4:00 AM
So I have been thinking about emergence recently
You know how classical computers can simulate classical mechanics like a rolling ball using only 0 and 1 states
and yet, that simulated ball never actually act like what a real ball does. In particular, it is intangible
So we have two things sharing the same features (to some order of resolution of course) and yet do not act the same within that context
Thus it seems successfully simulating all the classical properties of an object do not automatically give you the classical object itself
Why is that?

4:15 AM
dunno, pal
have you seen the latest veritasium vid?

4:55 AM
o yeah, that's surprising
what in abomination

One could guess it to be true in a blackhole

Well
If the speed of light has a directional dependence
Then it means looking further into space is the same as looking further back in time may not apply at all for some directions
and there is no way to test this because all the delays will be in the direction of the future light cone, so you cannot exploit that to produce a time paradox
In fact, if the speed of light has a directional dependence, then you can always reinterpret it as the speed of light having the same speed for the round trip, except that spacetime is curved in some way, meaning what we think to be minkowski spacetime like for our neighbourhood is actually not flat

5:23 AM
TIL Derek Alexander Muller is Australian

We Aussies are leading the next world transformation
(barring impotent governments)

and he's not a physicist

All systems of the old world will be broken
It's only 2 days away from that
The election, whoever wins, will be transformative

6:14 AM
Yo.

3 hours later…
9:24 AM
You can have SR with different directional speed
It's called the Reichenbach synchronization
Rather than the Einstein synchronization

10:03 AM
I see. Hmm, so it seems it is analogous to choosing reference frames and as long the same convention is used there will not be any difference in the interpretations of the results
assuming I read the synchronisation equation correctly that is

10:40 AM
that article ain't finished yet, alas

11:11 AM
How to position vector equation of projectile motion?

11:33 AM
What is the process of reopening a question on physics S.E.? In math S.E. you can ask for a reopening in the "CURED" room and for feedback in the "Constructive Feedback" room.
I think a question of mine has been erroneously closed. It has an answer already which is upvoted and directly contradicts the answer from the question it has been labeled a duplicate of. Furthermore, I have explained extensively in the comments and edit of the question why the linked is question is not a duplicate of this one.

12:04 PM
@user400188 The first time you edit a question after it is closed, it is automatically scheduled for reopen review.
3

What is the difference between polar coordinate and radial, tangential coordinate?

radial and tangential would usually describe more vectors than coordinates

@Slereah how do we calculate tangential, radial components at any point of curve using cartesian and polar coordinates?

they are the vectors stemming from the radial and angular coordinates
and vice versa for radial coordinates

12:21 PM
Also can we use trajectory equation of projectile motion as Position vector if we take both x,y on same side of equality and insert $R(x,y) = (\tan\theta\bf{x} -\frac{1}{2}\frac{g}{V_{o}^2\cos^2\theta\bf{x^2})\hat{\boldsymbol{\imath}} - \bf{y}}\hat{\boldsymbol{\jmath}}$
Ahh everything messed up.
.

1:02 PM
In the case of two vertically bodies of different mass connected by a string passes over frictionless pulley. Why Tension on the string is same on both side of the pulley?

1 hour later…
2:16 PM
@123 This question has been answered many times on physics SE - I suggest you take a look at the top results

Boom, boom and boom @123

@NiharKarve Pls share the link. I can not find it.
Thanks @Charlie
Yo @Charlie

2:34 PM
hi

I am still looking for equation of projectile in radial and tangential. If you have time to help me figure this out.
still a mystery for me

I don't know what that means

My question does not understand by you or solution what is need?

I don't really understand that either
I don't know what "equation of projectile in radial and tangential" means

I want to see in other way $a = a_{transltional} + a_{rotational}$

2:41 PM
You need to more precisely explain the situation you're talking about

@Charlie Ookay. But you and everyone helped in many ways.
Thanks

I don't even know what $a$ is here

Means $F_{Net} = F_{translational} + F_{rotational}$
After finding the net force break these two forces in translational and rotational components at every point.

What is "$F_{\text{translational}}$"?
I don't know what the "translational component" means
Are we talking about a projectile being thrown through the air on the surface of the Earth?

Yes this was related to projectile.
Basically i want to analyze equation how radial and tangential components of projectile behaves.
I learnt many many ideas from everyone here about projectile.
Can i start trajectory equation?

2:48 PM
@123 The tangent at every instant is the time derivative of position, i.e. velocity $\dot{r}(t)$. In the 2d description, there is a unique (up to sign and length) vector orthogonal to that, call it $\dot{r}^\perp$. Then you can split the force into components $F \cdot \frac{\dot{r}}{\lvert \dot{r}\rvert}$ and $F\cdot \frac{\dot{r}^\perp}{\lvert \dot{r}^\perp\rvert}$. Is this what you're looking for? If not, you'll have to describe a lot better what you want.

Ok, you need to explain what "tangential components of projectile" means

@ACuriousMind Thanks yes.
@Charlie i want to explain both radial and tangential of projectile.

But radial and tangential what of the projectile?

yes of projectile.

lol

2:51 PM
Why :D @Charlie due to silly question.
Who is acm?

It's not necessarily a silly question, it's just hard to decipher what you're actually asking imo P:
acm$\cong$acuriousmind

Oooo acm means @ACuriousMind
Aaa i got it.

I still don't get why you're trying to break down projectile motion into tangential and radial components. It seems more complicated than it's worth IMO

Yeah, also if we're talking about a projectile being thrown in a homogenous gravitational field the word "radial" seems misplaced

@JMac Problem is that either uniform circular motion explaination radial , tangential idea is wrong where they explain radial only changes direction not velocity as $a_c = \frac{v^2}{r}$ and tangential is responsible of changing velocity.

2:54 PM
@Charlie I was kinda talking to him about this yesterday. You technically could analyze the parabolic motion like you would circular motion; but I don't see the point when the radius and tangential velocity are constantly changing

I know this is correct there is some insight which still i don't understand completely

The question to me is probably just getting a bit lost in translation, which isn't your fault, but it does mean I can't attempt to answer it :P

Because in rectangular coordinate y-component changes velocity continuously which seen as direction change. I want it separate in radial and tangential.
:D @Charlie I know i became a long discussing on same topic. But @ACuriousMind given me equations i will do some calculation and will spend some time on it.
:P

@Charlie :P may be i wont...

2:59 PM
@123 I'm still wondering why you would want to do that... Like I'm just not sure if I see any benefit from doing it that way compared to just using rectangular coordinates and being able to easily find the path

Can we discuss new topic???
what is the problem when we consider non-inertial frame as reference frame?

What "problem" do you expect?

I have read in a book laws of physics are valid if inertial frame is reference frame.

Sure, at least the standard ones

I knew some basics may be wrong. If we consider non-inertial frame as reference frame we find force acting backward on an object but in reality there is no force on it actually frame is acceleration.
But in one book i read there are some cases where non-inertial frame considered as reference frame.
the force above called fictitious force. My understanding is correct?

3:04 PM
Non-inertial frames are fine to consider, what you're describing above is a called a pseudo-force

You can use non-inertial reference frames as long as you account for the non-inertial effects (usually as a pseudo-force)

If we are in non-inertial frame how do we exactly know there is some force acting on it or not. What if some force acting on it how do we encounter the actual force?
@JMac Aaah. Ookay. It means if we can exactly find how our own non-inertial frame acceleration . We can take it as reference. This is what you meant?

In the context of a generic vector bundle $(E,M,V,\pi)$, $\pi:E\rightarrow M$, why is it necessary to have the total space only locally look like $U\times V$ for some open region $U\subseteq M$? Is it common for the total space $E$ to have curvature or non-trivial topology? In other words why don't we just make $E=M\times V$ rather than having it only locally have this structure? Maybe this question is a bit too open ended

3:19 PM
@Charlie well, if you were just interested in the case $E = M\times V$, you wouldn't need to look into the theory of bundles :P

ah, naturally 3 lines beneath where I was sitting wondering about it :P

Doesn't Wilsonian renormalisation (with the hard cutoff) violate Lorentz symmetry?

@NiharKarve Yes.
but since the cutoff will not enter any physical quantities you compute, why would it matter?

Right, so what does 'violate Lorentz symmetry' mean?
In intermediate steps?

the canonical quantization procedure where we work with a Hamiltonian and asympotically free states as $t\to \pm\infty$ also "violates Lorentz symmetry" in that it is not manifestly invariant or properly covariant in its intermediate steps, but the results we compute - like scattering amplitudes - are

3:27 PM
Hmm - makes sense
On an unrelated note, when does the link to the h bar pop up on the website?

it's pretty random :P
but there is a persistent link to chat in the hamburger menu in the upper right

Yeah, it actually only turns up round about now for me
Ok last one, why does the Pauli matrix vector thing $(1, \sigma^\mu) have one dotted and one undated index? undotted* 3:43 PM @NiharKarve I'm not quite sure what you mean - what's the context here? @ACuriousMind Good Evening ACM yo yo yo everyone what's up! This might be incorrect, but I thought I saw$\sigma^\mu$described as$\sigma^\mu_{\alpha\dot{\alpha}}$Tensorial Contraction probably A tensor$K_{\mu \mu} ^\nu$can be contracted to$K^\nu$3:50 PM Yeah, this isn't that ok, no idea then, what's goin' here @Charlie The classical example is non-orientable manifolds Schwartz kind of pulls it out of a hat to justify all indices in$\tilde{\chi}^\dag\sigma^\mu\tilde{\chi}$being contracted A non-orientable manifold has a non-trivial tangent bundle yo yo yo 3:54 PM @Slereah oh ok, are these non-trivial cases of much interest in physics? I'm mostly reading about this because I don't know what a principle bundle is and it apparently comes up in gauge theory Sure or rather "is central to the study of" gauge theory :P there's plenty of non-trivial bundles in physics Related to the study of instantons and topological defects Are the spacetime tangent bundles non-trivial? Depends on which one! 3:55 PM hmm, instantons is a word I've only heard in passing :P :O$\mathbb{R}^n$has a trivial tangent bundle certainly Just to be clear, when we say "non-trivial" we mean topologically non-trivial right? We're not talking about curvature wait that requires a metric hmm in the context of bundles, the trivial bundle$\pi : E \to M$is the bundle where$E = F \times M$@NiharKarve In the chiral basis, the gamma matrices are$\gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}$With$F$the typical fiber 3:58 PM So this means the action of$\gamma^\mu$turns a left-handed (dotted) spinor into a right-handed (undotted) spinor and vice versa ok Oh boy, this is getting messy @ACuriousMind isn't there a negative sign here? in the bottom left So when you write the Dirac action in terms of Weyl spinors, you have that the$\gamma^\mu \partial_\mu$becomes a$\sigma^\mu \partial_\mu$where the$\sigma$changes the handedness, and this is why it has one dotted and one undotted index, so that whenever you contract it with a Weyl spinor, the result has the opposite handedness @Charlie I don't know what you mean ;) D: 4:03 PM so if I were to use a needlessly complicated analogy, it's like the tetrad e^\mu_a having one Greek and one Latin index because it converts from "local" to "global" I would say it is entirely unlike that (there is no geometry involved here as in the local/global distrinction) but if that analogy helps you I'm not gonna protest :P I mean, I would go for the actual transformation law, but it's not as intuitive for the Pauli vector as it is for say, momentum ^ No, the previous analogy (and I know it's stretching it a lot) was based on what it "eats" and what it "produces" Merely superficial ya go! @Azmuth If you've just returned to make pointless and annoying interjections, please stop. @NiharKarve yeah, in that respect they're similar in that they're producing a vector of a different "kind" than the one they ate Please don't ping currently chat-suspended users, either. 4:24 PM what?!! It's somewhat taunting/rude to address users when they cannot reply. The user you pinged is currently chat-suspended, as you can see from his chat profile. That didn't take long ... @JohnRennie Good Evening sir! :-) Is there any way to let all the responses to my question (even comments) to be notified to my email? 4:31 PM I think you can get the SE to mail you. There's a setting in the profile somewhere. Where? chat settings or which settings? You need to edit the URL and replace 1325 by the ID of your PSE account. ok Do you have an account on the PSE? I couldn't find you in the list of users. No, no account on pse, only AI and SO 4:34 PM I can't remember if the setting applies to all SE sites or if you need to set it on every site you use. ok, I'm doing that ri8 now Are job alerts good? And now I'm off for my weekly browse through the science blogs ... and to digest my lunch :-) Wow:) cool Don cooooooooool As Pomished Earlier: Blue Moon of 31st Oct Last Blue Moon of the Decade This one with higher EV Values... @Azmuth It seems like there are supposed to be other blue moons this decade? @JMac Yes, it happens in every 2.7 years. 4:49 PM Why did you say "last blue moon of the decade" then? Because last one of 2020... 2021 will start next decade.... right? Typically people refer to decades starting on the 0 year, like "the 90's" was from 1990 to 1999; so this decade would be from 2020-2029 the way most people use it hmmm... I thought counting starts with 1... So, first year as 2021 or sooo... :P Also, was the moon actually that blue for you, or did you like change the colour? I didn't hear anything about the blue moon actually appearing blue this year somewhere. @JMac Well, pedal back to 1 AD to see what happens :p 4:57 PM @JMac I did change the ACS Profile... but didn't tempered anything else... (It's to tell to capture which prominent color) blue moons aren't actually blue, see en.wikipedia.org/wiki/Blue_moon @Azmuth What is "ACS Profile"? it's not an astronomical phenomenon that turns the moon blue, you just made your camera record the usual color of the moon as blue :P @JMac similar to ISO, but they are used to change the exposure values to get different color. @Azmuth Okay so you basically changed the colour then... 5:00 PM @ACuriousMind Check ACS Value, I deliberately changed it to get more blue. @JMac Yes! @ACuriousMind It can be an atmospheric phenomenon, but I didn't hear anything about volcanoes or heavily particulate, so I figured that didn't add up. yes, there are blood moons tho for red colors @JMac well, the 2.7 years is exactly the cycle for "second full moon in a month" and that's what happened Oct 31, so what Azmuth posted is in fact just an ordinary picture of the moon What does "ACS" even stand for. I literally can't find anything about "ACS profiles" or "ACS Values" related to photography? @Azmuth yes, that's what I said. But why did you do that? Why did you take a normal picture of the moon and turn it blue? 5:02 PM blue moons aren't interesting Of what interest do you think an image you essentially just painted blue is to others? this is the neat one A wet moon (also called a Cheshire moon) is the visual phenomenon when the "horns" of the crescent Moon point up at an angle, away from the horizon, so that the crescent takes on the appearance of a bowl or smile. A wet moon occurs when the crescent Moon is low above the horizon and at a point more or less directly above the Sun's (invisible) position below the horizon. This in turn is determined by the positions of the Moon and Earth in their respective orbits, the inclinations of these orbits relative to one another and to Earth's celestial equator, and the observer's latitude on Earth. Wet moons... @ACuriousMind I did take normal picture. I didn't change it to blue @ACuriousMind So that I wouldn't have to say, it's blue moon to others :P XD If you messed with the colours of the picture, it's not really "normal". You literally changed camera settings until the moon appeared blue even though really looking at it, it wasn't blue... Can we measure forces of other frames if we are in non-inertial reference frame? 5:05 PM There's a difference between that and post-processing/editing yes @123 pseudoforce @123 all forces in every frame are the same, except for inertial forces @Azmuth Again, I can't find anything on what these "ACS Profiles" or "ACS Values" are. What are you talking about when you say you changed those? How do we know our own frame psuedo force? @JMac It must be white balance or (WB-ACS or WB) then. Check in manual or expert mode of your phone. @123 practically? made in india phones usually call it ACS tho Yes because our first target to know what force act on our own reference frame to make it non-inertial. 5:10 PM Good Night, it's 10:40 PM here :) @123 It's usually a theoretical tricks deployed to make calculations easy.. You can ask others in the room, it's easy thing... @123 You just create a situation where you would expect no force on something if you were in an inertial frame, and if there is a force, that's the pseudoforce. What I am thinking @ACuriousMind. How do we account this situation like in two busses moving with different acceleration. If a boy in one bus he is in non-inertial frame. How he calculate the other bus force. you can know if a force is an inertial force if the acceleration it causes is the same regardless of mass @123 by measuring the acceleration of the other bus, I'd say I'm really not sure what the difficulty is supposed to be here I don't understand the situation you said @SirCumference. Pls explaim 5:18 PM @123 if a force is proportional to mass, and therefore its acceleration is the same regardless of mass, then it's a "fictitious force" aka inertial force and you can get into a reference frame that doesn't contain that force Okay.. Hmm.. It means if we take inertial force as psuedo force for every other frame. This calculation makes our frame as inertial. Right? inertial force is just another name for pseudoforce/fictitious force one that i prefer tbh, since it mathematically works the same as a force, but is just explained by inertia in a different frame Yes @SirCumference How do we calculate psuedo force of our own non-inertial frame? There could be some other frame or object. gravity is also an inertial force, since it has a mass proportionality @123 you calculate any forces proportional to mass those are your inertial forces Exactly @SirCumference this was also in my mind we are also live in non-inertial frame. It means nature gives us this beauty gravitational force is proportional to mass. If this is not the case it is possible to calculate our own fictitious force? 5:26 PM i mean even without gravity we're on a rotating planet, so we commonly use a rotating reference frame but just approximate it as inertial Yes it is approximation @sir @123 you look for forces that are proportional to mass then you've calculated your own fictitious force i don't really see what the problem is we can calculate the coriolis force here on earth, and that's an inertial force What if there is no any force which is proportional to mass what about this case. then you're in an inertial frame, so there aren't any inertial forces Oooh I see. @SirCumference it means there is always forces which is proportional to mass. 5:29 PM @123 in a noninertial frame, yes some might say that's what defines a frame as noninertial now i'm no expert so i can't say if that's the most accurate definition Yes this will be my next question but for everyday use it might be helpful How many types of frames in physics? Inertial (constant velocity or no force) , non-intertial (at which inertial forces which is proportional to mass). Is there any other frames. not that i really know of someone else here might have a better answer Thanks @SirCumference for your help. I Google it there are only 2 frames inertial and no inertial. @ACuriousMind when you have time pls explain it further. It means psuedo force as in my example of bus each object has different mass but having same acceleration with the bus. It means it is proportional the mass. Excellent @SirCumference Am I correct? It is same behavior of gravity. Wow What is the difference between gravitational mass and inertial mass. Why we don't know how they are same? 6:02 PM @123 what do you want me to elaborate on? 3 hours later… 9:31 PM in Wolfram Mathematica, Oct 21 at 7:58, by Lukas Lang If you ever wanted to get something similar to Matlab's plot interactivity in Mathematica, there's now ResourceFunction["InteractiveGraphics"], with support for zooming, dragging, and data-tip placement: 2 hours later… 11:19 PM anyone have a sec to help me with the two dimensional heat equation on a disc with inhomogeneous BCs? 11:29 PM @psa If you have a question the rule is generally to just go ahead and ask, rather than asking if you can ask. Otherwise someone who can answer your question could come by in 30 minutes when you're gone and not be able to help guys why total mass is equal to$M=4\pi \int_{r=0}^{r=a}\rho(r)r^2dr$M=\int_{r=0}^{r=a}\rho(r)dr @Charlie yeah fair enough. alright, well, I'm looking to solve the heat equation$\dot{T} = \alpha\nabla^2 T$in 2D in polar coords on a disc$0 \leq r \leq R$with the weird BCs and IC$T(R, \theta, t) = T_0 \cos\sin(\theta) + \kappa t$and$T(r,\theta, 0) = T_i\$

I think it is total mass but that constant and r^2 is bit confusing to me
sorry I am in hurry so need to go

just a general starting point and intuition would be helpful here