The h Bar

John Rennie

09:09

@tatan Rutherford's model of the atom is wrong because it doesn't match observation. This is hardly surprising since quantum mechanics wasn't discovered at the time he formulated the model. However it does not violate the second law.

user228700

09:36

Hi, again :-)

user228700

I've a quick question about differentiation...hmm, perhaps I should ask at the M.S.E chat?

anonymous

@Kaumudi.H Depends

user228700

Wokay, I'll ask away before somebody brings up the Golden rule of chat again.

user228700

Sorry for the timeout. I'm typing my question now...

user228700

For functions like the following in which it is not possible to write an expression for $y$ wholly in terms of $x$:

user228700

09:45

$x^3+y^3=2xy$

JamalS

@dmckee I'm wondering if you can help a theorist like me with a problem propagating errors :) I have a spectrum which I am applying Wien's law to. To do so, I've applied a Hermite interpolation to the spectrum, and can use any number of algorithms to find the maximum. However, I'm not sure how to propagate the error of the interpolation ($\max_x |f''(x)|\prod_i^{n}(x-x_i)$ where $x_i$ and $f$ carry their own errors as well) through the method which finds the maximum. Any ideas?

user228700

How to find the range and domain of such implicit functions?

user228700

And I brought up differentiation because...well, never mind that just now.

anonymous

@Kaumudi.H Find value of y at x=0 and x=infinity and x=-infinity. You will get an approximate idea of the graph. Then find dy/dx and see how its sign varies over the number line.

John Rennie

@Kaumudi.H Use the chain rule

user228700

09:48

@anonymous I already suggested this (:-P )

user228700

yesterday, by Kaumudi. H

@JohnR: Whenever you've got the time, please do come up with a bunch of puns and then we'll all vote! (Using stars?)

user228700

@JohnRennie Yeah, no, never mind differentiation...that I can do using, yes, the chain rule.

John Rennie

$\frac{d}{dx} = \frac{dy}{dx} \frac{d}{dy} $

user228700

@anonymous Yeah, I guess that's the general way. Hmm, OK, thanks :-)

John Rennie

Oops, sorry you were asking about the range and domain ...

user228700

09:50

:-) NP, thanks. Although, hmm, so differentiating that function, I find that its derivative becomes undefined at a few points. What does this mean? How can the slope be undefined?

John Rennie

@Kaumudi.H what is the slope of 1/x at x=0?

user228700

Hmm, good point. Hang on...

anonymous

@Kaumudi.H At which points is it undefined ? Those points will have vertical tangent.

user228700

Gosh, OK, yeah, I'm dumb.

user228700

Thanks :-)

user228700

09:52

But it's interesting.

anonymous

The graph will have a loop

user228700

So the function may be defined at that point and still have an undefined slope but then we have cases like $1/x$ in which the function isn't defined at that point in the first place.

anonymous

@Kaumudi.H You studied limits and continuity ?

user228700

Yeah. (Studying...)

Balarka Sen

If it's not defined at a point there's no such thing as the slope there.

anonymous

09:55

You need to check the continuity of graphs

user228700

@anonymous For what?

anonymous

In 1/x there is a discontinuity

at x=0

Balarka Sen

@anonymous Actually it's a point where it's not defined. Not quite a discontinuity.

A "singularity"

user228700

@BalarkaSen Yeah, so what I was asking was if, for implicit functions, finding points where the slope is undefined is a way to find points at which the function itself is undefined...hence the relation to range.

Balarka Sen

I'd call a function discontinuous at $a$ only if it's defined at $a$.

anonymous

09:56

@BalarkaSen It is non removable discontinuity of infinite type.

Ramanujan

I think function is not defined at x=y=0

Balarka Sen

@anonymous I mean, some people do call it a discontinuity, but it's rather strange. I just think continuity should be a property of the function over it's domain (0 is not in it's domain of definition).

But I am not going to delve into semantics again

user228700

..?

Balarka Sen

I also don't know what infinite type means. It's a pole of order 1.

anonymous

@BalarkaSen Its a terminology issue. So lets leave it at that.

user228700

10:00

.__. Terminology? Does anybody know an answer to this:

user228700

4 mins ago, by Kaumudi. H

@BalarkaSen Yeah, so what I was asking was if, for implicit functions, finding points where the slope is undefined is a way to find points at which the function itself is undefined...hence the relation to range.

anonymous

@Kaumudi.H First of all the expression you gave is not a function.

$x^3+y^3=2xy$

isnt a function

user228700

How are u able to ascertain that..?

Balarka Sen

$y$ is - locally - a function of $x$.

anonymous

Put x=1 and see how many values you get for y

@Kaumudi.H

Ramanujan

10:04

@anonymous m.wolframalpha.com/input/?i=x%5E3%2By%5E3%3D2xy&x=0&y=0

user228700

Huh.

anonymous

A function cannot have two values of y for 1 value of x

Balarka Sen

The point is that away from the origin $y$ can be locally defined as a smooth function of $x$. This is the whole point of the implicit function theorem

user228700

Of course. I know that.

Balarka Sen

But as you pointed out, not globally so.

anonymous

10:07

@Kaumudi.H No that is not true. If slope is undefined that does not mean function is undefined at that point.

user228700

@anonymous I know, I know. Like u said, it may be that the tangent is normal to the x-axis and I was dumb enough not to have immediately realised this.

Balarka Sen

I don't even know what slope means for functions undefined at a point.

user228700

@BalarkaSen We're talking about what an undefined slope could mean for a point, not the other way around.

Ramanujan

@BalarkaSen :D

anonymous

@BalarkaSen Slope doesn't exists for those points. We look for f'(a+) and f'(a-) at those points

user228700

10:10

Thanks, guys.

Balarka Sen

@anonymous Those might not exist either; think 1/x at x = 0

Ramanujan

@BalarkaSen what do you think range for that function?

user228700

Oh, also, which are these "...terse posts suggesting that not everyone is happy with this (name change)" ?

anonymous

@BalarkaSen Yes, I agree. In those cases neither left hand derivative nor right hand exists

user228700

Last time I checked, almost everyone but Danu was either happy or didn't care.

anonymous

10:12

Or you can say they are infinite ...... f'(a)=infinity

(which doesn't make much sense though)

Balarka Sen

one heads to infinity and one heads to -infinity, so no.

f'(0) does not exist even in the extended reals

Actually you were correct. derivative of 1/x away from 0 is -1/x^2. so both derivatives are -infinity. So it should have been f'(0) = -infinity.

anonymous

Yeah :P I was thinking of $x^{1/3}$ all this time !

@BalarkaSen True. hyperbola :)

@BalarkaSen Have you learnt thermodyanamics ? I have some queries regarding Clausius inequality

DHMO

Am I doing something wrong, or is the maximum momentum of a particle with fixed mass when v=c/sqrt(2)?

Balarka Sen

what I know about thermodynamics can be written at the back of a postcard, so better ask someone else

DHMO

(sorry to interrupt)

anonymous

10:22

okay!

DHMO

never mind, I did something wrong.

anonymous

10:40

@JohnRennie Do you know in the Clausius Inequality $$\oint \frac{\delta Q}{T} \leq 0$$ what $\delta Q$ stands for ? And is $T$ the temperature of the system or the surrounding ?

John Rennie

$dQ/T$ is the entropy change $dS$. It's basically a statement of the second law of thermodynamics

anonymous

@JohnRennie I wanted to ask what $dQ$ stands for. Is it the heat absorbed by the system or the heat given out by the system ?

John Rennie

I think given the sign convention $dQ$ is the heat entering the system and $T$ is temperature of the system when the heat enters it.

There's quite a nice summary here

anonymous

Thanks. I'm reading it.

I read it. I'm talking an example. Please point out if I am making any mistake. Suppose the we have a Carnot's engine whose source temp is $T_1$ and sink temp is $T_2$. The heat absorbed from source is $q_1$ and heat released into system is $-q_2$ (since heat is being released so we take negative sign). So entropy change of the isolated system is $$\frac{q_1}{T_1}+\frac{-q_2}{T_2}=\Delta{S}$$ ? Right or wrong ? Secondly the condition for the process to be spontaneous is $\Delta{S}\leq0$. Right?

@JohnRennie

John Rennie

That looks OK, though I remind you that it's 30 years since I studied thermodynamics ...

In fact for a reversible engine that implies $\frac{q_1}{T_1} = \frac{q_2}{T_2}$ and I think that's correct.

anonymous

10:57

@JohnRennie One thing that I feel is strange is the fact that a spontaneous process requires the entropy change of the system to be negative or 0. Is this really correct? In that case an isolated system would violate the law that total entropy of universe always increases

John Rennie

Doesn't that just mean the engine is increasing the entropy of its surroundings?

anonymous

@JohnRennie I think $$\Delta S=\oint \frac{\delta Q}{T}$$ only true for reversible processes. For irreversible processes $$\Delta S > \oint \frac{\delta Q}{T}$$

@JohnRennie Yes, that seems correct :)

So it is not correct to say $$\Delta{S}\leq0$$ is the condition for spontaneous process.

John Rennie

For a closed system $\Delta S \ge 0$ for a spontaneous process.

Obviously because that's the second law.

anonymous

@JohnRennie That S is for the whole universe? Right?

I was talking about my previous assumption chat.stackexchange.com/transcript/message/34801647#34801647

14 mins ago, by anonymous

I read it. I'm talking an example. Please point out if I am making any mistake. Suppose the we have a Carnot's engine whose source temp is $T_1$ and sink temp is $T_2$. The heat absorbed from source is $q_1$ and heat released into system is $-q_2$ (since heat is being released so we take negative sign). So entropy change of the isolated system is $$\frac{q_1}{T_1}+\frac{-q_2}{T_2}=\Delta{S}$$ ? Right or wrong ? Secondly the condition for the process to be spontaneous is $\Delta{S}\leq0$. Right?

John Rennie

No, that's the change in entropy of the closed system

Well the Clausius inequality applies to heat engines doesn't it? And they work in a cycle so the net entropy change is zero.

I think the Clausius inequality means the heat engine cannot decrease the entropy of its surroundings i.e. it's a net exporter of entropy.

anonymous

11:11

Okay I think I have got it now! The Clausius Inequality is $$\oint \frac{\delta Q}{T} \leq 0$$. The other condition for spontaneous (isolated) process is $$\Delta S_{sys} \geq \oint \frac{\delta Q}{T}$$ (with equality holding for reversible processes). However it is not correct to conclude that $$\Delta S_{sys} \geq 0$$ for spotaneous processes.

@JohnRennie I don't think so. $\Delta S_{sys}$ for a heat engine may be negative however the total $\Delta S_{sys} + \Delta S_{surr}\geq 0$ for a spotaneous process.

John Rennie

Over the whole cycle the entropy change of a heat engine is zero. That's because entropy is a state function and the engine returns to its original state when the cycle is completed.

anonymous

@JohnRennie Do all heat engines complete a full cycle like Carnot's engine ?

John Rennie

Yes. A heat engine in the normal meaning of the term always works in a cycle.

anonymous

@JohnRennie Okay. I get it now. Thanks a ton =D!

DHMO

12:33

Can anyone explain to me, from a mathematical point of view, why p^2/2m is a better approximation of the kinetic energy than mv^2/2?

(p = γmv, γ = 1/sqrt(1-v^2/c^2), E^2 = (mc^2)^2 + (pc)^2)

(blue: p^2/2m, black: correct kinetic energy, red: mv^2/2)

ArtEze

12:53

Internet

The Internet is the global system of interconnected computer networks that use the Internet protocol suite (TCP/IP) to link devices worldwide. It is a network of networks that consists of private, public, academic, business, and government networks of local to global scope, linked by a broad array of electronic, wireless, and optical networking technologies. The Internet carries an extensive range of information resources and services, such as the inter-linked hypertext documents and applications of the World Wide Web (WWW), electronic mail, telephony, and peer-to-peer networks for file sharing...

DHMO

@ArtEze I tried searching it

ArtEze

@DHMO ¿internet?

John Rennie

@DHMO the exact equation for the KE is $$KE = \sqrt{m^2c^4 + p^2c^2} - mc^2 $$ Yes?

ArtEze

13:12

From Wikipedia.

DHMO

@JohnRennie yes

which is my black curve

John Rennie

OK. we rewrite this as: $$KE = mc^2\left(1 + \frac{p^2}{m^2c^2}\right)^{1/2} - mc^2$$ Is it obvious what I've done here or do you need more detail?

Kenshin

can anyone help in python?

DHMO

@Kenshin yes

Kenshin

yay

@DHMO I have a string that looks like this:

"AcesPercentage": 80.0,

"DoubleFaults": 3,

"DoubleFaultsPercentage": 50.0,

"FirstServePercentage": 47,

"FirstServeDividend": 40,

DHMO

13:18

@JohnRennie oh god why am i so stupid thanks

Kenshin

how can I use python to extract say the first server percentage from that string?

John Rennie

@DHMO you've jumped ahead of me?

DHMO

@Kenshin I have a strong urge to convert that to a dictionary lol

@JohnRennie yes

you've presented the key step

Kenshin

is that eays

DHMO

I automatically linked that to the first line and the last line

ArtEze

13:19

Aug 18 '13 at 16:12, by Dimension10

@Dilaton Of course... It's possible to use LaTeX everywehre, but only with the help of codecogs.

John Rennie

Cool. My work here is done :-)

DHMO

@JohnRennie oh, your work isn't done sorry

why is it a better approximation than mv^2/2?

Kenshin

@DHMO I will be looping through thousands of tennis matches and generating a text like above for each match

then I want to extract the key data from the text string

and save it and ultimately export to csv

ArtEze

LaTeX online. codecogs.com

John Rennie

@DHMO Well as you obviously anticipated the next step is to use a binomial expansion and we get: $$KE = mc^2\left(1 + \frac{p^2}{2m^2c^2} + O()^2\right) - mc^2 $$

DHMO

13:22

>>> s = """"AcesPercentage": 80.0,
...
... "DoubleFaults": 3,
...
... "DoubleFaultsPercentage": 50.0,
...
... "FirstServePercentage": 47,
...
... "FirstServeDividend": 40,"""
>>> s
'"AcesPercentage": 80.0,\n\n"DoubleFaults": 3,\n\n"DoubleFaultsPercentage": 50.0
,\n\n"FirstServePercentage": 47,\n\n"FirstServeDividend": 40,'
>>> eval("{%s}"%s)
{'AcesPercentage': 80.0, 'DoubleFaultsPercentage': 50.0, 'DoubleFaults': 3, 'Fir
stServeDividend': 40, 'FirstServePercentage': 47}
>>> eval("{%s}"%s)['FirstServePercentage']

@Kenshin

Kenshin

ty very much i will try that

John Rennie

@DHMO So we get: $$KE \approx \frac{p^2}{2m} + O()^2$$ i.e. the KE is slightly greater than $p^2/2m$. OK so far?

DHMO

@Kenshin it's probably cheating lol, since I used eval

make sure eval is enabled

@JohnRennie yes

Kenshin

cheating?

DHMO

@Kenshin well, if you consider eval as cheating.

Kenshin

13:24

I don't care about cheating I just want the code to work lol

DHMO

alright

ArtEze

@Kenshin I' m from StackOverflow in Spanish. :)

John Rennie

So if we use $p^2/2m$ we'll get a result slightly lower than it should be. The next step is simply to substitute $p = \gamma m v$ to get $$KE \approx \gamma^2 \tfrac{1}{2}mv^2 $$

@DHMO Yes?

RSerrao

Good afternoon all! I am studying thermodynamics and I got stuck in understanding something. Is there anyone able to help me?

DHMO

@ArtEze hola

@RSerrao not if you don't state your question

Kenshin

13:27

@DHMO I get this error: TypeError: %o format: an integer is required, not str

RSerrao

My question was already stated here physics.stackexchange.com/questions/305451/…

DHMO

@Kenshin what did you type?

ArtEze

@DHMO Hola, jeje.

DHMO

@JohnRennie yes

RSerrao

I may rephrase it

Kenshin

13:28

eval("{%loss_player}"%loss_player)

where loss_player is "s" in your example

DHMO

@Kenshin oh, I misled you

you are supposed to write eval("{%s}"%loss_player)

Kenshin

oh k

RSerrao

eval("{}".format(loss_player)) should work in Python 3 ;)

Kenshin

s is special?

DHMO

@ArtEze por que vienes aca?

John Rennie

13:28

@DHMO And $\gamma > 1$ so that means $p^2/2m > \tfrac{1}{2}mv^2$

DHMO

@RSerrao no, I want to make the string "{abc}" for the string "abc"

RSerrao

hmm ok :P I tried to guess it :P

Kenshin

@DHMO awesome it works thanks

ArtEze

@DHMO Porque vino Kenshin al chat en español.

John Rennie

@DHMO So we conclude that $$KE > \frac{p^2}{2m} > \tfrac{1}{2}mv^2 $$

Kenshin

13:30

I don't know why it works but it works

DHMO

@JohnRennie Could you tell me if I went wrong here?

> $E_{\mathrm k}$

> $\displaystyle = m_0 c^2 \left({\gamma - 1}\right)$

> $\displaystyle = m_0 c^2 \left({\left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5} - 1}\right)$

> $\displaystyle \approx m_0 c^2 \left({1 - (-0.5) \left({ \frac v c }\right)^2 - 1}\right)$

> $\displaystyle = \frac 1 2 m_0 v^2$

RSerrao

@Kenshin you may want to read about "string format specifiers"

Kenshin

@DHMO r u able to explain briefly why it works?

I have never used the "eval" before

DHMO

@Kenshin eval basically evaluates the given input as python code

since {"abc":1,"def":2} is basically python code for a dict

putting it in eval generates a dict

Kenshin

i see

cool

so I'm lucky the data I needed was in the dictionary format

DHMO

13:31

yes

that's why I said it was cheating

ArtEze

Genial animé Ranma y 1/2.

Kenshin

what would we do if the delimiters were different?

out of curiosity

John Rennie

@DHMO that looks OK. You got the correct result that the KE is approximately $\tfrac{1}{2}mv^2$

DHMO

@Kenshin I would use str.replace :p

Kenshin

lol

DHMO

13:32

@JohnRennie it looks like the same method to me

John Rennie

It's just that your approach oesn't help you understand why $p^2/2m$ is a better approximation.

RSerrao

@DHMO can you point me to anyone who can help me? ;)

DHMO

@RSerrao the guy above you

RSerrao

John Rennie?

DHMO

@JohnRennie ya, thanks

@RSerrao yes

Kenshin

13:33

thanks heaps @DHMO

RSerrao

@DHMO thanks ;)

@JohnRennie can you try and give me a hand with something?

John Rennie

@RSerrao I saw your question but my thermodynamics is pretty rusty. I mainly do general relativity these days.

DHMO

@JohnRennie wow, so I'm lucky

ArtEze

Pi

The number π is a mathematical constant, the ratio of a circle's circumference to its diameter, commonly approximated as 3.14159. It has been represented by the Greek letter "π" since the mid-18th century, though it is also sometimes spelled out as "pi" (/paɪ/). Being an irrational number, π cannot be expressed exactly as a fraction (equivalently, its decimal representation never ends and never settles into a permanent repeating pattern). Still, fractions such as 22/7 and other rational numbers are commonly used to approximate π. The digits appear to be randomly distributed. In particular, the...

John Rennie

@RSerrao I vaguely recall calculating the entropy change of stretched rubber bands - part of the eperiment that shows rubber bands contract on heating. I'd guess a quick Google would find you lots of articles discussing that experiment and they'd probably help.

@ArtEze why are you posting random links to Wikipedia?

ArtEze

13:38

@JohnRennie hehe, i don't know.

RSerrao

@JohnRennie thanks

DHMO

@JohnRennie So basically $E = \gamma m_0 c^2$ and $p = \gamma m_0 v$.

why?

John Rennie

Well $p = \gamma mv$ is the definition of relativistic momentum

DHMO

But you do see it in $\lambda = \dfrac h p$

John Rennie

That's the de Broglie wavelength. I'm not sure of the relevance of that ...

DHMO

13:46

Well, it shows that the $p$ does have some real-world significance...

John Rennie

Well yes, and $p = \gamma m v$ has real world significance too. For example the force exerted by a relativistic particle is the rate of change of relativistic momentum.

Anyhow, I have to go now. Back in a couple of hours.

DHMO

@JohnRennie so what exactly does $\gamma$ affect and why?

Ramanujan

Why are you people saying p=\gamma MV ? I only know p=mv

DHMO

@Ramanujan $p=mv$ is classical momentum

relativistic momentum is $p = \gamma mv$, and when $v \ll c$, $p \approx mv$.

heather

14:05

hello

DHMO

hi

Pissed off layman

14:25

yo

Wow! @JohnRennie the British pound took a dive after the Brext speech.

Pissed off layman

14:50

@anonymous The talk got cancelled :(

Probably health issues...

anonymous

:( I was suspecting something like that

Too bad

Wish he gets well soon and gives another speech some other day

@Pissedofflayman

Pissed off layman

Yup

heather

does anyone have any thoughts on having a Jon Skeet style post for John Rennie?

(see Jon Skeet Facts)

Pissed off layman

His new speech synthesizer is supposed to double his speed. @anonymous

John Rennie

@Pissedofflayman we haven't even had the Brexit speech yet! The fall in the pound on Friday was just due to the announcement of the speech. God help us when we actually have the speech!

heather

14:55

@JohnRennie, hello =)

John Rennie

@heather It was funny the first time, but I'm not sure it bears repeating.

Remember that Stack Overflow was the first ever Stack Exchange site, so Jon Skeet was the first big rep member the SE ever had.

I think all the good Jon Skeet type jokes have already been done.

anonymous

Make a Wiki page on John Rennie :D

@heather

John Rennie

: -)
They'd just close it on the grounds of "not famous enough"

heather

@JohnRennie "John Rennie can go faster than the speed of light."

anonymous

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