The h Bar

4 Thermodynamics Work done in Irreversible process, Isobaric Process

They are both huuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuge books :-(

user228700

Okay! I definitely don't have time to read either of them then! I'll have to resort to watching Hindi vids. like:

user228700

►

@JohnRennie Number 4: Doppler Effect - As the ambulance goes past the sirens are of shorter wavelength and hirer frequency, once the ambulance passes the sirens will have a lower wavelength, lower frequency, lower pitch ?

user228700

Please pray that I come out alive :P

10:04 AM

@CodeRatchet Looks good, with possibly a slight correction: as the ambulance moves towards you the frequency will be higher and as the ambulance moves away from you the frequency will be lower.

Though I suspect this is what you meant anyway.

@JohnRennie Yeah that's correct thanks

Can we tackle question 2 or 3?

2) You have filled a kettle with cold water and placed it on a gas stove to boil. Explain how heat exchange will occur in this situation.

3) Whilst being operated, one of the double insulated construction (DIC) equipment encounters a fault. Explain if the operator will suffer from shock or not

Q3 I can't answer because I don't know what double insulated construction (DIC) equipment is, and the question doesn't say what the fault is.

Ok leave that with me, question 2 would that fall into Energy Transfer?

Q2 seems awfully vague.

The water will get hot and eventually boil away.

So I suppose you'd say that heat from the burning gas heats the water then when the water reaches 100C the heat from the gas turns it into steam.

But it isn't clear to me if this is what the question means or not.

Ok so I'm starting to think question 2/3 are linked together

10:16 AM

@CodeRatchet it doesn't look to me as if Q2 and Q3 are linked ...

Ok, I think I have enough information to hopefully do some more research and give a better answer. Thanks for you help @JohnRennie appreciate it.

@CodeRatchet you're welcome :-)

user116211

10:32 AM

@JohnRennie hmm. Well, let me begin...

Good lunch? :-)

user116211

@JohnRennie Kinda... grilled veg sandwich ;)) My favourite.

user116211

With Heinz ;P

user116211

Anyways, ...

user116211

So, first he defines a line-element $\overline{\mathrm ds}^2$ based on kinetic energy of the system.

10:35 AM

OK

user116211

Namely $$\overline{\mathrm ds}^2 = \sum_{i\,=\,1}^Nm_i~(\mathrm dx_i^2 + \mathrm dy_i^2 + \mathrm dz_i^2)\tag{15.11}$$

user116211

Then he concludes that the co-ordinate system is Euclidean with coordinate axes being $\sqrt{m_ix_i}~~~ \sqrt{m_iy_i}~~~\sqrt{m_iz_i}\,.$

user116211

Well, I could digest that.

user116211

Then he transforms the variables to generalised coordinates $q_1,\ldots, q_n\,.$

user116211

He writes:

user116211

10:38 AM

> [...] the geometry remains Euclidean, although the line-element is given by the more general Riemannian form: $$ \overline{\mathrm ds}^2 = \sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ with $n= 3N\,.$

OK

user116211

Now, why does the geometry "remains" Euclidean? I didn't get that (since I'm not into differential geometry till yet, maybe).

user116211

Well, this is the part of the query but not the main one....

user116211

Then he considers a case of a system with given kinematical constraints.

I think his point is that (15.11) shows the space is flat, because we can choose global coordinates in which the space is flat everywhere.

user116211

10:42 AM

@JohnRennie sure, because that is Euclidean?

The curvature is coordinate independent, so it does not depend on the choice of coordinates.

user116211

@JohnRennie ohh, okay, got it!

user116211

@JohnRennie That's why even after transforming to the generalised ones, the geometry remains Euclidean, is it so?

Yes. Starting from a flat space it's possible to choose curved coordinates in which the metric doesn't look like a Euclidean metric. However the apparent curvature is due to your choice of coordinates and not a property of the space.

user116211

Going to the second case, he attacked the problem by firstly converting the coordinates to generalised curvilinear ones i.e. $q_1,\ldots, q_n\,.$

user116211

10:45 AM

@JohnRennie that's why he said, the line-element becomes general Remainnian but the geometry remains the same i.e., Euclidean...

> The last result and the remarks preparatory to
it have been aimed at understanding how one
might prescribe the vector z that arises in Eq.
(23) for any given practical situation. We have
found that the use of the generalized D’Alembert’s
Principle (which requires a speci5cation of the
vector C at each instant of time) yields a unique
characterization of Qc
ni, and therefore of Qc
. We
have thus obtained, within the framework of Lagrangian
dynamics, the explicit equation of motion
for systems with non-ideal constraints.

@MAFIA36790 for example, in the rest frame of an accelerating observer the metric is the Rindler metric and this doesn't look flat. It even has an event horizon!

What is lagrangian mechanics without the lagrangian appearing anywhere, sometimes I don't understand terminologies

But this is because the if we use the rest frame of an accelerating observer those coordinates are curved. Spacetime is flat.

user116211

@JohnRennie Then it must be that just because the coordinates are curvilinear doesn't have to mean the geometry is not Euclidean? good point.

user116211

10:47 AM

@JohnRennie ohh.

user116211

@JohnRennie This makes the thing clearer.... okay.

To look at the curvature you'd need to calculate the curvature invariants.

user116211

@JohnRennie Riemannian curvature tensor, I guess?

In GR they would be the Ricci scalar, Kretschmann scalar and so on. I don't know what the equivalents would be in this case, but they must exist.

user116211

@JohnRennie okay...

10:49 AM

The trouble with the Riemann tensor is that when you write it down its components will depend on the coordinate system you choose.

So it wouldn't necessarily be immediately obvious that the space is flat.

But isn't the riemann curvature tensor has to vanish if spacetime is flat?

But the curvature invariants are coordinate independent.

So you mean there are coordinate systems in flat spacetime where the riemann curvature tensor will not vanish?

user116211

@JohnRennie yes, that's the crux of the discussion, I suppose.

user116211

anyways, let me come to the second case....

user116211

10:54 AM

> [...] express from the very beginning the rectangular coordinates of the particles in terms of $n$ parameters $q_1,\ldots,q_n\,.$ these parameters are now the curvilinear coordinates of an $n$-dimensional space whose line element.... takes the form: $$\overline{\mathrm ds}^2 = \sum_{i, \, k\,= \,1}^n a_{ik}~\mathrm dq_i\mathrm dq_k\,.\tag{15.16}$$

user116211

> The $a_{ik}$ are here given the functions of the $q_i\,.$ The line element is now truly Riemannian not only because the $q_i$ are curvilinear coordinates, but because the geometry of the configuration space does not preserve the Euclidean structure of the original $3N$-dimensional space...

user116211

Notice that the line-element defined from $(15.11)$ and the one in $(15.16)$ look exactly the same.

user116211

They are in the general Riemannian form and defined on the generalised coordinates.

OK, though I must admit I'm not sure what he means by the geometry of the configuration space does not preserve the Euclidean structure of the original 3N-dimensional space

user116211

@JohnRennie My query!!!

user116211

10:57 AM

Why does the former line-element preserve the Euclidean Geometry while the latter develop geometry which does not preserve the Euclidean structure of the $3N$-dimensional space?

user116211

That's the point I'm not getting.

user116211

Why did the geometry developed from one remain Euclidean while the exact opposite thing happened in the latter case even though both used the curvilinear generalised coordinates and line-element in the general Riemanninan form and both looking exactly same?

user116211

@JohnRennie The former one does preserve that.... hmmm. Confusing ;/

Aaaaaaaaaaaaah ...

user116211

It spoilt my whole day; I'm stuck at this very point T__T

11:00 AM

Note that he says: Let us now consider a system with given kinematical conditions between the coordinates

user116211

@JohnRennie yes?

user116211

Degrees of freedom is not same in both the cases, I know that.

Does that mean the particles can interact with each other i.e. exchange momentum?

user116211

@JohnRennie He didn't say such thing.

I guess I just don't understand the terminology because I don't know what given kinematical conditions between the coordinates means.

user116211

11:02 AM

@JohnRennie He only says this:

user116211

> Let us now consider a system with given kinematical conditions between the coordinates.

user116211

That's it. Then he told there are two ways to attack the problem.

user116211

The one I discussed above is the second one.

user116211

One of the worst Sundays... how could I go ahead if I don't understand this? Damn ;/

Go back to (15.11)

user116211

11:07 AM

@JohnRennie okay

This form assumes all the particles move independently. That is, a movement $dx_i$ of one particle doesn't affect any other particle.

user116211

@JohnRennie yes.

If there were any interactions you'd get cross terms $dx_i dx_j$

And in that case the space would not be flat.

user116211

@JohnRennie sure.

user116211

Then?

11:09 AM

@MAFIA36790 I think so, though I wouldn't swear to it because I'm a fair way outside my comfort zone.

user116211

hmm. okay.

If this is what is meant by given kinematical conditions between the coordinates then such a system would not have a Euclidean space.

It would have a curvature that could not be transformed away by messing with the coordinates.

user116211

@JohnRennie Kinematical conditions mean constraints.

@MAFIA36790 ah, OK. I have no idea what constraints do to the geometry.

user116211

@JohnRennie They decrease the parameter $n,$ isn't it?