’ll be fine

Danu

Hi @ACuriousMind!

ACuriousMind

@Danu heya

How's it going?

Danu

pretty OK

Not super happy with my PhD progress so far, but I've gotten really into chess and am having a lot of fun with that :-)

How's your job?

ACuriousMind

Very nice so far, but then again it's only been two weeks

Danu

2:26 PM

What kinda stuff are you doing?

On a daily basis

Future Historian

@ACuriousMind. Er......yes.

I am.

Danu

Also a stupid question: Is it possible to have a (real) symmetric, positive-definite matrix $A$ such that $A_{11}=-(A)^{-1}_{11}$?

ACuriousMind

@Danu Currently mostly playing around with the new language until I've had proper training in it, not doing anything that'll land in production yet

Danu

Coding?

ACuriousMind

Sure, it's a job as a software developer after all ;)

Danu

2:28 PM

Okay

I'm not sure whether I knew that already

ACuriousMind

Also, drinking a lot of coffee :P

Danu

but aight

ACuriousMind

@Danu For a 2-by-2 matrix that would be $a = - d /(ad -bc) = - d / \det(A) $ so I don't see why that wouldn't be possible

Danu

You also need it to be positive definite

(and symmetric, which makes it a bit easier)

positive definiteness gives, for a 2-vector $(x,y)$, the condition $ax^2+2bxy+dy^2>0$ which means that $a,b,d>0$ all separately, right?

Maybe not $b$.. let's see.

I guess not $b$.

Mithrandir24601

@Danu This is really weird, but I was wondering this all of 3 days ago...

Danu

2:35 PM

Huh, haha

that's funny

Mithrandir24601

(came up with 'most likely not')

Danu

lol

I have an idea

maybe diagonalizability will help

ACuriousMind

What's so interesting about that condition that two people are wondering about it?

Danu

I wanna see if I have found a contradiction in a paper or not haha

ACuriousMind

lol

Danu

2:37 PM

I hope not

because it's my supervisors paper, and I'm supposed to work on it haha

Mithrandir24601

@ACuriousMind I've forgotten whether it came up looking at Pauli matrices in a curved background or a specific two-state vector that I found

ACuriousMind

@Danu Well, proving him wrong certainly is work!

Jake Rose

Is anybody free to try and help me with a probability question?

Danu

I have a conventions question too

When people write something like $ds^2= dz d \bar z$, what is the right hand supposed to read? $dz \otimes d \bar z$ or $\frac{1}{2}(dz \otimes d\bar z + d \bar z \otimes dz)$?

Balarka Sen

Quite sure the former

Danu

2:49 PM

hmkay

Balarka Sen

Hm, I guess if you have a Hermitian metric on a curve, you write that as $h = f dz \otimes d\bar{z}$. The corresponding Riemannian metric is $1/2(h + \bar{h})$ though...

Danu

exactly

cause if you have $d z\otimes d\bar z$ it's not symmetric in real coordinates

But I have a supposedly Riemannian metric that is written like this

Balarka Sen

Yeah well basically that's the real part of $h$

Danu

How am I supposed to interpret it?

symmetrize it myself, right?

Balarka Sen

So $ds^2$ is a Riemannian 2-tensor... so I guess you were right

You should interpret it like you said

Slereah

2:53 PM

It's the flat metric of the complex plane

Just pick the appropriate metric that gives you a flat plane

Balarka Sen

Yes, but it's a Hermitian vs Riemannian question

Danu

in the case I'm interested in it's of course not this simple

It's not obvious that the expression I'm given is not symmetric

it has many terms

Balarka Sen

i see

Danu

Sme of which are in real coordinates

some in complex

so hand-symmetrizing only parts of it is questionable, too

Captain Bohemian

4:17 PM

Does quantization just mean finding the representation of all quantities in terms of operators which can act on states in a certaiin Hilbert space?

CooperCape

allo

bolbteppa

In $H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$ if you set $x(t) = b \cos(\omega t)$ and form the dimensionless $a(t) = \frac{m \omega x + i p}{\sqrt{2 m \omega \hbar }} $ then we have $a(t) \sim b e^{-i\omega t}$ hmm

Seems like some way of saying you want strictly complex solutions

John Rennie

4:48 PM

I've just had a blue chip meal:

Literally! The chips were blue!

CooperCape

I've seen crisps like that as well..

I don't know why.

0celo7

dear god

John Rennie

A novelty thing I guess. They tasted no different to regular chips.

These possibly:

Adirondack Blue

The 'Adirondack Blue' is a potato variety with blue flesh and skin with a slight purple tint, released by Cornell University potato breeders Robert Plaisted, Ken Paddock, and Walter De Jong in 2003. The 'Adirondack' varieties are unusual because both the skin and the flesh are colored and have high levels of anthocyanins. This variety is good for boiling, baking, and mashing, and can be used for brightly colored salads. Unlike many blue potatoes, it does not turn grey after boiling. 'Adirondack Blue' was bred from N40-1 ('Chieftain' x 'Black Russian') x NY96 and is not under plant variety protection...

skullpatrol

Putting the blue into blue chips :-)

No ketchup?

Slereah

@JohnRennie are you back in the 90's

When colored food was an unexplicable trend

John Rennie

4:55 PM

@skullpatrol BBQ sauce! :-)

skullpatrol

Yummy.

Anonymous

They look like erm...blue worms

Anonymous

:D

Anonymous

But BBQ sauce sounds nice

John Rennie

It has to be said they don't look that great. They tasted OK though.

Venison sausages BTW.

skullpatrol

4:57 PM

That's quite a few sausages?

John Rennie

I was hungry ...

skullpatrol

Indeed.

John Rennie

I am no longer hungry :-)

CooperCape

did you weigh them though

I remeber you weighed your sausage rolls

0.86kg?

user55789

I have a shameless homework question!

But a hint would be sooo welcome

My current approach was: argue that the last term is symmetric and antisymmetric, but the problem is that my argument must be false, since both g hat and g_

John Rennie

5:01 PM

@CooperCape 600g (before cooking)

user55789

would satisfy this property in this line of reasoning...

Or I could use 1/D g_munu g^munu?

0celo7

5:30 PM

@user55789 should that be "conformal killing vector for $\hat g$"?

user55789

No, regular g

Or not? :P

0celo7

It doesn't make sense for it to be a killing vector for $\hat g$.

That would imply it's a Killing vector for $g$ too

user55789

5:52 PM

@0celo7 Why not?

I'm not sure either though

Actually, since the Christoffel symbols are to first order in the partial, I'm not so sure whether that's an obvious statement

I mean, the partial will act on both the exponential factor as well as $g$ itself, so you get extra terms

0celo7

6:12 PM

@user55789 $g$ is conformal to itself so if $\zeta$ is a Killing field for $\hat g$, well just take $\hat g=g$

user55789

@0celo7 Idk, wouldn't that imply that trivially the last part is zero?

I think we're side-stepping the intended thought process here

0celo7

@user55789 Yes, I'm saying there's a typo...

they want you to show that $\zeta$ is a conformal killing field for $\hat g$

user55789

So the intended question is

Well basically the same thing but swap g hat with g

0celo7

yeah, and the hint is telling you to be careful when you lower the index on $\zeta$

there's two metrics and the results are different

also the covariant derivatives will be different

if you want formulas that will help, check Appendix D of Wald

other than that, it's basically plug and chug

user55789

Ok just for reference

The implication I'm supposed to show is

$\nabla_{(\mu} \zeta_{\nu)} - \frac 1D \hat g_{\mu\nu} \nabla_\alpha \zeta^\alpha = 0 \Rightarrow \nabla_{(\mu} \zeta_{\nu)} = 0$

Right?

0celo7

6:22 PM

no, I just said that's what the typo in the problem implies.

user55789

so instead

$\nabla_{(\mu} \zeta_{\nu)} - \frac 1D g_{\mu\nu} \nabla_\alpha \zeta^\alpha = 0 \Rightarrow \nabla_{(\mu} \zeta_{\nu)} = 0$

(no hat)

0celo7

no

user55789

:D

Ok last try then I'm off to make tea

0celo7

You want to show that the first equation implies $\hat\nabla_{(\mu}\hat\zeta_{\nu)}-\frac{1}{D}\hat g_{\mu\nu} \hat\nabla_\rho\zeta^\rho=0$, where $\hat\zeta_\mu =\hat g_{\mu\nu}\zeta^\nu$.