The h Bar

user228700

9:00 AM

Okay so assuming that the box is at equilibrium at every minute step of this process, the spring force would equal the force applied by and that's how you wrote that integral.

user228700

But I don't understand what it means when we say "the spring does negative of the work done by us."

John Rennie

Work is the transfer of energy

user228700

Okay...

John Rennie

Support I lift you up then I have increased your (potential) energy by doing work on you.

But when I apply a force on you you apply an equal and opposite force on me

So that means you are also doing work on me

user228700

Riight. But am I changing your mechanical energy?

John Rennie

9:04 AM

If we define the work I do on you as positive that means the work you do on me is negative, because the forces are equal and opposite.

@KaumudiHarikumar don't worry about that for the moment.

The point is that the sign of work is just a convention.

user228700

@JohnRennie Okay, for the moment.

user228700

@JohnRennie Okay, I get that. But why? Why the convention?

John Rennie

> Why the convention?

The energy of an object can increase or decrease

Secret

Naively speaking, suppose we have a cannonball sized object that is indivisible (hence the whole cannonball has to be treated as basically a point particle), then the fact it is staying put on the floor in some reference frame means its momentum is well defined and is zero in that frame. Meanwhile, there is no fuzziness or uncertainty in where it is in space (because it just sat there on the floor), thus the uncertainty in position is also zero.
Therefore uncertainty principle is violated by such cannonball, which means the macroscopic object that obeys uncertainty principle as requested by…

John Rennie

If we define positive work as increasing the energy of an object then if we decrease the energy that would be negative work.

Negative work doesn't mean negative energy in some weird way, it just tells us which direction energy is flowing.

user228700

9:07 AM

Yeah, so when I'm doing negative work on you, I am decreasing my energy? Wait, what? I'm confused :/

John Rennie

No, if you're doing negative work on me you are decreasing my energy

If you are doing positive work on me you are increasing my energy

user228700

Okay, because your energy really is decreasing because you are increasing my energy?

John Rennie

Energy is conserved. If we are an isolated system and one of us increases in energy the other must decrease in energy

This is far simpler than you think. The phrase negative work confuses everyone because it sounds weird, but it isn't.

user228700

So I'm decreasing your energy while you're increasing mine, correct?

John Rennie

Yes

Go back to your spring, because that's a nice simple system.

user228700

9:10 AM

Okay, great.

John Rennie

Suppose we start with the spring compressed and the ball stationary, then we let go. The spring expands and increases the velocity (and therefore KE) of the ball.

So what happened to the energy of the ball?

user228700

It increased.

user228700

So the spring did work on it.

John Rennie

And what happened to the (potential) energy of the spring?

user228700

It decreased.

user228700

9:12 AM

Wait, quick question. This potential energy is the one given by $1/2kx^2$ no?

user228700

Okay, that didn't work. I'm sorry I'm so bad at MathJax. I'll eventually learn it.

John Rennie

@KaumudiHarikumar Yes. Though remember $x$ is the displacement from equilibrium. In this case you could argue that the length of the spring is increasing, but the distance of the spring from its equilibrium length is decreasing so $x$ is decreasing.

user228700

Yes, okay...

John Rennie

Anyhow, if you asked what work the ball did on the spring you'd have to say the ball did negative work on the spring.

user228700

I don't quite understand potential energy fully though. Is there something on the internet that I could read and learn this better?

user228700

9:16 AM

(Other than Wikipedia)

John Rennie

But all that means is that the force $\mathbf F$ and the displacement $d\mathbf x$ are in different directions so the product $\mathbf F \cdot d\mathbf x$ has a negative sign.

user228700

I mean, I can google, but do you know of anything that you've come across before?

user228700

@JohnRennie Okay...

John Rennie

I don't know of any articles specifically about potential energy, but it seems a simple enough concept.

One thing to note is that we can only ever measure changes in potential energy.

user228700

@JohnRennie ...which I have never really gotten my head around, sadly :-(

user228700

9:19 AM

@JohnRennie Okay..?

John Rennie

Well if I do some work $W$ in compressing a spring I know the potential energy of the spring must have increased by $W$.

i.e. the potential energy of the spring has increased by $W$

Potential energy has what we call a gauge freedom.

user228700

Okay, but...okay, if you were asked to define the potential energy related to any force, how would you..?

John Rennie

You can only ever define the change in potential energy caused by the force.

user228700

Okay..?

John Rennie

Suppose I'm on the ground floor of a building and you're on an upper floor 10m above me.

I'm going to say my potential energy is zero. That makes your potential energy $10mg$.

user228700

9:25 AM

Potential energy is defined like this in my textbook:

John Rennie

yes?

user228700

Sorry for interrupting. Give me one second. I'm terrible at taking photos :/

user228700

user228700

Sorry it took so much time! Like I said, terrible photographer right here :/

John Rennie

That's not really a definition is it? That's a lot of rather vague statements about potential energy.

user228700

9:32 AM

@JohnRennie EXACTLY.

John Rennie

If you have some system then there is an associated function $U$ that is a function of the configuration of your system.

user228700

Okay...

John Rennie

i.e. if the state of your system is defined by some variables, $q_1$, $q_2$, etc then $U(q_1, q_2)$ is this function.

This function $U$ has the property that if you do work $W$ on the system then $U$ increases by an amount $W$.

user228700

Okay, this is the definition. Okay...

John Rennie

And if the system does work $W$ on something else then $U$ decreases by that amount $W$.

user228700

9:35 AM

Okay.

John Rennie

This function $U$ is the potential energy.

user228700

Okay...Can you tell me the reason we defined it like this..?

John Rennie

Because that is the most general definition. It doesn't rely on any specific arrangement of springs or masses.

Probably the most important thing you need to know about $U$ is: $$\mathbf F = \nabla U$$

Secret

I think your book might have that "stuff rolling downhill" picture in mind (and also the work energy theorem) when they talk about potential energy, as when some object were being exerted by a force that is from the potential, it "rolls" to a configuration which has lower potential

Hmm...Johnrennie, am I making sense here in how I tried to understand how our usual definition of potential energy might coincide with the book?

user228700

@JohnRennie Didn't understand, I'm afraid...

user228700

9:38 AM

What are you doing to U?

user116211

Well, pedantically, $U$ is the work function while negative of work function is potential energy ie. $$V\equiv\textrm{potential energy}= -~U\,.$$

John Rennie

The symbol $\nabla U$ is called grad U.

user228700

@JohnRennie Okay..?

Secret

grad U measures how large and in what direction U is changing at a point

user116211

Anyways, @JohnRennie, I've something to ask you...

John Rennie

9:39 AM

See en.wikipedia.org/wiki/Del

Secret

It is a multivariable generalisation to the concept of slope

user116211

@KaumudiHarikumar spatial derivative of a scalar which turns out to be vector.

John Rennie

What I'm getting at is that if you want a precise definition of the potential energy it's actually quite technical.

user116211

Directional Derivative at the steepest direction.

John Rennie

Which is why we don't usually bother.

user228700

9:40 AM

@JohnRennie Oh, okay then...

user116211

@JohnRennie, i would be waiting for this conversion to get completed.

user228700

I'll understand it as you explained before going into this grad $U$ thing.

John Rennie

If we are working in one dimension $x$, e.g. like an extending spring, then the $\nabla$ operator is just $d/dx$.

user116211

Hey @yuggib.

user228700

@JohnRennie Okay...

John Rennie

9:42 AM

So we get: $$ F = - \frac{dU}{dx} $$

Secret

uh, you forgot the minus sign

John Rennie

Does this make sense (mathematically) so far?

user228700

@JohnRennie Okay. Wait, no negative sign? I've seen it with a negative sign before :/ Huh.

John Rennie

Oops, well spotted, it should have a negative sign.

user116211

@Secret There would be no negative sign provided $U$ is work function.

user228700

9:43 AM

@JohnRennie Okay. I mean, it sort of makes sense.

Secret

@MAFIA36790 Well, we tend to define potential instead of the work function, thus forces always point downhill

John Rennie

Now suppose we integrate both sides of this equation we get: $$\int_a^b Fdx = \Delta U $$

does that look familiar to you?

user228700

Yep...

user116211

@Secret sure.

John Rennie

... because $\int_a^b Fdx$ is just the work $W$ done in moving from $a$ to $b$

So this says the change in our potential energy is just equal to the work done. Which is where we came in.

user228700

9:47 AM

Yeah, ok...

John Rennie

The point of all this is that we define the PE in this slightly abstract way, and it works!

It behaves as we expect potential energy to behave.

user228700

Okay...

John Rennie

The function $U$ can have lots of different forms. For an electric field there is an electrical potential energy.

user228700

Yes...

John Rennie

For a complicated set of springs there is a potential energy that depends on the lengths of all the springs.

For a gravitational potential energy U depends on the positions of all the gravitating objects.

user228700

9:48 AM

Okay...

John Rennie

So the form of the potential energy can be very varied. However it is always true that $$F = - \frac{dU}{dx}$$

Secret

@johnrennie Is potential as "the tendency of an object to change its configuration to reduce it" a good word summary of the formula $F=-\nabla U$ or this is too vauge to capture the more advanced type of potential functions?

John Rennie

@Secret too vague. Vague statements like that are what confuse students in the first place.

Secret

ok

user228700

@JohnRennie Okay. Noted.

John Rennie

9:51 AM

So there you are, you now know how to define potential energy even if you're not entirely sure what the definition means :-)

user228700

@JohnRennie Yes, okay :-) Just one last doubt, if you don't mind...

John Rennie

Yes?

user228700

I am VERY confused about the work energy theorem :/

John Rennie

OK

user228700

Okay, so the problem is that on the website "Physics Classroom", this is the equation they gave:

user116211

9:54 AM

@KaumudiHarikumar You shouldn't be now that you know what potential is.

user228700

$W_C+W_NC+W_P=K_f-Ki$

user116211

@KaumudiHarikumar define the terms.

user228700

Where $W_C$=Work done on the body by the conservative forces, $W_{NC}$=Work done on the body by the non-conservative forces and $W_P$=Work done on the body by the pseudo forces.

user228700

And of course, $K_f-K_i$=Change in kinetic energy of the body.

user116211

@KaumudiHarikumar You should then use W_{NC} to get $W_{NC}\,.$

user116211

9:57 AM

@KaumudiHarikumar What's the thing that you didn't get?

yuggib

@MAFIA36790 hey

user228700

Okay, what I don't understand is, where does potential energy factor into all this? According to this equation, the net work done on the object changes the kinetic energy alone...

user116211

@yuggib: do you know anything about Nakahara M.: Geometry, Topology, Physics? It's a mathematical physics book ; that's only I know.

Secret

This is because $W_C=\int_a^bF dx = \Delta U$

That is, conservative forces can be derived from a potential

user116211

@KaumudiHarikumar I don't think you can as there is non-conservative force too.

yuggib

10:01 AM

@MAFIA36790 I don't know it

user116211

@yuggib hmm.

user228700

@MAFIA36790 I can't what?

yuggib

It seems to deal with those stringy/TQFT speculations

user116211

@yuggib yes, yes!

yuggib

I don't like those things

;-P

user116211

10:02 AM

@yuggib okay....

user228700

@Secret Okay..?

yuggib

More concretely, I also know very little about the whole topic

user116211

:32482122 I know that.

user228700

:32482122 Okay great, more misconceptions.

John Rennie

@MAFIA36790 sorry, I meant that for Kaumudi

user116211

10:03 AM

Those forces can be expressed with time-dependent work function.

John Rennie

@KaumudiHarikumar a typical example of a non-conservative force is viscous drag.

user228700

@JohnRennie Okay...

user116211

@yuggib sorry for asking then ;)

John Rennie

If you drop a ball in vaccum it accelerates so the change in KE is equal to the change in PE

If you drop a ball in water then it doesn't accelerate as fast because of the drag of the water so KE < PE.

user228700

Okay..?

John Rennie

10:05 AM

It looks as if energy has disappeared so we call the drag a non-conservative force.

user228700

Okay...

John Rennie

But what has actually happened is the missing energy has gone into increased kinetic energy of the water molecules.

user228700

Oh, right, and this sort of interaction doesn't happen when we're dealing with "conservative forces", yeah?

John Rennie

The trouble is that it's impossible to keep track of the umpteen gazillion water molecules the ball collides with, so we just say the energy has been turned into heat.

So a non-conservative force is actually conservative but (much) too complicated to keep track of in detailed way.

user228700

Okay..?

John Rennie

10:08 AM

But anyway, we started with $$W_C+W_{NC}+W_P=K_f-Ki$$ and I'm not sure where we got to ...

user228700

Yeah, that.

user116211

Anyways, I've postponed my study of Bourbaki in order to concentrate more on on-topics; I'm currently reading Linear Algebra by Hoffman and Kunze; this book is really an amazing piece of art @yuggib.

John Rennie

@KaumudiHarikumar Suppose we go back to my example of a ball falling in water

user228700

Sure...

John Rennie

Suppose the ball has a mass $m$ and falls a distance $h$ then the change in potential energy is $$\Delta U = mgh$$ yes?

Bernard Meurer

10:11 AM

@MAFIA36790 What's up?

user228700

Yes...

user116211

@BernardMeurer hey; studying....

Bernard Meurer

@JohnRennie How to use LaTeX on an email to a prof?

John Rennie

And the work done on the ball is: $$W_C = \Delta U = mgh$$ yes?

user228700

Yeah...

John Rennie

10:13 AM

So where has that work gone. If the ball were in vacuum then it would all go into the KE of the ball and we'd have: $$ W_C = \Delta U = mgh = KE_f - KE_i $$

But in water the ball collides with loads of water molecules and the ball does work on those water molecules. We can't keep track of each individual water molecule so we give up and just the ball does some work $W_{NC}$ on the water.

And now we get: $$ W_C = \Delta U = mgh = KE_f - KE_i + W_{NC} $$

user228700

Okay...

John Rennie

So that's how the potential energy relates to your equation.

user228700

Okay, now I understand...

John Rennie

I suspect you don't really, no-one does on their first attempt. Hopefully this will start to make sense after you've thought about it for a while ...

user228700

Yeah, I think I've got it now. I mean, I have to read this again but it's starting to make more sense now...

user228700

10:19 AM

I will definitely understand better now that I've understood the context, sort of...

John Rennie

All I can say is just keep working at it. All this stuff does eventually become instinctive with experience.

yuggib

@MAFIA36790 Yeah, Bourbaki can be read at a later stage

you'll appreciate it more

user228700

Okay. Thanks SO SO much sir. You have absolutely no idea how grateful I am :-)

user116211

@yuggib yeh, thanks for that; noted.

John Rennie

@KaumudiHarikumar you're welcome. I still have an hour free before I have to go work again, so I'll go answer some physics questions now :-)

user228700

10:21 AM

Also, before you leave, I just wanted to know how you, in your days, were able to understand all this so well...was it your profs. at school/university or books..?

user116211

@yuggib Presently, I'm feeling it kinda dull; but that's okay. The pace is slow; but I actually got to understand all the Criteria of Substitution ;)

John Rennie

@KaumudiHarikumar when I was your age I didn't understand it any better than you do. I only understand it better now because I've been using it for forty years.

So in forty years you'll understand it too - that's probably not very helpful :-)

user228700

@JohnRennie :P Okay. Thanks again sir! SO much.

user228700

Have a nice day!

Physics Meta

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I am very disappointed by my experiment on this site. Four days ago, I posted this question: Numerical simulation of the double-slit experiment including watching the electrons . After less that 24 hours, it was put [on hold] by 5 persons as being unclear. So I edited my question a first time in...

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discussion support

user116211

10:29 AM

@JohnRennie: Well, can I ask you now?

John Rennie

Yes, sorry, your question kind of got lost in the discussion about potential energy ...

Xasel

uh..I should wait then :0

user116211

So, $$\overline{\mathrm dw}= \mathrm dU$$ where $U$ is the work function and $\overline{\mathrm dw}$ is the work.

user116211

This means

user116211

$$\sum_i \mathbf F_i\cdot \mathrm dq_i = \sum_i \frac{\partial U}{\partial q_i}~\mathrm dq_i\,.$$

user116211

10:34 AM

Here, $U$ is taken to be time-independent.

user116211

Now, when $U$ depends explicitly on time $t,$ the above relation still holds, but...

user116211

Well, here Lanczos states:

user116211

> The above equation still holds, with the understanding that in forming the differential $\mathrm dU, ~t$ is considered as constant.

user116211

@JohnRennie: I didn't get what he meant by $t$ has to be considered constant. I mean if you want to write the differential of $U,$ how could you take $t$ constant?

user116211

Also, at another point he writes,

user116211

10:40 AM

> [...] a generalised force may have no work function, but still satisfy the law of conservation of energy, as for example the force which maintains the rolling.

user116211

@JohnRennie: I didn't get that also how without any work function, the force satisfies the conservation law. Also, in rolling, there is friction responsible for maintaining the rolling in some cases. Does that mean even friction maintains the conservation law? Hmmm.

John Rennie

I think that just means the equation becomes: $$ \sum_i \mathbf F(t)_i\cdot \mathrm dq_i = \sum_i \frac{\partial U(t)}{\partial q_i}~\mathrm dq_i $$

Both the force and the work function become time dependent, but at any given time $t$ the above equation still applies.

user116211

In Cartesian coordinates, the differential can be written as $$\mathrm dU=\mathbb{grad}(U(t))~\mathrm dx + \frac{\partial U}{\partial t}~\mathrm dt\,.$$

user116211

So, does that mean he is telling us to discard the last term? Why?

John Rennie

If you have $n$ parameters $q$ then your configuration space has dimension $n+1$

user116211

10:52 AM

@JohnRennie you are counting $t$ too?

John Rennie

@MAFIA36790 yes. You have $n$ dimensions for the parameters $q$ and one for time.

user116211

okay, then.

John Rennie

I would guess Lanczos just means you can foliate this space into $n$ dimensional subspaces of constant $t$

yuggib

@MAFIA36790 the foundational part has to be slow to be digested well...

John Rennie

Within each subspace dU/dt is zero because we've defined that subspace to have constant $t$.

To be honest I'm guessing because I haven't read Lanczos' book, but that seems a reasonable guess.

yuggib

10:57 AM

anyways, if you want a different approach on logic theories you should look for model theory (Chang-Keisler is a very nice book); and if you want to know set theory as it is nowadays formulated you should go for Jech's (huge) book (it is more fast-paced, but not exactly "fun" ;-P).

user116211

@yuggib sure; I've some treatises on First-Order Logic; let me complete them first; I wonder why they don't teach logic at the early undergrad classes ;/

user116211

@yuggib I'm currently following Jech too. It's intuitive till now.

yuggib

They don't because it is a discipline that developed later with respect to others

(at least with modern standards of rigour)

peterh

@JohnRennie I have a... layman's idea :-) Multiple time dimensions with non-deterministic time evolution (like QM), but with consistency constraints (i.e. time evolution is non-deterministic on t1 and on t2, but t1+t2 has to be the same as t2+t1). Do you know anything similar to this? The wikipedia is very cloudy in this sense.

user116211

@yuggib ohh.

Secret

11:06 AM

@peterh So you are basically saying that the state vector still evolves deterministically under t1 and t2, but the outcomes are probablistic like in QM?

user116211

@JohnRennie okay, even if I take this; what is the point of doing this?

John Rennie

I don't know. I would have to read the book to see what Lanczos is getting at. Sorry :-(

user116211

@JohnRennie okay. Should I ask it? Well, I have to read between the lines again before doing that....

user116211

Also, @JohnRennie

user116211

34 mins ago, by MAFIA36790

@JohnRennie: I didn't get that also how without any work function, the force satisfies the conservation law. Also, in rolling, there is friction responsible for maintaining the rolling in some cases. Does that mean even friction maintains the conservation law? Hmmm.

user116211

11:18 AM

I couldn't get the rolling example...

ACuriousMind

11:49 AM

@MAFIA36790 In pure rolling, all that the rolling friction does is continue to turn the wheel, whose velocity and angular velocity remain constant. This is because energy would only be dissipated by "ordinary" friction if the point of contact had non-zero velocity with the floor, which it by definition hasn't in pure rolling.

2

user116211

@ACuriousMind got that. I just could not remember in time that friction, in facts helps to maintain the rolling motion which commences when the point of contact has zero velocity.

Shing

um... not been here for a while... mind if I asked why Chris White seemed to be gone?

user116211

2 days ago, by ACuriousMind

@EmilioPisanty I'm pretty sure it's a (over?)-reaction to him getting banned on chat.

user116211

@Shing Let's just don't discuss about him; it wouldn't make him come back...

ACuriousMind

@Shing See here and here for some relevant chat parts and this message asking to avoid further speculation.

Shing

12:01 PM

okay, thanks guys

peterh

@Secret Yes, although I am not sure that the state could be best described by a vector. I would say, the state could evolve, but not so strongly non-deterministically, as in the QM. Only the summed non-determinism would result the probabilistic outcomes of the QM.

user116211

We're sorry you're disappointed with our service; your refund is being processed now. — Alfred Centauri 8 mins ago

user116211

Oh man ;P

John Rennie

@AlfredCentauri: I think your comment goes a bit far and you should delete it.

ACuriousMind

@JohnRennie Just flag comments you think are over the line, there's no need to tip-toe around comments

Enough flags will get rid of a comment even without moderator action

John Rennie

12:16 PM

@ACuriousMind I have done. But it Alfred sees this and deletes the comment that is all to the good.

ACuriousMind

12:26 PM

Hmmm, math.SE is so popular among people wanting to get their work done for them that they are explicitly on the lookout for questions from active contests oO

user116211

@ACuriousMind This is one of their weird common practices.

ACuriousMind

I find this very strange indeed: They help people with their homework questions not caring whether they are helping cheaters except when they are helping to cheat at a contest? How is helping to cheat on a homework assignment or an exam morally any different from helping to cheat on a contest?

That strikes me as even more of a mess than our homework policy :P

user116211

@ACuriousMind I never actually got what their policy to blunt HW questions actually is.

user116211

Most of the time, there is a sort of race who would post an answer first; rarely though sometimes homework questions are closed there on the grounds of research efforts.

user116211

But only God knows when that happens.

ACuriousMind

12:38 PM

I think it's...none? They have a policy (and a pre-generated close message) about giving context/motivation, but due to the high question volume it gets applied very inconsistently since not many users see any given question and the opinions about what constitutes "sufficient context/motivation" appear to vary greatly.

Loong

And they do this: math.stackexchange.com/…

I need 73000 NAA flags.

ACuriousMind

What the...? I've seen some of those answers but I didn't realize it was a pandemic

user116211

@Loong ;P

user116211

They are really weird like Stephen King's Haunted hotels.

ACuriousMind

That's a really weird comparison in itself :P

user116211

12:46 PM

Well, lately I was reading The Shinning....

user116211

Oh, yesterday, it was his birthday also, I see....

0celo7

@ACuriousMind I need help.

I'm a wuss

Clowns scare me and there's a clown epidemic in America

people are dressing up as clowns and mugging people...

user116211

Then you fear from redheads also, I guess.

user116211

@0celo7 Halloween is far away, isn't it?

ACuriousMind

@0celo7 what

0celo7

12:55 PM

Nothing to do with Halloween, it's a new "thing" the kids are doing

And it's creepy af

ACuriousMind

I, too, would be afraid of clowns mugging people.

But I'm also afraid of people mugging people, soooo...

0celo7

@ACuriousMind Nah, I'm a big guy. I wouldn't get mugged randomly. But I have no weapons against a clown.

Emilio Pisanty

yesterday, by David Z

@EmilioPisanty fair enough; I guess what I really want to ask is that everyone refrain from acting like they know what happened to Chris when they really don't.

just to stress @ACuriousMind's point. Also CC @MAFIA36790.

ACuriousMind

@0celo7 Well, I'm not a psychologist, so I can't do anything against your fear of clowns

0celo7

@ACuriousMind So, how do I cancel dinner plans w/ Reb's family? I don't want to be walking home from the parking garage at midnight.

I'm a giant baby, I know

ACuriousMind

12:59 PM

I'm assuming telling the truth is not an option? :P

0celo7

@ACuriousMind Truth? What's that?

ACuriousMind

@0celo7 Like "I'd love to come to dinner but I'm afraid of walking on my own that late at night"?

No need to get specific about the clowns

0celo7

@ACuriousMind It's clown specific. I usually have no problem walking through the seedier parts of town at night.

My choice of music helps me blend in, ofc.

ACuriousMind

Hm

Have you had dinner with them before?

If not, anything you say will seem like an excuse not to have dinner.

0celo7

Many times

ACuriousMind

1:04 PM

Okay, so the objective is to find a plausible but less embarrasing reason than "I'm afraid of clowns" to temporarily not have dinner with them?

(if you say no to this you'll have to explain to me in more detail why just saying "I'm afraid of clowns" isn't an option)

0celo7

The objective is for you to tell me if I see a clown, stab it and run.

That was my roommate's advice

ACuriousMind

lol, if your fear is so controllable that you could attack one and run away, then what's the problem to begin with?

0celo7

I'm not sure it is

@ACuriousMind Optics aren't good.

ACuriousMind

@0celo7 Only one way to find out: Tell your roommate to dress up as a clown and randomly attack you.

user116211

@0celo7 amazon.com/When-Clowns-Attack-Survival-Guide/dp/1607747030

0celo7

1:09 PM

@ACuriousMind ???

One of us would die in that situation

ACuriousMind

That would solve the problem either way, wouldn't it?

user116211

@0celo7 This is the infamous Aversion Therapy.

user116211

And this works.

0celo7

@ACuriousMind ????

youre a psycho

ACuriousMind

@0celo7 You're dead: You don't have to worry about dinner anymore. He's dead: You know you can take a clown.

0celo7

1:10 PM

@ACuriousMind you're insane

ACuriousMind

I'll see if I can come up with less drastic solutions while I run to the post office to ensure my continued enjoyment of universal healthcare.

brb

user116211

universal healthcare?

user116211

googling

Secret

@peterh But then what will nondeterministic nonrandom evolution even mean. If we cannot even predict the probability amplitudes wrt all t1 or t2 separately, how is it supposed to be described by an equation?

More importantly, even if you suppose there is a state that obeys two schrodinger equations for each variable t, and then demand consistency ,it has been shown to result in noninteracting fields

0celo7

1:33 PM

@ACuriousMind yeah I also don't want to have to pay for a cleric after a clown battle

the winner of a knife fight ends up in a coma...the loser in the ground

Secret

@Secret (Meanwhile, my higherspace enthusiastic friends (i.e. not actual scientists) basically get a similar conclusion via philosophical arguments that motion in multi time physics is going to need to obey more constraints than their single time counterparts

peterh

@Secret Wow! Thank you very much the link. I think maybe the time evolution on a single time axis could be maybe also random, but not so strongly random, as the QM. Only the combination of them would result the time-dependent schroedinger.

Secret

19

Q: Intuition for multiple temporal dimensions

It’s easy, relatively speaking, to develop an intuition for higher spatial dimensions, usually by induction on familiar lower-dimensional spaces. But I’m having difficulty envisioning a universe with multiple dimensions of time. Even if such a thing may not be real or possible, it seems like a go...

spacetime time dimensions

In addition to these, there's also a guy who worked on a 2 time model where one of the time dimension is thermal in nature (which I need to dig it up again in roder to recall the detials

ACuriousMind

@0celo7 I'm afraid the only realistic solution is to tell them the actual reason why you don't want to go.

user116211

Truth might be lethal....

0celo7

1:39 PM

@ACuriousMind Then I must go.

Secret

@peterh Ok, this stuff about thermal time is QFT level, thus I don't understand.

0celo7

If I get mugged or anally raped, I blame you...

user116211

@0celo7 you will not...

user116211

Otherwise, shout for the Dark Knight....

ACuriousMind

Well, you could dress up as a clown yourself to not get attacked...

Secret

1:42 PM

> time evolution on a single time axis could be maybe also random, but not so strongly random

So I am guessing for each t axis, you want something that has the property between that of probability amplitudes and classical trajectories. I don't know how will one come up of such a mathematical object

user116211

@ACuriousMind Can @0celo act as a clown?

Secret

and in order to obey the consistency condition you proposed, you want to combine whatever this mathematical object such that the classical trajectory part is cancelled or averaged out, leaving behind the probabilistic outcome...?

0celo7

You people have no regard for my well-being.

Secret

It is easy for me to think of something random, and something determinstic, but to think of somehting in between...?

Perhaps, one way to tackle this is as follows: You want some function $\phi (\vec{x},t_1,t_2)$ with some equation or rule such that it basically act as a seed for some kind of procedural generation process (because that is the only pseudorandom thing I know of) for each $t_i$. And after this, you want the $\phi$ that you obtained when you procedurally generate along the $t_1$ and $t_2$ axes respectively to combine (perhaps by adding, mutliplying i don't know) so that it produces the
schrodinger equation

Pseudorandom number generator. Otherwise I have no idea how to read this stuff without enough knowledge in computing. But this is one way to give the desired property of having the whatever it is evolving randomly but not totally randomly

Secret

2:17 PM

One possible way to implement this is to start with a seed $s=\begin{pmatrix}t_1 \\ t_2\end{pmatrix}$ and some kind of separable probability distribution $P(s)$ as the pseudorandom number generator such that $P(s)=p_1(s_1)p_2(s_2)$ where $s_1$ and $s_2$ are seeds with the $t_1$ and $t_2$ parameter unspecified. This will ensure that along the indivudal time axes, the result is basically speaking determinstic and pseudorandom,
but because there is no correlation with the result of any seed where $t_1$ and $t_2$ are interchanged, they are independent in some way in terms of their time evolution.

ACuriousMind

@EmilioPisanty thanks for asking this question - you have manage to thoroughly confuse me about the Hamiltonian formalism in this case!

What follows are my ramblings on a general underlying problem here:

Secret

and finally a word of caution: NO ONE will check whether the above actually make sense because multiple time physics is not very useful hence interesting to most of us since our world seemed to be fine with models of one time physics. So whatever you are doing, you are alone just like the other multiple time theorists who wrote those papers

ACuriousMind

By definition the canonical variables of the Hamiltonian formalism are the coordinates $x^i$ and the canonical momenta $\pi_i := \partial L/\partial \dot{x}^i$. Again, by definition, the Poisson bracket $\{f,g\}$ is defined by the derivatives w.r.t to these canonical variables, and Hamiltonian equations likewise involve those. The Hamiltonian formalism itself knows nothing about the kinematic momenta $p_i$ a priori.

But everything I can find mixes and matches the usage of $\pi_i$ and $p_i$ as the canonical momentum within the same computation. Your answer here just states in passing that $p_y$ is conserved since $y$ doesn't appear, but the proper Hamiltonian conclusion would be that $\pi_y$ is conserved since $y$ doesn't appear.

This page computes a non-zero $\partial H/\partial x_i$ although $H=\pi^2$ does not depend on $x_i$ in the partial sense of the proper Hamiltonian formalism

And it goes on like that

But to me, the only formally correct thing to state is that $\partial \pi_i/\partial x_j = 0$ because the Hamiltonian canonical variables are independent

In other words, the correct Hamiltonian formalism must have $\{x_i,\pi_j\} = \delta_{ij}$, but this means that $\{x_i,p_j\} = \delta_{ij} + \{x_i,A_j\}$ (modulo constants) and also $\{\pi_i,p_j\} = \{\pi_i, A_j\} = - \partial A_j/\partial x_i$.

0celo7

2:33 PM

Secret, this is not your...ACM what are you doing

Secret

(blah time out)

0celo7

Not you

ACM seems to be blogging

ACuriousMind

So, $\{H,p_j\} = \{\pi^2,A_j\} = \sum_i 2 \pi_i \{\pi_i, A_j \} = - 2\sum_i \pi_i \frac{\partial A_j}{\partial x_i} $

Secret

ACM: Just a small check, is the term that act as the generator of the infintesimal transformation always the term on the right entry of the poisson brackets (or commutators)?

0celo7

@ACuriousMind When am I allowed to Taylor expand?

I want to rigorously Taylor expand the coordinates of a curve in normal coordinates.

ACuriousMind

2:36 PM

For Landau $A_y = Bx$ and $p_y$ we then get $\{H,p_y\} = -2 \pi_x B = -2 p_x B\neq 0$, contrary to @EmilioPisanty's claim that it is conserved in Landau gauge.

I have no idea what's going on here.

@0celo7 Didn't I preface this by saying what follows are my ramblings on the Hamiltonian formalism for a particle in a magnetic field?

@0celo7 Rather tautologically if and only if the function is analytic

0celo7

@ACuriousMind oh, didn't see that.

@ACuriousMind not even in a small neighborhood?

What if the function is regular at the point of interest?

ACuriousMind

Sometimes you can get analyticity by some nice properties like the function obeying an elliptic PDE (no, I don't know anything about this, I just know this result)

0celo7

@ACuriousMind I think you're referring to Cauchy-Kovaleska.

Emilio Pisanty

@ACuriousMind That's a confusion with notation. In both the answer and the newer question I'm using $\mathbf p$ as the canonical momentum and $m\mathbf v$ (with no single symbol) for the kinematic momentum.

Secret

For example I know the general condition of the conservation of some conserved quantity Q is given by $\{Q,H\}=0$. When this is inverted into $\{H,Q\}=0$, it is saying the Hamiltonian is unchanged under the transformation generated by Q. So does that mean the generator is always whatever is placed on the right argument of the bracket?

ACuriousMind

2:42 PM

@EmilioPisanty Oh

Then disregard everything I said

Emilio Pisanty

@ACuriousMind fair enough

ACuriousMind

Should've gotten suspicious that the Hamiltonian would just have been $\pi^2$ in hindsight

Emilio Pisanty

It's a pain. There's just not quite enough letters to fit in all the concepts.

@ACuriousMind A Hamiltonian that's the square of the canonical momentum looks highly suspect to me. The kinetic energy is almost always the square of the kinematic momentum, and if this matches with the canonical momentum then cool, otherwise then not.

ACuriousMind

Yeah

Emilio Pisanty

@ACuriousMind But then I'm not completely sure what you mean by $\pi$ either, so I'll just leave it at: what I just said is fairly washy and don't take anything I say too seriously unless I've taken care to bolt every symbol to its corresponding concept.

@ACuriousMind I am glad about that though.

You'd think that question would be easy to answer but I couldn't come up with anything.

ACuriousMind

2:46 PM

Okay, now that I know that your question makes sense I'll try and follow my original idea to solve it ;)

Secret

I previously read about that $\pi$ often used to notate the canonical momentum for charge in magnetic field problems. I first saw something like this when it was introduced the ladder operators for angular momentum calculations back in my quantum courses

Emilio Pisanty

@Secret Yes. But you need to watch out each and every time: for every new text you read, make sure you check what symbol means what. Don't rule out the chance that some wacky source uses $\pi$ for kinematic momentum, because of reasons best known to whoever.

Secret

I will keep that in mind

ACuriousMind

Reading your post again I could have spared myself this issue had I only paid more attention to your usage of the words "canonical" and "kinematic" :/

Emilio Pisanty

@ACuriousMind cf. supra ;-)

I do look forward to what you can dig up on that one, though.