The h Bar

4:01 PM

"I have discovered a truly marvelous demonstration of the Riemann hypothesis that this stream is too low-res to contain"

John Rennie

:-)

barrycarter

Love it!

Or pretend that like you're explaining it then accidentally on purpose stab yourself with your pointer.

0celo7

@barrycarter you know any complex diff geo?

barrycarter

No, I don't even know what that means, Ocelot.

Obliv

@0celo7 and the proof to N = NP is.. cut stream

0celo7

4:02 PM

Complex differential geometry?

barrycarter

No, I got that. I was being a little facetious. I mean, no I don't know it.

Although maybe my people use a different word for it.

@Obliv We have to see the suicide, though.

We somehow have to merge a thoretical false proof with a bloody suicide.

Obliv

he accidentally tripped over the power cord and decided he was too depressed to show the proof. then he ended himself @barrycarter

barrycarter

@Obliv Wouldn't tripping over the power cord cut the video feed?

Obliv

yes thats why the stream is dead!

barrycarter

@Obliv But we'd miss the good part.

Obliv

4:05 PM

so is life

barrycarter

@Obliv But I like the idea... poorly shielded electrical cord... glass of water... some sort of metal. Just make sure the video camera's on a different circuit.

Physics Meta

0

Q: Additional chat room owners?

Since the moderators all have lives, we've been thinking about getting some additional eyes on chat in the form of room owners for The h Bar. In most chat rooms, ownership is meant to indicate who is "in charge" of a room and should be the point of contact for outsiders with questions about its...

6

barrycarter

And put the cam on a delay so no one comes to "rescue" him too fast.

Jiminion

@JohnRennie We'd just wait for it to evaporate, depending on how much you weigh....

barrycarter

Even black holes can get full.

skill patrol

4:08 PM

full of what?

John Rennie

I'm sure somone hereabouts did the calculation of whether a microscopic black hole dropped into the Earth would evaporate or consume the Earth.

barrycarter

full of Rennie, if you know what I mean.

@JohnRennie Perhaps I could help you with those calculations? :)

John Rennie

17

Q: Throwing a micro black hole into the sun: does it collapse into a black hole or does it result in a supernova?

What do we know about accretion rates of micro black holes? Suppose a relative small black hole (mass about $10^9$ kilograms) would be thrown into the sun. Eventually this black hole will swallow all matter into the star, but how much time will pass before this happens? Are there any circumstan...

supernova gravitational-collapse micro-blackholes

barrycarter

Does anyone want to get back to relativity, or was that too physics-y? :P

John Rennie

@barrycarter the trouble is you keep pressing points that seem irrelevant.

An object has a trajectory in spacetime,

and the form of that trajectory depends on the coordinate system you're using to plot it.

barrycarter

4:12 PM

@JohnRennie OK, then let's not continue. I'm building up to: relativity is stupid, but can be improved.

I just want to establish the first part right now.

Simultaneity is a hoax!

John Rennie

The length along a trajectory can be calculated using the metric, and in this way the trajectory can be parametrised using this length.

That length being the proper time, or proper length, which is the time shown by a clock carried by an observer following that trajectory.

0celo7

@ACuriousMind How does one actually take complex conjugates of complex forms?

barrycarter

Ocelot, a + bi -> a - bi

Flip the sign of the complex part.

0celo7

@ACuriousMind Like how would you take the complex conjugate of $\omega =\frac{\mathrm{i}}{2}h_{i\bar j}\mathrm{d}x^i\wedge\mathrm{d}x^{\bar j}$, the Kahler form

barrycarter

@JohnRennie OK, but the way "most" people do SR involves the Lorentz transform, right?

John Rennie

4:15 PM

@barrycarter That's the way it's taught to undergrads. If they go on to study relativity seriously they then have to unlearn all that stuff.

Well, not unlearn it, but understand the limited role that Lorentz transforms play.

barrycarter

@JohnRennie Well, OK, but I think that's my point then. The Lorentz transform is not the best way to do SR, though I can't think of a better way, but working on it.

John Rennie

The best way is to start from the invariance of the line element. Everything follows from that.

0celo7

@ACuriousMind And further, why does $\omega=\partial\bar\partial F$, where $\omega$ is a "real" form, imply that $F$ is a real function

barrycarter

My point is that simultaneity is an artifice.

@JohnRennie What does that mean? The speed of light is invariant?

John Rennie

For example:

5

Q: How do I derive the Lorentz contraction from the invariant interval?

Reviewing some basic special relativity, and I stumbled upon this problem: From the definition of the proper time: $$c^2d\tau^2=c^2dt^2-dx^2$$ I was able to derive the time dilation formula by using $x=vt$: $$c^2d\tau^2=c^2dt^2-v^2dt^2=c^2dt^2\left(1-\frac{v^2}{c^2}\right)\rightarrow d\tau = dt\...

homework-and-exercises special-relativity

barrycarter

4:18 PM

OK, that starts with stuff I don't understand.

A lot of the stuff I read starts with complicated and, in my mind, unnecessary equations and phrases.

You can derive SR from its postulates fairly easily.

0celo7

@ACuriousMind Hmm, a Standard Reference claims the Kahler condition needs an extra $\mathrm{i}$ in it.

ramsay

hello,
one question!
keep your pen on table in this way.......
and then give it impulse in the direction shown.
then my question comes, why does it go in parabolic path even we are exerting force in downward direction, centre of mass should move in vertical direction but it moves in horizontal as well as vertical direction, why?

John Rennie

@barrycarter Really? What don't you understand?

@ramsay the block exerts a force on the pen as well, and that force has a horizontal component.

ACuriousMind

@0celo7 $(\omega)^\ast = - \mathrm{i}h^\ast_{\bar{j}i} \mathrm{d}x^{\bar{j}}\wedge\mathrm{d}x^i$.

(I dropped the 2 because it's irrelevant)

0celo7

@ACuriousMind Yes, I think I get that now

I think it should be $\omega=\mathrm{i}\partial\bar\partial F$

barrycarter

4:23 PM

@JohnRennie The definition of proper time... that formula doesn't follow for me, I've never seen it before.

ramsay

@JohnRennie but pen makes a horizontal displacement and it seems a bit weird to me that normal force will cause a huge horizontal displacement. So, is that the only force resulting in displacement horizontally?

Obliv

@Ramsay and the block acts as a fulcrum for the pen if that helps.

barrycarter

@Obliv That would actually make the path a circle.

I assume the pen doesn't hit the side of the table at any point.

ACuriousMind

@0celo7 Conjugate the equation, and use that $\omega$ is real, this gives that $F$ is real.

ramsay

@barrycarter ya,we have to assume that condition

0celo7

4:26 PM

@ACuriousMind Does it though? Isn't $\overline{\partial\bar\partial}=\bar\partial\partial=-\partial\bar\partial$

John Rennie

@barrycarter the definition of the line element ds is the be all and end all of relativity, both special and general. All of relativity is founded upon this principle. Until you understand this you do not understand relativity.

barrycarter

@ramsay The circle argument seems sound to me, so I'm interesting in understanding this too

ACuriousMind

@0celo7 Oh, yes.

barrycarter

@JohnRennie What IS the line element 'ds'? I don't claim to understand GR.

ACuriousMind

Indeed, there's an $\mathrm{i}$ missing

0celo7

4:27 PM

@ACuriousMind So it's $\mathrm{i}\partial\bar\partial$ that's the Hermitian differential operator

John Rennie

Line element

In geometry, the line element or length element can be informally thought of as a line segment associated with an infinitesimal displacement vector in a metric space. The length of the line element, which may be thought of as a differential arc length, is a function of the metric tensor and is written ds or dℓ. Line elements are used in physics, especially in theories of gravitation (most notably general relativity) where spacetime is modelled as a curved Riemannian manifold with a metric tensor. == General formulation == === Definition using metric === The coordinate-independent defini...

barrycarter

@JohnRennie OK... what's the metric space associated with relativity?

0celo7

@barrycarter do you know Riemannian geometry

barrycarter

Ocelot, probably not.

0celo7

What kind of mathematician are you?

barrycarter

4:28 PM

In general, just ask me questions and see what happens.

0celo7

Logician?

barrycarter

Ocelot, not a very good one, apparently.

John Rennie

For special relativity the metric is diag(-1,1,1,1).

ramsay

@barrycarter do you know which force makes the rod rotate about centre of mass(COM) in the frame of COM ?

John Rennie

For general relativity the form of the metric is obtained by solving the Einstein equation.

barrycarter

4:29 PM

@ramsay That would be the torque you're applying.

@JohnRennie OK, wait, we're getting closer to mutual understanding. To me, a metric measures the distance between two points. Are you saying the points in relativity are events?

John Rennie

e.g. for a spherically symmetric static mass we get the Schwarzschild metric:

Schwarzschild metric

In Einstein's theory of general relativity, the Schwarzschild metric (also known as the Schwarzschild vacuum or Schwarzschild solution) is the solution to the Einstein field equations that describes the gravitational field outside a spherical mass, on the assumption that the electric charge of the mass, angular momentum of the mass, and universal cosmological constant are all zero. The solution is a useful approximation for describing slowly rotating astronomical objects such as many stars and planets, including Earth and the Sun. The solution is named after Karl Schwarzschild, who first published...

0celo7

@JohnRennie do you even know any GR

John Rennie

@barrycarter yes, exactly. The metric gives us a way of calculating the distance between any two spacetime points.

0celo7

"distance"

John Rennie

@0celo7 no, I fake GR and orgasms

barrycarter

4:30 PM

@JohnRennie OK, so by a spacetime point you mean (t,x,y,z) or something else?

0celo7

@JohnRennie It's waaaaaaaaaaay more complicated than that

John Rennie

I mean (t,x,y,z)

0celo7

Most spacetimes don't even have a notion of distance

Even lorentzian distance

John Rennie

@0celo7 don't all spacetimes have a metric?

barrycarter

@JohnRennie OK, so you're saying an event has a spacetime coordinate (t,x,y,z) but that coordinate depends on the observer? Or is there a fundamental coordinate system here?

0celo7

4:32 PM

A metric tensor does not give a distance in the "metric space" sense.

ramsay

@barrycarter sounds good :-)

John Rennie

If I have a curve in a spacetime, then the metric gives me a way to calculate the length of the curve. Doesn't it?

barrycarter

@ramsay But, apparently, there's more going on that I don't understand.

0celo7

@JohnRennie Yes, but that's not the distance between two points.

barrycarter

@JohnRennie A metric is between two points. You get length by integrating a metric.

Ocelot, damn, you're smarter than you look.

0celo7

4:33 PM

@JohnRennie In the Riemannian case, we have the following: (long equation is coming)

ramsay

@barrycarter like?

John Rennie

@0celo7 I never said it was

barrycarter

@ramsay Well, you asked why the pen went parabolic and not circular, right?

We have a math-physics language barrier :)

A metric is a binary function on a set S that yields a (usually real) quantity.

0celo7

$$\mathrm{dist}(p,q)=\inf\{L(c)\mid c:[a,b]\to M, c(a)=p,c(b)=q\}$$

ACuriousMind

@JohnRennie: Let me try to reformulate what 0celo7 is trying to say: The metric tensor gives a notion of length, but not of distance (the difference is that length is of curves, distance is between points).

0celo7

4:34 PM

@JohnRennie factually incorrect

4 mins ago, by John Rennie

@barrycarter yes, exactly. The metric gives us a way of calculating the distance between any two spacetime points.

ACuriousMind

You said: "The metric gives us a way of calculating the distance between any two spacetime points.", which is what he's taking offense at.

0celo7

@JohnRennie here $$L(c)=\int_a^b \sqrt{g(c',c')}\mathrm{d}t$$

barrycarter

Ocelot, care to explain that equation?

0celo7

@barrycarter we can define the length of a curve

barrycarter

The one before it

The second one I got.

0celo7

4:36 PM

The distance between two points is the infimum of the lengths of curves connecting the two points

where we take the inf over all piecewise diff curves connecting the two

barrycarter

Ocelot, so c() is a curve that goes from p to q?

0celo7

Yes.

barrycarter

OK, define what you mean by "curve". A curve in 4D space?

Obliv

does the statement: if and only if , mean that the condition is true when both "if" and "only if" are true? and when "if" is true but "only if" is false, the condition is false?

barrycarter

@Obliv The 2nd thing you said.

John Rennie

4:37 PM

@0celo7 I meant along a curve. Jesus, are all mathematicians worse than lawyers?

0celo7

A continuous map from a nondegenerate interval to the manifold

@JohnRennie I'm an engineer.

barrycarter

The mainfold being 4D space?

0celo7

Yes

barrycarter

Wouldn't the shortest curve then be a line?

0celo7

@barrycarter In flat space with no fucked up topological stuff, yes.

barrycarter

4:38 PM

So what's the fucked up topological stuff here?

ACuriousMind

@Obliv "A if and only if B" is the same as "If A, then B and if B, then A".

0celo7

@barrycarter If you remove a point, for instance.

John Rennie

@0celo7 all the engineers I know are basically friendly people who like to drink beer, tell bawdy jokes and fart. Few of them would be actively looking for ways to tell me I'm wrong.

barrycarter

A -> B && B -> A

0celo7

Example:

skill patrol

4:39 PM

@JohnRennie mathematians are the lawyers of science :P

3

0celo7

Take $\mathbb{R}^2$ and remove the origin.

ramsay

@barrycarter i am thinking why should it go circular? what reason you think?

0celo7

There is no shortest curve connecting (-1,0) and (1,0).

barrycarter

@ramsay Imagine there's no table, it's easy if you try... then you would have a circular arc around the fulcrum of the lever.

Ocelot, true, but how does this apply to relativity?

ACuriousMind

@JohnRennie Only some, but @0celo7 has this habit of going into pedantry and math dump mode whenever someone else talks about relativity :P

Obliv

4:40 PM

@acuriousmind what about "a+b if and only if f(a) = f(b)" does that mean I can say "c+d if and only if f(c) =f(d)"?

0celo7

@ACuriousMind I'm tired of physicists being imprecise about GR!

I can't help that I think geometry is cool

ACuriousMind

@Obliv Uh...you just renamed your variables.

0celo7

@barrycarter I was correcting something @JohnRennie said that is incorrect.

ACuriousMind

Both statements mean exactly the same.

Obliv

@acuriousmind can i say that if a,b,c,d are elements in set A

ACuriousMind

4:41 PM

Although neither is meaningful because "a+b" is not a statement that has a truth value.

Obliv

pretend + is just a relation

John Rennie

@0celo7 right, but you could have asked did you mean distance along a curve? rather that just saying that's wrong you stupid ass

0celo7

@JohnRennie Where did I insult you?

ACuriousMind

@Obliv I don't understand the question.

barrycarter

Mr Ocelot, if you teach me relativity correctly, I'll never make fun of you again.

0celo7

4:42 PM

@barrycarter Ok.

How much differential geometry do you know?

barrycarter

Possibly none.

0celo7

Real analysis, topology?

Obliv

okay @acuriousmind the truth is i'm trying to prove $\sim$ is an equivalence relation on A. it says $a \sim b$ iff $f(a) = f(b)$. people used the argument that $f(a) = f(a)$ since set equality is reflexive. does that mean you can say $a \sim a$ iff $f(a) = f(a)$?

oh and $f$ is a surjective map if that matters

barrycarter

Er, I know both ok, but could you try explaining and I'll ask if it gets too hard?

0celo7

@barrycarter amazon.com/Manifolds-Differential-Geometry-Graduate-Mathematics/…

This is a good intro to all the math you need for GR.

barrycarter

4:44 PM

Ocelot, erm, I said if you taught me relativity :)

ACuriousMind

@Obliv That's not really meaningful. $a\sim a$ iff $f(a)=f(a)$ is just a special case of the definition of the relation.

barrycarter

Not Mr Lee

0celo7

This chat is way too narrow for that shit

Obliv

@acuriousmind actually someone in math.SE chat told me that "Often, the words "for all $a$ and $b$" and similar are omitted in definitions"

barrycarter

Ocelot, we could find another venue, or I could just ask questions and see what happens?

ACuriousMind

4:45 PM

What you must say to prove reflexivity is: For all $a\in A$, $f(a) = f(a)$ since equality is reflexive, therefore $a\sim a$ for all $a\in A$.

0celo7

@barrycarter moment.

barrycarter

I think @Obliv is asking if the inverse image of a surjective function defines an equivalence relation?

ACuriousMind

@Obliv Yes. Why does it matter?

Obliv

i don't think so

John Rennie

@barrycarter: the point of all this is that if you think SR is all about Lorentz transformations you will never really understand it. That's why my enthusiasm for debugging your Lorentz transforming has been less than overwhelming.

ACuriousMind

4:46 PM

@barrycarter No, Obliv is still confused about how logic works, I think :P

0celo7

3

A: Question about basic formalism of GR and the metric tensor

When we talk about the geometry of GR, it is understood that the manifold of spacetime is not a Riemannian one, but rather a Lorentzian manifold. This means that the metric is not positive definite. With this understanding, we call $g(.,.):=\langle.,.\rangle$ an inner product as usual. This lack ...

Here is a good intro, I think.

Obliv

I'm just saying because if for all $a$ & $b$ isn't said then how can we apply the relation $\sim$ on $a \sim a$ iff $f(a) = f(a)$?

barrycarter

@JohnRennie But that's the point I've been making: LT is fairly dumb. But that's also how most people (incl me) do it, so it's good to understood.

John Rennie

@barrycarter: also note that the distinction between SR and GR is misleading. The SR metric is one of the vacuum solutions to Einstein's equation so SR is actually GR.

Slereah

It is the week end

Woo

0celo7

4:47 PM

@JohnRennie This is correct.

barrycarter

Ocelot, OK, can I ask you to simplify that?

John Rennie

:-) I love you man

Slereah

Well in SR spacetime is not dynamic, tho

SR is GR with $G \rightarrow 0$

0celo7

But it should be noted that people do talk about matter in SR and ignore the GR effects this should create.

Basically what Sam just said.

John Rennie

Yes, but it's common to ignore backreaction when working with metrics

0celo7

4:48 PM

The energy-momentum tensor decouples from spacetime geometry in SR.

Slereah

The old Physics Cube

ACuriousMind

@Obliv I don't understand the question

Obliv

give me a minute I'll probably just figure it out on my own.

Slereah

Classical mechanics without gravity is $c \rightarrow \infty, G \rightarrow 0, \hbar \rightarrow 0$

ACuriousMind

@JohnRennie It's common to ignore backreaction regardless of what exactly you're working with ;) "Background fields" are everywhere.

Slereah

4:49 PM

With gravity $c \rightarrow \infty, \hbar \rightarrow 0$

barrycarter

Ocelot, I need it dumbed down more. Can I ask some basic questions instead?

Slereah

QM is $c \rightarrow \infty, G \rightarrow 0$

SR is $G \rightarrow 0, \hbar \rightarrow 0$

0celo7

@barrycarter If you told me what exactly your background is, yes

Slereah

QFT is $G \rightarrow 0$

ACuriousMind

@Slereah Actually...$\hbar\to 0$ is a bit difficult :P

barrycarter

4:50 PM

Ocelot, it's a page from the swimsuit addition of Sports Illustrated...

0celo7

When you say you're a "mathematician" I expect "paracompact hausdorff manifold" to mean something to you

Slereah

Close enough

barrycarter

Ocelot, only some of those words make sense to me. Hausdorff... ok, that's the only one.

John Rennie

Cor, this is turning into a good discussion. This is exactly how the chat room should work!

barrycarter

Cor?

Slereah

4:51 PM

Maybe he's a mathemagician

0celo7

@ACuriousMind lol

ACuriousMind

I mean, it works often, but it gets kinda messy, especially when fermions and stuff are involved

John Rennie

urbandictionary.com/define.php?term=cor

Obliv

@acuriousmind I think the way I see it is: $a \sim b$ holds only if the image of $a$ through $f$ is equal to the image of $b$ though $f$. So I'm trying to make the connection that that logic is applicable for all elements in $A$.

0celo7

my QM prof was talking about the classical limit

Slereah

4:51 PM

Hm

ACuriousMind

@Obliv I don't understand the second sentence.

0celo7

I don't even think he said "limit", he just said when $\hbar=0$

barrycarter

@JohnRennie Oh, I knew cor blimey, but geez, I didn't think anyone actually still said it.

Slereah

Can you have classical fermions

barrycarter

Ocelot, so I can't just ask you questions? :(

0celo7

4:52 PM

Ask

Slereah

You can have spinor fields

ACuriousMind

@Slereah Yes and no. You can introduce classical fermionic d.o.f., but they don't describe actual classical systems

Slereah

Can you have them without SR though

0celo7

@Slereah No.

barrycarter

Ocelot, ok the basic thing in relativity is an event, correct?

John Rennie

4:52 PM

@barrycarter "cor" is still used occasionally amongst my age group, though usually ironically. I have never heard anyone say "cor blimey".

ACuriousMind

You only introduce them to give a notion of a classical fermionic theory that is then quantized

Obliv

@acuriousmind based on that information given to us in the first sentence, is it okay to make this connection between $\sim$ and the images of the elements that it is relating for all elements on $A$?

Slereah

What about "Zounds"

0celo7

@barrycarter Yes, an event is a point on the spacetime manifold.

barrycarter

@JohnRennie As long as it's not "cor blimey, guvnur"

ACuriousMind

4:53 PM

@Obliv I don't understand that question, either.

0celo7

It's nothing special, it's literally just a point.

Slereah

Couldn't 'ave been me guvnah I got sausage fingahs

John Rennie

Only Dick van Dyke ever said that

0celo7

@barrycarter do you know what a manifold is

barrycarter

Ocelot, ok, good. Now, when you spacetime manifold, is this a manifold with 4 dimensions?

Ocelot, I should but I don't know what a manifold is.

0celo7

4:53 PM

Yes

barrycarter

So, I can assign any event a set of 4 coordinates?

Obliv

like @acuriousmind $a\sim b$ if and only if $a$ is a banana and $b$ is an orange. can I say $c\sim d$ if and only if $c$ is a banana and $d$ is an orange?

0celo7

@barrycarter a manifold is a topological space where every point has a neighborhood homeomorphic to an open set in $\mathbb{R}^n$

Slereah

Proper british english

ACuriousMind

@Obliv Yes, you just renamed the variables, I already answered that question.

barrycarter

4:54 PM

Ocelot, OK, that still didn't make sense, although I think I know what you mean.

Ocelot, but OK, every event has a (t,x,y,z)?

0celo7

Why didn't it make sense

Obliv

@acuriousmind so the relation $\sim$ is simply defined by the conditions that proceed it? so if $a \sim b$ if and only if $f(a) = f(b)$ then $\sim$ is simply an equals sign?

barrycarter

I thought we were dealing with simple 4 dimensional space, but we're not?

0celo7

@barrycarter Let $M$ be a topological space. Then for any $p\in M$, there is an open $U\ni p$ such that there is a homeomorphism $\varphi:U\to U'$ where $U'\subset\mathbb{R}^n$ is open. Then $M$ is a topolgical manifold.

ACuriousMind

@Obliv Uh...no. If $f$ is not injective, $a\sim b$ is not the same as $a=b$.

Slereah

4:56 PM