The h Bar

4:47 AM

@Luyw see the Wikipedia article on the Young's modulus

The Young's modulus is a fundamental property of a material and it relates the tension to the fractional change in length.

For springs we have a corresponding equation:

$$ F = K \frac{\Delta\ell}{\ell_0} $$

Where $K$ is a constant that depends on the modulus but also on the geometry of the spring. It's quite a complicated thing to calculate, but as long as you keep everything except the length of the spring constant $K$ is constant.

So in the usual Hooke's law equation $F = k\Delta\ell$ the constant $k = K/\ell$.

John Rennie

5:03 AM

Oops, that should be $k = K/\ell_0$.

skullpetrol

5:36 AM

> Perhaps most notably, a change.org petition that addressed Microsoft, Google, and President Donald Trump as decision makers amassed approximately 143,000 signatures in three days.

@SirCumference

Despite rumors, Coolmath Games is not shutting down in 2020

skullpetrol

5:48 AM

Jul 23 '17 at 3:40, by The Raiders of Las Vegas

Bad idea #271: Dropping into the half-pipe on a Segway.

Scientist Smith YT

6:03 AM

@JohnRennie I have a question. I have a 5 watt blue laser. I want to find a formula to calculate the heat of the dot at a set distance. Or even a combination of formulas I can use to get the end calculation. What formula or set of formulas do that?

John Rennie

@ScientistSmithYT if you know the area of the dot $A$ then you know the intensity of the energy being supplied because it's just $I = 5/A$ watts per square metre.

But what you don't know is how fast the energy is being lost due to cooling.

At the equilibrium temperature the cooling rate must also equal 5 watts, i.e. the enegy being lost per second is equal to the energy being supplied per second. But to find out what this equilibrium temperature is you need to know how the temperature affects the cooling rate.

Scientist Smith YT

@JohnRennie Ok

@JohnRennie I do know for sure the output light power going out of the aperture is 5.16 watts. I rounded it to 5 watts.

John Rennie

@ScientistSmithYT to be honest, we'd normally answer the question by doing the experiment and measuring the temperature. It's too hard to predict theoretically because so many factors are involved.

If the object being heated is a perfect black body you can use Stefan's law to calculate the temperature. That would give you a guideline.

Stefan's law tells you the rate of heat loss per unit area, $j$, is given by $j = \sigma T^4$ where $\sigma$ is a constant.

So if the area of your dot is $A$ the rate of heat loss will be $\sigma A T^4$ and just set that equal to 5.16 watts to get the temperature.

Stefan–Boltzmann law

The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time j ⋆ {\displaystyle j^{\star }} (also known as the black-body radiant emittance) is directly proportional to the fourth power of the black body's thermodynamic temperature T: j ⋆...

But this will only be a guideline as heat can be lost by other mechanisms such as conduction and latent heat of melting or vaporisation as well. So the real temperature will be lower than this.

Scientist Smith YT

6:23 AM

@JohnRennie Ok thanks, I'll just take it as a guideline.

Gyromagnetic

6:54 AM

@JohnRennie riddle me this: I have a large pot of boiling water, and I throw a chicken egg inside it... effectively boiling it in X minutes. Then I decide to throw an ostrich egg, whose radius is 3 times larger. How much longer will it take to cook?

Prof. threw this at me yesterday, i found it a good question ;)

John Rennie

You can estimate it by assuming that the thermal conduction inside the egg doesn't matter and all that matters is the heat flow through the shell. The mass of the ostrich egg is 27 times greater (3^3) than the chicken egg so the thermal capacity is 27 times greater.

I wonder if you could approach the problem using dimensional analysis ...

skullpetrol

is the answer 27X @Gyromagnetic?

John Rennie

6

Q: Dam Son's problem: how long does it take to boil an ostrich egg?

The egg boiling problem is a seemingly simple problem posted from the blog of UChicago's physicist Dam Thanh Son. A chicken's egg has a length of 5 cm and takes 6 minutes to boil. An ostrich's egg has a length of 15 cm. How much time does it take to boil an ostrich's egg? (Photo for il...

Novalium Company

@kylecampbell It's a joke - Noether = No ether.

2

skullpetrol

niice play on words

Gyromagnetic

7:10 AM

mass and thermal capacity are 27 times larger yes

but the time to cook the egg is not 27X

John Rennie

@Gyromagnetic But the shell surface area is a factor of 9 larger so for a given temperature gradient 9 times as much heat per second flows through the ostrich egg shell. That reduces the difference to a factor of three.

Gyromagnetic

true, but the answer is not 3X

skullpetrol

questions like this beg the response "when am I ever going to use this?" :-)

John Rennie

@Gyromagnetic And finally the temperature gradient is the temperature difference divided by the radius, and the ostrich egg has a radius three times greater than the chicken egg so the thermal gradient is factor of three lower.

Gyromagnetic

exactly

the time is 9X

that last term is the funkiest one imho

John Rennie

7:14 AM

@Gyromagnetic or in general the time is proportional to $r^2$

Akash. B

@JohnRennie I have a question

John Rennie

@Akash.B morning :-)

Gyromagnetic

@skullpetrol , "when am I ever going to use this?" :-)... next time you boil an ostrich egg, ofc! haha :)

Akash. B

Morning

Gyromagnetic

@skullpetrol also good conversation at dinner parties hahahaha

Akash. B

7:16 AM

@JohnRennie earth is a giant magnet right?

skullpetrol

true dat @Gyromagnetic

Novalium Company

mOrNiNg

Gyromagnetic

@Akash.B among other things

Akash. B

So if I move a iron piece randomly

John Rennie

@Akash.B yes. The outer core is liquid iron and the flows in it cause the magnetic field.

Akash. B

7:20 AM

it should produce electricity

@JohnRennie right?

Gyromagnetic

well, earths magnetic field is quite homogeneous

and to produce current you need d flux/dt

but in principle yea

Akash. B

If a conductor cuts a magnetic field it should produce electricity

@Gyromagnetic change in flux?

Gyromagnetic

true, if you rotate a loop in earth constant field, you would get an AC current

John Rennie

@Akash.B this is exactly how an electric generator works.

Gyromagnetic

but if you just move it around you wouldnt

John Rennie

7:26 AM

When you rotate the generator you are, in effect, waving a conductor (the wire) around.

Gyromagnetic

yeaaaaah, but you get what im saying, you can also wave wire around and get nothing

but i think thats beside the point atm

skullpetrol

Let @Akash.B finish asking his question @Gyromagnetic :-)

Gyromagnetic

@Akash.B yeah sorry :(

Sir Cumference

7:49 AM

@skullpetrol YO WHAT

skullpetrol

yup, looks like a vicious rumour

Sir Cumference

here i was wondering how i'd continue learning math

skullpetrol

stick with what works for you

Sir Cumference

@skullpetrol i've been kidding about learning from there :P

the site's for elementary school math

skullpetrol

:-)

Happy World Emoji Day!

Slereah

8:13 AM

morning

Luyw

good morning

Gyromagnetic

mornin'

Slereah

the heat death of the universe is poorly named

it doesn't die of heat at all

quite the opposite

Gyromagnetic

it also can't die

John Rennie

World Emoji Day?

Good grief, that actually exists.

@Slereah it's the death of heat not the death of the universe

The universe and Cthulhu never die.

The difference between them is that Cthulhu is real.

Luyw

8:25 AM

@JohnRennie hey, I still have a question on springs. When we cut a spring in $n$ parts, the force will still be the same since $\Delta l/\ell_0$ won't vary, but won't the stiffness increase by a factor of $n$? So shouldn't the force also increase by a factor of $n$?

John Rennie

suppose you cut a spring into $n$ pieces, then two things happen:
1. the stiffness of each subspring increases by a factor of $n$
2. the extension of each subspring deceases by a factor of $n$

Luyw

but also $\ell_0$ is now $\ell_0/n$ which cancels the $n$ in $\Delta l/n$

I'm using $F=k\Delta l/\ell_0$

John Rennie

Suppose I take a 1m spring and I stretch it to 1.1m so the extension is 0.1m and $\Delta \ell/\ell_0 = 0.1$

Luyw

if you cut it in $n$ parts you'll have $F_n=k_n\Delta l/\ell_0$ assuming the equal forces you have $k\Delta l/\ell_0=k_n\Delta l/\ell_0$

John Rennie

Now I cut the spring in half to give two springs with an original length of 0.5m that are stretched to 0.55m i.e. the extension is 0.05m

Luyw

8:32 AM

wait so it appears $k_n=k$?

John Rennie

So for the two springs $\Delta \ell/\ell_0 = 0.05/0.5 = 0.1/1$ i.e. $\Delta\ell/\ell_0$ hasn't changed.

Luyw

yes I am cool with that already

what have I done wrong in the last equation?

John Rennie

You need to be absolutely clear what constant you are taking about. When we talk about the force constant of a spring we usually mean the constant $k$ in $F = kx$. This constant $k$ depends on the original length of the spring so it changes if we cut the spring in half.

Slereah

@JohnRennie But

strange eons

Also I don't actually care for Cthulhu

John Rennie

If you write the equation as $F = K x/\ell_0$ then this constant $K$ is independent of the length of the spring and it does not change if you cut the spring in half.

Slereah

8:36 AM

I'm all about the Great Race of the Yith

look at those dorks

John Rennie

@Luyw I suspect what you are doing is mixing up the two constants when you say:

> won't the stiffness increase by a factor of $n$?

The constant $k$ will change but the constant $K$ will not.

Luyw

But $F=(\frac{K}{\ell_0})\Delta l$ cut spring in $n$ parts, $F_n=(n\frac{K}{\ell_0})\frac{\Delta l}{n}$

oh yeah

my bad xD

John Rennie

Yes :-) The $n$s cancel!

Luyw

yes and writing the new force as $F=k\Delta/n$, we consider $k$ as being $n$ times more than that before cutting

ok

thank you thank you

John Rennie

:-)

Slereah

9:42 AM

How does one prove that a ball falls down using string theory

Luyw

9:56 AM

I thought string theory isn't taken seriously anymore, it is not the case?

John Rennie

@Luyw I suspect Sam was joking :-)

He knows perfectly well how to derive general relativity as the low energy limit of string theory.

The fact that string theory gives us GR for free is one of the reasons it became so popular.

Although if I understand correctly any theory that includes supersymmetry as a gauge symmetry inevitably reproduces GR so it's less of an amazing achievement than you might initially think.

Luyw

Yes yes, totally understanding you

Captain Bohemian

10:23 AM

publications of string theory look the most inscrutable to me

students in my previous lab said string theory is more harmonious to people with background in quantum field theory while loop quantum gravity is more harmonious to people with background in general relativity

Gyromagnetic

string theory is a swamp... that must be drained

Captain Bohemian

10:41 AM

it's like some research groups designate their research area as string theory but don't actually work on string theory or do not only work on string theory; they work on various theories of gravity and quantum fields, like supergravity, higher derivative gravity, supersymmmetry gauge field theories, higher spin theories, bigravity, etc.; the reason they use string theory to designate their research area probably because these theories they are involved with can be derived from

string theory at certain limits.

Secret

youtube.com/user/keeroyz/videos

Slereah

11:26 AM

@JohnRennie you may be overestimating my power

@CaptainBohemian maybe putting string theory in the names gets you more grant money

John Rennie

Really. The string spectrum includes the states of a massless spin 2 QFT, and a spin 2 QFT produces a force like gravity. Perhaps neither you nor I understand all the details (I certainly don't) but the general principle is straightforward.

Slereah

Well I know it in the broad sense

Knowing how to show it

That's a different matter

Also spin 2 stuff doesn't necessarily produce GR

You can add arbitrarily many terms to the Lagrangian

And indeed must for renormalization purpose

Although I'm not sure if adding terms in a non-trivial way produce proper QFTs

Since it makes the stress energy tensor not locally conserved in a classical theory

Not sure what that would do for a QFT

Unstable vacuum???

John Rennie

I recall reading a paper discussing whether a massless spin 2 QFT was equivalent to GR, and the conclusion was that unless you are being pedantic yes it is.

Slereah

Slim chances of me not being pedantic

Slereah

12:00 PM

One thing I'd like to do on my site (one of the many things, of course) is just doing solutions of simple problems in various theories

like free particles in Newtonian mechanics/Hamiltonian/Lagrangian/relativistic/GR/QM/QFT

etc

And the same for free fall, two particles in orbit, etc etc

Akash. B

@JohnRennie so if I just keep on moving an iron piece randomly in atmosphere it should produce an electric current right?

Slereah

Although rly, free particles in string theory is quite awful already

bloody boundary conditions

PM 2Ring

1:03 PM

@Akash.B You'll get some current, but the Earth's magnetic field is rather weak, so you need a long piece of metal to get useful current. And you shouldn't move the metal randomly, you should move it in a path that maximizes the crossings of the magnetic field lines. This is impractical on the Earth's surface, but it's feasible to do it in orbit:

Electrodynamic tether

Electrodynamic tethers (EDTs) are long conducting wires, such as one deployed from a tether satellite, which can operate on electromagnetic principles as generators, by converting their kinetic energy to electrical energy, or as motors, converting electrical energy to kinetic energy. Electric potential is generated across a conductive tether by its motion through a planet's magnetic field. A number of missions have demonstrated electrodynamic tethers in space, most notably the TSS-1, TSS-1R, and Plasma Motor Generator (PMG) experiments. == Tether propulsion == As part of a tether propul...

Ryan Unger

1:19 PM

@Slereah yo

have you ever seen the normalization $G=(n-1)\omega_{n-1}T$

for $n+1$ dimensions

Slereah

What

What is even $\omega$

volume form?

Ryan Unger

area of $S^{n-1}$ sphere...

Slereah

oh

I see

I have not but I guess it's supposed to give you the proper Newtonian formula in the classical limit that way?

Ryan Unger

it's probablt there to normalize flux somehow

Slereah

yeah

Usually factors of the EFE are either nothing or just SI units

or "some factor"

not a whole of in between

Ryan Unger

1:23 PM

this is the most complicated factor I've seen

@Slereah hmmm

there's an explanation

you know what the constant in the ADM mass should be so that it gives the right result for Schwarzschild

but since that's vacuum, you don't know what the constant in the EFE should be

but you want the ADM mass to be the integral of the energy density

so pretending the ADM mass integrand is the scalar curvature of the Cauchy hypersurface, that fixes the constant in the EFE

because $R"="\mu$

Slereah

Ah yes

Ryan Unger

there's a factor of 1/2 there

Slereah

I tried to prove Newton's formula from Schwarzschild once

without ever assuming Newton at all

it was a tricky thing

Ryan Unger

so the ADM mass is $$\frac{1}{2(n-1)\omega_{n-1}}\int (g_{ij,j}-g_{ii,j})n^j$$

Slereah

Hard to keep track of where all the $c$'s should go

Ryan Unger

1:33 PM

and the integrand is $R$ modulo lower order terms

uhhh that's not true sorry

you integrate by parts first

then it's an integral of $R$ over the bulk modulo lower order terms

so you want $$\frac{1}{c_n}\int R=\int\mu,$$ which fixes the overall constant

where $\mu=T(\nu,\nu)$

Slereah

$\nu$ timelike I s'ppose

Ryan Unger

its the normal to the Cauchy hypersurface

Slereah

sounds timelike enough

you know I wonder if there's like

A dumb paper on string theory

like just something practical

like "scattering cross section of strings"

Captain Bohemian

2:14 PM

it's like a lot of recruitment systems use their native languages, so if you want to read their information, you need to invoke google translate.

Balarka Sen

3:06 PM

2

bolbteppa

3:18 PM

Zizek Debates Peterson: "Where Are These Postmodern ‘Marxists’?"

►

This was the moment of moments regarding JP

Balarka Sen

Zizek & JBP was good entertainment

skullpetrol

Albert Einstein - Riding on a Beam of Light

►

bolbteppa

Apparently it's still an open question whether LQG even has GR as it's classical limit, that can't be...

skullpetrol

IMHO, Peterson is a good social psychologist

Ryan Unger

3:36 PM

One of the great stable geniuses of our time, skull.

skullpetrol

Socially, yes.

Gyromagnetic

3:58 PM

no JBP posting guys, we might have to lock you up in the hbar-gulag

Ryan Unger

why

Gyromagnetic

just kiddin, its fine ;) <3

Balarka Sen

It's like, no, it's not.

Gyromagnetic

some ppl in my work group have very strong feelings about him hahaha

Balarka Sen

It's not.

Gyromagnetic

4:09 PM

academia is full of commies

Balarka Sen

Postmodern neo-Marxists, to be precise. And you have to be precise about that, bucko

Otherwise you will end up in deep, deep, trouble. And that's no joke.

Ryan Unger

cant tell if Balarka is serious

Balarka Sen

im doing a jbp impression

Gyromagnetic

hahaha a good one also

Ryan Unger

clean your room Balarka

that's step one

Gyromagnetic

4:12 PM

thats the belly of the whale

roughly speaking

Ryan Unger

has JBP done an analysis of toy story 4

Gyromagnetic

wut? rly hahahaha

Ryan Unger

its a question

molyneux probably has

he has actually gone full cult leader in the last few years

Balarka Sen

@RyanUnger has to

bolbteppa

He always was with his de-fooing

Gyromagnetic

4:17 PM

i wonder how things will play out... i have the feeling everything is very volatile atm tbh

John Rennie

@Slereah tricky? It's straightforward isn't it?

Ryan Unger

don't you get it from writing out the radial geodesics in coordinates or something

John Rennie

The geodesic equation gives you:

$$ {d^2 r \over d\tau^2} = - \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) u^t u^t $$

Ryan Unger

where is equation (1)

John Rennie

Approximate $d\tau = dt$, $u^t = c$ and $2GM/c^2r \ll 1$ and the result drops out.

Balarka Sen

4:24 PM

$$\text{Here} \tag{1}$$

John Rennie

@RyanUnger oops, copy paste error :-)

skullpetrol

$$\text{Pre-Here} \tag{0}$$

Slereah

4:36 PM

@JohnRennie not terribly hard but I tend to lose some factors of $c$ along the way

Doing it in natural unit's a breeze tho :p

skullpetrol

natural-units-rule

Physics Meta

7:03 PM

0

Q: Ask to have a link to denounce the abuse of power of some moderators

Referring to the previous question Can we report moderators for abuse of power? I ask, kindly, il link of the tool to report abuse of power by some moderators. Thank you.

support specific-question

tpg2114

7:27 PM

@ACuriousMind Question for you (and possibly follow-ups based on the answer) -- are you familiar with how faculty positions work in Belgium? (Is it similar to the German system?)

ACuriousMind

I have no idea how it works in Belgium, I'm afraid (and I wouldn't be sure about Germany, either, to be honest :P)

tpg2114

Heh, okay.

There's a faculty position open that intrigues me... but, I know how the US system works in terms of teaching/research loads, funding, tenure, etc..

But I have no idea how the various countries in Europe operate

ACuriousMind

Tenure definitely varies markedly between Germany and the rest

Funding variations are more determined by (sub)field than country, I think

tpg2114

I guess that's true in the US also -- the way aerospace engineering faculty need to work for money is different than math or philosophy or whatever else.

ACuriousMind

Since I didn't end up as a lecturer, I have no idea about teaching loads at all. Some lecturers taught much more than others, but I don't know whether that had anything to do with requirements or just personal preference

Cang Ye

11:32 PM

Hi, John good night. I was having some trouble to access the site yestoday night, I am in Asia and then went to bed, Sorry

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