The h Bar

@Semiclassical In a graduate class that uses complex analysis but isn't a complex analysis class, do you think I can just write "Laurent series computed with Mathematica"?

Semiclassical

What series are you having to compute?

0ßelö7

@Semiclassical like $(z^2-c^2)^{-1/2}$

I need to do residues

Semiclassical

well, that particular one wouldn't have residues

0ßelö7

it should have a simple pole, no?

Semiclassical

hmm

at infinity, you mean?

it's pretty easy to do by hand if you do $z=1/w$.

0ßelö7

1:08 AM

no....when $z^2=c^2$...

Semiclassical

that's not a simple pole.

0ßelö7

it's 1/0 there and the denominator scales linearly

Semiclassical

$\frac{1}{\sqrt{z^2-c^2}}=\frac{1}{\sqrt{z-c}\sqrt{z+c}}$

near $z=c$, that's $\frac{1}{2c}\frac{1}{\sqrt{z-c}}$.

and similarly near $z=-c$

it's got branch points at $z=\pm c$ and a pole at infinity.

0ßelö7

hmmmmm

Semiclassical

plus, the only place where the denominator scales linearly is at infinity

$\sqrt{z^2-c^2}$ scales as $\sqrt{z-c}$ near $z=c$.

0ßelö7

1:15 AM

well this is no bueno because the lift is given by the residue of that thing :<

need to think more

Semiclassical

note that while you don't have a residue except at infinity, you do have a nonzero contour integral if you integrate around a circle containing $z=\pm c$.

0ßelö7

Oh, this thing isn't meromorphic so residue doesn't apply?

Semiclassical

Not sure what you mean by that.

Oh, I see.

0ßelö7

to apply the residue theorem you need a meromorphic function, I think

Semiclassical

Yeah.

0ßelö7

1:20 AM

and this has a branch cut so it's not (?)

Semiclassical

Simple version of this: Suppose you've got $\sqrt{1-z^2}\,dz$.

0ßelö7

k

Semiclassical

Suppose you integrate around the branch cut $[-1,1]$

0ßelö7

I still don't really understand those. Why is the branch cut there?

Semiclassical

my brain isn't working like it should, but note that it behaves like $\sqrt{-z-1}\sqrt{z-1}\approx \sqrt{-2}\sqrt{z-1}$ near $z=1$

and since $\sqrt{-z-1}$ is analytic away from $z=-1$, it'll have a convergent Taylor series in the vicinity of $z=1$. Hence the powers of $z-1$ that show up in the product are $1/2,3/2,5/2,\ldots$.

But I feel like this misses the point somewhat.

0ßelö7

1:31 AM

@Semiclassical Yeah, and the function I have is actually $z/\sqrt{z^2-c^2}$

this doesn't have a limit as $|z|\to\infty$ which is pretty bad

Semiclassical

Well, note that $$z(z^2-c^2)^{-1/2}=z^2(1-c^2/z^2)^{-1/2}=z^2\left(1-\frac{c^2}{2z^2}+ \frac{c^4}{8z^2} +\cdots\right)$$

so it's not analytic at infinity, but it doesn't have a residue there.

0ßelö7

@Semiclassical wolfram gives the expansion as $z+$ stuff

Semiclassical

where? near zero you're right.

0ßelö7

yeah

Semiclassical

I was doing it at infinity

0ßelö7

1:44 AM

oh

ah, yes

I am not pleased with this. It's supposed to go to zero at infinity

Semiclassical

oh. I made an error above: factoring out the z^2 from inside the square root should cancel the z out front

(it's a power of -1/2, not 1/2)

so the z^2 out front in the rest of the terms should be ignored.

0ßelö7

so that's the expansion at Re z = + infty?

Semiclassical

$|z|=\infty$.

0ßelö7

that doesn't sound right because the limit as $z\to-\infty$ is $-1$

Semiclassical

Why? $\sqrt{z^2-c^2}\to +\infty$ as $z\to-\infty$

doh

0ßelö7

1:49 AM

but $z\to-\infty$ in the numerator

Semiclassical

yeaah

I suspect the right way to do this is to consider $\dfrac{z}{\sqrt{z^2-c^2}}dz$

0ßelö7

This is really fishy. I need to find the far-field velocity to compute the lift, but there's no far-field velocity!

Semiclassical

and let $w=1/z$

Actually

maybe I should be writing this as $$\frac{z}{\sqrt{z^2-c^2}}=\frac{z}{\sqrt{z^2}}\left(1-\frac{c^2}{z^2}\right)^{-‌1/2}$$

0ßelö7

ugh also the proof of the lift theorem assumes the velocity's integral is given by the residue theorem

Semiclassical

point being, $\frac{z}{\sqrt{z^2}}\to +1$ along $z\to\infty$

0ßelö7

1:53 AM

well, I don't understand fluid mechanics

I'm gonna fail

Semiclassical

this is conformal mapping business?

0ßelö7

yeah

Semiclassical

What are you actually supposed to compute, some contour integral?

0ßelö7

I conformally mapped a flow around a disk onto the exterior of an ellipse, and need to calculate the lift

Semiclassical

hmm. my knowledge is too limited to offer anything much

0ßelö7

1:57 AM

but the result is strange...and it doesn't really make sense in the case when the ellipse is just the circle again

it should just be the same flow but...scaled?

and there's no $|z|\to\infty$ limit of the velocity, so it's not a well-defined far-field type flow

Semiclassical

Did the flow around the disk have well-defined far field flow?

0ßelö7

Yeah, it's $U(1-a^2/z)$ for a disk of radius $a$

so it goes to $U$ at infinity, which is $U\mathbf i$ in real coordinates

Semiclassical

you may want to recheck the conformal transformation itself, then

because somethhing sounds fishy

0ßelö7

msme.us/2010-1-5.pdf what I did matches up with this

namely (11) for the potential, and then I differentiate that

Semiclassical

hmm, but the profiles they get do look to have good far-field flow

0ßelö7

2:03 AM

@Semiclassical By inspection you can see that the derivative of (11) is $\propto C_1+C_2 z/\sqrt{z^2-c^2}$

Semiclassical

what are a,b here?

0ßelö7

major and semimajor axis

Semiclassical

hmm

if a=b, then the square root term will cancel out

so if a=b then C2=0

0ßelö7

I get $$U\frac{1}{2}\frac{2bz}{(a-b)\sqrt{z^2-c^2}}$$

it doesn't cancel out unless I'm missing something

Semiclassical

if a=b then that's not well-defined

0ßelö7

2:06 AM

yeah

well if $a=b$ then there's some other issues

Semiclassical

yeah

I wasn't reading that right.

0ßelö7

to get to this form they had to multiply and divide by stuff

Semiclassical

What is $c$ here?

0ßelö7

$a^2-b^2=c^2$

Semiclassical

oh, $a^2-b^2$

Right. So $c\to 0$ as $a\to b$

and then $\frac{z-\sqrt{z^2-a^2+b^2}}{a-b}$ acts a bit weird as $a\to b$

0ßelö7

2:09 AM

so I have $$w(z)=\frac{U}{2}\left\{z+\sqrt{z^2-c^2}+\frac{(a+b)^2}{z+\sqrt{z^2-c^2} }\right\}$$ before simplifying

there

that works as $a\to b$

Semiclassical

oh, that's $U(\frac{a+b}{2})=U\cdot (a+b)/2$?

0ßelö7

yeah

Semiclassical

ugh

0ßelö7

$U$ is the velocity of the circular flow

Semiclassical

sure

I mean, the $z=x+i y$ in there makes sense

0ßelö7

2:14 AM

@Semiclassical Hmm?

Semiclassical

well, if you only had $w(z)=Uz=Ux+iUy$

right.

0ßelö7

there might be another derivative in there, I'm confused now

I might find the professor, this isn't nice

Semiclassical

I feel like we're making this more complicated than necessary somehow

but me trying to help you is a pretty futile endeavor given how little I remember

0ßelö7

@Semiclassical I mean the guy gave us that transformation in class and said to use it on the homework

but the answer is clearly nonsense because the far-field flow is wrong

which you confirmed

Semiclassical

@0ßelö7 well, the version you wrote here doesn't look entirely absurd

for large $z$, it behaves as $w(z)=(U/2)(z+\sqrt{z^2})$

which if you can justify interpreting as $2z$, gives $w(z)=Uz$ as expected

0ßelö7

2:19 AM

@Semiclassical in that note they multiplied and divided by $z-\sqrt{z^2-c^2}$ to get rid of the denominator

Semiclassical

sure, they rationalized it

0ßelö7

yeah, that's the PhD math word

or...highschool math :P

@Semiclassical hmm

so the derivative should again be $U$ (for large $|z|$), but it isn't

Semiclassical

it's something to do with those square roots

0ßelö7

@Semiclassical the derivative is $$\frac{z\sqrt{z^2-c^2}+z^2-(a+b)^2}{\sqrt{z^2-c^2}(\sqrt{z^2-c^2}+z)}+1$$

that lone $z$ out front causes issues again

@Semiclassical Look at (18) of that note. That $y$ on top also causes issues! Where did they come up with those plots

I am misunderstanding something crucial here

Semiclassical

What I'm trying to do is use their $w(z)$ to make an actual stream plot in mathematica

0ßelö7

2:29 AM

@Semiclassical the streams are level sets of the function $\psi=\mathrm{Im}\, w$

Semiclassical

yeah, mathematica doesn't like those level sets

this seems like a more readable source: math.nyu.edu/faculty/childres/chpfour.PDF

see in particular example 4.6 on the fourth page

alas, it seems to duplicate the results of the other source

0ßelö7

"This insures that that infinity maps by the identity and so the uniform
flow imposed on the circular cylinder is also imposed on the ellipse."

I think our calculations directly contradict that, no?

Semiclassical

actually, here's something weird

If I write the square root in mathematica as $\sqrt{z-c}\sqrt{z+c}$ instead of $\sqrt{z^2-c^2}$

then I think I get the right contours

0ßelö7

@Semiclassical if you post the code I'll plug it into MMA

Semiclassical

Show[ContourPlot[
Im[1/2 (z + Sqrt[z - Sqrt[3]] Sqrt[z + Sqrt[3]] + 9/(
z + Sqrt[z - Sqrt[3]] Sqrt[z + Sqrt[3]])) /.
z -> x + I y], {x, -3, 3}, {y, -3, 3}],
Graphics[{Disk[{0, 0}, {2, 1}]}]]

that's with a=2, b=1

0ßelö7

2:38 AM

Nice!

Semiclassical

I'm not entirely surprised: I've run into this sort of thing before where mathematica has trouble with branch cuts

0ßelö7

@Semiclassical but isn't the branch cut inside of the ellipse?

Semiclassical

yep.

from focus to focus

0ßelö7

@Semiclassical that's something at least

Semiclassical

thing is, the trouble mathematica has isn't just at the branch cuts

to see what i mean, replace Sqrt[z - Sqrt[3]] Sqrt[z + Sqrt[3]] with Sqrt[z^2-3] and see how much trouble it runs into

0ßelö7

2:41 AM

@Semiclassical So can this plot $w'(z)$?

Semiclassical

the trouble, I think, is that along the imaginary axis one has $z^2-3<0$

so it ends up jumping in its definition of $\sqrt{z^2-3}$ along the imaginary axis

and thus, suffering

0ßelö7

@Semiclassical lol, ok

Semiclassical

that simplifies to $$w'(z)=\frac{z^2+\sqrt{z-\sqrt{3}} \sqrt{z+\sqrt{3}} z-6}{z^2+\sqrt{z-\sqrt{3}} \sqrt{z+\sqrt{3}} z-3}$$

in the case of $(a,b)=(2,1)$

oh, nice

Show[StreamPlot[{Re[w], -Im[w]} /.
w -> (-6 + z^2 + z Sqrt[-Sqrt[3] + z] Sqrt[Sqrt[3] + z])/(-3 +
z^2 + z Sqrt[-Sqrt[3] + z] Sqrt[Sqrt[3] + z]) /.
z -> (x + I y), {x, -3, 3}, {y, -3, 3}],
Graphics[{Disk[{0, 0}, {2, 1}]}]] @0ßelö7

blah, should've put the $z$'s in front of the square roots there. (i c/p'd from mathematica)

0ßelö7

o.o

Semiclassical

Lesson of the day: "g***** branch cuts"

0ßelö7

2:49 AM

so...where do the z's go?

this plot looks really good

Semiclassical

what I meant re: z's was that I should've written $z\sqrt{z-\sqrt{3}}\sqrt{z+\sqrt{3}}$ not $\sqrt{z-\sqrt{3}}\sqrt{z+\sqrt{3}}z$

as for wtf is going on...hnnng

0ßelö7

why does that matter?

Semiclassical

aesthetics

0ßelö7

lol

Semiclassical

harder to miss the $z$ out front

0ßelö7

2:58 AM

@Semiclassical how do I get these to plot farther out?

like -100 to 100

Semiclassical

change {x,-3,3},{y,-3,3}

to whatever bounds you want

0ßelö7

ah yeah

@Semiclassical yeah well I must be retarded. I just plotted the thing that should diverge and it seems fine

I don't get it!

Semiclassical

g***** branch cuts, man

I have memories of having to deal with those issues during prior research.

Not fond memories.

0ßelö7

@Semiclassical I feel for you

@Semiclassical you know this is symmetric so there still probably won't be any lift .-.

Semiclassical

yep

0ßelö7

3:11 AM

I need to add a vorticity term

Semiclassical

welp

have fun with that

0ßelö7

gonna play a video game and talk to a guy in my class tomorrow

this is too hard

Semiclassical

good call

0ßelö7

thanks for your help

Semiclassical

One thing I can confirm: $\frac{dw}{dz}$ has a series expansion of the form $1+C/z^2$ for large $z$

0ßelö7

3:13 AM

well that's good

Semiclassical

so no pole at infinity

which seems consistent with the symmetry guaranteeing zero lift

0ßelö7

I just need it to be constant at infinity

Semiclassical

yeah, it's definitely got sensible far-field flow

0ßelö7

the lift is given by $$-\rho |\mathbf u_\infty| \hat n\,\mathrm{Re}\oint_{\text{boundary of the body}}w' \,dz$$

Semiclassical

right

0ßelö7

3:15 AM

where $\hat n$ is the perpendicular to the far-field flow

Semiclassical

so difference of integrals of $dw$ along the top boundary vs. bottom boundary

and since there's no pole at infinity, and it's analytic away from the cut, one can just stretch that contour and contract it to zero at infinity

hence, zero lift.

to get it nonzero, hmm

0ßelö7

@Semiclassical Well I learned how to give lift to a flow around a disk

Semiclassical

I think it should be enough to change the far field behavior

0ßelö7

one adds a swirly thing

Semiclassical

yeah, sounds right

what I have in mind is that if you were to rotate the ellipse slighty, then I think---hmm, maybe no

0ßelö7

3:18 AM

$$w(z)=U\left(z+\frac{a^2}{z}\right)+\frac{\Gamma}{2\pi i}\log z$$

Semiclassical

I'm trying to think whether that breaks the symmetry enough

0ßelö7

@Semiclassical I think I should just apply the conformal transformation to that guy

Semiclassical

yeah

godspeed :)

0ßelö7

the result will be horrific

Semiclassical

actually, that's not quite true

0ßelö7

3:19 AM

my avatar nicely sums up my reaction to this task

Semiclassical

the new term is $\log(z+\sqrt{z^2-c^2})$, right?

But I think that's an inverse trig function

0ßelö7

@Semiclassical with a $1/2$ in there, but I guess that can come out

@Semiclassical huh

Semiclassical

hrm. I may be wrong

0ßelö7

maybe inverse hyperbolic

Semiclassical

yeah, that's what I meant.

ahah, check equation (1): mathworld.wolfram.com/InverseHyperbolicCosine.html

0ßelö7

3:22 AM

it's almost inverse cosh

@Semiclassical that works when $c^2=1$...

can one scale that away?

Semiclassical

hm. no wonder I wasn't getting that to work

no, I don't think so

This probably is some statement about what happens as $c\to 1$

0ßelö7

Huh. It's $c=0$ that's exceptional.

Semiclassical

though I can't quite see what that represents.

Yeah.

I mean, $c=1$ really shouldn't be anything special.

So idgi

0ßelö7

Since I have to figure this out tomorrow I should probably do my programming tonight

damn homework

@Semiclassical I don't know what's worse, Fortran or banch cuts.

Semiclassical

3:50 AM

doing branch cuts with Fortran :P

0ßelö7

@Semiclassical there's a special place in hell for people who do that

Semiclassical

yes, the place in which people doing conformal mappings in the 70's had to live :P

John Rennie

5:09 AM

@BalarkaSen Go on then :-)

Balarka Sen

6:16 AM

@JohnRennie I had this in mind.

John Rennie

Eloy. Gosh that takes me back.

There were a lot of those trippy bands around back in the day. Mostly forgettable it has to be said.

Balarka Sen

Yeah psychedelic rock seems to be a common trend back in 70s/80s

John Rennie

Is that a skunk or a honey badger? Or something else?

Balarka Sen

Sloth :P

John Rennie

I must admit the exact meaning of the meme escapes me :-)

Balarka Sen

6:20 AM

I originally created it as a response to someone on the math chat who uses spacey backgrounds with sloths floating over them as gravatar.

But I figured you'd appreciate the Eloy reference.

John Rennie

In principle I listen to bands like Eloy because I'm a cool hip dude. In practice a little Eloy goes a long way. I have a large collection of albums from Eloy, The Enid, The Orb, etc, etc that get listened to at least once a decade.

These days I tend to listen to the weird shit that you lot post about :-)

Balarka Sen

I actually quite liked "Ocean". Admittedly I haven't listened to any other album of theirs

Hah. I am really into weird songs these days.

John Rennie

I guess my view is that there's lots of interesting stuff around that I haven't heard. Albums that are several decades old need to be really good for me to listen to them instead of the interesting new stuff. There are albums like that, but Eloy aren't high on that list :-)

I'm currently listening to last night's BBC Radio One rock show. OK 90% of it is average, but there is some really good stuff there that is good to hear.

Balarka Sen

Yeah I understand that. My musical experience is really very shallow compared to yours so I tend to fiddle around and listen whatever as long as it's not mainstream.

I don't know a lot of good rock music of the modern times.

It's mostly hip hop that's stealing the show, which I can't for the life of me appreciate

John Rennie

That's because I've been listening to crap music for nearly 50 years :-)

Balarka Sen

6:29 AM

That edit!

John Rennie

:-)

One of those oops moments :-)

I think it's good to have a varied taste. I like to listen to new stuff because that's how you find out about interesting new bands. But it also means you end up listening to some really awful stuff. You need to play some of your old favourite albums from time to time to stay sane :-)

Balarka Sen

Hahah

Mithrandir24601

@BalarkaSen does classical music count as non-mainstream?

Balarka Sen

Probably, yes. I don't listen to classical music very much, but I should.

Dodsy

6:45 AM

@BalarkaSen try the goldburg variations.

Balarka Sen

I'll have a look at some point in my life.

I have 0 background on classical music

Mithrandir24601

:) I have a new favourite piece of piano music - Beethoven's Appassionata :)

Dodsy

I'll have to give it a shot.

I like the gymnopedies by erik satie.

my favourite beethoven is moonlight sonata 3rd movement.

Anonymous

@JohnRennie Do you happen to know about leakage inductance and stuff related to that? I'm stuck...I don't know why leakage factor $\sigma=1-k^2$

Balarka Sen

I don't enjoy Beethoven; I prefer Bach way more. But that's just me

Dodsy

6:49 AM

The goldburg variations are bach.

Sid

...I don't listen to music. :/

Should I?

Dodsy

eh, I'd probably sooner go deaf than blind.

John Rennie

@Blue Leakage factor is useful flux divided by total flux isn't it? Or something like that.

@Sid only if you enjoy it.

Anonymous

@JohnRennie I think so, but I don't know how to derive that formula

John Rennie

What's $k$?

Anonymous

6:51 AM

$k$ is the coupling factor

Anonymous

Inductance

In electromagnetism and electronics, inductance is the property of an electrical conductor by which a change in current through it induces an electromotive force in both the conductor itself and in any nearby conductors by mutual inductance. These effects are derived from two fundamental observations of physics: a steady current creates a steady magnetic field described by Oersted's law, and a time-varying magnetic field induces an electromotive force (EMF) in nearby conductors, which is described by Faraday's law of induction. According to Lenz's law, a changing electric current through a circuit...

Anonymous

"The coupling coefficient is a convenient way to specify the relationship between a certain orientation of inductors with arbitrary inductance. Most authors define the range as 0 ≤ k < 1, but some[11] define it as −1 < k < 1. Allowing negative values of k captures phase inversions of the coil connections and the direction of the windings."

Mithrandir24601

@Dodsy it's brilliant, but not as fun to play as appassionata

@BalarkaSen in that case, the fantasia and fugue in g minor is astounding

John Rennie

@Blue Isn't $k^2$ the ratio of the output power to input power? And the flux coupling will be related to the power.

Mithrandir24601

I nabbed a chordal progression from that in A level music and my teacher thought it was jazz influenced

Balarka Sen

6:57 AM

hah

Anonymous

@JohnRennie I'm not sure how you got $k^2$ is the ratio of input and output power. We know $k^2=M^2/(L_1L_2)$

Anonymous

wikimedia.org/api/rest_v1/media/math/render/svg/…

John Rennie

$k$ is the ratio of the actual output/input voltage to the ideal input/output voltage, and power is proportional to voltage squared.

Balarka Sen

@JohnRennie I don't know what kind of weird music you like/enjoy, and it seems people have really been making comparisons of this with Captain Beefheart's TMR, but you might check out Scott Walker's "Bish Bosch"

Anonymous

$P=VI=V^2/R$. Power is proportional to $V^2$ only if the resistance is constant, isn't it?

Balarka Sen

7:02 AM

It's an epic freakshow, but I kind of am enjoying it.

Anonymous

Here we don't know about resistance

John Rennie

@BalarkaSen I like some Captain Beefheart. Doc at the Radar Station is very good and I recommend it. Ice Cream for Crow is good ish. His first album Trout Mask Replica is the one everyone raves about, but it's pretty weird. For the hard core Beefheart enthusiast only.

Balarka Sen

Thanks, I'll have to check these out!

Anonymous

Or perhaps we take inductive reactance instead...

John Rennie

@Blue remember that $k$ isn't the ratio of the voltages, it's the ratio of the ratio of the voltages i.e. $$ k = \frac{V_{in,actual}/V_{out,actual}}{V_{in,ideal}/V_{out,ideal}} $$

So the resistance cancels out.

Anonymous

7:09 AM

Okay, so $k^2 = \frac{P_{in,actual}/P_{out,actual}}{P_{in,ideal}/P_{out,ideal}}$. Right?

Anonymous

Then what does $1-k^2$ give....

Anonymous

$1-k^2$ should give (useful flux)/(total flux)

Anonymous

I'm not sure how we get that from there ^

Anonymous

Let's write like $k^2$ = (actual power gain)/(ideal power gain)

Anonymous

So, $1-k^2$ is (power gain lost)/(ideal power gain)

Anonymous

7:15 AM

Actually $1-k^2$ should give (flux lost)/(total flux)

John Rennie

There's something screwy about the definition because in the ideal case $k=1$, which means $\sigma=0$.

I wonder if the $k$ in your equation is defined differently to the usual $k$

Anonymous

In ideal case there is no flux leakage so $\sigma=0$ seems fine

Anonymous

Short-circuit inductance

Short-circuit inductance is the inductance when one of the primary winding or the secondary winding of the transformer is short-circuited and measured from the other winding. This value is often called informally as leakage inductance of the transformer. But the term leakage inductance which is in the electromagnetic literature is defined as an inductance caused by a magnetic flux (leakage flux) which is interlinked with one winding of a primary winding or a secondary winding and does not interlink with the other winding. So a confusion is caused by called short-circuit inductance as leakag...

Anonymous

I don't think so

Anonymous

"Assuming that the coupling coefficient is k"

John Rennie

7:19 AM

If I'm honest I'm not familiar with the subject. I'm just trying to throw out ideas that might be useful.

Anonymous

I think we're close to the answer. We just need to understand the meaning of $1-k^2$

Abcd

@JohnRennie Please let me know how to put limits of integral. I am writing an answer and ...

(using mathjax)

I want the format.

Anonymous

\int_{a}^{b}

John Rennie

\int_{a}^{b} gives $\int_{a}^{b}$

Abcd

Thank you :)

I am supposed to write dm/dt as $\dfrac{dm}{dt}$, right?

Or is there any other way?

Anonymous

7:24 AM

That's fine

John Rennie

Some people will moan that you should use the upright $\mathrm d$

Anonymous

Or just write $\dot{m}$

Abcd

@JohnRennie How to do that?

And can you tell me how to switch to next line in mathjax?

Anonymous

@Abcd Right click on the d there and go to "Show Math as" and click "Tex commands"

John Rennie

$$ \frac{\mathbf d m}{\mathbf d t}$$ but I never bother about the \mathbf

Abcd

7:29 AM

@Blue Done.

How to switch to next line?

John Rennie

\\ gives a line break in MathJax $$ line 1 \\ line 2 $$

Anonymous

@JohnRennie Does the square of flux have any physical significance? $\phi^2$

Anonymous

Here they define $k$ as ratio of linked flux and total flux

Abcd

$\int dv_{v_i}^{v_f}$

I want limit over integral sign.

Not happening.

Anonymous

\int_{v_i}^{v_f} dv

Abcd

7:33 AM

yes

@Blue Isn't the integral sign too short here?

Anonymous

@Abcd Use \$\$ \$\$ instead of \$ \$ make it bolder

Anonymous

$$\int_{v_i}^{v_f} dv$$

Abcd

Oh, yes.

Anonymous

$k=\frac{\phi_p}{\phi_s}$

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A: Change in momentum for a rocket

Thrust force $R$ on rocket is given by $M\dfrac{\mathrm dv}{dt}$ = $- v_{rel}\dfrac{\mathrm dm}{dt}$ where $\dfrac{\mathrm dm}{\mathrm dt}$ is the rate of fuel consumption and $ v_{rel}$ is the velocity of rocket relative to the ejected mass. Multiplying both sides by $dt$ we get $$\mathrm dv...