« first day (2312 days earlier)      last day (2620 days later) » 
01:00 - 20:0020:00 - 00:00

8:00 PM
@Slereah Are you here?
I need a notation for antisymmetrization but w/o one of the indices in the middle
 
mb
 
like $\mu-\nu$ antisym of $G_{\mu\sigma\nu\rho}$
$\sigma$ should not be touched
 
$A_{[b|c|d]}$ is the standard notation
 
Ah, yeah. Thanks
 
for antisymmetrization on $bd$
 
8:03 PM
Hmm. I need to symmetrically antisymmetrize an antisymmetrization.
$G_{(\mu|[[\sigma||\nu)|\rho]]}$?
@Slereah does that make sense?
 
a bit hard to read
though when there's a lot of indices it gets hard
you know what notation is easy to read for antisymmetrization?
Penrose notation.
 
Proof?
 
8:16 PM
very nice notation
 
idk if serious or not
 
Well for that aspect, penrose notation is usually clearer
 
Ok.
Almost there.
 
bad point
 
10 indices
is that a PSE record?
 
8:37 PM
Condensed notation!
 
8:49 PM
@Secret I think I was just saying that any normal object would be hotter than this absolute-zero object, and thus there would be at least some heat flow from any other body to the object. The object could only be approximated as having constant temperature if we consider it to be an extremely large "reservoir" which could take in any amount of heat. While that's used in thermodynamics in some cases, it's not realistic in most applications, and isn't the case here.
 
9:01 PM
0
A: On the Uniqueness of the Riemann-Christoffel Tensor

0celo7Note that we're working in a normal coordinate system, so your equation for the second derivative of the metric simplifies to $$\tag{$1$}\frac{\partial^2 g_{\mu\nu}}{\partial x^\xi \partial x^\rho}=\frac{\partial\Gamma^\sigma{}_{\mu\rho}}{\partial x^\xi}g_{\sigma\nu}+\frac{\partial\Gamma^\sigma{}...

@Slereah lookie
I think I have the record for # of indices in a post.
@BalarkaSen see above for the horror of general relativity
 
@0celo7 lol, I can't believe you wrote all of that out!
 
@BenNiehoff It wasn't that bad. I didn't make any mistakes
Everything worked on the first try
you should read it
 
well, hopefully the internet points you get should be worth it :)
 
@BenNiehoff I wanted to make sure Weinberg is consistent
Btw, there's a more general proof for the uniqueness of the Einstein tensor
 
more general how?
 
9:15 PM
The einstein tensor is the only tensor that is linear in the second derivatives and has zero divergence
 
you mean as being the only thing built out of curvatures which is divergence-free?
ah
 
yeah, basically
It only works in four dimensions. In higher dimensions you get other terms
 
you mean there are other divergence-free tensors?
are they symmetric and rank 2?
 
I think so. Let me get out the book to double check before I say something wrong.
 
that sounds suspicious to me
 
9:20 PM
@BenNiehoff The theorem is not formulated as I remembered. I'm thinking...
Lovelock's Theorem is a theory of general relativity which says that from a local gravitational action which contains only second derivatives of the four-dimensional spacetime metric, then the only possible equations of motion are the Einstein field equations. The theorem was described by British physicist David Lovelock in 1971. == Statement == The only possible second-order Euler-Lagrange expression obtainable in a four-dimensional space from a scalar density of the form L = L ...
That's the statement in 4 dimensions
 
ah, yes, I do know people who have worked on so-called "Lovelock gravity"
where they add extra terms
 
the proof is extremely hard btw
it uses jet bundles
@BenNiehoff apparently, yes
I might make understanding this paper a long-term goal
@BenNiehoff In any case, I have a much better description of the Riemann tensor if you want to hear it
 
ok, sure
 
@BenNiehoff How well versed in differential geometry are you? Do you know what a local isometry is?
 
I'm pretty well versed
 
9:26 PM
Do you know the Frobenius theorem?
 
vaguely...it has to do with whether first-order systems are integrable?
 
Yeah.
Let me find the statement
 
or whether a spray forms surfaces, etc.
what's your description of Riemann?
 
@BenNiehoff The Riemann tensor being zero in an open set is a necessary and sufficient condition for the open set to be locally isometric to Euclidean/Minkowski space.
 
yes, of course
it's the integrability condition to find coordinates in which the metric tensor is diag(1,1,1,1...)
 
9:30 PM
Yep
 
was that all, though?
 
There's an even better trick using Jacobi fields
@BenNiehoff Yeah
I figured if you were asking about Weinberg you wouldn't know any geometry, frankly.
You can use Jacobi fields to show that the exponential map is a local isometry when there's zero curvature.
 
oh, I thought you were in the room when I was explaining that I'm a string theorist and I do differential geometry for a living :P
 
@BenNiehoff I'm a mathematician.
Maybe I was here...I remember a string theorist being here.
 
ah, maybe you didn't agree that what I do is differential geometry ;)
 
9:32 PM
:p
So is what I wrote on the main site what you're looking for?
 
what I'm looking for?
I didn't ask the Weinberg question
 
wait what
I need more sleep
 
lol
 
Ok then
Back to...hmm. I guess thermo homework.
 
so, there was a question earlier about the meaning of the Einstein tensor, which got me thinking about easy ways to interpret the various curvatures
the scalar curvature is of course the deficit solid angle for geodesic balls
and the Ricci tensor measures how geodesic balls deform into ellipsoids
 
9:35 PM
Ricci controls the volume expansion of geodesic balls
 
the Riemann tensor must control higher-order deformations of geodesic balls, like twisting, etc.
 
Twisting? Which tensor appears in the Raychaudhuri equation?
 
I don't remember, but GR gives you a deceptive idea of these things
because of the time/space split
so I want to think just in Euclidean signature
 
We usually say Riemann controls geodesic spread because the Jacobi equation
 
well yes
ah, I remember
 
9:39 PM
Remember?
 
the Riemann tensor controls how the cross section of a geodesic ball deforms into an ellipse as you travel along a geodesic radius
which is another way of saying Jacobi fields of course
 
that's what you said for Ricci curvature
My interpretation comes from $\sqrt g \sim 1-\frac{1}{6}R_{ij}x^ix^j$
 
ah, yes
the Ricci tensor measures how the area of the cross section changes as you travel along a geodesic radius
whereas the Riemann tensor describes how the shape changes
 
those are the kinds of things you hear, but I'm not sure anyone has written them down
 
no, I don't think most people think about it often
 
9:48 PM
What do you mean by "cross section"
 
take a geodesic ball centered at O. Draw a geodesic ray from O, and slice perpendicularly to the ray
 
slice with what?
more geodesics?
 
yeah, I know
what you really have to do is take the orthogonal subspace to your ray at O and parallel transport it
that's just the collection of Jacobi fields
I'm assuming the Levi-Civita connection, of course
so the orthogonal subspace will stay orthogonal :)
 
that's the best connection
 
eh, torsion is pretty neat, too
took me a long time to understand what it actually measures, since all the books just ignore it
and I actually had to use a metric-incompatible connection once :)
 
9:55 PM
I have absolutely no idea what it does
I understood it for an hour once
Then I forgot
Never seen it since
 
so, take a look at the operator, built out of the Riemann tensor, that describes Jacobi fields: $R_{abcd} X^b X^d$
it's a symmetric matrix in ac
 
Riemann tensor. We're working with LC?
Or general connection here
 
assume LC for now
So, as a symmetric matrix, it can describe ellipsoidal deformations of the orthogonal subspace along geodesics
but it cannot describe twisting
torsion is precisely the thing that gives you twisting
 
@BenNiehoff That's precisely what happens in Raychaudhuri's equation.
But with the signature, who knows if that's actually what's happening.
 
Raychaudhuri's equation is on a Lorentzian manifold
so, he's only looking at a slice through the orthogonal subspace :)
 
10:00 PM
There's an $R_{abcd}X^bX^d$ term in it is what I'm saying
I've got to go
cya
 
10:14 PM
@0celo7 Ugh what a horrible jumble of symbols
 
10:25 PM
Hi everyone... anyone here understand Reynold's Transport Theorem really well?
 
Hello
 
I wonder if there's physics papers for really obvious facts
Like who do you cite if you want to say "objects fall on earth"
(if they have sufficient density)
 
If I numerically solved the SEQ to get $x-y$ pairs for the wavefunction and I plot it and it looks like this, what might have gone wrong? The problem here is that my wavefunction goes below $y=0$. This is for the infinite square well, so my energies should always be above $y=0$, right?
 
@BalarkaSen it's beautiful
@Slereah how much pressure is 2MPa
 
10:45 PM
About 20 atmosphere?
 
@Slereah the saturation temperature is 212C
must be pretty high pressure.
 
212 Coulombs isn't a temperature
 
Celsius
485 K
 
There is a specific K in unicode for Kelvins
 
...why
 
10:53 PM
KK
u can see the difference
 
nope
 
I'm guessing it's easier to parse for computers
 
@Slereah have you been reading Munkres or nah
 
not today, no
if u want me to do math every day you can hire me
I am available for a thesis
 
in constructive set theory?
 
11:01 PM
 
What?
 
@0celo7 as long as money is involved, sure
 
will you compute scattering cross sections?
in QCD
 
I've done it before
though it was effective QCD
 
5th order?
 
11:07 PM
Nah, boring old first order
in the $O(N)$ sigma model
For $N = 2$ and $N = 3$
I don't even know if fifth order for the sigma model would be interesting
There's some assumptions to reduce QCD to the sigma model, i'm not sure the results will be that great at 5th order
 
Knock yourself out
 
It would be pretty horrible
I don't know how to vary the full Riemann tensor
squared!
@Slereah The issue is that its variational derivative is a total divergence, right?
That's what we mean by "doesn't contribute to the EoM"?
 
11:23 PM
That is usually the case, yes
 
Straumann says the derivative is $$(RR_{ab}-2R_{ac}R^c{}_b-R^{cd}R_{acbd}+2R_a{}^{cde}R_{bcde})-\frac{1}{4}g_{ab}‌​(R_{cdef}R^{cdef}-4R_{cd}R^{cd}+R^2)$$
lol
 
11:38 PM
@dmckee What does "enthalpy out" mean?
I don't know what $h_{out}=u_{out}+(Pv)_{out}$ is supposed to mean
specifically the $(Pv)_{out}$ piece
 
Enthalpy is energy that is available for use in a constant pressure environment. Enthalpy out is enthalpy removed from the system. It is useful in constant pressure contexts.
Because the system is held at constant pressure some energy added won't go to raising the temperature but (generally) to increasing the volume occupied by the system. As energy flows out some will come from the work the environment does on the system to decrease the volume occupied.
So $(Pv)_out$ is a measure of work involved in changing the volume occupied by the system, and at this point the matter of sign conventions rear's it's ugly head.
 
yeah, but how do I compute the flow work?
I guess $P$ is not changing
but what about $v$?
I have water in a tank that's being emptied. I'm not sure if this is constant pressure
 
@0celo7 $v$ will generally change, and you have to know enough about the system to find that.
 
For all I know it explodes when the valve is opened
@dmckee then...how do I evaluate it?
 
@0celo7 If it is exposed to atmosphere than it is under (at least roughly) constant pressure.
 
11:45 PM
@dmckee Must be one hell of an atmosphere. The water is at 200 celsius and is saturated
the tables tell me that means the pressure is 1555kPa
 
@0celo7 You have to know enough about the system.
 
@dmckee could you please be more specific?
 
No. You have to figure out the volume change from the specifics of the problem you are working.
 
volume change?
I don't know what this symbol means!
What is $(Pv)_{out}$?
 
For that matter you have to show that the pressure is (or is not) constant from the specifics of the problem.
$P$ is the pressure.
 
11:47 PM
yes
 
$v = V/N$ is the volume per particle. Or they might mean $v = V/n$ is the molar volume.
 
so is $v$ the total volume that has left?
@dmckee it's volume per unit mass
 
Usually it would be the current volume of the system.
@0celo7 OK. Good.
 
@dmckee at the final state?
that doesn't make much sense
we're taking $h_{out}$ out of the system
shouldn't it depend on what left?
 
@0celo7 Slow down. Why are they interested in $Pv$ in the first place?
Answer ... because it is related to work.
 
11:49 PM
@dmckee The first law gave me $$Q_{in}=\frac{m_1}{2}(h_{out}-u_1)$$
I need to calculate $h_{out}$ somehow
 
or because it is involved in the Legendre transform between internal energy and enthalpy.
 
@dmckee I suspect this is not constant specific volume
so how do I evaluate $v_{out}$?
 
@dmckee Legendre transforms is on the list of things I need to take a day to understand for real.
 
01:00 - 20:0020:00 - 00:00

« first day (2312 days earlier)      last day (2620 days later) »