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Joshua Lin
Joshua Lin
2:20 AM
If I'm grading N papers with N large; but I don't have a rubric yet and I want to grade everyone an average of B+ with one standard deviation as one letter grade or something - I recall that the 'best' way to do similar kinds of problems (efficientlyish) is to grade sqrt(N) papers with no rubric as a guideline, then use the sqrt(N) to calibrate my grading to get a guideline. What is this kind of principle called again?
https://en.wikipedia.org/wiki/Secretary_problem
wait this is what I was thinking about
maybe not so applicable to grading
 
vzn
vzn
2:51 AM
@MoreAnonymous cf Bringing balance to the universe: New theory could explain missing 95 percent of the cosmos phys.org/news/2018-12-universe-theory-percent-cosmos.html
 
uhoh
uhoh
0
Q: What are the different kinds of stability that apply to planar periodic orbits, and what do they mean?

uhohThis is a question about terminology related to orbit stability. I had wanted to ask about stability of orbits described in the paper Three Classes of Newtonian Three-Body Planar Periodic Orbits Milovan Šuvakov and V. Dmitrašinović Phys. Rev. Lett. 110, 114301 https://arxiv.org/abs/1303.0181 h...

classical-mechanics stability-in-odes stability-theory bifurcation celestial-mechanics
 
 
2 hours later…
Avnish Kabaj
Avnish Kabaj
4:49 AM
@JakeRose Research in what field?
That's what I meant
 
 
4 hours later…
Slereah
Slereah
9:01 AM
morning
 
Slereah
Slereah
9:45 AM
You know I think the worst possible thing in GR was the introduction of the word "black hole"
A lot of confusion in GR seems to stem from people confusing the Schwarzschild, Kerr, Oppenheimer, Vaydia etc metrics together
 
Slereah
Slereah
10:07 AM
it's not really a good idea to regroup such different metrics so casually
 
GodotMisogi
GodotMisogi
@Slereah I think I remember watching some Neil DeGrasse Tyson video in which he uses "black hole" as an example for keeping terminology simple in physics compared to fields such as law
 
Anshuman Sinha
Anshuman Sinha
10:33 AM
why does vortex formation takes place in kitchen sink ?
 
John Rennie
John Rennie
10:44 AM
@AnshumanSinha vortices start because angular momentum is conserved.
 
Anshuman Sinha
Anshuman Sinha
11:10 AM
@JohnRennie then why is it not there during the start of the flow ? or is it always there ? could you share some relevant link
 
John Rennie
John Rennie
@AnshumanSinha in real flows there is always a non-zero angular momentum because it's impossible to get the flow completely radial. Since angular momentum is conserved the sideways velocity of the water increases as it approaches the drain, so even a tiny non-zero angular momentum causes a significant rotation near to the drain.
And once a vortex forms it drags on the water flowing in from the tap and the vortex self amplifies.
 
Anshuman Sinha
Anshuman Sinha
@JohnRennie flow radial or towards the z_axis ( i mean along the axis of the tube at the bottom of the sink ? why does it self amplifies? i.e why does it drags water ?
what will happen if I put an objext over the free surface of water in the kitchen sink; will it still form vortex ?
 
John Rennie
John Rennie
By radial flow I mean flow directly towards the centre of the drain
 
Anshuman Sinha
Anshuman Sinha
@JohnRennie what will happen if I put an objext over the free surface of water in the kitchen sink; will it still form vortex ?
 
John Rennie
John Rennie
@AnshumanSinha Do you mean something like a piece of wood floating on the water?
 
Anshuman Sinha
Anshuman Sinha
11:24 AM
yes
 
John Rennie
John Rennie
Generally speaking the vortex will still form. You could of course try the experiment for yourself. I assume you have a sink and something that floats to hand.
 
Anshuman Sinha
Anshuman Sinha
can you find me a proper paper where it's being discussed!
I've tried searching it myself
but the papers are more of related to natural vortex
 
John Rennie
John Rennie
Offhand I don't know of any such article. My guess is that anything that discusses vortex formation is likely to be quite technical.
 
Anshuman Sinha
Anshuman Sinha
like cyclones, where Coriolis force is a major force
could you explain why the vortex starts from the surface of the water in the sink, rather than starting from the bottom hole
@JohnRennie
 
John Rennie
John Rennie
I'm not sure what you mean when you say the vortex starts from the surface of the water. The vortex has one end at the surface and the other end in the drain. Is that what you meant?
 
Anshuman Sinha
Anshuman Sinha
11:33 AM
no the end is not at the bottom, but the top of the vortex is at the free surface
from 0:14 onwards
@JohnRennie
 
John Rennie
John Rennie
@AnshumanSinha you mean the way the surface dips downwards towards the drain as the vortex builds up?
 
Anshuman Sinha
Anshuman Sinha
yes
 
John Rennie
John Rennie
@AnshumanSinha Remember that the vortex is forming in the water, but you can't see it because the water is clear.
As the vortex forms it lowers the pressure due to the Bernoulli effect. That pulls the surface downwards and that is what you are seeing.
 
Anshuman Sinha
Anshuman Sinha
11:48 AM
oh yeah ! you are right! Maybe if i use some kind of oil or color water, then I'll get the effect correct
 
John Rennie
John Rennie
@AnshumanSinha Typically we put mica flakes in the water so we can see the flow.
 
Anshuman Sinha
Anshuman Sinha
okay.
can you explain this formation more mathematically ? I mean with the help of vortex stretching and turbulence
@JohnRennie
 
John Rennie
John Rennie
As a general rule explaining any flows mathematically is hard. The flow is described by the Navier Stokes equations and they are fiendishly difficult to solve analytically except in the simplest cases. I doubt you'll find any simple explanation for vortex formation.
 
Anshuman Sinha
Anshuman Sinha
complex explanation would help me greatly! I'm in my 4th year of engineering
Navier stokes is an everyday event! I just wanted to get the staring points where I can build the mathematics by myself.
 
John Rennie
John Rennie
If you search the literature there must be loads of stuff on vortex formation because it is so fundamental. Fluid mechanics isn't my area so I can't help.
 
Anshuman Sinha
Anshuman Sinha
11:57 AM
okay
thanks
Your edit is asking for definitions, which your course material should provide. Usually such questions should be asked in chat. — GodotMisogi 2 days ago
@JohnRennie did you find an answer to this problem
-2
Q: Recurrence relation in thermal speeds

Anshuman SinhaProve the recurrence relation for averages of even powers of the thermal speend in an equillibrium gas https://imgur.com/a/igTPsJ2 Please refer the above picture for the relation. Is the thermal speed here means rms velocity of the gas ? What does m stands for ?

thermodynamics molecular-dynamics
 
newUser
newUser
I think this is an interesting problem physics.stackexchange.com/questions/461480/…
 
John Rennie
John Rennie
@AnshumanSinha that's not something I know about. Sorry.
 
Anshuman Sinha
Anshuman Sinha
okay, thanks anyways
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
12:41 PM
hello?
is it correct to say:
I = Z Z* ?
where I is the intensity in optics
and Z is the complex wave?
 
Emilio Pisanty
Emilio Pisanty
12:59 PM
yo @ACuriousMind race you to the rep-cap
(sorry, I should stop. I just can't help it =P.)
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
hrllo?
*hello?
is what i said true?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor well, it isn't too wrong
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
well I tried googling the equation but I couldn't find any evidence that its true
 
Emilio Pisanty
Emilio Pisanty
for one "intensity" can mean a number of things
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but in my home work, it gave the equation as such
I was wondering that this equation is applied to that specific problem or its a general case
 
Emilio Pisanty
Emilio Pisanty
1:05 PM
you probably mean irradiance
Irradiance
In radiometry, irradiance is the radiant flux (power) received by a surface per unit area. The SI unit of irradiance is the watt per square metre (W/m2). The CGS unit erg per square centimetre per second (erg·cm−2·s−1) is often used in astronomy. Irradiance is often called intensity because it has the same physical dimensions, but this term is avoided in radiometry where such usage leads to confusion with radiant intensity. Spectral irradiance is the irradiance of a surface per unit frequency or wavelength, depending on whether the spectrum is taken as a function of frequency or of wavelength...
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
because I know the equation: I = |E part of EM wave|^2
 
Emilio Pisanty
Emilio Pisanty
but it's worth thinking carefully about what exactly you mean
i.e. whether it's a quantity that's over a unit solid angle illumination or not
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
I mean the definition of intensity is just Power over unit are aright?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor often, yes. But there's a number of radiometric quantities and you need to be careful in distinguishing them (so see in particular this table).
@WantingtobeanAndroidDevelor and, separately, saying "Z is the complex wave" is a very squishy statement and it could mean a number of different things
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
I thing in the question it said "complex space amplitude" of which I don't really know the meaning of
I just assumed that its just an amplitude of the wave
or at least the real part is the amplitude of the wave
 
Emilio Pisanty
Emilio Pisanty
1:08 PM
@WantingtobeanAndroidDevelor so you're asking us to confirm a quote and you cannot recall the specific wording of the quote?
that doesn't tend to go very well
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
I mean that isn't so important though right?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor whether it's right or wrong?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
because I want to know if its a general statement or a specific statement
 
Emilio Pisanty
Emilio Pisanty
I dunno, that's up to you
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
Question 2: Fabry-P´erot etalon interferometer
 
Emilio Pisanty
Emilio Pisanty
1:10 PM
@WantingtobeanAndroidDevelor as a general statement, it is indeed true that there exist relevant intensity measures I in optics, and relevant complex-valued phasor quantities Z that describe the wave amplitude, such that I=ZZ*
as a specific statement, it is almost certainly incorrect
i.e. it is almost certainly missing a constant or geometrical factor of some sort
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
A Fabry-P´erot etalon interferometer has a transmitted spatial complex amplitude of
Z =
aT
1−Re−iδ .
 
Emilio Pisanty
Emilio Pisanty
but saying what that factor is depends on the details of what each symbol is taken to mean
@WantingtobeanAndroidDevelor that is unreadable.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
ok sorry
 
Emilio Pisanty
Emilio Pisanty
Use MathJax for rendering
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but thats basically a fraction
 
Emilio Pisanty
Emilio Pisanty
1:11 PM
instructions are here meta.stackexchange.com/a/220976
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
$Z = \frac{aT }{1−Re−iδ }$
...
I don't know why its not working but you get the idea right
Show that the intensity is given by
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor it's working on my system. You need to enable ChatJax to get it to work. Link in my previous comment.
@WantingtobeanAndroidDevelor ... where $a$, $T$, $Re$ and $\delta$ are, respectively...
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
$I = ZZ^* = \frac{(\frac{(aT)^2}{(1-R)^2})}{1 + F sin^{2}2(δ/2) }$
right?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor OK.
What about it?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
so as I said the question is basically irreverent to what I want to know
but is $I = ZZ^*$ true in general case in optics?
 
Emilio Pisanty
Emilio Pisanty
1:17 PM
@WantingtobeanAndroidDevelor it's irrelevant to the question you've been given
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
or is it just true for the Fabry-P´erot etalon interferomete
 
Emilio Pisanty
Emilio Pisanty
the question is defining $I = ZZ^*$
and asking you to perform further manipulations on it
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but don't I need to explain why $I = ZZ^*$ if its specific to the problem?
 
Emilio Pisanty
Emilio Pisanty
It doesn't mean that the $I$ that they've defined corresponds explicitly to an energy flux per unit area or any other specific measure of energy flux
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
because I really don't know why $I = ZZ^*$ can be true I can see why $I = |Re[Z]|$ can be true
 
Emilio Pisanty
Emilio Pisanty
1:20 PM
@WantingtobeanAndroidDevelor I don't think so. If you're in doubt you should check with your instructor, but the question as written is just asking you to perform specific manipulations on $ZZ^*$.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but he says "Show that the intensity is given by" meaning I is intensity right?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor why would $I = |Re[Z]|$ be true? That one is certainly wrong.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
because in lecture he defined intensity of light as the square of the E part of EM wave
which is generally true so,
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor I can't speak to what the person asking the question meant by it or what they are looking for. I can provide the standard reading on the question. If you're in doubt about whether it applies, the only person who can settle that is your instructor.
@WantingtobeanAndroidDevelor so why do you think that $I = |Re[Z]|$, which doesn't have the square, work at all?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
well Im not interested in how to solve the question
oh sorry I meant $I = |Re[Z]|^2$
 
Emilio Pisanty
Emilio Pisanty
1:22 PM
@WantingtobeanAndroidDevelor In that case, for one, you can drop the absolute value - it's made moot by the square
but more importantly, what happens if $Z$ is pure imaginary?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
then intensity is zero
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor why so?
what does it mean for $Z$ to be purely imaginary?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
can I have a second
 
Emilio Pisanty
Emilio Pisanty
... and more importantly, what does $Z$ mean in the first place?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
here I was looking at this
9
Q: Complex numbers in optics

IsaacI have recently studied optics. But I feel having missed something important: how can amplitudes of light waves be complex numbers?

optics electromagnetic-radiation waves polarization complex-numbers
 
Emilio Pisanty
Emilio Pisanty
1:24 PM
@WantingtobeanAndroidDevelor There's no $Z$ in there, if you want to link there you need to be more specific.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
The answer says that there is no imaginary part to a wave, its just convenient for doing some of the math
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor which answer?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
the one with most upvote
Im guessing its convenient when you want to do reflection by mirror by multiplying z by i and stuff
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor That doesn't mean that $\mathrm{Im}(Z)$ can be disregarded at will
Basically, this is how you do it
when you say
> $Z$ is the complex amplitude of the wave
what you mean is
(something of the form)
> The electric field of the wave, as a function of time, is given by $E(t) = \mathrm{Re}\left[ E_0 Z e^{-i\omega t} \right]$
where $E_0$ is some pre-defined constant
the intensity that corresponds to that has the form
$$I = k \left< E(t)^2 \right>$$
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
can I have a second to turn on my mathjax plugin
 
Emilio Pisanty
Emilio Pisanty
1:28 PM
where the angled brackets denote time-averaging over one period, and $k$ is some constant whose value and physical dimensions depend on the details of what you're doing
when you do that, you get the following
$$I = k \left< E(t)^2 \right> = \frac{k}{T} \int_0^T E(t)^2 \mathrm dt$$
for $T=2\pi/\omega$ the period of the wave
with $$E(t) = \frac{E_0}{2}\left( Z e^{-i\omega t} + Z^* e^{+ i \omega t} \right)$$
so therefore
$$I = \frac{k}{T} \int_0^T E(t)^2 \mathrm dt= \frac{k}{T}\frac{E_0^2}{4} \int_0^T \left( Z e^{-i\omega t} + Z^* e^{+ i \omega t} \right) \left( Z e^{-i\omega t} + Z^* e^{+ i \omega t} \right)\mathrm dt$$
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
ok
sorry
mathjax is working now
 
Emilio Pisanty
Emilio Pisanty
$$I =\frac{k}{T}\frac{E_0^2}{4} \int_0^T \left[ Z^2 e^{-2i\omega t} + ZZ^* + Z^*Z + (Z^*)^2 e^{+2i\omega t}\right] \mathrm dt$$
here the time-dependent terms integrate to zero
and you're just left with the time-constant terms
$$I = k \frac{E_0^2}{2}ZZ^*$$
so, some constants in front
as I said before
but basically $I= ZZ^*$
so
why was your initial guess of $I = |\mathrm{Re}(Z)|^2$ wrong?
well, if you look at the electric field
$$E(t) = \mathrm{Re}\left[ E_0 Z e^{-i\omega t} \right]$$
having a purely-imaginary $Z$ just implies a time delay
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
Im reading from top right now
 
Emilio Pisanty
Emilio Pisanty
i.e. a $\sin(\omega t)$ time dependence instead of $\cos(\omega t)$ for real $Z$
that cannot change the intensity.
@WantingtobeanAndroidDevelor QED.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
give me a min to read through top Im sorry
Ive got some questions already though
if $E(t) = \mathrm{Re}\left[ E_0 Z e^{-i\omega t} \right]$, then why can we say
$$I = k \left< E(t)^2 \right>$$ and not
$$I = k \left< Re(E(t))^2 \right>$$?
 
Emilio Pisanty
Emilio Pisanty
1:38 PM
@WantingtobeanAndroidDevelor the second expression is superfluous
if you take the real part of a real number then you get the real number
why do you think the second one is needed?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but first term is oh....
so <sin^2(x)>= <cos^2(x)>
woops
sorry what I meant is
<sin^2(x)> = <sin(x)cos(x)>
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor what?
@WantingtobeanAndroidDevelor try this for further background
15
Q: What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$?

FedUserI've read that plane wave equations can be represented in various forms, like sine or cosine curves, etc. What is the part of the imaginary unit $i$ when plane waves are represented in the form $$f(x) = Ae^{i (kx - \omega t)},$$ using complex exponentials?

quantum-mechanics waves complex-numbers linear-systems plane-wave
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
ok thanks but why would you say they are superfluous?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor (this is dead wrong, by the way. $\langle \sin^2(x) \rangle = 1/2$, while $\langle \sin(x) \cos(x) \rangle = 0$.)
2 mins ago, by Emilio Pisanty
if you take the real part of a real number then you get the real number
$E(t)$ is explicitly real
saying $\mathrm{Re}(E(t))$ doesn't do anything.
You just get $\mathrm{Re}(E(t)) = E(t)$.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
Sorry I meant
<RMS[sin^2(x)]>= <RMS[sin(x)cos(x)]>
where RMS is taking the root mean square
 
Emilio Pisanty
Emilio Pisanty
1:43 PM
why is that relevant here?
I mean, it's also wrong (it's off by a factor of four), but that's beside the point.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
I skim read the link but he also says its doesn't really play a role
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor OK.
so.
How about you sit.
then you read carefully
instead of skimming
and then you come back and ask?
It's all in the text I wrote above.
I have very little to add, beyond "see above".
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
well do you have the time?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor To do what? To repeatedly go over material that you would get on your own if you sat down to read carefully?
I don't, I'm afraid.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
well thanks
everything in our analysis is linear, so it will automatically work for sums? but isn't the trig linear as well?
 
Emilio Pisanty
Emilio Pisanty
1:52 PM
@WantingtobeanAndroidDevelor everything is linear except for the quadratic parts
which are basically everything
so.... no.
> isn't the trig linear as well?
I have no idea what you mean there
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
what do you mean by linear here? linear is the same meaning of linear in linear algebra correct?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor yes, obviously.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
so the operation that he says to be a linear operation is taking the square of it self?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor what are you talking about?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
In my mind, the definition of linear is that the equally same contribution is made towards a variable by components of another variable no matter the magnitude
So he is saying, you can add two waves together by adding the complex representation, multiply by the result of the complex conjugate and then take the real part?
 
Emilio Pisanty
Emilio Pisanty
2:00 PM
@WantingtobeanAndroidDevelor I have no idea what you're on about. I can't answer without knowing what context you're trying to work in, or how it relates to what you're saying, so I won't try.
@WantingtobeanAndroidDevelor who is "he", and where is he "saying" this?
unless you provide clear context, it's impossible to help you.
We cannot see what's going on inside your head.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
the one who answered the question you gave me on stack exchange
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor ... you changed the topic, but you couldn't be bothered to say so?
which question, and which answer?
i.e. give an explicit link
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
well Im still trying to understand the answer to my question
but you gave me a link
15
Q: What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$?

FedUserI've read that plane wave equations can be represented in various forms, like sine or cosine curves, etc. What is the part of the imaginary unit $i$ when plane waves are represented in the form $$f(x) = Ae^{i (kx - \omega t)},$$ using complex exponentials?

quantum-mechanics waves complex-numbers linear-systems plane-wave
that I should understand
 
Emilio Pisanty
Emilio Pisanty
and if this "he" is just me, then why are you talking about "he"?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
in fact "he" is you
I didn't realize you wrote the answer
 
Emilio Pisanty
Emilio Pisanty
2:03 PM
@WantingtobeanAndroidDevelor ...
.....
..............
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but the context is your answer to the link you gave me
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor it took you more than one full page of chat transcript to provide context. This doesn't make it very easy for other people to help you.
@WantingtobeanAndroidDevelor This is discussing other people's use of the notation
it means that the notation works IF everything that follows is linear
if it doesn't, then the analysis does not work
in your case, the stuff that follows from the introduction is not linear
more specifically, it's quadratic
so it is not appropriate to replace the real-valued physical field $E(t)$ with a complex-valued stand-in
instead, you need to start from the top and perform a careful analysis of what does happen
there's a useful, recent reference for that careful analysis
in this chat transcript
starting here:
41 mins ago, by Emilio Pisanty
Basically, this is how you do it
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
so you are saying I can't use f(x)=Aei(kx−ωt) where amplitude = Re|f(x)| can't be used for finding the intensity because calculating intensity is not a linear method??
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor No, that's not what I'm saying.
I'm saying you can't use $f(x)=Ae^{i(kx−ωt)}$, where the amplitude is $f(x)$, for finding the intensity because calculating intensity is not a linear method
see the difference?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
well I see the difference in the text, specifically the definition of amplitude
but I don't understand why would I think the amplitude of wave as a complex number?
 
Emilio Pisanty
Emilio Pisanty
2:14 PM
@WantingtobeanAndroidDevelor well, some people do
you included
@WantingtobeanAndroidDevelor "Z is the complex wave"
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
so you wanted me to say Re(Z) is the wave?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor No.
The wave is $\mathrm{Re}(Ze^{-i\omega t})$ (or something of that form)
"the complex wave" is a term that's too vague to mean anything specific
it would be accurate to describe $Z$ as the complex amplitude of the wave, though.
"phasor amplitude of the wave" would also work.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
sorry Im just doing a bit of calculation before I ask
I assumed $Z = e^{i\omega t}$ but I'm guessing I assumed wrong
what is Z?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor that's up to you to say.
You brought it up.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
I see
 
Emilio Pisanty
Emilio Pisanty
2:25 PM
You bring up a symbol, you need to define it.
that's why I kept asking that same question from the very beginning.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
I just thought Z is usually used to denote complex number
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor All sorts of symbols are used for all sorts of purposes in all sorts of different contexts.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
the only way in which a complex number can do anything to do with wave is $Z = e^{i\omega t}$
 
Emilio Pisanty
Emilio Pisanty
You still need to define them when you bring them up.
Particularly when prompted to do so.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
he says Z is transmitted spatial complex amplitude
 
Emilio Pisanty
Emilio Pisanty
2:28 PM
... where "he" is ... ... ... ... ... ?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
my lecturer
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor so then it can't be time-dependent, can it?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
why? mustn't it be time dependent?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor an amplitude?
where have you seen amplitude used to describe an oscillating quantity?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
oh I see sorry
so its a standing wave
 
Emilio Pisanty
Emilio Pisanty
2:30 PM
@WantingtobeanAndroidDevelor what?
no
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
why?
 
Emilio Pisanty
Emilio Pisanty
why would that be implied by the above?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
because the only way in which a light wave to be not be dependent on time is for it to be a standing wave correct?
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor standing waves do depend on time
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but the amplitude doesn't>
*?
 
Emilio Pisanty
Emilio Pisanty
2:34 PM
@WantingtobeanAndroidDevelor the amplitude of a way is typically understood to be a number $A$ such that the wave signal $f(t)$ is given by $f(t) = A \cos(\omega t +\varphi)$
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
yes
 
Emilio Pisanty
Emilio Pisanty
... so why would you think that the amplitude can depend on time?
@WantingtobeanAndroidDevelor (note that you can edit your previous messages by pressing the up key)
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
sure
but its a wave
I see so Z is not actually a wave
its just a constant
its the amplitude over position
which is a wave
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor if by $Z$ you mean this
11 mins ago, by Wanting to be anAndroidDevelor
he says Z is transmitted spatial complex amplitude
then yes
if by $Z$ you mean something else, then $Z$ means whatever that something else is.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
ok
I see
so what did you want me to take away from physics.stackexchange.com/questions/316135/… ?
so is the statement I = ZZ* where Z is complex amplitude of a wave correct?
 
Emilio Pisanty
Emilio Pisanty
2:43 PM
@WantingtobeanAndroidDevelor with a number of caveats, yes
but it really doesn't sound like you understand the caveats.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
so this complex amplitude only make sense when I define wave as $Re(Ze^(iwt))$
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor as a general statement, yes
but frankly, why wouldn't you include an explicit statement linking the objects you've introduced (say, a complex amplitude $Z$) to the physical fields involved?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
ok
it feels a bit awkward to make the imaginary part of the wave function contribute towards the wave's displacement by having an imaginary part as the amplitude
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor You're confusing the imaginary part of the complex wave function $Ze^{-i\omega t}$ with the imaginary part of the complex amplitude $Z$.
They're extremely different things.
The imaginary part of the complex wave function $Ze^{-i\omega t}$ doesn't contribute to anything.
The imaginary part of the complex amplitude $Z$ is completely different, and it contributes every bit as much as the real part.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
but when you times the complex part of the complex amplitude with the complex part of $e^{-i\omega t}$ don't you get a real number that contribute towards the displacement of the actual wave?
 
user351417
2:59 PM
@EmilioPisanty Are you going to ask that thing about Dimethyl Hydrazine's melting point on Chem SE, or could I do that? I'm quite interested to find out the answer, and I've only been able to find a non-handwaving description of a difference between the two forms which suggests that the unsymmetrical form should have a higher melting point (which is not the case).
 
user351417
(sorry if that question sounds a bit forceful/impatient; I'm not sure how to phrase that :P)
 
Emilio Pisanty
Emilio Pisanty
@Chair if you want to ask, go ahead.
 
user351417
Cool, I'll put together some sources and do that. Thenx
 
user351417
I'll ping you if there're any answers.
 
Emilio Pisanty
Emilio Pisanty
@WantingtobeanAndroidDevelor The real wave is $\mathrm{Re}(Z e^{-i\omega t})$. In which way to you envision the imaginary part $\mathrm{Im}(Z e^{-i\omega t})$ contributing to the real wave?
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
3:04 PM
I meant so that $Im(Z) \times Im(e^-{i\omega t}) = Real$
 
Slereah
Slereah
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
@EmilioPisanty Well thanks for dedicating a large amount of time in answering my question, I think I now understand why I = Z Z* and I have a much better understanding about waves represented as complex numbers, really cleared some stuff up for me. I was lucky to have you answer my questions thanks
 
vzn
vzn
← lol 1sthand knowledge of physics mafia/ omerta :P
@Slereah its a work in progress and there have been decades of paradigm shifts in the area, some theories superceding prior ones. a well written historical ref/ survey could help sort it out. yeah it would be nice to see one. maybe black holes for dummies :P dummies.com/education/science/physics/…
@AnshumanSinha wikipedia is often a great place to start en.wikipedia.org/wiki/Vortex
 
vzn
vzn
3:42 PM
lol once got dinged for calling n00bz n00bz... who blithely assumed it was derogatory? :P
1 1
 
Anonymous
4:07 PM
@WantingtobeanAndroidDevelor To get an upright font you should use \text{} to enclose Im and Re, like \text{Im} and \text{Re}. Alternatively, you could enclose them with \mathrm{}.
 
Wanting to be anAndroidDevelor
Wanting to be anAndroidDevelor
k
thx
 
Anonymous
@WantingtobeanAndroidDevelor Re(Ze^(iwt)) renders it as $Re(Ze^(iwt))$. Write it as \text{Re}(Ze^{iwt}) to get $\text{Re}(Ze^{iwt})$ i.e. with curly braces. Anyway, it's good that you're trying to learn! :)
 
Anonymous
2586
Q: MathJax basic tutorial and quick reference

MJD(Deutsch: MathJax: LaTeX Basic Tutorial und Referenz) To see how any formula was written in any question or answer, including this one, right-click on the expression it and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next p...

discussion faq mathjax reference
 
Emilio Pisanty
Emilio Pisanty
@Blue Not quite. The primary choice is \mathrm
unless it's actual text
 
Anonymous
@EmilioPisanty I don't think there's any difference in the rendered output? But yes, I do think the convention is \mathrm{}.
 
Emilio Pisanty
Emilio Pisanty
4:16 PM
@Blue the rendered output is indeed the same, but the rendered output is not the only thing at play here.
MathJax and LaTeX are semantic markup languages
i.e. the markup is saying additional things about the content beyond just how it should be rendered
things like "this is text"
or "this is mathematical notation that should be typeset in upright roman font"
 
Anonymous
@EmilioPisanty Ah, that is indeed true!
 
Emilio Pisanty
Emilio Pisanty
So if there are two ways to get the same output, but one of them is semantically wrong and the other is semantically correct, you use the latter.
 
Anonymous
Indeed. I suggested using \text{} as its easier to remember for a beginner. It's easy to get confused about which to use for upright units, and which to use to upright text and which to use for these cases: Re and Im. I certainly wouldn't use \text{} for that in formal writing. But I get your point, yes.
 
Emilio Pisanty
Emilio Pisanty
@Blue sure
I mean, everything is up to a point
and if you're really being pushy about semantic markup, then you really ought to be using $\operatorname{Re}(z)$ ($\operatorname{Re}(z)$) instead
but I'd tend to discourage teaching the use of \text{} to newcomers
 
Anonymous
Hehe. That's there too. :P
 
Emilio Pisanty
Emilio Pisanty
4:23 PM
it's liable to create all sorts of misconceptions that rear their heads down the line
 
Anonymous
Right. It's better to learn the correct syntax from the beginning itself. I did have several misconceptions. as my learning of TeX was very haphazard (mostly included picking up bits and pieces from LaTeX SE every now and then). But staying around on SE has corrected a lot of them. :)
 
Anonymous
For example, the other day I was a bit uncomfortable about the large spacing between a and mod b in $$a\mod b.$$ So I tried to correct it using \text{}. Really shouldn't do that!
 
Anonymous
58
A: Writing mod in congruence problems without leading space

WernerThe spacing issue is an obvious indication that it's not meant to be used that way. However, you could wing your own, depending on the sophistication you're after: \documentclass{article} \usepackage{amsmath} \newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)} \begin{document} \begin{align*} a &\e...

 
Anonymous
4:38 PM
Anyway, I'll keep your advice in mind. "MathJax and LaTeX are semantic markup languages" is a very convincing point to change my habits for good. Thanks! :)
 
Emilio Pisanty
Emilio Pisanty
@Blue good thing :-)
@Blue huh
\pmod, eh?
that one's news to me
 
Anonymous
It's the one with parenthesis!
 
Anonymous
I learnt it from Mithrandir. :P
 
Emilio Pisanty
Emilio Pisanty
and I just put this in a paper I submitted last Friday
oh well
 
Anonymous
Heh. Well, you learn things everyday. :P
 
Emilio Pisanty
Emilio Pisanty
4:41 PM
indeed 
  \def\pod#1{\allowbreak\if@display\mkern18mu\else\mkern8mu\fi(#1)}
  \def\pmod#1{\pod{{\operator@font mod}\mkern6mu#1}}
well, those are definitely fancier than straight user-defined spacing
though then again it prints well enough, so who cares
@Blue if you're looking for a good TeXing exercise, finding those definitions is a good one ;-)
 
Anonymous
4:55 PM
Yep. Lots of useful ones out there!
 
Anonymous
3
A: How to typeset the following table in MathJax?

Willie WongSuch complex tables are not supported by MathJax, and it is outside their intent: it is supposed to be a software package to help the presentation of Mathematics on the internet, not a generic typesetting language to replace HTML. As I answered here, tables are potentially supported by MarkDown...

 
Anonymous
Reminds me. There is a workaround to typeset such complex tables, using array, in MathJax albeit it's not without flaws.
 
Anonymous
5:08 PM
Oh, robjohn has already covered it.
 
Anonymous
147
A: MathJax basic tutorial and quick reference

axblountArrays It is often easier to read tables formatted in MathJax rather than plain text or a fixed width font. Arrays and tables are created with the array environment. Just after \begin{array} the format of each column should be listed, use c for a center aligned column, r for right aligned, l for...

 
Anonymous
That is one place where \text{} really comes handy. I remember creating this table for a PSE question.
 
Anonymous
\begin{array}{|c | c | c | }\hline \dim M & \# & \\ \hline 1 & 1 &
\text{Morse-Radon theorems} \\ 2 & 1 & \text{Morse-Radon theorems}
\\ 3 & 1 & \text{Morse-Radon theorems} \\ 4 & \text{uncountably
infinite} & \\ 5 & \text{finite} & \text{surgery theory} \\
6 & \text{finite} & \text{surgery theory} \\ \vdots & \text{finite} &
\text{surgery theory} \\ \hline\end{array}
 
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