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2:43 AM
Has anyone ever participated in a private beta before?
 
 
2 hours later…
4:58 AM
@ACuriousMind @AaronStevens @JMac @DavidZ @peterh-ReinstateMonica Thank you for your response!
@JMac The time due for ending the suspension is the same on both the sites. And it is very unlikely that both the suspensions would have occurred at the same time. Moreover I cannot dig up any inappropriate activity by the user on Chemistry SE (which is expected, because all the mess would have been cleared). So I suppose this was a case of very closly timed suspensions on both the sites.
 
5:35 AM
@user1271772 I did, multiple times. In a time, I was very active on the a51
 
6:10 AM
@FakeMod You know, on Chemistry we have moderators, too.
 
6:56 AM
I was reading about about QFT from daily physics magazine I have a doubt, if the universe behaves in a way that contradicts the predictions of QFT with weak gravity (extremely unlikely), or if information is destroyed by black holes (almost inconceivably difficult to test, even with Planck energy accelerators) or by mapping the landscape, and showing it doesn't match our universe (incredibly computationally infeasible).
 
7:25 AM
@YuvrajSingh... whether QFT can be applied to gravity at all is unknown.
In weak gravitational fields we can write down a quantum field that we think would work as an approximation. This is what we call an effective field theory.
The question is whether it is useful to try and describe gravity using an effective quantum field theory. And the simple answer is that no-one knows.
 
@Loong Yes I do :) So, now I suppose that both the suspensions were manual and intentional. Thanks!
 
Some of physicist try to relate QFT with gravity.
@JohnRennie
 
@YuvrajSingh... the reason we do this is that we understand QFT very well, so if we can show that gravity can be described by a QFT then that means we understand quantum gravity.
But whether this is going to work or not no-one knows at the moment.
@YuvrajSingh... the attempt to write gravity as a quantum field is called asymptotic gravity:
Asymptotic safety (sometimes also referred to as nonperturbative renormalizability) is a concept in quantum field theory which aims at finding a consistent and predictive quantum theory of the gravitational field. Its key ingredient is a nontrivial fixed point of the theory's renormalization group flow which controls the behavior of the coupling constants in the ultraviolet (UV) regime and renders physical quantities safe from divergences. Although originally proposed by Steven Weinberg to find a theory of quantum gravity, the idea of a nontrivial fixed point providing a possible UV completion...
 
8:03 AM
@JohnRennie Good Afternoon sir...
Why $g_{\mu \nu} \delta g^{\mu \nu} = - g^{\mu \nu} \delta g_{\mu \nu}$?
Are tensor products commutative?
in Torsion free Conditions?
 
@AbhasKumarSinha Don't know, sorry.
 
@JohnRennie okay...
@JohnRennie Sir, Can I multiply $g^{\mu \nu}$ both sides of Field Equations and get a single equation to solve instead of multiple ones?
 
There are ten independent values in the metric, and no matter what you do you are always going to need ten equations for those ten independent values.
 
okay....
 
I guess the question is whether if you take one of the scalars, e.g. the Ricci scalar, can you write a single equation for the Ricci scalar?
I don't know the answer to that, but I would guess that it is impossible.
 
8:10 AM
@JohnRennie yes...
@JohnRennie :54032223 What we solve Field Equations, what we solve exactly? :P I wrote Stress Energy Tensor, but, now I dun know what to solve $g$ or $R$
I forgot that there are two other tensors apart from that...
For example in Schrodinger's equation, we solve, $\psi$ then what we solve here?
 
We don't really solve the field equations in the sense that you think of solving a simple linear differential equation. Typically they are solved by guessing the general form of the metric using properties like symmetry.
 
@JohnRennie means?
 
Then once you have the general form you can write the Ricci tensor and scalar and write down the ten equations. Finally play with them until you manage to guess the exact form of the metric.
@AbhasKumarSinha Google for the derivation of the Schwarzschild metric as this is a good example. The spherical symmetry restricts the form the metric can take so we can guess a starting point that we can take forward.
 
@JohnRennie Okay... Thanks sir :-)
The facial expressions of mice
This is lit! XD
 
:-)
 
8:16 AM
hehehe....
@JohnRennie :-))))
@JohnRennie Yes, that's what I needed, that answered all my questions... :-)
The Schwarzschild solution describes spacetime under influence by a non-rotating massive spherically-symmetric object. Of the solutions to the Einstein field equations, it is considered by some to be one of the simplest and most useful. == Assumptions and notation == Working in a coordinate chart with coordinates ( r , θ , ϕ , t ) {\displaystyle \left(r,\theta ,\phi ,t\right)} labelled 1 to 4 respectively, we begin with...
 
@JohnRennie Well, more computationally inclined physicists do solve them like that! It's just not what you tend to see in a textbook or what you could do by hand. Don't fall prey to theorist myopia ;)
 
I was about to say I don't know of any analytic solutions that were arrived at by directly solving the ten independent equations, but solving with an ansatz is a direct solution really so I'm not sure my statement is fair.
I assume numerical solutions are done that way, but I confess I know nothing about the way numerical GR is done.
My guess is that numerical GR is mind manglingly complicated.
 
All numerics is pretty complicated, honestly, unless you can use one of the solvers that already exist out of the box for your particular problem
The crucial achievements there are materialized not only in papers (the algorithms used) but also in the software libraries that actually implement them
 
8:33 AM
morning
There are pretty complicated existing GR solutions
ie the Mixmaster universe
which is named after a blender
 
I thought it was discovered by Charles Mixmaster
 
Discovered by DJ Mixmaster
 
I would have said that the mixmaster metric was pretty simple, but in the sense that the Navier Stokes equations are simple.
Its general form is simple.
 
@Slereah Sounds like an advertisement for an up-and-coming pop star :P
 
Mixmaster is mostly well-known because it's one of the rare spacetime solution with no Killing vectors
 
8:42 AM
So...a friendly spacetime?
 
Oh it's a Bianchi universe
 
No Killing sounds nice.
 
neat
 
@Slereah Wait, Wiki says Bianchi geometries always have three Killing vectors
 
Hm
They are homogeneous, after all
I think it may be Bianchi for specific functions $L$
 
9:06 AM
I was so disappointed when I discovered Killing was a person's name.
 
What were you expecting?
If it can make you feel better, the Wave of Death spacetime is exactly what it says
 
A Killing vector is a mystical object that slays all who attempt to compute its norm.
 
No that's just the Navier-Stokes equation
 
What is a null singularity? A half hearted Google has failed to uncover a Dummies guide to null singularities.
(in the context of the wave of death spacetime)
 
9:21 AM
spacelike/timelike/null singularities are singularities where curves of that type stop
 
Aha, so at a null singularity null geodesics are incomplete?
 
yes
 
OK, I can see why you wouldn't want one sweeping through your region of spacetime.
Though you would no longer be worrying about coronavirus.
 
wave of death spacetime is just when the universe ends
Alarming
 
9:37 AM
Hey what is the first thing that comes to your mind when I say f(x)?
Google f(x) and you will be surprised!!(well at least I was).
lemme know what was your reaction ;)
 
@HrishabhNayal Oh nooooooo!!! We are doomed!
 
10:31 AM
@Slereah Infinite things are a big problem in the QFT, too, but somehow the reality remains always finite.
At least, what we can measure from it.
 
11:08 AM
-1
Q: Using Lagrangian equation

Yasir SadiqIn classical mechanics we have a concept of generalized coordinates. Say my generalized coordinates are (x,y). My doubt is ,Is it legal to write the position vector in any vector basis say polar basis but having components which are functions of x, y and then use the Lagrangian equation?

This may be the worst thing I've ever seen
 
11:18 AM
Auch, it seems like I have misunderstood the method I am implementing for my Master's thesis, and now the programming I have done the last two months is kind of useless
 
 
1 hour later…
12:18 PM
@Slereah Don't worry, there's more... I do exist.
naked Singularity.... Eh... I dun feel good at the book's intentions...
@Slereah Those are kids' books, become real men... Read real books!
 
12:33 PM
@Secret Are you okay with General Relativity? I've a quick doubt.
 
depend on level
 
@Secret Why $g^{\mu \nu} \delta g_{\mu \nu} = - g_{\mu \nu} \delta g^{\mu \nu}$?
Also, Is tensor product commutative?
 
no they do not commute
 
@FakeMod Haha lol
 
@Secret why?
 
12:38 PM
also I think you incur a minus sign when you raise and lower index, I don't recall the rules completely
 
@Secret why? Why? '-' sign for raising and lowering...
 
Pretty sure at this point SE forgot about Astronomy when it graduated all those beta sites
 
Just use the Leibniz rule
 
@SirCumference No, Astronomy first, so to pay respect, I add $\kappa term in Einstein Field Equations for the Expansion of Universe...
@Slereah means?
@Slereah why sign changes for changing the indices?
 
$$\delta (\delta^\mu_\nu) = 0 = \delta (g^{\mu\sigma}g_{\sigma\nu}) = g^{\mu\sigma} \delta g_{\sigma\nu} + g_{\sigma\nu} \delta g^{\mu\sigma}$$
 
12:42 PM
@Slereah So, that's true only for $g_{\mu \nu}$ tensor?
@Slereah Btw, very professional argument... +1
 
It's true for any tensor and its inverse
As long as $T$ has an inverse, $T \delta T^{-1} = -T^{-1} \delta T$
It's true for any form of derivative and inverses, really
 
@Slereah Also, one more question, why they don't multiply $g^{\mu \nu}$ both sides in Einstein Field Equations and get the DE in just one equation?
 
$$D(1) = 0 = D(f f^{-1}) = f D(f^{-1}) + f^{-1} D(f)$$
@AbhasKumarSinha Because that equation would be underdetermined
 
@Slereah means?
 
The metric tensor has 12 components, you can't solve it in one equation
 
12:48 PM
@Slereah So, that's why I'm converting all 12 component into 1.....
 
lol that reminds me back in 3rd year when I tried to do exactly that, and I end up spending the next 3 hours looking up how to find the eigenvalues of a tensor and such thing does not really exists
 
@Secret k.... I'll look at it then...
@Slereah Another Question.... Why tensor products are non commutative?
 
Why would they not me
 
@Slereah means?
 
why do you think the tensor product wouldn't be commutative
 
12:51 PM
@AbhasKumarSinha that's cosmology :P
 
It's defined by an ordered pair
 
which i guess is astronomy
 
There's no particular reason for it to be commutative
 
Tensor products are defined ontop of cartesian products, and $A \times B$ and $B \times A$ are not identical in general
 
@SirCumference Okay.... But, Astrophysicist read Cosmology because it's space and time and spacetime...
@Secret K... I guess I assumed commutative and messed whole equations... I'll have to redo everything.
 
12:54 PM
@AbhasKumarSinha what
 
@SirCumference Does celestial mech comes in astrophysics?
 
plenty
 
dat's it...
 
celestial mech is very different from cosmology
 
@SirCumference just same buddy.... It's their propaganda to make them look different.
 
12:58 PM
what
 
@SirCumference That's a secret... don't tell anyone... That Astrophysics and Cosmology are same.. but it's physicist's propaganda to make them look different.
 
$\text{wut}_{\alpha\beta\delta}^{\sigma\phi}T^{\beta_{\sigma}}_{\delta} = F_{\varnothing^{\varnothing}}_{\varnothing \alpha}$
$\text{wut}_{\alpha\beta\delta}^{\sigma\phi}T^{\beta_{\sigma}}_{\delta} = F^{\varnothing^{\varnothing}}_{\varnothing^{\alpha}}$ is my response to the above comment
 
@Secret instructions unclear, accidentally proved Schwarchild's Conjecture.
$\varnothing$
$\otimes$
cool thing :-)
@Secret I guess you've added extra 't'
 
1:33 PM
@Slereah Ew. It's defined by its universal property!
 
Anyone wanna play chess/cricket?
Schwarchild solution is for flat spacetime...
@JohnRennie Sir is that huge thing for flat spacetime?^
 
@PM2Ring I think once upon a time moderators were able to merge accounts on their own, but they removed that feature after one merge too many between accounts that didn't actually belong to the same person and thereby exposed PII of one user to another
 
@AbhasKumarSinha the Schwarzschild spacetime isn't flat ...
 
@JohnRennie $T_{\mu \nu} = 0$ should be flat na?
 
nope
 
1:41 PM
@AbhasKumarSinha The word is 'no', not 'na', and no, vanishing energy-momentum is not the same as vanishing curvature.
 
@Secret You mean spacetime isn't flat even after the stress energy tensor is 0?
@ACuriousMind Why?
 
It's 'vacuum' ($T=0$) but it's not 'flat' (vanishing Riemann tensor)
 
$\mathbf{T} = \mathbf{0}$ only means it is one of the vacuum solutions, where there is no stuff in spacetime
but spacetime can still be e.g. gravitational wave like for example
 
@Secret if there's no stuff, then it should be flat?
@Secret gravitational wave without matter? insane!
 
@AbhasKumarSinha That's simply not what flat means. 'Flat' means vanishing curvature (=Riemann tensor)
 
1:42 PM
@AbhasKumarSinha you might think so, but you would be wrong :-)
 
@AbhasKumarSinha Exactly analogous to classical EM where you also have EM waves propagating in vacuum as an allowed solution
Less 'insane', more 'unsurprising'.
 
@ACuriousMind so what it has to do with spacetime?
There is no matter, but spacetime isn't flat... Only matter make curves in spacetime? true now?
 
@AbhasKumarSinha I'm saying it is analogous - just like EM waves propagating in vacuum without charges are allowed solutions of Maxwell's equations, gravitational waves propagating in vacuum without mass is also an allowed solution of Einstein's equations. You need charge/mass to generate the waves, but not for them to propagate.
 
> Only matter make curves in spacetime? true now?
No
 
@JohnRennie Then what other thing?
OMG... I thought only matter causes curvature in spacetime because of gravity....
 
1:46 PM
In the Schwarzschild solution $\mathbf T = 0$ everywhere except at $r=0$ where it is undefined.
Yet this is a curved spacetime.
 
@JohnRennie This isn't in assumption list of Schwarchild Solution wiki... (I may be wrong...)
 
It is possible to associate a mass with the curved spacetime. This called the ADM mass. But pinpointing exactly where this mass is located is impossible.
 
(note that a common variant of Schwarzschild is that the solution is valid up a certain radius and then you have a massive body sitting there instead of going all the way to the event horizon)
That is, the actual solution for a massive body that's not a black hole is Schwarzschild in the exterior of the massive body, and something else for its interior.
 
If $T = 0$ isn't a flat spacetime, then what is flt spacetime? $g \neq 0$ and $R = 0$ is curved.... :-(
 
@AbhasKumarSinha the common way out if this is to say that the mass is located at the singularity, though note that while this makes intuitive sense it is mathematically meaningless.
 
1:50 PM
@AbhasKumarSinha I already told you - the definition of 'flatness' is 'vanishing Riemann tensor'.
 
@JohnRennie where is singularity?
 
At $r=0$
 
@ACuriousMind $R=0$ cannot be curved? you mean?
@JohnRennie So, curvature starts from $r=0$?
 
That depends on whether by $R$ you mean the Ricci scalar, the Ricci tensor or the Riemann tensor :P
 
@ACuriousMind So which R you mean exactly?
 
1:51 PM
(another reason to lose a few more words about what you're writing instead of just posting equations without elaboration, by the way)
1 min ago, by ACuriousMind
@AbhasKumarSinha I already told you - the definition of 'flatness' is 'vanishing Riemann tensor'.
What exactly is unclear to you about me saying the Riemann tensor?
 
The trace of a non-zero tensor can be zero
 
@ACuriousMind prove it.
 
@AbhasKumarSinha Be more polite.
This is not a way to treat people who are offering up their free time to answer your questions.
'Can you prove that?' or 'Do you have a reference for that?' are perfectly fine and acceptable questions. 'prove it.' is a rude command.
 
I'm amused...
So, T=0 doesn't mean matterless.... Cool...
 
@AbhasKumarSinha have a look at:
In general relativity, a vacuum solution is a Lorentzian manifold whose Einstein tensor vanishes identically. According to the Einstein field equation, this means that the stress–energy tensor also vanishes identically, so that no matter or non-gravitational fields are present. These are distinct from the electrovacuum solutions, which take into account the electromagnetic field in addition to the gravitational field. Vacuum solutions are also distinct from the lambdavacuum solutions, where the only term in the stress–energy tensor is the cosmological constant term (and thus, the lambdavacuums...
 
1:58 PM
@JohnRennie You will agree that $R_{\mu \nu} - \frac12 R g_{\mu \nu}$ on RHS is purely a mathematical thing... nothing to do with reality... it's abstract... But, where does Gravity entered here?
atleast it's derived without Gravity into consideration...
 
19
Q: How does "curved space" explain gravitational attraction?

Mason WheelerThey say that gravity is technically not a real force and that it's caused by objects traveling a straight path through curved space, and that space becomes curved by mass, giving the illusion of a force of gravity. That makes perfect sense for planetary orbits, but a lot less sense for the expr...

 
@JohnRennie Does Schwarchild Solution work for blackholes? you told about singularity thing....
 
The Schwarzschild geometry is an idealised black hole. It wouldn't perfectly describe a real black hole but it would be an excellent approximation to a real black hole.
 
@JohnRennie okay... does it work for points moving in vaccum? (mass points)?
 
Can you clarify what you mean by that?
Do you mean a point mass i.e. a single point with infinite density?
 
2:02 PM
@JohnRennie finite density... single point mass...
 
How can a point mass, i.e. zero volume, have a finite density?
 
@JohnRennie yes
@JohnRennie electron
 
An electron isn't a point mass because it is always delocalised over some region of space.
 
@JohnRennie okay... then it can be approximated for electron?
 
An electron would have a Schwarzschild radius smaller than the planck length, so it's unclear whether GR could describe the geometry near such a small object.
 
2:07 PM
@JohnRennie so black-holes are even smaller...
 
The problem with GR is that everything is complicated.
Consider a ball of gas collapsing under its own gravity. This would be a reasonable model for a black hole forming from some collapsing star.
 
@JohnRennie woops.... It's harder than QM. I completed whole QM volume in a week and I'm struggling 1 Page for GR for weeks...
@JohnRennie so there's mass here... how can schwarchild work here?
 
With a few not too horrendous simplifications we can describe this process using a geometry called the Oppenheimer-Snyder metric.
 
@JohnRennie very hard name....
 
It goes as you'd expect, the ball shrinks under its own gravity and ultimately shrinks to an arbitrarily small size.
 
2:10 PM
@JohnRennie Okay... So, no mass here?
 
Basically it shrinks to a point, but the problem is that the geometry becomes undefined for a point mass, so GR can describe the evolution as long as the radius is greater than zero by even the tiniest amount, but it cannot descrbe what happens when the ball radius becomes zero.
 
@JohnRennie okay....
 
The Schwarzschild geometry is the $r=0$ limit of the OS geometry, but with the point $r=0$ removed. i.e. it covers the whole spacetime except for the point $r=0$.
 
But, this answer is insane! physics.stackexchange.com/a/222433 You did Newton's laws using GR!! That's enormous amount of skills for it!!
@JohnRennie okay.....
 
So it's a bit of a cheat. The Schwarzschild geometry only works because it specifically excludes the point where the geometry goes screwy.
 
2:14 PM
So, first learning... - I don't need Stress Energy Tensor for Blackholes....
@JohnRennie What are scalar fields?
 
In the OS geometry the mass is obviously collapsing towards $r=0$, so in the limit it makes sense to say all the mass is in a point mass with infinite density at $r=0$.
 
@JohnRennie OS Geometry seems difficult... need some time to digest information...
 
So by analogy we'd say a Schwarzschild black hole with a mass M has that mass as a point mass at $r=0$. But the point $r=0$ is excluded from the Schwarzschild metric, so where then is the mass?
 
@JohnRennie doesn't exist in manifold.
@JohnRennie Why we require scalar fields? (what are those? gravity, EM wave or what?)
 
A scalar field is just a scalar that is a function of position in spacetime. For example the good old electrostatic potential is a scalar field.
 
2:22 PM
@JohnRennie How to write lagrangian of a scalar field of Gravitational field?
 
@AbhasKumarSinha in GR?
 
@JohnRennie you mean? Gravity is different in GR?
 
Gravity as in Newtonian gravity doesn't exist in GR. It emerges as a low energy approximation.
 
@JohnRennie So what's GR version of Gravity?
 
In GR the motion of particles is calculated using the geodesic equation. I guess this is the equivalent to Newton's gravitational force.
 
2:26 PM
okay....
 
The problem is that you're trying to understand GR by asking radom questions about it, and that's a terribly inefficient way to learn it.
It will mainly seem to you that GR is a hodge podge of bizarre and random facts.
 
2:55 PM
@JohnRennie I mean, it is a general theory ;)
 
@AaronStevens you don't realise how weird GR is until someone unfamiliar with it starts quizzing you :-)
 
 
1 hour later…
4:15 PM
@peterh-ReinstateMonica Wow! I didn't know that! Why haven't you committed to any proposals recently?
 
@user1271772 Area51 is currently basically dead and it's highly unlikely that SE will launch any new sites, cf. meta.stackexchange.com/a/344244/263383. If you want to start to follow proposals there and expect anything to come of it you're probably several years too late.
 
4:49 PM
@user1271772 Recently not. A51 was always disliked at the company. They did not update it since a decade. Meanwhile, they've continuously said, that they can't update it, because it is so old. ;-) In theory, the A51 works, but only very rarely are new sites created ( stackexchange.com/sites#newest )
@user1271772 Btw, it is a like a beta, the differences: 1. rep levels are much lower than even on the betas 2. the site is invisible on the stackexchange.com
@user1271772 On the Engineering SE, I created the first question. :-) I could see a completely empty SE site! For some seconds only. It was clear that I must create a question on the spot, so I could not create a screenshot.
@user1271772 But the private beta phase lasts only for some days, and then we lose the wonderful privileges that we got.
@user1271772 Btw, the also the Physics SE was created on the A51 by Tobias Kinzler. He is a physicist from Switzerland.
Now the SE has no idea, what could they do to this whole network. My impression is, that if they could change the past, they should have not ever created the SE network et al. Or they had created only IT-related SE sites.
Maybe superuser, serverfault would exist, but no more. At the time, I've read from some SE bosses a very strong urge that "we don't divide the SO", i.e. they view the site network as the SO, with some extras.
 
5:58 PM
Ancient vedic quantum mechanics
 
@Slereah Is that in Tamil?
 
It is in the official language of India
The Queen's english
 
It's taking forever to load
 
It is about in part the Vaiśeṣika Sūtra, which is in Sanskrit
 
Is it related to the Kama Sutra?
Is it the same Sutra?
 
6:02 PM
It is not.
Sutra roughly means "book"
Kama sutra is the book of lovin'
 
I see
StackOverflow is the sutra of programmin' ?
 
I wouldn't call it a book, but why not
 
Finally the PDF opened: P R I Y A D A R S H I
 
Although there is a book of StackExchange
that guys gathers posts of SE and puts them in book forms
Which is an odd business model
I hate this PDF so much
Getting nationalism into science history
 
6:32 PM
@Slereah where do you get this stuff from? :P I am quite sure that you wouldn't be intentionally searching for that...
 
@FakeMod It's actually the subject of the 11th highest voted meta thread on physics meta. physics.meta.stackexchange.com/questions/10581/… Featuring an answer by the author of the books himself.
 
Bam
Time flies
 
@JMac Well no, I was talking about the Sanskrit/ancient Vedic stuff...
 
@FakeMod Oh my bad, I see that now
Anyways that is an interesting read if you haven't seen it
 
@JMac It's more about the history of science rather than science itself. So a good read, but not for me :)
 
6:39 PM
I was thinking in more of general interest of the SE system and what some people apparently do with the information
 
@JMac Oh my bad, I was talking about that ancient Vedic thing :D
It seems as if we both are consistently focused on different things... :)
 
6:56 PM
lol yeah talking right over each other I guess. At least it was all civil talking over each other.
 
 
3 hours later…
10:11 PM
@JMac Random thing I noticed: we have about the same number of posts in the hbar
 
10:26 PM
@AaronStevens That's weird, that's weird... kinda weirding me out.
 
10:44 PM
@JMac Did you know that Lincoln’s secretary was named Kennedy?
 
 
1 hour later…
11:45 PM
Why does the integral in 2.16 evaluate to $\rho(\mathbf{r})$ and not $\rho(\mathbf{r’})$? Isn’t the spike of the Delta function at $\mathbf{r’}$?
 
@schn The Integral is over the primed coordinates. So the spike picks out when $r'=r$
 
@AaronStevens Then it makes sense, thanks. Since $\mathbf{E}(\mathbf{r})$, I thought it would be over $\mathbf{r}$.
It's from Griffith's Introduction to Electrodynamics, by the way.
 
@schn Yeah, the variable you are integrating over will go away since it is a dummy variable.
 
@AaronStevens Thanks!
 

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