The h Bar

12:17 AM

@0celo7 u there

12:34 AM

argh what was I doing wrong ._. Can someone confirm that this was the right thing to do: Given a point P(1,-2,3) and the equation of a plane x+2y-4z = 1 find the distance from the point P to the plane. I started by finding another point on the plane then drew a vector from it to P. Then I projected this vector onto the normal vector of the plane

I checked wolfram alpha and this is the correct procedure but when I picked different points on the plane to make sure the distance was correct, I got different results ;(

I tried 2 different points like 4-5 different ways and my brain hurts from thinking about it now

$\vec{P_pP} = (1,-2,3) - (1,0,0)$ was one, then $\langle 1,\frac{-3}{2}, 3 \rangle$ was another

Then you take the dot product between this and the normal vector then divide by the magnitude of the normal vector, right?

I just checked online using a calculator and my original point on the plane yielded the correct result

Sir Cumference

2:13 AM

Wait, holy crap

Is it possible to only change your name on one SE site?

0celo7

2:30 AM

@Obliv yes

0celo7

2:56 AM

@ACuriousMind Help!

I think I proved something, but it hurt

vzn

3:25 AM

@BernardMeurer cool, sounds like a smartphone built by/ for power users. am a bit amazed myself how much this stuff has evolved in ~1decade o_O

user228700

5:01 AM

Hi everyone :-)

John Rennie

Morning

DanielSank

@JohnRennie hi

user228700

@DanielSank: Hi. @JohnRennie: Morning sir.

John Rennie

@Kaumudi morning :-)

user228700

@JohnRennie: I feel a little guilty for attacking u with questions every morning :-P So much that I almost feel like not asking u about that car problem that I told u about last night.

John Rennie

5:12 AM

That's OK. I have an hour before I have to do any real world and I have my morning coffee to hand, so I'm good :-)

user228700

OK :-)

John Rennie

But I think I need to be clearer exactly what you are asking

user228700

Yes, OK. We have a car that is moving on a curved flat road.

John Rennie

OK

user228700

The task is to analyze the force that is responsible for the curved motion of the car.

John Rennie

5:15 AM

OK, sounds easy

user228700

This curved motion can apparently be brought about by the force of friction, or by banking the road or both.

John Rennie

Yes

user228700

I understand the case of banking the road but I'm very confused about the case in which the frictional force is responsible for the curved motion.

John Rennie

In both cases there must be an inward force of $mv^2/r$. Yes?

user228700

Firstly, we're dealing with static friction, and not kinetic friction and secondly, how in God's name does friction act in the direction of the inward radius for a car that's moving in a curved path?

user228700

5:18 AM

@JohnRennie Definitely.

John Rennie

Suppose the car was driving ina straight line on the side of a hill i.e. the hill goes up to the left of the car and down to right of car, but the car itself is moving at a constant level.

What stops the car from sliding sideways down the hill?

user228700

@JohnRennie U mean on one of those roads paved on the side of a hill? The road is mostly straight, no? ie. The force of gravity is mostly normal to the car and doesn't have a sideways component to make it slide down, no?

John Rennie

We're dealing with an idealised hill, so it's a smooth inclined plane.

The car moves in a straight line at constant altitude.

user228700

@JohnRennie OK...

John Rennie

So if you put a ball on a point on the cars path the ball would start rolling down the hill

user228700

5:22 AM

Yes.

John Rennie

So what stops the car sliding sideways down the hill?

user228700

The engine of the car turns the wheels and the frictional force acting on them pushes the car upward? (I'm guessing...)

John Rennie

Suppose the car is stationary. What stops the car sliding down the hill?

user228700

@JohnRennie Definitely, the frictional force applied on the car by the road, acting upwards.

John Rennie

Now imagine the car is moving very slowly e.g. 1mm per year. Does the car now start sliding down the hill?

user228700

5:25 AM

@JohnRennie Uh, no.

John Rennie

OK, now speed up to 1 metre per year. Does it now slide?

user228700

No...I mean, I dunno. It's moving so slowly that I would assume that the static friction continues to act on the wheels, as in the case in which the car is just stationery.

John Rennie

1 metre per hour, 1 metre per second, why should the car suddenly start sliding sideways at some transition speed?

user228700

@JohnRennie What do u mean "sideways"?

John Rennie

The car is driving on an inclined plane that slopes up to one side and down to the other, but the line along which the car is driving is at a constant level.

user228700

5:29 AM

@JohnRennie Yes, I got that...

John Rennie

So sideways means the the car slides down the slope normal to its direction of travel

user228700

@JohnRennie Right. I wasn't entirely sure if this is what u meant. Was making sure.

John Rennie

Obviously, the point I'm making is that the friction between the tyre and the road stops the car sliding sideways, and the speed the car is travelling shouldn't affect this.

user228700

@JohnRennie And this frictional force acts up, pointing toward the hill..?

John Rennie

Yes

user228700

5:32 AM

@JohnRennie And we gather this...how?

John Rennie

If the hill angle is $\theta$ then there is a sideways force on the car of $mg\sin\theta$.

So unless a frictional force is acting the car will accelerate sideways down the hill at $g\sin\theta$ m/s$^2$.

user228700

But this force acts up the plane, no?

John Rennie

The force due to gravity acts down the slope normal to the car's direction of travel, and the frictional force acts up the slope normal to the car's direction of travel.

If the car is stationary this is a simple problem of the type you've probably done a hundred times.

user228700

John Rennie

Yes, that looks good.

user228700

5:40 AM

But it doesn't seem to me like the frictional force is acting toward the hill, it looks like it's acting up the plane...

John Rennie

Maybe we need to clarify what you mean by the frictional force is acting toward the hill. I took this to mean up the plane i.e. towards the top of the hill.

i.e. to the right in your diagram, just as you have drawn $f_r$

user228700

Ah. OK then. I kept thinking this:

DanielSank

Καλιμερα, @JohnRennie

John Rennie

@DanielSank Calamari to you too :-)

DanielSank

Gee, I wonder if I swear in foreign languages here if that counts against the be-nice policy...

@JohnRennie HAHAHAHAHA

That's the beauty of it, scientists can phoenetically read Greek.

John Rennie

5:43 AM

My niece is studying (ancient) Greek at school. At age 14!

DanielSank

@JohnRennie wow

John Rennie

I didn't realise they taught it these days.

DanielSank

I'm trying to scrub the rust off my my modern Greek cogs.

Duolingo is amazing.

I've been doing Russian there as well.

user228700

Prepare ur eyes for the blinding beauty of my drawing:

John Rennie

:: drum roll ::

user228700

5:45 AM

user228700

Oh my God, that looks so shitty >.<

DanielSank

@Kaumudi I am blind!

The beauty... my eyes cannot handle it!

@Kaumudi how could you do this to me?!

Valter Moretti

@vzn I have read your message right now, sorry

user228700

Lol. I'm truly sorry :-( As u can see(or...not), I suck at this.

user228700

@JohnRennie: Did u understand my horrible drawing?

John Rennie

5:49 AM

Is that clearer?

user228700

@JohnRennie Whaat? MUCH clearer.

user228700

OK, I think I understand the point u were making before.

DanielSank

@JohnRennie !

John Rennie

Maybe I've chosen a bad way to approach this, but my point is that if the car doesn't slide to its left, down the hill, that must be because the tyre road friction stops it sliding.

user228700

@DanielSank Aren't u supposed to be blind? -_-

user228700

5:53 AM

@JohnRennie Right, yes...

John Rennie

@Kaumudi so you concede that tyre-road friction can stop the car sliding sideways?

user228700

@JohnRennie Yessir.

user228700

But the car is on an incline!

user228700

What about the case when it's moving on a flat road on a curved path?

John Rennie

@Kaumudi In this case we have an external sideways force (due to gravity) and the tyre-road friction opposes that force and prevents the car from accelerating. Yes?

user228700

5:55 AM

Yes.

user228700

When the car is on the flat road, there is no external sideways force, so why does the friction start acting radially inward?

John Rennie

On a level road the driver turns the front wheels so the wheels now align along an arc of a circle.

user228700

@JohnRennie OK...

John Rennie

I can draw another of my amazing diagrams if that would help :-)

user228700

@JohnRennie :-) Truly amazing indeed. But I have a diagram here: (I didn't draw it, don't panic :-P)

John Rennie

5:58 AM

@DanielSank Google Draw. For simple diagrams it's the dogs bollocks (another use of the word bollocks :-)

DanielSank

@JohnRennie o_O

Are dog's bollocks... good... or... well... dog's bollocks?

John Rennie

dog's bollocks means really good, though quite why escapes me :-)

@Kaumudi This might be of interest ...

3

A: Direction of friction when a car turns

This is my attempt to illustrate what happens when the car wheel is turned: Focus on the bit of the car tyre marked with a red spot, and the bit of the road marked with a green spot. If we could look at the contact patch between the tyre and the road we'd see something like the rectangle I've ...

user228700

John Rennie

@Kaumudi If you make the car body invisible, so you can see the tyres it would look like (pause while John draws a diagram) ...

user228700

@JohnRennie Reading this...

user228700

6:04 AM

What do u mean by contact patch?

DanielSank

@JohnRennie British English is a silly language.

John Rennie

@Kaumudi the contact patch is the area of the tyre that is in contact with the road. The tyre deforms under the weight of the car so we get a roughly rectangular area of tyre that is flattened against the road.

This is my attempt to show how the car turns in a circle ...

user228700

@JohnRennie OK...

user228700

"The red spot on the tyre is scraped across the road surface. It's this lateral motion of the tyre surface across the road surface that causes the frictional force that turns the car."

John Rennie

The driver turns the front wheels so both the front and back wheels are aligned on a tangent to the circle.

user228700

6:08 AM

I don't understand this bit...

John Rennie

Look at my diagram above.

user228700

OK...

John Rennie

If the car were to go straight ahead it would scrape the front tyres sideways.

Because the front wheels have been turned inwards

user228700

Yeah, OK...

user228700

Wait, why? I can imagine the situation and yes, that's exactly what happens, but which force turns the front wheels that way?

John Rennie

6:11 AM

@Kaumudi the driver does, by turning the steering wheel.

Actually, perhaps this isn't obvious if you don't drive ...

user228700

Right. OK...

user228700

(I can't drive cars/motorized bikes yet)

user228700

Alright, so the wheels keep on turning like that and back wheels keep moving forward?

John Rennie

In most cars the back wheels are fixed and don't turn. That's why I've drawn the car slightly displaced to the right. The idea is that the plane of rotation of all the wheels is tangential to the circle.

@DanielSank True, and that's what makes it such fun :-)

user228700

"The plane of rotation of the wheels is tangential to the circle"; I don't understand this by looking at your diagram. (I'm very sorry on behalf of my dumbness)

DanielSank

6:17 AM

@JohnRennie I'm disappointed that Google translate can't go from English to Cockney rhyming slang.

John Rennie

@DanielSank Hmm, to what extent Cockney rhyming slang really exists, and to what extent it's a creation of the media, is an open question.

user228700

@DanielSank You're not blind -__- My poor(albeit, silly) drawing.

John Rennie

I suspect it genuinely exists, but has been exaggerated by the media.

DanielSank

@JohnRennie Is that didgeridoo?

Or just your rack and pinion?

John Rennie

:: John very nearly really does fall off his chair laughing ::

DanielSank

6:20 AM

:)

John Rennie

I think Monty Python did a sketch along those lines ...

@Kaumudi: anyway, back to tyres ...

DanielSank

@JohnRennie Forest fires.

::ducks::

John Rennie

@DanielSank Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrghhhhhhhhhhhhhhh!

DanielSank

::dies laughing::

user228700

:'-(

DanielSank

6:21 AM

Oh let some guys have their fun.

user228700

@DanielSank Sure sure :-)

user228700

OK, should I come back in awhile after all the fun has been had?

DanielSank

I think we're done.

We've had our midday sun.

user228700

:-D How are u so good at this?!

user228700

@JohnRennie: Is it time for u to start ur work now?

DanielSank

6:23 AM

@Kaumudi Language is one of my "things".

John Rennie

In about five minutes ...

user228700

@DanielSank :-D I see.

John Rennie

A wheel will roll in the direction it's pointng.

user228700

Hmkay...

John Rennie

So in my diagram above the rear wheels will want to roll to the right, while the front wheel will want to roll down and to the right. Does that make sense?

user228700

6:25 AM

Yes, it does.

John Rennie

But of course the rear and front wheels are attached to the same car so they can't move independently.

user228700

Right.

John Rennie

The result is that the front of the car moves down and to the right and it pulls the rear wheel after it.

user228700

Okay..?

John Rennie

So the car ends up driving in a circle

user228700

6:27 AM

And what about the frictional force..?

John Rennie

The wheels want to roll in the direction they are pointing because to move in any other direction requires them to slide sideways a bit.

And that's where the friction comes in.

user228700

OK..?

John Rennie

Imagine you are holding a car tyre, or anything else the same shape.

user228700

OK.

John Rennie

You can roll it in the direction it's pointing very easily. Yes?

user228700

6:30 AM

Yes.

John Rennie

But sliding it sideways, normal to the direction it's pointing, is hard. Yes?

user228700

Yes, relatively.

John Rennie

And that's because rolling forwards doesn't involve any sliding so there is no frictional force opposingthe motion.

user228700

OK...

John Rennie

But moving the tyres sideways means you have to slide the tyre surface over the floor surface, and there is a frictional force that opposes this motion.

user228700

6:32 AM

Yeah, OK...

John Rennie

So the only way a wheel can move without sliding is in the direction it is pointing.

user228700

Yes, OK...

John Rennie

So when the car driver turns the front wheels inwards the car cannot move straight ahead without the front tyres sliding over the road.

user228700

Yes...

John Rennie

And the friction opposes that sliding

user228700

6:34 AM

Hm, OK...

John Rennie

That's why the car turns round the circle

It's the only motion possible that doesn't cause the tyres to slide

user228700

Okay! This makes more sense now. Thanks so much sir :-) And sorry for the trouble.

John Rennie

Good timing - now I have to get to work for half an hour or so ...

DanielSank

@JohnRennie Pfffft. "Work".

John, Nethack is not "work".

John Rennie

@DanielSank Checking server event logs. Not the most exciting thing I've ever done ...

DanielSank

6:43 AM

Hmph.

Shing

Hi everyone

DanielSank

Hi, everybody.

Shing

I am reviewing E&M. I always think Purcell was the best when it comes to undergrad text..... however, he almost never deal with dirac delta function in his book...

user228700

9:23 AM

Hi :-) I've got a quick question; how is the angular frequency for simple harmonic motion defined..?

John Rennie

@Kaumudi the frequency is in units of cycles per second

And each cycle is $2\pi$ radians

So $\omega = 2\pi f$ radians per second

user228700

I've been searching and searching but all I've seen is $\omega=2\pi/T$.

John Rennie

If $T$ is the period then $f = 1/T$

user228700

@JohnRennie We've just defined each cycle to consist of $2\pi$ radians?

John Rennie

A simple harmonic oscillator is closely related to circular motion.

user228700

9:26 AM

Yes, OK. I know that bit.

John Rennie

And one cycle of a SHO is equivalent to one rotation round a circle

And one rotation around a circle is an angle of $2\pi$

user228700

@JohnRennie Right. OK, thanks! :-)

John Rennie

That was easy :-)

user228700

:-P Yep.

user228700

10:15 AM

Can u help me with a bit of a homework-tsy question..?

John Rennie

Yes, ask away!

@Kaumudi Hello?

user228700

10:54 AM

Oh, crap, I'd assumed that u were away and left :/ My phone doesn't give me notifications on time.

user228700

And now u're really away. I'll wait!

Slereah

12:42 PM

I guess before I try rebuilding QFT from scratch, I should finally understand how to do Klein Gordon from the Wightman axioms

Still not sure

Physics Guy

12:52 PM

@Slereah What do you mean by "do the Klein gordon from the Wightman Axioms" ? Do you mean you want to derive the Klein Gordon Equation from the Whightman axioms ?

Slereah

Well the solution, not the equation

Qmechanic

1:31 PM

3

Q: Are dimensions a fundamental property of the universe or an emergent result of other physical laws?

Are spacetime dimensions a fundamental property of the universe or an emergent result of other physical laws?

Primarily opinion-based? Unclear what you're asking?

Slereah

1:41 PM

Not opinion based but a bit vague

And odds are without answer

0celo7

@Slereah How is the research going?

Slereah

Do you mean hilbert spaces or jobs

Jim

But if a car crashes on the road and no one is around to hear it, does it make a sound? — Jim 1 min ago

Welcome to philosophysics.SE

0celo7

the first

is Jen a troll?

Jim

@0celo7 Unless explicitly acting like a troll, I treat everyone as if they are not a troll

Slereah

1:56 PM

I am investigating Klein Gordon in the Wightman axioms to see what I really need to transfer

Also thinking aboot it, I will lose separability of the Hilbert space

Since the cardinality of the basis will be like

$\Bbb R^\Bbb N$

0celo7

god I hate point set topology

Slereah

That's like... $2^{\aleph_1}$?

0celo7

so damn messy sometimes

Slereah

I dunno

I mean there's an equivalence on that set so it might be smaller

I dunno

Let's see

Tho then again the cardinality of the field doesn't dictate the cardinality of the basis

Oh apparently hyperreal numbers also have the cardinality of the real

yuggib

@Slereah no, it is $(2^{\aleph_0})^{\aleph_0} $

Slereah

2:07 PM

yeah apparently that's $\approx 2^{\aleph_0}$

yuggib

That should be $2^{\aleph_0}$ if I remember cardinal arithmetics

Slereah

I wonder if there's a constructible bijection

yuggib

Anyways, that is if you want to consider the vector space over $\mathbb {R} $

@0celo7 point set topology is the interesting topology

0celo7

@yuggib you should read this little note once it's done

If $X$ is a locally Euclidean Hausdorff space, then the following are equivalent:

(1) $X$ is second countable.

(2) $X$ is metrizable and has a countable number of connected components.

(3) $X$ is $\sigma$-compact.

(4) $X$ is paracompact and has a countable number of connected components.

(5) $X$ is separable.

I am proving this

yuggib

Sounds ~~boring~~ very interesting

0celo7

2:13 PM

it is boring, don't get me wrong

but someone should write it down

Slereah

I'm sure someone did

0celo7

find it

yuggib

Probably someone else already did

0celo7

wtf Latex

@yuggib local compactness + Hausdorff is OP

0celo7

2:32 PM

@yuggib do you know anything about locally compact topological groups?

Slereah

I know that they are compact, locally speaking

They also have a group structure

0celo7

whoa

I never thought about it like that

Slereah

Really makes you think

yuggib

@0celo7 yeah, I know a bit of harmonic analysis on the abelian ones

And some representation theory

But not so much

There is a classical book by loomis and mackey iirc

And a modern one by Rudin

0celo7

2:48 PM

@yuggib Is any topological book generated by an open neighborhood of the origin or do you need local compactness?

I can't find where one would need local compactness in the proof

But Wolf seems to think one needs it

yuggib

Mmmh I think that is true for banach spaces

Also metric probably

Not sure about locally convex ones

0celo7

@yuggib Banach spaces?

yuggib

So I think local compactness is not necessary

0celo7

viewed as an abelian topological group?

yuggib

Yes

For example

But I'm not 100% sure

0celo7

2:52 PM

Well, yeah, its' true

any Banach space is generated by its unit ball

so just shrink that to fit inside any neighborhood of the origin

I'm fairly sure that works

hmm, is that true?

the generated by its unit ball thingie

@yuggib what does scalar multiplication mean in the topological group context?

yuggib

@0celo7 is not important

0celo7

@yuggib well it kinda is...

yuggib

@0celo7 yes it is true if I recall correctly

0celo7

my ball statement is a homeomorphism from the unit ball to the whole space

Slereah

@0celo7 has huge balls

0celo7

2:57 PM

yes, I often get compliments

Sir Cumference

Does anyone know if it's possible to get a page or two from George Lemaître's "The Beginning of the World from the Point of View of Quantum Theory" without having to pay for it?

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