The h Bar

user228700

11:02 AM

Where does the $40 \rho G$ come in? :-/ @Mew

Mew

yeah

it's bettter to do it this way:

let Pt2 be the point in the FIRST tube 40cm up to the mark

user228700

Okay...

Mew

now do you agree that Pt= 40pg + Pt2

user228700

Yeah. I mean, OK, wait.

Mew

40pg = pressure due to water between Pt and Pt2

user228700

11:08 AM

But, but ._.

Mew

you don't get that Pt = 40pgh + pt2?

remember Pt is the bit on the first tube that intersects wtih th eline

and Pt2 is 40cm above

user228700

No ;_; I get it but then I don't.

Mew

ok let me know when you get it

user228700

Huh? I thought Pt was the one on the other tube!

Mew

Pt = tube on the left

Pt2 = tube on the left (but 40cm above the line)

user228700

11:10 AM

Ohhhhh.

Mew

Pt = Pt2 + 40pgh

user228700

OK, yeah, I get that.

Mew

cool

now let us define Pt3 as tube on the right (but in line with Pt2)

now do you agree that Pt3 = Pt2

and then define Pt4 as tube on the right (but in line with the line)

user228700

@Mew Can u tell me why? It's easy for me to convince myself and then completely disagree 2 mins later.

Mew

Pt4 = Pt3 + 40p1gh

So now we have Pt4 = Pt - 40pgh + 40p1gh

and so clearly it was incorrect to say Pt4 = Pt which we said earlier

and since Pt4 = Pa + pgh1

and Pt = Pa + pgh2

Pa + pgh1 - 40p1gh + 40pgh = Pa + pgh2

user228700

11:14 AM

:-/ Can u tell me why Pt3=Pt2?

Mew

cos they are both at the same point (40cm above)

and they are the same liquid

user228700

(This is the stuff I actually have trouble with. I can write the equations but I don't understand/unable to make out where the pressures are equal. How dyou make that out?)

Mew

ok

I dunno the meaning

I just know how to get the answer

@JohnRennie what's the meaning?

user228700

No, but how are u able to figure out where the pressure is equal?

Mew

by seeing what gives the right answer

user228700

11:16 AM

-_____-

Mew

^_^

user228700

Awesome.

user228700

@JohnRennie: ? _/\ _

Mew

but i bet i can solve any question

even without the answer

by using my approach above

John Rennie

@Kaumudi yes?

user228700

11:17 AM

How to figure out where the pressure is equal?

John Rennie

@Kaumudi what pressures? Bear in mind I haven't been following the conversation so I don't knoiw what pressures you're specifically referring to.

user228700

Oh, right, OK.

user228700

Um, this:

user228700

40 mins ago, by Kaumudi

user228700

In a system like that ^ how to figure out at which the points the pressures are equal?

John Rennie

11:22 AM

I'm not sure I'd approach that problem by trying to work out where the pressures are equal ...

koolman

@Mew I have again have a problem with entropy

Mew

what @koolman?

@JohnRennie is it not correct though that for a given continuos liquid, at a given height the pressure would be the same?

user228700

@JohnRennie Huh? Really?

Mew

@Kaumudi there are many ways to solve this

Electrodynamist

@JohnRennie Can you help me with this? Weinberg writes that the property that distinguishes Lorentz Transformation is that it leaves the proper time invariant. I struggle to understand this. To me, the important property of LT is that it leaves the components of the metric tensor invariant. Any GCT would leave the proper time invariant as it has no free indices by definition - it's a scalar. Is there any justification to Weinberg's claim?

koolman

11:25 AM

We know S = q/T , according to this if we increase the temperature the entropy decreases but @Mew according to me increase in temperature would increase disorderness hence entropy increases

user228700

Wokay. That's a big question. I'll come back later then, after trying some more.

user228700

Thanks, u guys :-)

John Rennie

@Electrodynamist yes, I think I agree with you. The proper time is an invariant in general relativity, so that property isn't unique to special relativity since it's true in curved as well as flat spacetime.

Mew

@koolman but this is for a given Q

koolman

@Mew what will be the effect of q on temperature

Mew

11:29 AM

So the higher the temperature, the less a given additional heat will increase the entropy further

@koolman Q is the heat energy enetering the system

@koolman so technically $\Delta S = \frac{\Delta Q}{T}$

where $\Delta Q$ is the heat added to the system

So think of two scenarios

John Rennie

@Electrodynamist I don't know where Weinberg writes the stuff you cite so I can't comment on what he means.

Mew

1. T is low for a given $\Delta Q$

and 2. T is High for a given $\Delta Q$

John Rennie

@Kaumudi what is the question? Are you trying to calculate h1 and h2?

koolman

@Mew why there is less additional heat

Mew

In scenario 2, temperature is already high, so adding a further $\Delta Q$ doesn't increase entropy as much

@koolman I said in both scenarios assume the same heat

i.e $\Delta Q$

Electrodynamist

11:32 AM

@JohnRennie Ok. Could it be that we take a tensorial representation for granted but Weinberg might mean something about the adequacy of a tensorial expression itself? Btw, he has written this in his book "Gravitation and Cosmology: Principles and Applications of General Theory of Relativity" (Part 1 - Preliminaries, Chapter 2 - Special Relativity).

koolman

@Mew if we are doing same amount of work in suplying heat

Mew

@koolman mate I said the heat is the same in both scenarios. The heat is $\Delta Q$ being added to the system in both scenario 1 and 2

Now the question is, what will increase entropy more?

ADding heat to a hot system, or adding the saem amount of heat to a cold system?

Because a hot system is already random, adding more heat doesn't increase the entropy as much as adding heat to a cold system

koolman

Ohk @Mew

John Rennie

@Electrodynamist I've got that book somewhere. Let me see if I can dig it out ...

koolman

@Mew but if we want to know change in entropy in two processes

1) when temperature increses

2) when decreases

In first case the change in entropy would be positive

Am i correct @Mew

Mew

11:36 AM

generally as temperature increases, entropy increases yes

and that's because to increase temperature you add heat, $\Delta Q$

so to work out the entropy, if temperature isn't fixed you have to evalute the integral:

Electrodynamist

@JohnRennie It is on page 26-27 in the print that I've got. It's 1972 print - seems like the first edition as the edition is mentioned nowhere.

John Rennie

@Electrodynamist Ok he's saying that if we take the metric to be the Minkowski metric then the Lorentz transformations are the only ones that preserve proper time. That rules out e.g. Rindler transformations because they change the metric. So yes, as you say, the key point is that the metric remains Minkowski.

Mew

$\int \frac{dQ}{T(Q)}$

koolman

So @Mew we cannot answer my question without having proper values of Q and T

Mew

@koolman in the entropy formula you have $\Delta Q$ but you don't have $\Delta T$, you only have $T$

so the entropy formula doesn't tell you how entropy changes as T changes

it only tells you how entropy changes as Q changes, (at a particular temperature)

koolman

11:40 AM

@Mew but we can entropy change between two state having different temperature

Mew

of course

Secret

HDE's answer give a very concise example that illustrates the major reason I learn physics

Mew

@koolman but the formula isn't easy to evalulate in that case

koolman

@Mew for different temperature

Mew

yes

of course entropy changes with temperature

but the formula doesn't show how at first glance

the formula only shows how entropy changes with heat

to work out how it changes with temperature one must find a formula for Temperature in terms of heat and evaluate the integral

Remember the formula is $\Delta S = \frac{\Delta Q}{T}$ NOT $S = \frac{\Delta Q}{T}$ So the formula shows change in entropy only. It tells you nothing about how entropy itself depends on temperature or not

@koolman one can show using certain assumptions that $S_{final} = S_{initial} + C_p ln\frac{T_{final}}{T_{initial}}$

and thus clearly as Temperature increases, entropy increases

koolman

11:51 AM

@Mew how you got this formula

Mew

@koolman you can see a derivaiton from the entropy equation here: everyscience.com/Chemistry/Physical/Entropy/g.1312.php

@koolman here is a simple argument. Suppose in a special case we have $\Delta Q = \Delta T$

that is, as

1 unit of heat is added, the temperature then increases by 1 unit

Then the formula for entropy would become $dS = \frac{dT}{T}$

and thus $\Delta S = \int \frac{dT}{T} = lnT_{final} - lnT_{initial}$

koolman

Ohk

Mew

Thus $S_{final} - S_{initial} = ln\frac{T_{final}}{T_{initial}}$

Now because in reality $\Delta Q = Constant * \Delta T$,

the formula is instead

$S_f - S_i = C_p ln \frac{T_f}{T_i}$

So as we can see, if temperature increases from T_i to T_f, the entropy increases too

koolman

Ohk

Mew

so common sense prevails after all

koolman

11:59 AM

I am little bit confused . But most of the doubt is cleared

Mew

Just remember the entropy formula tells how entropy CHANGES as HEAT is ADDED to a system at a PARTICULAR temperature

and adding heat to a hot system increases entropy less than adding same heat to a cold system

koolman

So finally @Mew we can say increase in temperature increases entropy but change in entropy is less than when decreases

Mew

huh

Entropy decreases if temperature decreases

and Entropy increases if temperature increases

koolman

Ohk

Now I got it

Thanks

Mew

If your talking about temperature changes u have to use the ln (Tf/Ti) formula

koolman

12:04 PM

Yeah

Mew

note the limitations of the ln (Tf/Ti) formula. That is this assumes a constant heat capacity.

that is, the derivation of that formula assumes $\Delta T$ is proportional to $\Delta Q$

Sir Cumference

So...why aren't the mods active on weekends?

koolman

Yes sure i will remember it

@Mew I have another question

Mew

yep

koolman

We know entropy is a state function

Mew

12:08 PM

yep

koolman

But @Mew for reversible process it is zero and for irreversible process it is non zero .

So it should be path dependent

Mew

nope

koolman

For deltaS surrounding

@Mew why

Mew

well what you said isn't inconsistent with it being a state function

it can still be path independent and be 0 for reversible processes and negative for irreversible proceses

Sir Cumference

@Mew Do you like helping people?

Mew

12:10 PM

@SirCumference no why?

Sir Cumference

@Mew Ya mind helping me with psychology?

@SirCumference sure

@Mew wait really

go ahead

@Mew ok

gimme a sec

koolman

12:11 PM

@Mew sorry mew i could not understand this

Mew

@koolman if a system goes from state A to B, then it doesn't matter what path it took to go from A to B, the entropy changes will be the same

Sir Cumference

@Mew can ya explain the social penetration theory?

koolman

@Mew but as i said it is different for reversible and irreversible paths

Mew

@SirCumference relationships get deeper as they develop

@koolman no you don't have reversilbe and irreversible "paths", you have reversible and irreversible state changes

Sir Cumference

@Mew all right, that's surprisingly simpler than i thought

Mew

12:14 PM

@koolman if state B has higher entropy that state A, then a transition from A to B will always be irreversible regardless of the path taken

koolman

@Mew why

Mew

@koolman because entorpy is a state function

it doesn't depend on path

Entropy is a measure of randomness

IF state B is more random than A, it doesn't matter the path

it just is more random than A

just like temperature is path independent

koolman

Oh yeah

Mew

So if one path from X to Y is reversible, all paths from X to Y are reversible

and if one path from M to N is irreversible, all paths from M to N are irreversible

koolman

I see

Sanya

12:17 PM

@Mew ?!?!?!?!?!

yes?

I don't understand

understand what

or wait, maybe that's just a language problem

koolman

@Mew but can we have X toY reversible and Y to X irreversible

Mew

12:18 PM

@Sanya what language problem?

@koolman if X to Y is reversible, then Y to X is also reversible

Sanya

I can compress a gas from (p, V, T) to (p', V', T') reversibly (in theory) or irreversibly

meaning "in quasi-equilibrium at all times" or not

the path will depend on how I compress

but I can get from the initial state to the end state in both modes

can't I?

koolman

@Mew ohk

Secret

0

Q: Could there be a causal relation between the horizon problem and entanglement?

Usually the horizon problem is solved by an inflation of the expanding universe. But if space-time consist of virtual particles they could be entangled no matter what their distance is? Or perhaps the other way around that when you break the entanglement spacetime get ruptered and causing an ex...

big-bang

Quantum gravity experts needed

(My interpretation: What if an answer is wrote using Sean Caroll's model of emergent spacetime via quantum entanglement)?

Having said that, I have not fully read that paper yet

Sanya

"But if space-time consist of virtual particles" <<< Aren't virtual particles internal lines in Feynman diagrams and thus not real?

having said that, I have no strong background in QFT and none in General Relativity :D

Secret

yeah, virtual particles (if I recall Johnrennie and Acuriousmind's conversation correctly), are basically ways to expand the interaction of the quantum fields into a series

koolman

12:27 PM

@Mew is @Sanya is correct

Sanya

@Secret yep, that's what you do with Feynman diagrams, series expansions

and then you see internal lines where one particle goes from one vertex into another and has no connection to the outside

heather

hello everyone

Sanya

which I always thought were the virtual particles that are basically "nonreal" as they are "non measureable"

Mew

hi @heather

@koolman so entropy isn't a state function?

Sanya

@heather hey heather

koolman

12:31 PM

@Mew No it is

I asked that as you have not replied to sanya

Mew

So how do you go from A to B irreversibly one path, and reversibly another path?

unless the entropy decreases at some stage along the irreversible path and then increases again.

I'm only considering paths however where entropy is increasing

koolman

Ohk

Mew

I imagine you could take a path from A to B that is irreversible if entropy decreases along that path then increases again back to 0

Sanya

en.wikipedia.org/wiki/Second_law_of_thermodynamics#Entropy @Mew @koolman

Secret

A brief discussion on how to detumble satellites, from a company start up from my uni

Mew

12:35 PM

@Sanya so the "irreversible" paths you speak of ones where entropy decreases then increases again?

Sanya

@Mew sorry, I'm not following you

Mew

@Sanya suppose you can go from state A to state B by following a reversible path

we thus know the entropy change is 0

heather

we're creating a "four-year plan" to sketch out what we want to take during highschool

Sanya

@heather math, physics, IT? :D

@Mew yep

Mew

@Sanya and also we know that the entropy change at any small bit of the path is also 0

since entropy can never decrease

Sanya

12:38 PM

sure, for a reversible path, dS=0 anywhere

heather

@Sanya yep. They've got AP Physics I and II, AP Chemistry, AP Calculus I and II, AP Programming, Virtual Reality, Web Design I and II, Digital Electronics...=D

Mew

@Sanya so how is it possible to follow an "irreversible" path from A to B?

heather

and I think they might have a few artsy type classes, but I'm not sure =P

Sanya

@heather sounds like a good future ... and yeah, Chemistry is actually cool up to a certain level too

Mew

and still maintain dS = 0 (as required by 2nd law of thermodynamics)

Sanya

12:39 PM

@Mew this is not required by the 2nd law

heather

@Sanya, yeah, I'm super excited. But also nervous because highschool. And my mom's like "oh my goodness STOP GROWING!!" =)

Mew

@Sanya yes it is, because I can "pause" the process at any time and consider just this change

Sanya

the 2nd law states that for irreversible processes, $dS \geq 0$

Mew

no

heather

0

Q: How many dimensions does electricity have?

My six year old daughter asked me this morning, 'how many dimensions does electricity have ?' What would be the best answer bearing in mind the age !?

electricity dimensions

Mew

12:40 PM

@Sanya the 2nd law states that all processes dS > 0

>=

when considering system and environment

Sanya

@heather highschool, so nice ... brings back memories ::wipes away tears::

Mew

The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an isolated system, will always increase over time.

Sanya

or stay constant

yup

Mew

correct

So again back to my question

so how is it possible to follow an "irreversible" path from A to B?

heather

@Sanya, yeah, I may or may not have the course handbook up even though it is an in-school thing to fill out the plan =P

Mew

12:42 PM

and still maintain dS = 0 (as required by 2nd law of thermodynamics)

Sanya

@Mew this is where we disagree

Mew

But we already stated that we can have a reversible path from A to B

where dS = 0 along the whole path

thus we know the net entropy change between A and B is 0

thus any path from A to B must have dS = 0, or have dS +ve at some points and dS -ve at others (given entropy is a state function)

but given dS cannot be -ve (by 2nd law)

dS >= 0 and thus dS = 0 (given entropy of A = entropy of B)

Sanya

@Mew no - see the definition of entropy as a state function

the entropy difference between two states A and B is the entropy difference assuming a reversible process between A and B

Mew

@Sanya so you are saying the entropy of B actually does depend on the path to get there?

Sanya

@Mew no, because reversible processes are, as far as I remember, path independent

Mew

12:48 PM

@Sanya yes then we can conclude what I stated earlier

user228700

@JohnRennie Yes, basically.

Mew

that if an irreversible path takes you from A to B then all paths from A to B are irreversible

Because otherwise there is a reversible path from A to B

John Rennie

@Kaumudi Ah you're back!

Mew

and we have the scenario I described where you cannot define the irreversible process

Sanya

@Mew no - reversible processes are path dependent

user228700

12:49 PM

@JohnRennie Yes, I'm sorry. I didn't realize that u'd pinged me :-/

Sanya

irreversible processes are not

both can go from A to B

so to speak, I can divide the set of paths from A to B into two classes

Mew

@Sanya and is the Entropy the same in B in both cases?

Sanya

yeah, the entropy state function

the entropy production along the paths is different

Mew

but the entropy production must always be >=0 for any dS along any path

and if we say that dS is strictly greater than 0 at any point, then the total entropy of B must be greater than A

this then forbids any reversible path

Sanya

we are talking of two entropies

Mew

12:51 PM

two entropies?

user228700

@JohnRennie: Dyou have to go? Shall I come back another time?

John Rennie

@Kaumudi Suppose you ignore the middle bit of the equipment (the bit above the dashed line) and just calculate the pressure difference between the two sides. Are you OK with this approach so far?

Sanya

one is $\frac{\delta Q}{T}$, the other is the state function given by only connecting states with reversible processes

they do only coincide as long as we restrict ourselves to reversible processes

user228700

@JohnRennie Okay, hang on...

Mew

@Sanya I'm thinking entropy as the number of possible microstates that can represent a macrostate

user228700

12:53 PM

What dashed line? Which two sides?

John Rennie

@Kaumudi There's only one dashed line in your drawing ...

Mew

and for a small path dS, comparing the state entropy before and after

user228700

Oh crap, I'm looking at the wrong diagram!

user228700

@JohnRennie I think so.

Sanya

@Mew that's both only a good idea for equilibrium/reversible thermodynamics

I think

Secret

12:55 PM

I suspect, the answer to that question may be twofolds:
1. The dimensions of electromagnetic phenomenon (which includes electricity as a subset) is 3 spatial + 1 temporal dimensional. (I am pretty sure there are formal papers worked out by some guys in the physical community some time ago that has similar results as what I am going to say) Some of the 4D enthusiastics I mentioned a long time ago found that in all dimensions other than 3 + 1, electromagnetism in the form of maxwell equations breaks down and becomes something a bit weird (including magnetic fields having two components due t…

John Rennie

heather

0

A: How many dimensions does electricity have?

Here's an idea for what you maybe could say: Well, there are kind of two "types" of things in the world. First, there are physical objects, like you, me, this house, and so on (here she might chime in with the toaster, or her doll, or something). These physical objects have the property of dimen...

^would anyone mind looking at this?

user228700

@JohnRennie ::Embarrassed::. Yes...

John Rennie

OK suppose we igniore the middle tube then, $P_1 = P_a + \rho g h_1$ and $P_2 = P_a + \rho g h_2$. Yes?

user228700

Yup.

Mew

12:57 PM

@heather I've solved human motivation

John Rennie

@Kaumudi So the pressure difference is $P_1 - P_2 = (P_a + \rho g h_1) - (P_a + \rho g h_2)$.

Mew

@Kaumudi I can verify that for a given line, for the SAME continuous liquid, the pressures are equal (as I used to calculate earlier)

user228700

Oh, crap, sorry.

Secret

Q: Maxwell's equations of Electromagnetism in 2+1 spacetime dimensions

What would be different in the theory of electromagnetism if instead of considering the equations of Maxwell in 3+1 spacetime dimensions, one would consider 2+1 spacetime dimensions?

electromagnetism dimensions maxwell-equations

heather

@Mew, really? how?

John Rennie

12:59 PM

@Kaumudi Now ignore the tanks at the left and right and consider only the tube in the middle above the dashed line. Still OK with this approach?

Mew

@heather easy by considering the limbic system of the human brain

Transcript for