The h Bar

Silly Goose

00:58

wow ive never seen no one in Hbar

123

03:24

Hello Everyone...

What happened if direction of angular momentum is perpendicular to the direction of torque?

123

03:45

I created simulation for conical pendulum. Where pivot point B is taken as origin. Angular momentum about B is $L_B$ can be seen at point B and $L_B$ is constant is magnitude changing direction. $L_B$ has two components $L_z$ which is constant is magnitude and direction both has no torque. But $L_B$ in the radial direction $L_r$ constant in magnitude but changing direction. which is shown at M in radial direction.

and Torque $\tau_B$ is perpendicular to $L_r$ . I my point of view torque and angular momentum perpendicular to each other which causes rotation. This is why origin at B show circular motion.

123

04:17

Pls see the link video imgur.com/mJ2SGQS simulation of conical pendulum. Taken origin B pivot point . $L_r$ radial angular momentum which is radial component of $L_B$ and $\tau_B$ torque about origin B.

Both $L_r$ and $\tau_B$ perpendicular.

Pls help me what happened if angular momentum and torque perpendicular.

123

04:31

Hi @JohnRennie Sir

John Rennie

Hi :-)

123

I need your help keenly

Few topics still unclear to me.

Need your precious time

John Rennie

I'm a bit busy at the moment. Sorry :-(

123

I know you are busy these days. Pls share the time when you are available. I will be there

nickbros123

04:47

@123 you answered it yourself, the angular momentum vector rotates (or rather the radial component of the angular momentum vector)

123

05:43

@nickbros123 I want what is the formula or procedure which shows if angular momentum and torque are perpendicular to each other then object rotate.

Physics Meta

0

Q: How do we deal with interdisciplinary questions?

As the borders between various disciplines are slowly vanishing with the advent of interdisciplinary areas like Biophysics, Environmental and climate studies, and Complex systems, how do we deal with such questions in Physics SE? Usually, I try to answer questions that can be reasonably explained...

naturallyInconsistent

@123 This demand is wrong.

nickbros123

06:18

@123 in this particular strong constraint one can show it but this is not a general rule The thing is the angular momentum

The angular momentum has 2 components along z and r

And considering everything inside the expression constant, u can differentiate the r hat unit vector itself but cannot differentiate the z hat

In the case when your radial distance, velocity, angle of cone, none of it changes u can show this. To obey this there are a set of constraint relations on the angular velocity of rotation and angle of cone

123

07:50

@naturallyInconsistent why. There must be some reason or statement at which we can say if angular momentum and torque perpendicular to each other then object rotate in the direction of torque.

naturallyInconsistent

08:03

@123 No because an object's instantaneous rotation is to do with its angular momentum and nothing to do with its torque. The torque is to do with its angular acceleration.

Silly Goose

how do i formally think about taking direct sums of unitary operators represented as matrices acting on hilbert space?

for example, I have a representation of $SU(N)$ acting on $\mathcal{H}_S \otimes \mathcal{H}_\mathcal{E}$. ok. these matrices look like sums of $U_1 \otimes U_2$, which act on vectors like $\psi_1 \otimes \psi_2$. But you can then actually also consider this as an $N \times N$ matrix acting on an $N$ dimensional state.

What if I want to only look at $U \in SU(N)$ such that $U$ is block diagonal. I.e. $U = U_1 \oplus U_2$. Then can one consider the direct sum of representation of $SU(N)$ on $\mathcal{H}_1 \oplus \mathcal{H}_2$, which makes elements of $SU(N)$ look like $U_1 \oplus U_2$? And then simply consider $U_1 \oplus U_2$ as $N \times N$ matrices acting on $N$ dimensional states?

where $\mathcal{H}_1 \oplus \mathcal{H}_2 \cong \mathcal{H}_S \otimes \mathcal{H}_\mathcal{E}$

and $U_1, U_2 \in U(N)$ with det$U_1$*det$U_2$ = 1.

ACuriousMind

08:21

@SillyGoose why are the states $N$ dimensional?

a representation of $\mathrm{SU}(N)$ can have many other dimensions

naturallyInconsistent

@SillyGoose There is a reason why we teach QM using angular momenta. It provides a simple finite-dimensional system that we can combine in various ways. The usual combinations are $U=U_1\otimes\mathbb I+\mathbb I\otimes U_2$ and so forth, leading to whatever you might want to have later on.

ACuriousMind

It's perhaps not directly related, but the interplay between direct sums, direct product and tensor products is generally the following: Given two groups $G,K$ with representations $(H_G,\pi_G), (H_K,\pi_K)$, we have that the Lie algebra of $G\times K$ is $\mathfrak{g}\oplus\mathfrak{k}$, and $(H_G\otimes H_K,\pi)$ is a representation of $G\times K$ such that $\pi(g,k) = \pi_G(g)\otimes \pi_K(k)$ and $\mathrm{d}\pi(T_G,T_K) = T_G\otimes 1 + 1 \otimes T_K$

123

@naturallyInconsistent what is the role of radial angular momentum and tangential torque here. Pls explain. How to think these with origin at pivot point B.

naturallyInconsistent

@123 I can explain, but I don't think our language differences will sustain the communication. What is the point of you trying to do this? Why can you not just do the simple case of taking the centre of the circle of rotation to be the origin and thereby not have to deal with this at all?

Amit

@123 You may want to read up on precession that can be torque induced. Having the torque at right angles to the angular momentum is a special case of that

123

08:34

I am emphasizing this because i believe no matter if i change the origin and also change of origin change our parameters. But motion must be consistent and valid with new results. This is i believe that's why i wanted to understand my example with new origin.

naturallyInconsistent

@123 This is not sufficiently English to be transmitted but it is obvious that the result will be the same in any choice of origin and only the mathematical details differ.

123

@naturallyInconsistent Also new way of thinking give me confidence in understanding the physics more and more better way. Pls explain it.

@naturallyInconsistent Pls explain does not matter it is mathematical or english.

ACuriousMind

The whole point of knowing on an abstract level how physics is independent of the choice of frame is that we don't have to contort our brains to think through different frames in every single instance of every single situation :P

123

@Amit The gyroscopic precession is in mind. That's why i said when direction of angular momentum and torque perpendicular then object rotate in the direction of torque. I have taken this idea from precession.

Amit

@123 But it doesn't have to be exactly perpendicular

123

08:41

@Amit In my example with origin B both $L_r$ and $\tau_{B}$ are perpendicular. How to understand conical pendulum using these two information?

@ACuriousMind You can help me in understanding this. You all have great knowledge. That's why i need help from all of you. ACM you are the most experienced what i think.

Silly Goose

err sorry i mean ato talk about $SU(N)$ represented over an $N$ dimensional hilbert space

@naturallyInconsistent I am looking for block diagonal combinations which are not of the form $U_1 \otimes I + I \otimes U_2$ i believe

ACuriousMind

@SillyGoose so you are just literally looking at the unitary operators on a N-dimensional Hilbert space

Amit

@123 Just calculate the vectors

ACuriousMind

I'm a bit confused why we'd couch that in rep theory language

123

@ACuriousMind If you will guide me i will change the frame according to this in simulation for better understanding.

Silly Goose

08:46

hm yes i suppose well this is another thing i am confused about

ACuriousMind

@123 I don't know what you want; I said that I don't see the point in trying to view every physical situation from "randomly" chosen frames

the whole point of us having freedom of choice is that we can choose the one frame where the physics is easiest to understand and deduce there what happens

Silly Goose

really what is happening is we are representing $SU(2^n)$ over $\mathcal{H} \cong \otimes_i \mathcal{H}_i$ where dim$\mathcal{H}_i$ = $2$. Then, $U \in SU(2^n)$ are generated by generalized gell-mann matrices. But when writing this into code, one simply actually computes the tensor products of matrices, resulting in a $2^n \times 2^n$ dimension matrix. so yes eventually it is just the unitaries acting on $2^n$ dimensional hilbert space

123

@ACuriousMind Whenever humans thought in a new way , it open us a way to new physics.

ACuriousMind

@SillyGoose what do you mean "we are representing"

isn't $\mathcal{H}$ just $2^n$-dimensional and you're looking at the unitary matrices on it

I mean, sure, technically the fundamental representation is a representation, but if there are no other representations in play I don't understand why we talk about "representations" at all

123

@Amit I don't find any formula to calculate this. Because simple equations gives us only magnitudes of vectors not direction and position of quantities. Position vectors gives me vector magnitude and direction not position. And i don't find any way of calculation in physics at which can take care of vector position also along with magnitude and direction.

Amit

08:53

@123 To think in a new way it may be better not to rely on someone else's explanation

123

Pls share the procedure at which we can have vectors position, magnitude and direction.

ACuriousMind

@123 What are you talking about? The solutions to the equation of motion $\vec F = m\vec a$ are trajectories $\vec x(t)$ that are very much the position vectors "along with magnitude and direction"

Silly Goose

hm well something must encode the tensor product structure of the space i guess is what i am thinking

ACuriousMind

@SillyGoose sure but what does that have to do with the $\mathrm{SU}(2^n)$?

123

@Amit Yes but my knowledge is very limited . We always need a good great teachers.

Amit

08:54

@123 what do you mean? $\mathbf{\tau} = \mathbf{r}\times\mathbf{F}$

Silly Goose

also the meaning of direct product of groups seems to be representation dependent. e.g. $SU(2) \times SU(2)$ in this context can mean the set of all local unitary operators, $U_1 \otimes U_2$. but it seems to also be able to mean all operators of the form $U_1 \oplus U_2$?

Amit

@123 The teachers are the block

123

@Amit Yes , using this formulation we know magnitude only not the direction and position of vector.

Amit

@123 cross product gives a vector!

123

@Amit sorry. I mean this does not defined position of torque.

ACuriousMind

08:56

@SillyGoose By definition, $\mathrm{SU}(2)\times \mathrm{SU}(2)$ is just tuples $(U_1,U_2)$ where $U_i\in\mathrm{SU}(2)$. Your local unitaries are of the form $U_1\otimes U_2$, so there is a map from $\mathrm{SU}(2)\times \mathrm{SU}(2)$ simply by $(U_1,U_2)\mapsto U_1\otimes U_2$

Silly Goose

but shouldn't there also be a map $(U_1, U_2) \mapsto U_1 \oplus U_2$ then?

ACuriousMind

I don't know what $U_1\oplus U_2$ is supposed to mean since we're in a tensor product not a direct sum

Silly Goose

$U_1 \oplus U_2$ = $[[U_1, 0], [0, U_2]]$

the array is meant to be interpreted as a matrix. so it is the usual (?) notion of direct sum of matrices

ACuriousMind

@SillyGoose that only works by the very accident that $2+2 = 4$ and $2*2 = 4$

otherwise the direct sums and tensor products do not have the same dimension and this makes no sense at all

also, in what basis are you writing that down?

Silly Goose

why do direct sums of matrices of different dimensions make no sense?

ACuriousMind

08:58

what are these blocks relative to

Silly Goose

eveerything is in the spin-z basis

123

@ACuriousMind Does solutions of F = ma give us the position of velocity , acceleration vector? Is it applicable on every possible motion?

Silly Goose

I am interested in the more general case $U_1 \oplus U_2$ where $U_1 \in SU(2)$ and $U_2 \in SU(2^{n-1})$ where $U_1$ is $2 \times 2$ and $U_2$ is $2^{n-1} \times 2^{n-1}$

ACuriousMind

@SillyGoose If you have the case of SU(3), then the tensor product $H\otimes H$ of the 3-dim fundamental is 9-dimensional, but the direct sum is 6-dimensional

Silly Goose

which is a matrix on $2^n$ dimensional hilbert space

ACuriousMind

08:59

so $U_1\otimes U_2$ and $U_1\oplus U_2$ make no sense on the same space

@123 of course, that's what the equation of motion is for!

Amit

@123 choose some origin

Silly Goose

hm wait

ACuriousMind

@SillyGoose No, $U_1\oplus U_2$ is a $2 + 2^{n-1} \neq 2^n$ matrix

Silly Goose

i see

123

@ACuriousMind I will definitely try F = ma for projectile motion and in cartesian system. and circular motion in polar coordinate system. To see does these solution gives me the positions of velocity and acceleration vector.

ACuriousMind

09:02

I don't know what you mean by "try"

how did you get the trajectories thus far if not by solving the equations of motion???

Silly Goose

okay hm... so actually i should consider $U_1 \oplus U_2$ where $U_2$ is a $2^{n}-2$ square matrix...

123

@ACuriousMind Try means solution on paper first then simulate using these. It gives me more power in simulation.

ACuriousMind

@SillyGoose Why are you "considering" these direct sums to begin with

@123 I meant that I have no idea how you have been solving anything so far if you don't know that solving $F=ma$ is the equation of motion...

Silly Goose

i am trying to figure out the $SU(2^n)$ stabilizer subgroup of a particular initial state

i think i wrote it down concretely but i was trying to figure it out on a more abstract level

123

@ACuriousMind I found trajectories using formulas. I shared you my calculation picture for projectile motion. Pls see and guide me what is the problem in that

ACuriousMind

09:05

@SillyGoose yeah but I have the feeling that something went wrong during your abstraction :P

Silly Goose

oh yes

ACuriousMind

@123 and where did these formulas come from?

Silly Goose

xD

123

@ACuriousMind Pls see the formula in pictures for cartesian system.

Silly Goose

but in the more concrete analysis, it was just looking for $U \in SU(2^n)$ such that $[U, \rho] = 0$. Which for this particular $\rho$ is all block diagonal matrices ($2 \times 2$) $\oplus$ ($2^{n-2} \times 2^{n-2}$) as described above.

ACuriousMind

09:06

@123 yeah, that's what I mean: where did you get e.g. $x(t) = v_0 gt - 1/2 gt^2$ from?

123

I have calculated these formula from textbooks and on my own.

ACuriousMind

I mean if you're just using the trajectories you find somewhere instead of understanding that they generally are solutions to the differential equation $F=ma$, then this explains a lot of the strange communication problems you've been having

123

@ACuriousMind This comes from second equation of motion. $s = V_i t + \frac{1}{2} a t^2$

Silly Goose

which i thought should then be the group $U(2) \times U(2^n -2) \cap SU(2^n)$

but then this is the confusion about what $\times$ of groups really means and if it can represent direct sums of group elements as well as tensor products of group elements in different scenarios

because ultimately i wanted to look at the Lie algebra of this Lie group, so i thought this would be a way there. but the concrete way got there anyways. but doing it more abstractly would still be a good check i think

123

@ACuriousMind Aaaaaaahhhh... I see. I must notice this. I know how to do solution using integral. I didn't worked on it intensely because i find it difficult to first know the solution.

ACuriousMind

09:09

@123 I see...you're trying to do classical mechanics without actually having learnt classical mechanics! You're trying to run before you can walk: If you do not understand that everything in classical mechanics comes down to solving $F=ma$ - and that this holds in every frame once you know about fictitous forces - then this explains why you're having so much trouble

Silly Goose

but i mean if you look in say Hall's book on Lie theory, it presents the direct product of groups $G \times H$ with the associated Lie algebra $\mathfrak{g} \oplus \mathfrak{h}$. and the book provides the particular representation of this Lie algebra $\pi(X, Y) = X \otimes I + I \otimes Y$, i.e. what you see in say combining two spin-1/2 particles. but this is not what i am interested in doing

123

@ACuriousMind But finding solutions of the differential equations is very complicated. Let me try this 1st in projectile motion. Can you pls help me how can i find solution of F = ma in circular motion. I know this formula $\vec{R}(t) = A cos(\omega t) + A sin(\omega t)$

ACuriousMind

sorry, you're really just asking me to teach you the entirely of classical mechanics, that's a bit too much to do in chat

Silly Goose

and then a "direct sum" of groups doesn't really seem to mean what i mean by it... based on the wikipedia page

naturallyInconsistent

@SillyGoose Identity matrices are diagonal, so that that form is necessarily block diagonal of sorts; it is not going to span all possible block diagonal forms, but it is sufficient to become block diagonal.

123

09:16

@ACuriousMind I will consult to the book. I have read these in books. But i just wanted to know. If i differentiate above position vector with respect to time. Then it will give me velocity vector magnitude and direction not position. This is the problem i wanted to know.

ACuriousMind

@123 I don't understand what you mean by "magnitude and direction not position"

Silly Goose

well i'd like to generate all possible block diagonal $2^n \times 2^n$ special unitary matrices. where the first block is dimension $2 \times 2$

ACuriousMind

the velocity vector $\dot{x}(t)$ is "attached" to the point $x(t)$ if that's what you mean by "position"

123

Let me explain. If i differentiate $\vec{R}(t)$ it will give me velocity magnitude and direction of velocity vector. But this velocity vector have position tangent to the curve. This position is not mentioned in this type of formulation.

Silly Goose

these matrices generically look like $e^{-i\varphi}M_1 \oplus e^{i\varphi}M_2$ where $M_1$ and $M_2$ are special unitaries of the proper dimension. So then I was just trying to figure out what this subgroup of $SU(2^n)$ is. Which it msut be a subgroup because it is a stabilizer

123

09:19

@ACuriousMind Yes. But differentiating position vector does not give me position where particle moving.

ACuriousMind

@SillyGoose you already said the right thing, this subgroup is just $\mathrm{SU}(2)\times \mathrm{SU}(2^n -2)$

Silly Goose

oh

ACuriousMind

@123 but the position is $x(t)$! I don't understand what information you think is missing.

Silly Goose

@ACuriousMind but then in this context writing down what representation one means is important right?

123

@ACuriousMind Let me share you the picture

Silly Goose

09:21

because one might take this to mean operators of the form $U_1 \otimes U_2$ instead of, correctly, $U_1 \oplus U_2$

123

ACuriousMind

@SillyGoose I mean, $U_1\otimes U_2$ makes no sense in this context since that would have dimension $2\cdot (2^n - 2) = 2^{n+1} - 4 \neq 2^n$

Silly Goose

hm i see. so if we make clear that we are representing $SU(N)$ as unitary acting on $N$ dimensional hilbert space, i suppsoe it is unambiguous

123

This is circular motion. The vector u is velocity vector sit at origin O. which i calculated using differentiating position vector $\vec{R}(t) = A cos(\omega t) + A sin(\omega t)$

ACuriousMind

@123 I mean, that's just you choosing to draw the vector at the origin and not at the point $R$. That's a problem of your drawing, not of physics.

123

09:24

The u velocity vector is at origin , it should have position where object moving.

naturallyInconsistent

@123 The velocity vector is supposed to start at R, not at O

123

@naturallyInconsistent Yes exactly...

velocity vector should maintain its position at R using physics formula. This is what i am saying position

ACuriousMind

but you obviously know where the point R is - you drew it! Why do you not just draw the velocity vector as starting there?

naturallyInconsistent

If you study this in even greater detail, the vector space of velocities is NOT the same vector space as Cartesian position vectors. The origin in the vector space of velocities is NOT the same as the origin in your Cartesian position vectors. You can just move the origin of the velocities to where R ends, and that will be what we mean.

ACuriousMind

in formulae, the derivative $\dot{x}(t_0)$ of some $\vec x(t)$ is always attached to the point $\vec x(t_0)$

there really is not much of a mystery here

123

09:29

I always shift velocity vector to position R. It is very tedious task always. But physics formulation does not take care of the position. Do you know does this support by physics or not

ACuriousMind

(but yes one can take this as a starting point of a discussion of tangent vectors and whatnot, but that would be overkill)

@123 there is a lot of formal mathematics one can do here to formulate properly what is happening with the velocity vector

the "correct" solution involves switching to a differential geometry viewpoint

but this would be extreme overkill for the level of mathematical sophistication you're at

at your level, I think

2 mins ago, by ACuriousMind

in formulae, the derivative $\dot{x}(t_0)$ of some $\vec x(t)$ is always attached to the point $\vec x(t_0)$

is really all you need ot know

123

@ACuriousMind It means it is possible. I have tried this using matrices.

naturallyInconsistent

We are trying to tell you that, essentially, what you have learnt is already correct; you are complaining about a tediousness when, actually, we are already protecting you from a lot more nonsense that is beyond your current abilities to handle. Do not ask for even more than that! It is so easy to just shift the vector over!

123

@naturallyInconsistent Thanks for notifying good point.

If possible i wanted to know in conical pendulum at origin B we have two vectors perpendicular. Radial angular momentum vector $L_r$ constant in magnitude changing direction. and Tangential torque vector $\tau_{B}$ . Is it possible to understand the motion of conical pendulum using these two vectors?

Relativisticcucumber

@ACuriousMind i swear i posted a reply to this but i cant find it? am i dreaming

123

09:37

Relativisticcucumber

@ACuriousMind ok my reply was like i have seen the form of solving for stationary points of action where we use E-L equations to get EOM, but this method of varying the action has always been rather confusing to me. namely because it relies on $\delta$ which i have never been able to understand how to use and [...]

[...] havent been able to find a suitable resource on. thus im rather confused about what is meant by "the Einstein-Hilbert action $S_{EH}[g]$ is an action on the space of all metrics on a manifold, the dynamical variable is the metric itself". so in the first case of wanting to solve for the path of noninteracting and free particles, i just took the given lagrangian for granted and used that to do the normal E-L analysis, [...]

[...] so i thought in that case the dynamical variable would be proper time, but i guess i am misunderstanding something.

and sorry if this is replying twice and im blind. feel free to ignore.

ACuriousMind

@Relativisticcucumber I'm not quite sure what you mean because when you derived why the E-L equations are stationary points you should have also seen some "$\delta$s"

the proper mathematical context here is calculus of variations btw

Relativisticcucumber

okay so you mean the method of varying the action is the same thing as E-L formalism?

ACuriousMind

yes!

you derive the E-L equations by varying the action and assuming the action has the specific form of being the integral of a Lagrangian

Relativisticcucumber

okay i will look into the derivation for E-L again and then i think it will become more clear i hope

ACuriousMind

09:50

and by "dynamical variable" I mean "the variable we're trying to solve for"

Relativisticcucumber

@ACuriousMind oh

okay i see

thanks

ACuriousMind

which is "the path of the particle" in the geodesic action and "the metric" in the E-H action

Relativisticcucumber

okay okay

Relativisticcucumber

10:14

@ACuriousMind okay so is it like we express a lagrangian as $\mathcal{L}(q,f(q),\dot{f(q)})$ which resembles the general format $\mathcal{L}$(parameter, dynamical variable, time derivative of dynamical variable)and so in the first case i mentioned, we have $\mathcal{L}(\tau, x(\tau), \dot{x(\tau)})$ and for the second, we have $\mathcal{L}(\tau,g,\dot{g})$ and the variational principle says that when we vary the action wrt the dynamical variable, the quantity vanishes? [...]

[...] i still dont quite understand the E&M analogy given above but i guess i do see why these would be two different actions generally speaking.

im not sure $g$ should be a function of $\tau$ though, so im a bit hesitant on that front

ACuriousMind

10:29

@Relativisticcucumber it's a function of $x$ (point in spacetime), not $\tau$

Relativisticcucumber

okay that makes more sense.

ACuriousMind

you can do everything you do with "non-field Lagrangians" $L(t,f(t),\dot{f}(t))$ also for "field Lagrangians" where the $f$ depends not on one-dimensional time but multi-dimensional spacetime

Relativisticcucumber

i see

tangential to that discussion, i have one more question. related to something slereah said yesterday, carroll says "the interactions of matter fields to curvature are minimal: they do not involve direct couplings to the riemann tensor or its contractions". we see in the einstein equations, though, that the stress energy tensor is related to the curvature, so what does he mean by "interactions of matter fields to curvature"?

ACuriousMind

@Relativisticcucumber The point is that the interaction terms with the metric/curvature in the action are purely through covariant derivatives, not some sort of term that would involve a product/contraction of the Riemann tensor with some matter fields

Relativisticcucumber

okay and is this physically significant?

ACuriousMind

10:42

it's interesting because it gives a very general pattern for how to couple arbitrary matter fields to this theory - just replace all derivatives by covariant derivatives in the original matter Lagrangian, add it to the E-H action, done

Relativisticcucumber

@ACuriousMind ack wait so you said that the Einstein-Hilbert action is on the space of all metrics, and there is no particles involved. so could we say that the einstein equations are the "equations of motion" that we use to solve for a metric? i think im muddling concepts because you say there is no particles, but if there is no matter at all, wont the stress-energy tensor be zero?

i thought the point of einstein equations is that they relate curvature and mass/energy so we usually input a metric?

ACuriousMind

@Relativisticcucumber well you can add stuff to it! The E-H action $S_\text{EH}[g]$ is an action for the metric. Now you have some matter action $S_M[\phi]$. You replace all derivatives by covariant derivatives to get some $S'_M[\phi,g]$ (this depends now on the metric because the covariant derivative depends on the metric), then you get a joint action $S[\phi,g] = S_\text{EH}[g] + S'_M[\phi,g]$

the stress energy in the EFE is precisely related to the matter part $S'_M$ there

when there's no $S'_M$, then you get the vacuum EFE with $T=0$

Relativisticcucumber

:ooooooo

okay great that makes sense

er wait one more thing -- so this means there are metrics that minimize action? so is it possible to use this to find metrics? im kind of confused how we could do this. maybe we could input T?

but prob impossible to solve. ok but still hypothetically if we solve EFE, we get metrics that are stationary points of action?

ACuriousMind

@Relativisticcucumber Yes, the solutions to the EFE minimize the corresponding action

Relativisticcucumber

10:57

interesting. i assume in practice its impossible to solve the EFEs though?

ACuriousMind

depends on what you mean by "impossible" :P

plenty of known vacuum solutions

123

Is it always the case direction of angular momentum is the direction of axis of rotation?

What happens if direction of angular momentum not in the direction of axis of rotation?

ACuriousMind

also plenty of work on numerically solving the EFE

@123 what do you mean by "axis of rotation" if not the axis of angular velocity/momentum?

123

@ACuriousMind angular momentum. I am confused about the directions and their role in different directions of rotational parameters. Like angular velocity, angular momentum, torque.

Slereah

one way you can see the minimization of the action at work is if you consider some very restricted classes of metrics

ACuriousMind

11:05

@123 I understand you are confused but I don't understand what you are confused about. you did not answer my question: What is your definition of "axis of rotation" and how is it different from "axis of angular velocity"?

Slereah

For instance if you consider only the class of maximally symmetric metrics, where the curvature is constant, then the set of all metrics in that class is one dimensional

Only parametrized by the curvature constant

In which case you can just treat the variation problem as just an ODE

the so-called minisuperspace method

123

@ACuriousMind Axis of rotation is a straight line through all fixed points of a rotating rigid body around which all other points of the body moves in circle.

I have read direction of angular velocity $\omega$ is in the direction of axis of rotation. Is it always the case.

Slereah

So if you have your maximally symmetric metric $g_R$; with a one dimensional configuration space in $R$, then your action will be something like $$S = \int_U R \sqrt{-g_R} d^n x$$

So your action is just $$S = R \mathrm{Vol}_{g_R}(U)$$

Your extremizing metric will be the value of $R$ at which the volume of some region of space is extremized

Which you can show is flat space

What's the volume of a ball in some space of arbitrary constant curvature, I forget

Relativisticcucumber

@Slereah interesting -- why do we need a maximally symmetric metric for this?

Slereah

Scalar curvature

In the mathematical field of Riemannian geometry, the scalar curvature (or the Ricci scalar) is a measure of the curvature of a Riemannian manifold. To each point on a Riemannian manifold, it assigns a single real number determined by the geometry of the metric near that point. It is defined by a complicated explicit formula in terms of partial derivatives of the metric components, although it is also characterized by the volume of infinitesimally small geodesic balls. In the context of the differential geometry of surfaces, the scalar curvature is twice the Gaussian curvature, and complete...

@Relativisticcucumber You don't need to, it's just a very nice model to work with

Since it's one dimensional

As opposed to infinite dimensional

ACuriousMind

11:11

@123 My point is that this question really only makes sense if you have a definition of "direction of angular velocity" that's not just the same as "axis of rotation"

123

Means direction of angular velocity is normal to the instantaneous plane of rotation. There are so many terms in rotational dynamics. Center of rotation, plane of rotation, axis of rotation.

Relativisticcucumber

@Slereah ok i see hm interesting

Slereah

The volume of a ball in a space of constant curvature like that is about $\mathrm{Vol}_R(B_r) \approx \mathrm{Vol}_0 (1 + \frac{R}{6(n+2)} r^2)$

123

@ACuriousMind By definition direction of angular velocity is normal to the instantaneous plane of rotation or angular displacement.

Slereah

So the action in that case will be $$S \approx \mathrm{Vol}_0 (R + \frac{R^2}{6(n+2)} r^2)$$

ACuriousMind

11:15

@123 that's not a definition of angular velocity

that's just some property it has

what is your definition of angular velocity?

123

@ACuriousMind rate at which angular displacement changes with respect to time is angular velocity

Slereah

Hm

Am I doing this right

Does this extremize at $R = 0$

ACuriousMind

@123 and what problem do you have in determining whether or not that is the same as the "axis of rotation"?

you have a definition of "axis of rotation", you have a definition of "angular velocity", you can just apply them and look at whether they're the same

123

@ACuriousMind My question about direction of angular momentum not about direction of angular velocity.

ACuriousMind

also angular velocity is a vector so your "rate at which" definition is still incomplete; it's no wonder you keep getting confused about all the terms if you don't straighten out what the terms are

@123 but you also have a definition of "angular momentum", don't you?

123

11:21

@ACuriousMind Yes i have

ACuriousMind

your question is extremely straightforward to answer if you just write down all the quantities involved: The axis of rotation is parallel to angular velocity $\vec \omega$. Velocity and momentum are related by $\vec L = I\vec \omega$, where $I$ is the inertia tensor. So this equation directly gives you the necessary and sufficient conditions on $\vec \omega$ and $I$ for $\vec \omega$ and $\vec L$ to be parallel.

it's just a bit of linear algebra

123

@ACuriousMind Yes i know this rule. I am confused because i have seen the situation where direction of angular velocity and direction of angular momentum are different. as in the case of gyroscope.

ACuriousMind

@123 and why are you confused if you know this rule?

according to $\vec L = I\vec \omega$, the two will not in general be parallel when $I$ is not a multiple of the identity matrix

again, that's just straightforward linear algebra; the problem here is not some physical interpretation or whatever, you can read off directly from this equation that $\vec L$ and $\vec \omega$ will not point in the same direction in general

123

You can see my conical pendulum example. In this example direction of angular velocity and direction of angular momentum and direction of torque all are directed differently.

ACuriousMind

@123 of course, because torque is again only related to $\vec \omega$ via $\vec \tau = I\dot{\vec\omega}$

there's no reason anyone would expect any of these three quantities (torque, ang. velocity, ang. momentum) to be aligned for general $I$ just from looking at the defining equations

so I don't understand where the confusion here comes from or what the confusion even is

you seem to think we should expect them to point in the same direction, but there is simply no reason to expect that

123

11:27

@ACuriousMind I see.. This is the point. Here rotational inertia changing direction

@ACuriousMind This is what i think all these should be parallel. But when i saw they directed differently i got confused and have no rule to understand and calculate them

ACuriousMind

yeah, but why did you think they should be parallel? again, the fundamental equations are $\tau = I\dot{\omega}$ and $L=I\omega$ and none of these imply they should be parallel!

123

Pls explain my picture what is the role of $L_r$ radial ang. momentum and torque at origin $\tau_{B}$

ACuriousMind

no, I don't know what you mean by "role" and I don't know why you refuse to say why you think they should be parallel

we can't help you if you don't tell us what the (wrong) reason was you thought they should be parallel so that we can correct the underlying misconception; if you don't have a reason then you need to rethink your approach to physics entirely :P

123

@ACuriousMind Because i was not notified rotational inertia is tensor not a scalar. I considered rotational inertia as just scalar. then scalar multiple of vector is vector and direction is same. As in F = ma , direction of F and a is parallel. I was doing this way before you realized me rotational inertia is tensor.

ACuriousMind

@123 okay so just like with linear kinematics the problem here is just that you keep trying to do very general rotational kinematics when you only know a specific simplification

what you need to do is pick up a mechanics text that treats rotational kinematics in full generality and with an inertia tensor

talking about some specific pictures in this chat won't teach you kinematics

and we really can't replace an entire course/textbook for you

123

11:36

@ACuriousMind But textbooks and google won't address this kind of informations.

Slereah

looks like the full expression of the volume in constant curvature is

ACuriousMind

@123 I'm sorry, but that's just not true: I know for a fact that there are many mechanics textbooks that treat both linear and rotational kinematic in the generality I'm talking about.

that there are no textbooks that teach you about the inertia tensor is just a ridiculous claim

123

@ACuriousMind Can you pls share book name if you remember.

ACuriousMind

Jul 7 at 14:45, by 123

@Amit Yes i am reading. Kleppner kolenkow, symon mechanics , Tylor mechanics and others

123

In my picture we treat bob as point particle. There is no problem with the rotational inertia of the point particle. We only changed origin.

ACuriousMind

11:39

according to you, you're already reading Kleppner & Kolenkow

but now I find that hard to believe because they certainly introduce the inertia tensor in chapter 7

123

@ACuriousMind Yes the picture and simulation i created which is from K&K

I am at chapter 6. This is the picture from chap-6

ACuriousMind

yeah, so you need to just read the full book before you claim the book isn't teaching you anything :P

almost all physics texts always go from the simple to the more complicated situations

but you seem to just have read the part about the simple situations and then you try to apply it to the complicated situations

that's not how it works

but it explains why you're having so much trouble

123

Ooookay.. I have read the whole book 2 times. First time i have read without solution of example problem. 2nd time more carefully. Now i am solving examples.

@ACuriousMind You are right.

If i treat pendulum bob as point particle there should be no problem with the rotational inertia. We just changed the origin from A to B

ACM give me your brain so i can save my time and energy.

I will definitely read the chapter-7 more carefully this time. Thank you very much i learned a lot from all of you. And you notified my mistakes and important points to think.

Slereah

@ACuriousMind When mathematicians say "Riemann surface" am I being tricked into thinking they mean "a Riemannian manifold in two dimensions" but they actually mean "A one dimensional complex manifold"

ACuriousMind

11:56

@Slereah Yes

Slereah

Dagnabbit

ACuriousMind

that is exactly what's going on!

Slereah

What's the word math people use to talk about the moduli space of a regular old surface

ACuriousMind

what exactly do you mean by "moduli space" there?

Slereah

The moduli space of Riemannian metrics

ACuriousMind

11:57

that's one of these words I hate because everyone "knows" what they mean but there's like a dozen different definitions

Slereah

I mean it in the physicist sense of the SUPERSPACE

Metrics up to diffeomorphisms

Conformally equivalent metrics are not counted as the same here

ACuriousMind

I'm not sure there's a name for that space other than "space of Riemannian metrics"

Slereah

Hopefully google will not think I mean a Riemannian surface because I see that gag coming

ACuriousMind

also this sounds like one of these things where you quickly run into nLab-style generalized spaces because I don't expect this "space of metrics" to be a nice finite-dimensional space in any sense in general

Slereah

according to nlab it is the concretification of the internal hom of the space of metrics, if it helps

ACuriousMind

12:00

called it

Slereah

@ACuriousMind It is very much not

It's not even like locally Euclidian

because of the weird behaviour near isometries

Although fortunately I am reading Feyenabend currently so with some luck I will have realized the foolishness of science soon enough

He is claiming that science and magic are about as equivalently valid

although I suspect him of being a little bit of a provocateur

Making faces at your empiricism

ACuriousMind

"Feyerabend" is always a very funny name to me because in German it means both the time when one stops working and the time spent off-work in general

Slereah

Is that his real name or is it a cool nickname

ACuriousMind

so using Feyerabend's work to get out of science work is rather fitting

Slereah

Mr. Lazy Bone

ACuriousMind

12:03

@Slereah I assume it's a real name because the modern German spelling is Feierabend

Amit

Feyerabend may have said some provocative things mainly to be noticed

Slereah

I mean he makes some good points but for the most part they're not that new ideas

ACuriousMind

the word used to mean "the evening before a holiday" (where holiday=holy day, really), so people born on the evening before a holiday probably got that name

Slereah

Same method that got us the name Sabbatai

"Scientists don't actually conform to the scientific method" and "We can't separate scientific measurements from the theory that we have" isn't really a brand new idea

ACuriousMind

still needs to said every once in a while so the nerds don't get too cocky

Slereah

12:07

still plenty of neat things he does bring up though

Amit

It's hard to stand out in the shadow of Popper, he was his student

Slereah

apparently the old thermodynamic theory with Clausius entropy and such is in contradiction with the better statistical mechanics theory (obviously), but the actual proof of why one theory work and not the other could not actually be proven in the old theory

ACuriousMind

I'm not surprised thermodynamics is also a philosophical nightmare :P

Slereah

For instance if you consider Brownian motion as a counterexample, unless you can make some unrealistically precise measurements, you could never actually prove that Brownian motion violates the laws of thermodynamics

So that if you had Brownian motion without the framework of statistical mechanics, it probably wouldn't be able to disprove it

Claiming that without statistical mechanics we would just consider brownian motion as just a weird behaviour unrelated to thermodynamics

Amit

This sounds like a direct attack on Popper, that new theories don't always arise from refutations of the old ones, he may be correct

Slereah

12:14

Not really I guess?

Amit

Idk the details well enough

Slereah

Popper would basically say the same thing, you have a given paradigm and all the facts are interpreted within that paradigm

Amit

But it is imaginable, some model may simply work better, maybe easier to calculate with, even before you find the refutation of the old one

Slereah

Popper does bring that up

Amit

Oh okay

Slereah

12:16

Saying that such ideas do occasionally show up, but if there are no crisis going on in science, they're unlikely to make any waves

Amit

I see

Slereah

I think basically all philosophers of science do say "Actually real science doesn't operate by any particular method" but Feyerabend says "This is good actually"

Amit

I heard that he encouraged some of his students to "go back" to magic?

Slereah

That would be in line with the book certainly

I guess you should at least attempt it, in the name of empiricism

Check the stars before doing an experiment

Observe flocks of birds

Read the entrails

Amit

Lols, some people get rubbed the wrong way by that. I mean I think he may have had good intentions, but this kind of stuff makes it easy to misread him/twist his ideas

Slereah

12:26

I think it mostly makes it easier to dismiss him as just being a contrarian

but then again it was the 70's

Amit

That's the other side of it yeah

Slereah

That was the kind of vibe people had back then

Amit

Lol, drugs and anything goes

Slereah

You just had to freak out the squares

Amit

youtu.be/8GrVlLYgeZ8 -- right at the beginning of this interview, he admits he chose to express himself in "as extreme as possible" manner

Slereah

12:31

yeah you can tell :p

@ACuriousMind Did people try to freak out the squares in the 70's in Germany

Or did they stick with their German efficiency

Amit

Lol, I suppose jacobians are for freaking out the squares

Also it depends whether you're talking east or west Germany, maybe

ACuriousMind

@Slereah you mean the 1968ers?

it was a pretty rough time as far as I understand

over the 70s we even had several kidnappings and assassinations of prominent politicians and industrialists by the RAF

(this is all West Germany)

Slereah

We also has the 1968ers in France :p

also some terrorism but that was mostly connected to the Algerian war I think

Relativisticcucumber

12:59

why is it valid to say that $R$ vanishes in a vacuum? i know that if the stress energy tensor vanishes, then the related curvature should vanish, but why can there not be some background energy and/or curvature? since gravitational waves can travel in vacuum doesnt that mean there is something there?

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