The h Bar

Manas Dogra

05:26

Recently I came to know that classical trajectory of a particle given initial conditions is not unique..Initial conditions fix the arbitrary constants of integration which we get by integration the equation of motion...What extra things fixes.i.e. determines uniquely the trajectory in reality?

I mean if we get a set of trajectories as solution even after setting the initial conditions then what determines which trajectory the particle will follow in reality.

Korra

Am I missing any?

Slereah

05:52

@ManasDogra Initial position and momentum is enough to fix it in classical mechanics

Since it's just a second order ODE

Or alternatively, initial and final position, if you're feeling fancy and can see the future

Then it's

\begin{equation}
\vec{x}(t) = \frac{\vec{x}_b - \vec{x}_a}{t_b - t_a} t + \frac{1}{2} \left[ \vec{x}_a + \vec{x}_b - \frac{\vec{x}_b - \vec{x}_a}{t_b - t_a} (t_a + t_b)\right]
\end{equation}

Manas Dogra

So I thought until I read a paper by Dhar
https://aapt.scitation.org/doi/10.1119/1.17411

Slereah

But $\vec{x}(t) = \vec{v}_0 t + \vec{x}_0$ will fix it for $\ddot{\vec{x}}(t) = 0$

yeah but he's talking about weird systems

Probably more Norton's Dome stuff

In those cases it's possible that there are no unique solutions given initial conditions

Manas Dogra

@Slereah Yes that's what I was talking about...If we can make such a potential in reality what will determine which trajectory the particle would follow?

Although it's not practical to make a singular potential

Slereah

Well in reality we can't make systems with arbitrary forces

And also quantum mechanics makes it not an issue in the end

IIRC the QM version of Norton's dome isn't a problem

Manas Dogra

What about $F=\frac{1}{x^3}$? with $x=0=\dot{x}$ at $t=0$ It has three solutions even after putting the boundary conditions..

Slereah

05:59

I mean if you have discontinuous and divergent forces in the first place

It's gonna be weird anyway

there's plenty of known solutions like that

Manas Dogra

sorry $F=x^{1/3}$

Slereah

ie the Painlevé non-singular collisions for instance

Which just uses gravity

Although it's harder to prove

Manas Dogra

So we can ignore those kind of things for all practical purposes...

bolbteppa

@ManasDogra "In the case of singular forces, for which some of the higher derivatives do not exist at a given point"

Slereah

I mean if you use weird equations you get weird results

ACuriousMind

07:32

@ManasDogra Note that scenarios like Norton's dome are only pathological for certain points in the space of initial conditions (in the dome's case, for something sitting motionless at the exact top)

but in reality things are not points without extent - no "real" thing can be like that,real things have extent and hence will never hit only that single point with their state

Norton's dome is a philosophical problem (i.e. what does causation mean?), not a practical problem (i.e. how do we solve equations of motion in practice?)

Slereah

Well you can construct problems like that for most physical laws if you're loose enough with it

ACuriousMind

sure - whenever you have a differential equation with some function as input, choosing that function to be pathological will likely mess up the uniqueness properties of the solutions at some points

Korra

08:08

@ACuriousMind can you pretty please take a look at my question?

ACuriousMind

08:40

@Korra I have no idea what the question is. (I saw your message, I just don't know what the image depicts or what "am I missing any?" is supposed to mean)

Korra

09:04

@ACuriousMind it's the Feynman diagrams

And I am trying to work out the symmetry factors

So far I concluded that the factor is 4

Ryan Unger

12:02

Does a generic dome not have the Norton's dome property?

Slereah

Only if its derivatives diverge at some point I guess

or whatever the full condition is for Lipschitz continuous functions

ACuriousMind

@RyanUnger depends on what you mean by "generic"

Ryan Unger

I clearly mean whatever so that the statement is true

the usual meaning

ACuriousMind

you get Norton's dome-like phenomena exactly when the force resulting from the shape of the dome is not Lipschitz continuous

Ryan Unger

Are you sure that every non-Lipschitz function has this property though

Norton's dome is C^\alpha right

Slereah

12:14

I dunno if it's a sufficient condition for that no

Ryan Unger

I would believe that lots of C^\alpha vector fields have unique integral curves. I also believe that lots of them don't

Slereah

would be hard to have a dome shaped like a Dirichlet function

ACuriousMind

@RyanUnger probably not every

in any case, I think the "generic" dome many people would write down would be some sort of polynomial, and you don't get issues with that :P

Slereah

The usual one is basically $F = \sqrt{r}$

Ryan Unger

@ACuriousMind I clearly mean Baire generic in some Holder space

Slereah

12:20

it's my favorite shape of domes

Love it when the church has a Baire generic shape

Ryan Unger

its modern art

Slereah

Can't even find an easy example online of what's a good function to talk about Baire generic

Emilio Pisanty

5

Q: What is a $0\times0$ or $0\times3$ matrix?

In the comments to another question, the following exchange was noted: ... wait until you see a 0×0 matrix. and ... or worse, a 0×3 matrix! What are these things? Do they have a name or any special proprieties? Where are they used?

2

what

Ryan Unger

@Slereah what

do you know what baire generic means

Slereah

I don't rly

It's one of those words you rarely see

Ryan Unger

12:27

the baire category theorem is easy you should know about it

Slereah

Maybe but it's not one of those things that come up often?

Ryan Unger

uh, ok

if you say so

Slereah

3 PSE results for "Baire"

Emilio Pisanty

@Slereah math.stackexchange.com/search?q=baire

Slereah

Different tribe

I know it is a math term, it's just not common in physics

Vincent Thacker

12:41

@EmilioPisanty I learn something new everyday

Ryan Unger

physicists not using baire is not surprising

well, it is actually

arent they supposed to read reed and simon

Slereah

They are not

Someone repost that xkcd comic about specialists

ACuriousMind

@RyanUnger no, they're supposed to say "it's somewhere in Reed and Simon" when someone asks for proofs

Slereah

the QFT works in mysterious ways

Ryan Unger

citeseerx.ist.psu.edu/viewdoc/…

this might be relevant

Slereah

12:53

thanks

Emilio Pisanty

@VincentThacker I just added my take there

Slereah

13:33

Greg Egan published a new book and he's too ashamed to say it's about wormholes

Nazmul Hasan Shipon

14:18

Here is a circuit with two battery $V_1$ and $V_2$, you can clearly see the direction of the electric current $I_1$ and $I_2$. Can anyone please tell me, what will be the direction of $I_1$ if I reverse the terminal of the $V_1$ battery?

And, will the direction of $I_2$ change for it?, here $V_1$, $V_2$ has no fixed voltage.

John Rennie

14:38

@NazmulHasanShipon Hi Nazmul. Whenever you have a circuit with two batteries I suggest using the principle of superposition to calculate the currents. This is by far the easiest way to do it.

Vincent Thacker

@NazmulHasanShipon It depends on the values of V1 and V2.

Nazmul Hasan Shipon

14:55

@JohnRennie I don't know about principle of superposition. Can you give me a source where I can learn about it? I am high school student, about to complete.

John Rennie

@NazmulHasanShipon Have a look at this Khan Academy article.

And this is also a good article. Possibly better than the Khan Academy one.

Nazmul Hasan Shipon

@VincentThacker But I have to show diagrammatically what would the direction of currents after reversing the terminals of the voltage source $V_1$. Can @JohnRennie help me about it? Note that V_1 and V_2 isn't given.

John Rennie

@NazmulHasanShipon suppose we flip V₁ then the circuit looks like this:

OK so far?

Nazmul Hasan Shipon

yes

@JohnRennie

John Rennie

Suppose we ignore battery 2 and just draw in the currents we will get from battery 1. Then the circuit looks like this:

@NazmulHasanShipon OK so far?

Nazmul Hasan Shipon

15:04

yes

John Rennie

Now ignore battery 1 and draw the currents we expect for battery 2:

@NazmulHasanShipon Yes?

Nazmul Hasan Shipon

Yes, @JohnRennie

John Rennie

What superposition says is that to get the current when both batteries are present we just add the currents from the two diagrams, so we get this:

Now, if you look at the left part and the centre part the currents are in the same direction, so when we add them the total current will be in that direction i.e. down on the left and up in the middle. Yes?

Nazmul Hasan Shipon

Yes, @JohnRennie

cOnnectOrTR 12

Hi, @JohnRennie if two charges are present one inside a closed surface and one outside. Then we say net flux is due to inside charge only. Charge outside contribute to zero flux. The reason stated is field enters from one side and pass through other side.

my question is

These charges field will get superposed. So after superpose

tion

John Rennie

15:15

@NazmulHasanShipon But on the right side the red and blue currents are in opposite directions. Which way the total current flows will depend on which is bigger, the red or the blue current, and we can't tell that without knowing the values of the voltages and resistances.

@cOnnectOrTR12 I'm starting my lunch now. I'll be back in about an hour.

Nazmul Hasan Shipon

@JohnRennie okay @JohnRennie but

cOnnectOrTR 12

Ok! can we differentiate between the two fields and say it’s field just passes through the surface so it’s flux is zero.

Nazmul Hasan Shipon

@JohnRennie will the original direction of $I_2$ change for it?

cOnnectOrTR 12

This is the question. Answer it after your healthy lunch

John Rennie

16:05

@NazmulHasanShipon If we take the original diagram, before the battery was reversed, and draw in the directions of the currents we get (red is battery 1 and blue is battery 2 as before):

ACuriousMind

John's "lunch" being healthy would be a new occurence ;P

John Rennie

Now if I₁ is upwards then on the left branch the red current from V₁ must be larger than the blue current from V₂.

Now reversing V₁ swaps the direction of the red currents but it does not change their magnitude. If we look at the diagram with V₁ swapped:

It suggests that on the right most branch the red current will be larger and the net current will be upwards.

So swapping V₁ reverses the direction of both I₁ and I₂.

But I'm not absolutely sure about this as it may still depend on the resistances present. I'd have to do the calculation to be sure.

> Hi, @JohnRennie if two charges are present one inside a closed surface and one outside. Then we say net flux is due to inside charge only. Charge outside contribute to zero flux. The reason stated is field enters from one side and pass through other side.

My question is these charges field will get superposed. So after superposition can we differentiate between the two fields and say its field just passes through the surface so its flux is zero.

@cOnnectOrTR12 There is only one electric field, and it can be described as the sum of the fields from the two charges.

If we write the field as the sum of the two fields then:
- the field of the inside charge has a non-zero total flux through the surface
- the field of the outside charge has a zero total flux through the surface

@ACuriousMind my thousand words on the matter :-)

Traditional English sausage butties!

ACuriousMind

mhhh...the sausages look a bit strange, did you bread them or something?

John Rennie

I sliced them lengthways before frying them - it makes them flatter so they work better in rolls.

2

Nazmul Hasan Shipon

@JohnRennie @JohnRennie, does that mean V_1 > V_2 ?

PM 2Ring

16:23

5

Q: Unpinning the accepted answer from the top of the list of answers

Today we changed the way we sort answers on Stack Overflow. We no longer pin the accepted answer (with the green checkmark) to the top of the list of answers. By default, we now sort strictly by votes (descending order by highest score), and the accepted answer's order in the list is based on its...

discussion answers announcements answer-quality

I don't know if we need this on Physics.SE, but I guess we should consider it.

John Rennie

@ACuriousMind The rough appearance is because they were made from coarsely ground meat and where the inside surface has browned it looks very granular. They were exceedingly good though.

@NazmulHasanShipon I think it probably means V₁ > V₂ but I would have to sit down and do the calculation to be sure.

ACuriousMind

@JohnRennie I see, sounds indeed good!

Nazmul Hasan Shipon

@JohnRennie can you do it for me please, I can wait?

ACuriousMind

@PM2Ring at least worth thinking about

John Rennie

@ACuriousMind For the rolls I used a trick I picked up from a YouTube video. You sprinkle grated cheese onto a hot frying pan, wait for it to melt and brown then put the roll down on it. Then when you lift the roll up the cheese sticks to it and you get an instant toasted cheese layer on the roll.

ACuriousMind

16:29

@JohnRennie and here I was thinking you had just browned the bun

I was almost concerned - just bread and sausages didn't sound like a Rennie lunch ;)

satan 29

XD

John Rennie

I spread barbeque sauce on the rolls before adding the toasted cheese layer just to make it extra healthy.

satan 29

@JohnRennie make sure to have a diet coke too

John Rennie

@NazmulHasanShipon It's a messy calculation, but I had a quick look and I don't think we can definitely say V₁ > V₂.

Nazmul Hasan Shipon

@JohnRennie Okay @JohnRennie , thank you so much for doing it for me.

John Rennie

16:37

For example if R₃ is very large then the current through it will be very small even if V₂ is also very large so I₁ could still be upwards even though V₁ < V₂

cOnnectOrTR 12

@JohnRennie but this means we can tell and differentiate between the fields even after superposition. In other words there is one field after superposition. So how can you tell whose contributing to the flux because that can be done only by determining and distinguishing between the field lines of the two charges.

Nazmul Hasan Shipon

@JohnRennie I get it, thank you :) .

John Rennie

@cOnnectOrTR12 The flux is ∫E.dA where E is the total field and E = E₁ + E₂ where E₁ and E₂ are the fields due to the charges Q₁ inside and Q₂ outside. OK so far?

@NazmulHasanShipon you're welcome :-)

cOnnectOrTR 12

Seem right. Yeah

John Rennie

And if E = E₁ + E₂ we can write the integral as:

∫E.dA = ∫(E₁ + E₂).dA = ∫E₁.dA + ∫E₂.dA = Φ₁ + Φ₂

where Φ₁ and Φ₂ are the fluxes from the charges Q₁ inside and Q₂ outside.

cOnnectOrTR 12

16:42

Yes

John Rennie

So while there is only one field we can still calculate the fluxes separately and add them to get the total.

In this case we'd get Φ₁ ≠ 0 and Φ₂ = 0

cOnnectOrTR 12

How is phi2 zero

Nazmul Hasan Shipon

@JohnRennie but if R_3 is very large, how does it affect the current of V_2 ?

John Rennie

@cOnnectOrTR12 Φ₂ is the flux through the surface due to the charge outside the surface.

So it has to be zero.

@NazmulHasanShipon All the current that V₂ produces has to flow through R₃ before it splits to flow through the left and right branches. Yes?

Nazmul Hasan Shipon

yes

@JohnRennie

cOnnectOrTR 12

16:52

So charge outside will have + and - flux?

John Rennie

@NazmulHasanShipon And if R₃ is much larger than all the other resistors the current from V₂ is approximately given by I = V₂/R₃. Yes?

@cOnnectOrTR12 Every field line from the charge outside has to enter the surface then leave it again, so the number of field lines from the outside charge entering the surface has to be the same as the number of field lines from the outside charge leaving the surface. Yes?

cOnnectOrTR 12

Ok!

So?

Here is the confusion

John Rennie

So the total flux through the surface from the outside charge is zero.

Nazmul Hasan Shipon

@JohnRennie yes

John Rennie

9 mins ago, by Nazmul Hasan Shipon

@JohnRennie but if R_3 is very large, how does it affect the current of V_2 ?

cOnnectOrTR 12

16:56

How can you talk about field lines from individual charges if it has been superposed?

John Rennie

That's why R₃ affects the current from V₂

@cOnnectOrTR12 Because electric fields superpose.

You can take the field lines from the two charges separately, calculate the fluxes separately, then just add them to get the total.

Nazmul Hasan Shipon

@JohnRennie but I was wondering if that would affect the current of V_1 too.

John Rennie

No, because the current from V₁ splits. Some flows through the middle branch where R₃ is, but some can flow through the right branch where R₄ and R₅ are.

cOnnectOrTR 12

You mean take one charge find it’s flux1 then take another flux2. Add them and you will get net flux from charge inside.

John Rennie

So even if R₃ was infinite there would still be a current from V₁ because it could flow through the right most branch of the circuit.

@cOnnectOrTR12 the sum will give you the total flux from both charges, but the charge outside has a net flux of zero so in fact this is equal to the flux from the charge inside.

Nazmul Hasan Shipon

17:01

@JohnRennie so if V_2 > V_1, I_2 direction would be unchanged.

John Rennie

i.e. Φtotal = F₁ + F₂ but F₂ = 0 so we end up with Φtotal = F₁

Nazmul Hasan Shipon

@JohnRennie according to this picture

John Rennie

@NazmulHasanShipon Yes

56 mins ago, by John Rennie

cOnnectOrTR 12

Yes we can do that. And I was confused how could you talk about field lines passing through when it’s superposed. How can you map out individual lines when it’s already superposed

John Rennie

@NazmulHasanShipon I₂ is always downwards regardless of what the two voltages are.

Nazmul Hasan Shipon

17:04

@JohnRennie why is that?

cOnnectOrTR 12

Thanks I was grinding on this for days

John Rennie

@NazmulHasanShipon the red and blue arrows show the contributions to the current from V₁ and V₂. Yes?

@cOnnectOrTR12 OK :-)

Nazmul Hasan Shipon

yes

John Rennie

And if both arrows point in the same direction the total current must also be in that direction. Yes?

Nazmul Hasan Shipon

@JohnRennie but why should it point in the same direction?

@JohnRennie here it point upwards

@JohnRennie here downwards

John Rennie

17:08

@NazmulHasanShipon That's with V₁ reversed.

I'm talking about the diagram with V₁ in its original direction

Nazmul Hasan Shipon

@JohnRennie oh yes, I was talking about the reversed case.

John Rennie

In the reversed case I'm not sure what will happen.

Nazmul Hasan Shipon

so in the reversed case if V_2 > V_1, I_2 direction would be unchanged or we are not sure?

okay.

John Rennie

2 hours ago, by John Rennie

I'm not sure.

Nazmul Hasan Shipon

@JohnRennie okay :) .

cOnnectOrTR 12

17:11

When we find the E due to a charged conductor (eg a thick plate)it comes as E=sigma/e0. It has been derived by taking a Gaussian surface one end inside the conductor and other outside. We take charges on one side and not on other side. But when we talk about field due to thin sheet we draw Gaussian surface cylinder with both ends on both sides. Why have we ignored the charges on other side of the thick conductor. Does it not spread on both surfaces in case of thick conductor?

John Rennie

I need to go now.

cOnnectOrTR 12

18:34

Why don’t we find flux of the system of two charges, one of them inside the closed surface ?

why do we calculate flux separately?

ACuriousMind

@cOnnectOrTR12 you've been asking more or less the same question for days now here. What different answers than those you've already gotten do you expect?

If you are not satisfied with the explanations you've received so far, you really need to be more specific what you're dissatisfied with

At this point I don't see why I should try to answer your question when you'll just come back tomorrow and ask the same question again with no indication that you retained anything from the previous conversations

cOnnectOrTR 12

But no one answered this. Only recently I was told you have to find flux separately. And I am asking just why? Because now after superposition we should be finding the flux of resultant field from the system of charges

ACuriousMind

you don't have to find the flux separately, it's just easier

cOnnectOrTR 12

18:49

So even if we find the flux of resultant we will get the same net flux from inner charge?

And it’s the same thing

@ACuriousMind can you tell me how do you do that? Is it complicated?

ACuriousMind

didn't John explain that already here:

2 hours ago, by John Rennie

And if E = E₁ + E₂ we can write the integral as:

∫E.dA = ∫(E₁ + E₂).dA = ∫E₁.dA + ∫E₂.dA = Φ₁ + Φ₂

cOnnectOrTR 12

Total flux due to resultant field= flux1+flux2=flux1

Can you answer my charged conductor question ?

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