has anyone here used openems? writing some code using it and getting funky results, and their documentation is...lacking a bit

has anyone found galois theory to be useful in any encounter with physics?

@SillyGoose Probably not applying via polynomials themselves, but there are plenty of similar things. For example, would you consider finding symmetric solutions to physics problems along those ideas as Galois theory? I mean, in quantum computing there is a lot of ansatz that are simply guessed from a mix of Galois theory and Fourier analysis. Or would you accept that tensorial expansions that group the polynomial expansions into fixed $\ell$ angular momentum groupings, as Galois theory?

@SillyGoose I've seen it used

Though it is pretty niche

@Jakobian en.m.wikipedia.org/wiki/Big_Bang this picture says the duration of inflation is about 14 billion years. time is likely undefined in the initial quantum fluctuation region due to quantum gravity

big bang assumes a boundary in the past, so i dont think it allows for infinite time until the present

but maybe spacetime can be modeled as an open set. then we can add a boundary in the past extended to a quantum fluctuation region. and the integral of proper time from the present to the boundary could still go to infinity

maybe "big bang + infinite time until the present" is possible theoretically, but the actual calculation gives 14 billion years as the time from the end of quantum fluctuation to the present

If there was infinite time to the boundary it wouldn't be a singularity

That would just be the timelike/null past infinity boundary

oh

i was thinking something like : we take Minkowski spacetime, map it diffeomorphically to a bounded open set, and add a boundary in the past which we call Big Bang

this allows for infinite time from the boundary but it doesnt have singularity, yeah

That's just conformal boundary construction

how do we get that our universe has a singularity at in the past? do inflating spacetimes have singularities in the past?

We don't

That's just assuming the FRW model with current observations

oh

we also sometimes assume that the singularity part is governed by quantum gravity. the 14 billion estimate does not include the singularity region, right? because things are ill defined there

14 billion must be the age of the "well defined universe", excluding the infinite curvature part

or maybe the calculation is complete because the singularity is just a point, so one doesnt need to exclude any ill defined region

@RyderRude the singularity part is assumed. And only in certain cosmological models. Eternal inflation, for instance, doesn't have a curvature singularity. The 14 billion years is found by doing the age integral and integrating back to when the scale factor was zero (or really close to it)

hi guys

Where am I doing wrong?

@Jim oh

so the GR calculation is exact... but it's just that GR cant be expected to hold in the early part

@Jim does FRW fit data best over eternal inflation and others

@RyderRude FLRW is just a metric we use. It is generic, so yes it works well. The Lambda-CDM model fits the data the best, but eternal inflation is a model for a region of time when we have no data. It's really just saying "sure, that model is good, but let's say what if in this unknowable period of time"

@naturallyInconsistent Yes, they are so confusing. But what is the real answer to this problem? It seems a real contradiction to me.

@Pizza Everything is correct except that ,h=$frac{l}{2}cos(\theta)$ ,note that $\theta$ is the angle with the vertical wall not the horizontal one,hence height of COM initially is h=$frac{l}{2}cos(\theta)$

@Arjun the perpendicular distance from the center of mass of the rod to point A is: $\frac{l}{2}\sin(\theta)$?

For the tension in the cord, the distance to point A is: $l \cos(\theta)$?

@Pizza what I'm trying to say is in the initial potential energy part of your solution..the height of COM above the ground has been incorrectly taken as $\frac{l}{2}\sin(\theta)$,it should instead be $\frac{l}{2}\cos(\theta)$

Also on a second look,how did you conclude that $v_x$ of com =0?

@Arjun okok , but you confirm that

to find A. The tension of the string and the magnitudes of the normal reactions of floor and vertical wall that acts on the ends of the rod (SOL: 𝑁1 = 𝑚𝑔 = 78. .5𝑁; 𝑁2 = 𝑇 = 22.6𝑁)

N2 = T

N₂ 𝓁 cosθ = mg ¹⁄₂𝓁 sinθ

I have trouble with trigonometry. If it was sin before, shouldn't it be N2 • l • sin , now?

Since there are no other horizontal forces other than the normal force by the vertical wall,I don't think $v_x$ of com =0 when the rod hits the floor

vx = 0 Is a part of the solution

vy = -2.26

If I make your modification, I find this solution for vy

so its correct

@pizza lemme think for some time..

Okok

I don't understand why we use N2 • l • cos , if it is horizontally it should be sin?

@Pizza Think about it ,initial horizontal component of COM's velocity is 0 and COM would be displaced in the horizontal direction when the rod hits the ground, Now there is no other horizontal force than the rightward pushing normal reaction from the vertical wall,hence $V_x$ COM is NOT zero when it hits the ground

@Pizza In this solution you took A as a point that's at rest and evaluated the kinetic energy of the rod about it,but this will only be true if A is momentarily at rest ,but I don't think it's the case

So was point A done well?

@Arjun mm wait let me think

@Pizza Always keep in mind that when you are evaluating moments about a point ,you're taking the CROSS product of two Vectors,here moment of N about A would be by the cross product rule: $N(l/2)sin(90-\theta)$=$N(l/2)cos(\theta)$ since angle between N vector and vector BA is $90-\theta$

Ah okok

@Arjun anyway back here

Since the center of mass moves both horizontally and vertically, we must consider both components of velocity

Yes

so for vx = omega • l/2 • sin(theta) and vy = omega • l/2 • cos(theta)?

@Pizza Why did you equate both the moments!? How do you know angular acceleration about A is zero??

@Arjun can you check problem solving room pls

@Pizza ohk

@Bml There is no problem. Just consider every answer that talked about centrifugal to be wrong. Then you will be left with only correct answers covering centripetal properly.

@Jim thanks for sharing