Hi All.

Hello @SirCumference

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

@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"

ah yes, thank you

(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?

what

@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.

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

Ah, OK, don't know.

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

@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

@bolbteppa that's a great example

@schn what are you plotting?

@NiharKarve m.imgur.com/3hDAEVt

The boxes are supposed to be squares.

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

You could just manually position it

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?

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

Alright, thanks.

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

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

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

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...

@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

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

@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

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

@JohnRennie accuweather.com/en/videos/…

Hi, I'm taking my electromagnetism final next week

Would anyone be interested in private lessons, to help me revise for the exam?