The h Bar - 2016-09-13 (page 1 of 3)

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user228700

1:57 AM

@JohnRennie Hello sir! :) In the comments to your answer to this question: physics.stackexchange.com/questions/61864/… you have mentioned that emissivity is a function of the wavelength. How so? What did u mean..?

user228700

Also, why is emissitivity equal to absorptivity?

Sir?

Wow, assuming zir gender

user218912

2:37 AM

what?

user228700

@0celo7 Huh? Zir? What do you mean?

Hi guys, I'm an undergrad (physics) about to graduate and I want to specialize in high energy physics in grad school but most of my research experience is in condensed matter physics. I have some exposure to QFT in a condensed matter context as well. How would potential advisors view this?

user218912

@0celo7 how does $\hbar$ appear in the canonical commutation rules?

@IceLord I am not helping you with homework.

user218912

the course didn't even begin yet dude

user218912

2:45 AM

I'm asking this for myself

user218912

:/

sigh...why does $\hbar$ appear at all

user218912

yes why

I'm asking you

why does $\hbar$ appear in QM

user218912

dunno has something to do with quantization.

2:47 AM

figure it out

user218912

idk

user218912

just tell me please

user218912

nvm I figured it out

3:06 AM

Whoa

What happened to 0celo's pic?

I am reborn.

3:56 AM

@KaumudiHarikumar to take the easy question first: emissivity is equal to absorptivity becuase emission/absorption is a reversible process. Emission is just the time reverse of absorption and vice versa.

There's also a thermodynamic argument. Suppose you had two objects, one with high $e$ and low $a$ and the other with low $e$ and high $a$. In the same environment the low $e$ high $a$ object would get hotter than the high $e$ low $a$ object.

Hi, everybody.

So you could use the two objects to drive a heat engine and you've create a perpetual motion machine.

@DanielSank Morning.

@JohnRennie evening.

Morning, evening.

@KaumudiHarikumar because they're both a measure of how strongly the electrons in the material interact with the surrounding electromagnetic field.

...I wonder if that's a meaningful explanation...

Here's a related post

9

Q: Why is black the best emitter?

Why are emitters colored black better emitters than other colors? Why is white a worse emitter?

thermodynamics electromagnetic-radiation blackbody

user228700

4:15 AM

@JohnRennie Yes, OK.

@KaumudiHarikumar as for the emissivity, materials don't interact with radio waves in the same way they interact with visible light, x-rays etc. So it shouldn't be surprising that the emissivity will change for the different types of EM radiation.

user228700

@DanielSank Hi :-) I read the answer and wow, is it thorough. That cleared up that part of my question, thanks.

@JohnRennie I have discovered a truly wonderful proof of Riemann's theorem

Is that different to Riemann's hypothesis?

@JohnRennie Riemann's theorem is that Riemann curvature = 0 implies locally flat.

user228700

4:18 AM

@JohnRennie What do you mean by "it will change for different types of E.M radiation"? How does the definition of emissivity incorporate this dependence? I don't get it :/

@0celo7 I must admit that seems obvious to me as a mere physicist ...

@JohnRennie Please explain.

@KaumudiHarikumar we've already agreed that emissivity = absorptivity. Yes?

user228700

@JohnRennie Yep, when at thermal equilobrium, no?

@KaumudiHarikumar and the absorptivity of materials changes with wavelength, otherwise they wouldn't look coloured.

user228700

4:21 AM

@JohnRennie Okay...

@0celo7 oh hang on, Riemann curvature is probablky some technical term that I don't understand. I assumed it meant the Riemann tensor was zero but I guess that's wrong.

@JohnRennie That's what it means.

@0celo7 Well if the Riemann tensor is zero that obviously means space is flat. What am I missing?

Obviously?

Why is it obvious?

Yes, parallel transport round a curve leaves a vector unchanged.

4:23 AM

@JohnRennie That's a highly nontrivial theorem, but ok. So?

@0celo7 that's what we mean by flat.

@JohnRennie No, by flat we mean we can map it isometrically onto a piece of flat space.

There is a map from the manifold to $\Bbb R^n$ that carries the Riemannian metric onto the Euclidean one.

@0celo7 I knew this was going to be more complicated than I thought ...

@KaumudiHarikumar therefore emissivity must change with wavelength as well

@JohnRennie How do you go about proving Riemann = 0 implies parallel transport is path independent?

@0celo7 it just is.

4:30 AM

uhhh

4:41 AM

@0celo7 definition, right?

@DanielSank Which definition?

The Riemann tensor is either a mess of covariant derivatives or a mess of Christoffel symbols

neither one gives you parallel transport without some work

user228700

4:56 AM

@JohnRennie OK, thanks :-)

@KaumudiHarikumar Black body radiation generally assumes there are no specific interactions.

Materials will have electronic excitations at certain wavelengths and at these wavelengths they interact very strongly with the light.

Ooops.

Found an error :^)

That looks like a load of balls to me :-)

user116211

What's that?

user116211

0

A: Why can't echoes be heard inside a room?

Furniture absorbs sound so no echo is heard

5:00 AM

I'll fix it while I'm pretending to pay attention in QM tomorrow.

user116211

Comment or answer?

::puts various geometry texts into bag::

The idea is correct though.

user228700

@MAFIA36790 Hey, thanks for editing my question!

@JohnRennie Crap. I have to take a better look at how one solves $x''=0$ :P

user116211

That's my work; nothing to worry.

5:07 AM

user image

@JohnRennie I take it back, this is correct.

It's so clean.

I take credit for the proof, the idea of using Jacobi fields is not mine though.

user116211

Title of the Day:

user116211

-1

Q: It is in Mechanics

A particle m can slide without friction at the outer surface of a sphere of radius R .The particle was released from the top from rest .Determine the limiting angle of contact .The reference at the ground.

homework-and-exercises newtonian-mechanics

user116211

i don't know why they tag such questions with classical mechanics

user228700

6:37 AM

@MAFIA36790 😂

@MAFIA36790 Somewhere I've already heard about it

@MAFIA36790 Actually, echoes can be heard, but we are highly accustomed to it. In acoustic, there is a significant effect that in high-end sound studios there is no echo -> the recordings sound empty and false. There are sw ways to generate echo for that.

7:00 AM

Is it possible to change the parent site for an SE account? For historical reasons my account is linked to the ServerFault Stack Exchange but since I hardly ever visit it these days I'd prefer Physics to be my parent site.

user116211

@JohnRennie yes.

user116211

Go to the user profile.

@MAFIA36790 I can't see anywhere in the profile I can change the parent site ...

user116211

@JohnRennie You meant the SE chat account, didn't you?

Aha, thanks got it!

user116211

7:08 AM

@JohnRennie \o/

user116211

7:24 AM

Ah! @JohnRennie, you changed your bio also; I see.

Yes. I started out on Stack Exchange when I was working full time as a network manager and I only posted on Server Fault, so my profile was tailored to that.

have to clear some concepts-

is this doe physics?

for*

@Adi Yes, we discuss physics hereabouts :-)

electric potential energy - The energy expended in moving enough charge from one point A, to another point B in a circuit so that the potential difference becomes 0

@Adi that seems an odd definition.

7:28 AM

potential - The work done per charge to produce a push on the electrons at a certain point in the circuit. It can differ from point to point in a circuit

its now a definition

its just me trying to grasp this concept'

potential difference - The difference between potentials of 2 points in a circuit

i had a LONG discussion with this guy last night to understand the concept - chat.stackexchange.com/transcript/45262

The potential has a gauge freedom so it isn't well defined. The ony well defined quantity is potential difference. That is, we can unambiguously measure the energy needed to move a test charge between two points so we can unambiguously determine the potential difference.

When people talk about potential they almost invariably really mean potential difference.

so i am wrong?

i am talking in very loose terms here btw'

Take any two points in a circuit. If the energy needed to move a test charge $Q$ is $E$ then the potential difference between those two points is $V = E/Q$.

Does this make sense so far?

yes

OK so potential difference is well defined and easy to understand.

7:40 AM

yes

but am i right on potential?

its the work done to push the charge

No, I don't think that's a good definition of potential because it isn't clear to me what it means.

To specify the potential we just choose some reference point.

Then the potential is just the potential difference from that reference point.

user116211

@JohnRennie, one question for you; I'm bit confused....

?

potential diff. and potential are 2 very diff. thing

things(

@Adi no they aren't

user116211

For conservation of angular momentum to be valid, there must be the strong form of the third law.

7:43 AM

@Adi Why do you think they are different?

user116211

To do this, the potential must be dependent on the distance.

user116211

Goldstein writes:

@MAFIA36790 err, yes I suppose so ...

becasue steeven tole me so

chat.stackexchange.com/transcript/45262 that guy

user116211

He considers the potential $V_{ij} = V_{ij}(|\mathbf r_i- \mathbf r_j|)\;.$

7:46 AM

That's the potential difference between the two points $r_i$ and $r_j$

the symbols arent translating for mr

me

@Adi the chat room doesn't have MathJax enabled.

what should I do then?

17

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

@Adi Theres no problem, I know MathJax well enough to read your equations.

7:48 AM

no but i need to read your equations, and I cant

user116211

> The two forces are then automatically equal and opposite. $$F_{ji} = -\nabla_i V_{ij} = \nabla_j V_{ij} =- F_{ji}\;. $$

user116211

My question is: how does this follow automatically by defining the potential as $V_{ij} = V_{ij}(|\mathbf r_i- \mathbf r_j|)\;?$

@Adi Oops, I got a bit mixed up between the conversations.

its ok

user116211

@Adi math.ucla.edu/~robjohn/math/mathjax.html

7:51 AM

@Adi Suppose we know the potential difference everywhere in a circuit. We take this potential difference relative to some reference point e.g. the battery anode.

yea, relative to 2 referenc epoints

points

So relative to the anode the potential at any point is the energy needed to move a unit charge from the anode to that point.

we take the potentials of 2 deff. points and subtract them

@Adi hang on, how do you know what the potentials of 2 deff. points are?

i thought potential was the work dont at that point /charge

idk

user116211

7:53 AM

@MAFIA36790 typo: $\mathbf F_{ji} = -\nabla_i V_{ij} = \nabla_j V_{ij} =- \mathbf F_{ij}\;.$

If we take the anode as our reference point then the potential difference relative to the anode is the work needed to move a unit charge from the anode to that point. Let's call this potential difference $V(x)$ where $x$ is some variable that tells us where the point is.

potential is the work done/charge at a point in the conductor to push that charge. The electric force which pushes the charge becomes lesser and lesser as we move away from the terminal fro where the force is being originated...thats what I know

user116211

@JohnRennie: Any clue what Goldstein meant by saying that the forces being equal and opposite automatically follows from the definition of $V_{ij}\;?$

@Adi that's a meaningless statement. The phrase work done/charge at a point doesn't mean anything.

work done on every charge at a distance from the terminal

to push the charge forward

7:57 AM

You can measure the work done to move a charge from some reference point to that point and that's well defined.

due to the electric force

@Adi No, that's not a useful definition.

that work done/charge decreases as the distance between the charge and the terminal increases

im not telling you if it is right. I am telling what I know as I just learned it yesterday

@Adi Go back to my statement:

4 mins ago, by John Rennie

If we take the anode as our reference point then the potential difference relative to the anode is the work needed to move a unit charge from the anode to that point. Let's call this potential difference $V(x)$ where $x$ is some variable that tells us where the point is.

Does this make sense?

ok

yes'

8:00 AM

So for example $x$ could be what you mean by the distance between the charge and the terminal.

my definition of electric potential is not a useful but is it correct to some extent?

yes

@Adi No, I'm afraid not.

user116211

What I got!!!

user116211

4

Q: Strong Newton's third law of action and reaction: Mathematical Interpretation

According to the strong law of action and reaction for internal forces(Goldstein): "$F_{ij}=-F_{ji}$ and the forces lie along the direction joining the particles." Now consider the statement If those internal forces are conservative we can associate the internal forces with a potential of th...

newtonian-mechanics forces potential

:(

8:02 AM

If we take the anode as our reference point then at the anode $x = 0$ and $V(x) = 0$ because the potential difference between the anode and itself is zero.

user116211

And as I guessed, it is Qmech who answered ;))

As we move round the circuit $x$ increases and $V(x)$ changes. It could increase or decrease depending on exactly what is in the circuit.

But everywhere in the circuit $V(x)$ tells us the potential difference between the anode and the point $x$.

yes true

So, are you happy this makes sense and explains exactly what potential difference is?

yes

but isnt that what i said?

8:05 AM

OK. Now for the potential.

Let's call the potential $U(x)$

Suppose the potential is defined as:

U(x) = V(x) + C

?

where $C$ is a constant. Don't worry about the value of $C$ just accept that it is a constant.

oh ok

So the potential at the anode would be $U(0)$

true

wait

anode is where the charge starts moving from or where it ends up?

8:08 AM

The anode is where the charge starts from.

oh....but shouldnt the potential at the anode be A LOT

you know, its the starting

Potential of anode - POtential of cathode = potential difference of the battery

so how can potential of anode = 0

wait

its now 0

x = 0, so the distance from the anode is 0

oh ok

U(0), clear

I think you've got it. My potential difference $V$ is zero at the anode because I've chosen the anode as the starting point.

yes

But I've defined the potential as $U = V + C$ so the potential $U(0)$ is not zero at the anode.

In fact at the anode the potential has the value $U(0) = C$.

Because $U(x) = V(x) + C$ and $V(0) = 0$

yes

true

8:13 AM

Now I just guessed this equation for the potential, so the question is have I guessed right? Is this the correct definition for the potential?

What do you think? Does this definition make sense?

but at anode the potential is at its highest level

here its lowest

as potential difference is 0

@Adi why do you say the potential is highest at the anode?

If we take the anode as zero then the potential difference for the cathode is +v where v is the battery voltage.

So the potential of cathode is the potential of the anode + v i.e. the potential of the cathode is higher.

Isn't it?

because Potential of anode - Potential of cathode = potential difference of the battery

and if potential of anode is not the highest the potential difference on some point of the conductor can become negative

and the PD of cathode is lowest

@Adi other way around, isn't it?

Ah, OK, this is because of the rather silly way we define currents.

In reality a current is electrons flowing from the anode to the cathode.

But we define conventional current as flowing from the cathode to the anode.

Do you remember that I defined the potential as $V = E/Q$ ?

8:19 AM

shouldnt then the potential at the anode be the highest as the electric force s higher

the more the elctrons move away from the anode the lesser the force, and so the lesser the potential

as work/charge will be lesser

Be careful about electric force as that has a precise definition that I think you've misunderstood slightly.

negative repels negative positive repels positive

electromagnetic force i think

Yes, but remember that an electron charge is negative

so?

is the anode positive or negative?

negative i think

When all this stuff was first formulated back in Michael Faraday's time people though electric current was carried by positive charges so it flowed from the cathode to the anode.

And because things flow down a potential gradient that means the potential of the cathode was defined to be higher than the potential of the anode.

That's why we call the cathode positive and the anode negative.

Obviously positive > negative

8:25 AM

?

the electrons will experience greater force from the anode as the anode is is negative

and more force = higher potential, thats what i know

Force is a vector and can be positive or negative.

Work = force times distance, so that means work can also be positive or negative.

Mmm, I suspect this is just confusing you further.

have a look at this prntscr.com/chfeky

In any case it doesn't really matter. Can we just forget about all this stuff for now and go back to where we started?

thats what that guy told me

The trouble with work is that it can mean two things:

1. the work done on an electron

2. the work done by an electron

8:29 AM

he was talking about the work done on electrons

These two definitions are equal and opposite i.e. one is positive and one is negative

by the force of electromagnetism

@Adi that's definitely not true, by the way

what is not true??

the things that Steeven told?

@Adi OK let's take your statement:

3 mins ago, by Adi

he was talking about the work done on electrons

8:33 AM

yes

An electron flows from the anode to the cathode because it's being pulled by a force (the electric force).

That means the force is doing work on the electron.

i.e. the work done on the electron is positive.

And if potential is work done on a unit charge that must mean the potential is increasing.

no, its being repelled by a force

the negative charge of the cathode

anode*

not cathode

Whether the force is pulling or pushing doesn't matter. Either way it's doing work on the electron because it's accelerating it.

yes

but the force decreases as the electron moves away from the anode

So as I said above the work done on the electron is positive

8:36 AM

he told me that the cathode too has a negative charge but the charge at the anode is morwe negative

more negative*

Are we agreed that the work done on the electron is positive

how is work done postivie or nrgative?

If something is accelerating work is being done on it because its energy is increasing

and?

So work is being done on the electron.

8:43 AM

YEA

when is the work negative?

This is a metter of terminology to some extent, but if we are doing work on something we are increasing it's energy. The energy change is +ve so we say the work done on the objct is +ve.

user116211

@Adi When the object loses energy.

Now suppose the energy of the object was decreasing. We generally say this means work is being done by the object.

But we can also say that negative work is being done on the object.

how?

oh ok

the work is +ve

!!hang g

If we do positive work on something we increase its energy. So by analogy if we do negative work on something we decrease its energy.

8:47 AM

oh

We are getting miles away from your original question. If you're interested to discuss all this stuff that's fine by me, but you might want to take a step back to talking about potential and potential difference.

ok back to topic

Where we'd got to is that I said the potential difference relative to the anode at some distance $x$ along the circuit is given by $V(x)$.

Where $V(x)$ is the work done on the electron.

And I suggested we could define the potential as $U = V(x) + C$.

yes

Now suppose we use this definition for the potential to calculate the potential difference between the anode and the point $x$.

8:52 AM

how is PD work done on the electrons

@Adi oops, $V(x)$ is the work to move a unit charge not one electron.

yes

wait brb

k bak

but how is PD thw work done on electrons? its the diff. between potentials of 2 points

user116211

@Adi Work-kinetic Energy Theorem.

Potential is short for "potential energy". The potential difference between two points is the energy needed to move a unit charge between those two points.

user116211

@yuggib: o/

8:57 AM

When we say potential this is always a measure of energy

i thought potential was work/charge

Potential is work per unit charge so potential = W/1 because a unit charge is 1

yes

idk can i understand this..im 15

10th grade

This sort of stuff even confuses undergraduates, so you're doing fine :-)

i am following till now but..

...

i need to go

maybe next time

9:03 AM

OK :-)

am I even close to understading it?

I think you're closer than you suspect. You're getting confused about the definition of work, but that confuses everyone.

whats the flaw in my explanation of potential?

And in any case it isn't central to the argument.

@Adi Potential is just defined by $U = V + C$ as I said earlier.

And $C$ is an arbitrary constant.

oh

9:05 AM

That means potential doesn't have a unique definition because we can choose any value for C

This is quite common in physics and it's known as a gauge freedom.

But it does mean that you have to be careful talking about potential.

On the other hand potential difference is always well defined.

user116211

@Qmechanic: Ah! Point noted; sorry for that; wouldn't happen in future; Thanks for making me aware of that :)

@Adi the message I replied to (that more force = higher potential) is not true

It should be steeper the potential = more force

@DavidZ: I suspect that's hard to understand for a tenth grader.

Perhaps so, but that doesn't make it true

9:13 AM

We forget how much basic stuff we've learned over the years and how much we take for granted.

There would be a lot of background discussion needed to put your statement in context.

@MAFIA36790 \o

Think of it this way using newtonian gravitational potential which we are most familar with since little. A potential is defined to be something such that the force is given by how steep the potential is at a point. That is for 1D it is given by the slope of the potential at some point. (or in calculus $F(x)=-\frac{dV(x)}{dx}$, and for higher dimensions $\vec{F}(\vec{x}=-\nabla V(\vec{x}))$)

Now imagine the potential is like climbing a hill. Using our daily experience, if you place a ball on a steeper hillside the ball will roll down quicker than when placed on a not so steep place. This…

@MAFIA36790: your question kind of got lost in the potential energy discussion

user116211

@Secret use \nabla

user116211

@JohnRennie ah; never mind; I have found a relevant post that discusses it; I'm still reading that.

user116211

9:19 AM

5

Q: Strong Newton's third law of action and reaction: Mathematical Interpretation

According to the strong law of action and reaction for internal forces (Goldstein): $\mathbf F_\mathrm {ij}=-\mathbf F_\mathrm{ji}$ and the forces lie along the direction joining the particles. Now consider the statement If those internal forces are conservative we can associate the int...

newtonian-mechanics forces potential

I hate message expiry and I just noticed an ugly typo

user116211

@Secret Just ping your typo post writing the amended note.

bleh...

@Secret Typo: $\vec{F}(\vec{x})=-\nabla V (\vec{x})$

@Adi ^ See above and the message it is replied to

Meanwhile I am working on some rather hardcore derivation stuff on Lebniz integral rule

user image

user116211

9:26 AM

@Secret Saw that in the maths room.

And the issue is that the 3 terms underlined diverges and still trying to figure it out

I have not post this as a question yet because I am nto sure what question I want to ask about it other than how to resolve this divergence (which is not my major aim)

Meanwhile the only answer to the question that leads to the above one is not really answering the question

0

A: Exploring the properties of the limit when Lebniz rule fails

Define a function $G$ of two variables $r$, $s$ by $$ G(r,s) = \int_{a}^{r}f(x',s)dx'. $$ Under some reasonable circumstances, $G$ is a differentiable function of 2 variables, and its partial derivatives $G_1$, $G_2$ are $$ G_1(r,s)=f(r,b),\;\;\;G_2(r,s)= \int_{a}^{r}f_2(x'...

I explictly said I am not interested in how to evlauate it using lebniz integral rule, yet that answer gave me an evaluation based on lebniz integral rule

@MAFIA36790 : No problem.

user116211

Qmech comes and goes.....

user116211

o/

3 hours earlier, I was reading the rest of Susskind's classical mechanics. When he get to the section of the definition of the Hamiltonian, I then started to wonder about the interpeetation whether the Lagrangian is actually describing the worldsheet of the system in phase space

9:32 AM

Any stat mechanics here?

This is because the hamitonian is the total energy of the system, and if the lagrangian has no explicit time dependence, then the hamitonian will be conserved. This suggest any trajectory in phase space over the course of time would not look very different in 'shape' and should resemble some kind of prism with the height in the time dimension

So in a sense, while the lagranginan will always have indirect time dependence due to how $q$ and $\dot{q}$ have a time dependence themselves, if we consider all trajectories that are allowed by that lagrangian (thus in some sense listing out all the dependence of the trajectories with time), then overall we are getting some kind of static strructure in phase space

which remains unchanged in shape as it move through time

What's new here?

A simple example to illustrate that will be the lagrangian for circular motion which has the form:
$$L = T - V = \frac{1}{2}mr\dot{\theta}^2$$

where $r$ is constant. Plotting $L(\dot{\theta},r,t)$ should give a cylinder shaped graph

However, one might argue that this is a convenient example because of the circular symmetry, thus I need to check the most general case to see if this interpretation holds. But before then, I will read the rest of the chapters of susskind

user228700

10:16 AM

@JohnRennie Wow! You should get paid for all this great teaching you're doing here! :o

@KaumudiHarikumar I don't think my attempt this morning was very successful :-(

user228700

10:56 AM

@JohnRennie Oh :-( [Haven't read it all yet] Well, it was a tough task and you gave it your best.

user228700

11:51 AM

Also, what do you mean when you say you work part time as a computer nerd?

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