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123

04:26

Hi All.

123

05:16

Hello @SirCumference

3 hours later…

Nihar Karve

08:15

@ACuriousMind re the Callan-Symanzik equation, does $\beta = \mu\frac{\partial\lambda}{\partial\mu}$ have the \mu up front to make it dimensionless

ACuriousMind

08:45

@NiharKarve that's a good observation, that's probably it

well, less making it "dimensionless" and more taking the scale dimension out

so that $\beta$ has the same dimension as the coupling $\lambda$ it is "running for"

Nihar Karve

ah yes, thank you

2 hours later…

Physics Meta

10:21

Q: Fixing the closed-question notices following changes to the software

About a year ago, the post notices on closed questions changed across the entire SE platform. This changed a number of aspects of the Q&A engine software's behaviour and how it communicates meta information about threads to different classes of users, but one particularly important change is that...

6 hours later…

Nihar Karve

16:02

(warning: this may not be a well-founded question) - which of the (equivalent) Einstein equations is more useful in practice: the one with R or the one with T?

Ryan Unger

what

John Rennie

@NiharKarve What two versions?

R is normally the Ricci tensor/scalar and T is the stress-energy tensor, so both R and T appear in the EFE.

Nihar Karve

the "normal" one is the one with the Ricci scalar, but you can rewrite it so that there's the stress-energy trace on the other side

$R_{\mu\nu} = \kappa(T_{\mu\nu} -\frac12 Tg_{\mu\nu})$

obviously they're both really just the same thing, and I know you don't usually go the whole hog and solve the EFE directly, but I was just wondering

John Rennie

Ah, OK, don't know.

schn

16:34

Anyone good at plotting in Python here? I’m looking to plot a rather simple diagram, which I have sketched.

bolbteppa

@NiharKarve here's a basic example: in solving for Schwarzschild, if you use $R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 0$ then the resulting ODE's are straightforward to solve, but you need to do more work and compute $R$ and the whole thing is big enough as it is. If you use $R_{\mu \nu} = 0$ you then need to play with the resulting coupled ode's to eliminate variables which is not ideal and if you did it on your own would you even spot it

Nihar Karve

@bolbteppa that's a great example

@schn what are you plotting?

schn

17:01

@NiharKarve m.imgur.com/3hDAEVt

The boxes are supposed to be squares.

schn

17:17

@NiharKarve Almost have it. Do you know of a way to add labels beneath specific intervals on either axis?

Nihar Karve

You could just manually position it

schn

Say I would like to label the x-axis with an x, but also with a label beneath the interval with the boxes. Is that possible?

Nihar Karve

Yep, use xlabel for the main one and plt.text for the intervals

schn

Alright, thanks.

1 hour later…

M. Çağlar TUFAN

18:26

Hello

How do I express the gray area in terms of a and b vectors?

I tried many things but I couldn't simplified my result to a and b vectors

I end up with a and b vectors' projections on x and y axes

danielunderwood

18:39

Try a cross product? Not sure how it looks like a cross product geometrically, but the components certainly look like one

fqq

Are you looking for a basis-independent expression? the area is defined using the two vectors and the x,y axes, it probably does depend on the basis

bolbteppa

What's wrong with $( a \cos \theta_a \times b \sin \theta_b) - (b \cos \theta_b \times a \sin \theta_a)$ where $a$ is the norm of $\mathbf{a}$ and $\theta_a$ is the angle of $\mathbf{a}$ with respect to the positive x-axis etc...

M. Çağlar TUFAN

19:08

@bolbteppa yes, I find the same result but I'm not sure that this result is in terms of a and b vectors, it doesn't include neither a or b vectors...

@danielunderwood Yes, I tried to find a cross product that can give the area but I can only get the result when do cross product of vectors' projections

bolbteppa

It is in terms of the vectors a and b, not sure what you mean

M. Çağlar TUFAN

@fqq Sorry, I don't know what "basis-independent expression", it only asks me to find the grey area in terms of a and b vectors

@bolbteppa I mean it's not like $2\vec{a}\times\vec{b}$ etc. We have no $\vec{a}$ or $\vec{b}$ in our result.it's

bolbteppa

You can trivially attach this to a unit vector built from $\mathbf{a}$ and $\mathbf{b}$ if you want
$$\{ ( a \cos \theta_a \times b \sin \theta_b) - (b \cos \theta_b \times a \sin \theta_a) \} \frac{\mathbf{a} \times \mathbf{b}}{||\mathbf{a} \times \mathbf{b}||}$$

M. Çağlar TUFAN

Hmm, it kinda looks in terms of a and b vectors now :D

I guess we can't transform the result more, I don't know if there is any other solution to this problem

@bolbteppa Thanks for your help, I will use the result you have given

bolbteppa

This is what one does to express the surface area of a surface $\mathbf{r} = \mathbf{r}(u,v)$ as a vector when doing surface integrals, $d \mathbf{S} = \mathbf{n} dS$ where $\mathbf{n}$ is a complicated vector normal to the surface built from tangent vectors to that surface
$$\mathbf{n} = \frac{\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}}{|| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}||}$$