The h Bar

I will outline how you got your reuslt, please be patient

1)

user228700

Patient=Me. Please go on...

you assume Q = mcT, (2) then you go dQ/dt = mdC(T)/dtT + mC(T)dT/dt

this is the equation you arrived at before

user228700

Yup.

the error in the reasoning is because step (1) was wrong

it is false that Q=mcT

user228700

10:01 AM

Okay, I understood that.

instead it is true that dQ = mcdT

user228700

Okay...

Thus all we need to do is divide both sides by dt

dQ/dt = mcdT/dt

whether c(T) is time dependent or not doesn't change the above equation

temperature dependent*

user228700

@JohnRennie: Can we divide like this?

yeah, it looks suspiciously like a deriviative...

10:04 AM

of course

no

it would be incorrect to apply the "d" operator to both sides without applying the product rtule, but you can certainly divide both sides by a small quantity, dt

@KaumudiHarikumar Mew's terminology is a bit misleading. We are differentiating both sides with respect to time, not dividing by dt.

John we are dividing both sides by a small time element dt

there just happens to be a small change in temperature element on the numerator

and thus the result is a derivative

but the mathematical operations are sound

user228700

@JohnRennie Okay, but since it's already $dt$, how exactly do we differentiate that wrt time?

I think differentiate is the wrong word here

because we are not applying the differntial operator in this step

we are merely dividing by a small element of Temperature

i mean time

So you mean it is actually something like this?
$$\frac{\Delta Q}{\Delta t}=mC(T(t))\frac{\Delta T}{\Delta t}$$

10:06 AM

correct

user228700

But dividing isn't the same thing as differentiating...

user228700

So we're not differentiating?

Suppose I have a small element of change in velocity dv

and I divide by a small element of time dt

the result is dv/dt which is accelewration

the act of dividing by dt wasn't differentiating

but it resulted in the final answer being a derivative

user116211

@Mew Acceleration is the limiting value of that fraction.

yes

and d implies the limit

If i use delta instead of d that's fine

but then i would have to write the limit signs

user116211

10:08 AM

@Mew there is difference between $\rm d$ and $d\;.$

but if I use d the limit sign is implied

user228700

@JohnRennie: What is the truth?!?!

user116211

@KaumudiHarikumar what truth? New question today? Or still thermo?

I'm afraid I've completely lost track of the thread ...

user228700

No, this only :P Everyone's disagreeing so needed his expert advice.

10:10 AM

John, we are up to dQ = mC(t)dT

and I say we can divide both sides by dt

which gives dQ/dt = mC(t)dT/dt

No

Yes

user228700

:'-(

user116211

@Mew What about the case when you work with the grad operator? How can you do stuffs like this then?

What we can say is $$\frac{dQ}{dT} = C$$ I'm leaving out the $m$ for convenience

10:11 AM

Yes, but we want the time variation

@Mew For the rate of heat change equation, if we use what @Johnrennie said to "differentiate wrt t, then we will still end up with product rule because $$\frac{d(\Delta Q)}{dt}=m\frac{d}{dt}(C(T(t)\Delta{T}))$$, only if we divide by $\Delta (t)$ instead we recover the correct form of the equation?

And the chain rule tells us that $$\frac{dQ}{dT} \frac{dT}{dt} = \frac{dQ}{dt}$$

And we know $$\frac{dQ}{dT} = C$ so we get $$C \frac{dT}{dt} = \frac{dQ}{dt}$$

which proves my result

There's your answer with no suspect dividing by infinitesimals

Wait, so $$C(T(t)) \frac{dT}{dt} = \frac{dQ}{dt}$$ is a definition?

user116211

10:13 AM

0

Q: Position eigen functions

Confusion about Dirac notation.When represent basis kets of position state by dirac delta function,when by matrices.These are supposed to be eigen function right ?

quantum-mechanics

user116211

Why do they use vague pics like that ;((

i could have proven my method works from first principles (in the same way one can prove the chain rule is correct), but it's not necessary

user228700

Wait, but this $C$ is different from the specific heat!

user228700

This $C$ is just a constant...

C is just the heat capacaity = mc

10:14 AM

no

C can be temperature dependent

user228700

Oh no! The symbol that @JohnRennie used...

user228700

Never mind, I'm an idiot. Understood.

I still don't get how we can ignore the implicit time dependence in C(T) unless we accept that $$C(T(t)) \frac{dT}{dt} = \frac{dQ}{dt}$$ is a definition, and not derived from $dQ=CdT$

user228700

@Secret Yeah, neither do I actually.

Secret, the definition is $dQ = CdT$

or

user228700

10:17 AM

But you know what? Never mind! I'm just gonna stick to the FIRST explanation that @JohnRennie gave, saying $C$ may be considered a constant for exceedingly small temperature changes.

$\frac{dQ}{dT} = C(T)$

Now using the chain rule

user228700

That's probably exactly how my textbook reasoned it.

user228700

Thanks a TON!

\frac{dQ}{dt}\times \frac{dt}{dT} = C(T)$

$\frac{dQ}{dt}\times \frac{dt}{dT} = C(T)$

Multiple by sides by the reciprocal of the derivative $\frac{dt}{dT}$ to give

$\frac{dQ}{dt} = C(T)\frac{dT}{dt}$

I prefer dividing by dt which is mathematically fine, but the above works for those who prefer more rigerous mathematics

What is Q written in full arguments in the dQ/dT=C(T) equation? is it Q(T) or Q(t) or (because of chain rule) Q(T(t))?

Otherwise I can see how it holds now

10:23 AM

It would be Q(t)

hard to say

since Q is only meaningful when considering the change dQ

It should be noted that the usual convention is a lower case 'c' to refer to "specific heat capacity" and an upper case "C" for heat capacity

Back in my physical chemistry course and thermodynamics course, we never encountered many cases where the equation dQ/dt=CdT/dt is used (we mainly used dQ=CdT), which is why it naturally cause me to have a lot of questions. I am glad that you guys are here as otherwise I might have said the wrong thing for this question

The difference is C = mc, that is capital "C" or just "heat capcity" already includes the mass of the sample

yup

by the way:

@mew You can tell Joan to beef up her model to QFT level as there's a comment about the approach

Jan 30 at 15:20, by ACuriousMind

@Secret Well...I have a feeling that trying to model the T-violation in some particle experiment with non-relativistic quantum mechanics instead of QFT or at least relativistic quantum mechanics is not really a sensible approach.

that might help fill in some potential holes in her model that others might have pointed out

Yeah

I was potentially going to do a research topic on QFT applied to this paper

But the more I learned of QFT the harder it seemed to apply to the approach in this paper

That said I never got around to learning QFT that well

Do you know QFT?

None, because I am still trying to beef up my quantum and classical mech more. I am not ready yet but I am eager to learn it

because QFT promise even stranger ways to think about things

Recall that in classical mech, we have well defined trajectories and nearly all dynamics can be formulated in terms of hamitonians and lagrangians, then we entered quantum where things can superimpose and interfere, trajectories no longer well defined, we can only talk about probabilities and then states can be correlated when they interact (entanglement).
Going further into QFT, you cannot even say when and where fields interact and how it interact with itself, only that "it interacts somewhere"

I like the weirdness and non intuitive viewpoints presented by these models, because it makes one to think outside the box more

For example, telling anyone next to you that time may be discontinuous will surely make them go O_O

yasar

10:41 AM

Newbie Question: I remember there was a formula that calculates relative speed of objects that move near the speed of light. How can I find that formula?

i'll find it for you

@yasar See here

hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

man you beat me

fark

yasar

Thanks a lot :)

no worries mate

11:34 AM

it would be great if someone could tell the fault in this derivation

user116211

11:48 AM

@OsheenSachdev Could have used spherical coordinates.

user116211

$$\mathbf v= \dot {\mathbf r} = \dot r\mathbf{e_r} + r~\dot \theta\mathbf{e_{\theta}}\;.$$

user116211

$$\therefore ~~\mathbf a = \dot{ \mathbf v} = \left[\ddot r - r\dot \theta ^2\right]\mathbf{e_r} + \left[r\ddot \theta + 2~\dot r\dot \theta\right]\mathbf{e_\theta}\;.$$

user116211

@OsheenSachdev Now, find any fault by comparing the above general result with your case.

The error is dr/dt does not equal v in your notation

@MAFIA36790 I don't know how to work with spherical coordinates so this won't help

12:00 PM

@OsheenSachdev in your notation, v is tangential velocity, but dr/dt represents the change in the radius over time. In most simple circumstnaces the radius is fixed thus dr/dt is 0.

@Mew dr/dt is change in position vector per unit time so it should be v

user116211

@OsheenSachdev Are you working for circular motion? If so, then $\dot{\mathbf r}= 0\;.$

@OsheenSachdev if r is the position vector then w X dr/dt is equivelent to w multiplied by dr'/dt where r' is the radius

and dr'/dt is 0 as before

@Mew how is w X dr/dt is equivelent to w multiplied by dr'/dt ?

user116211

@OsheenSachdev Remember velocity is $\dot{\mathbf r}$ and not $\dot r\;.$ They are different things.