The h Bar

danielunderwood

00:10

Are there two distinct definitions of functionals? I've always heard of functionals as something that maps a function to a number, but it seems there may also be a definition that relates to 1-forms. Are they the same thing?

ChoMedit

00:47

Functional is actually, the map from the vector space to the underlying field, in mathematical view. Therefore, 1-form is functional since it's a function from the vector space of smooth functions on manifolds to real number.

@ACuriousMind Thanks. To study about the statistical method using in papers, is it enough to see papers and applications? Because choosing some textbooks etc. seems redundant..

danielunderwood

Ahh I never thought of the space of functions being a vector space for whatever reason

ChoMedit

Oh it's not a VS of smooth functions. It's tangent space.

I realize that this is not an answer why two functionals are called "functionals". Maybe they are originated from same sense, like a function of function(Wikipedia said).

danielunderwood

01:15

It sounds like the vector space definition covers both cases I think. Most of the places I've run into functionals just mentioned them as "functions of functions" or "map a function to a number" rather than defining them fully

rob

04:50

21 hours ago, by Swapnil Das

Is chemical physics physics or chemistry?

Lots of people have this question. Here is how you tell participants in these three fields apart:

1. A chemist is a person who makes a sample, and buys a laser to study it with.

2. A physicist is a person who builds a laser, and buys a sample to study with it.

3. A physical chemist is a person who buys a sample, buys a laser, and publishes a paper.

There! Now you know.

Al Nejati

lol

John Rennie

Speaking as a physical chemist I would have said a physical chemist makes the sample themselves and builds the laser themselves. In my experience physical chemists are very hands on characters.

Al Nejati

so a physical chemist spends four years building their own lab equipment and that becomes their PhD thesis?

:D

John Rennie

That's both in academia and industry, though admittedly the larger budgets in industry mean we buy a lot more.

@AlNejati actually, that's not so far from the truth :-)

Al Nejati

well it actually sounds like fun tbh

I would much rather be doing that than my current job

John Rennie

05:04

Well I have to say I enjoyed the three years of doing my PhD in physical chemistry more than I've enjoyed any other period of my life (so far)

Al Nejati

it's funny isn't it, you look back on your phd and it seems like the best time of your life

John Rennie

Because it was :-)

Al Nejati

yep

it's just that you don't realize it at the time!

Anonymous

06:28

@rob Small advice: It isn't worth your time trying to engage with this user. Your answer seems totally on point to me.

Anonymous

He will keep editing your answer to suit his agenda. Don't worry. We'll roll it back later (and possibly ask a mod to lock it).

rob

06:41

@Blue I have that impression from communication in other channels. Watch me win the argument by disengaging now.

Anonymous

:)

Anonymous

Okay, maybe I shouldn't import the drama

David Z

Yeah probably not, there are better ways of dealing with it

Kyle Oman

08:32

On Friday, a spokesperson for the Royal Swedish Academy of Sciences, which awards the Nobel Prize in Physics, told Motherboard that while it does not endorse the attitudes displayed in the video, "Professor Gérard Mourou has been awarded the Nobel Prize because of his groundbreaking contributions to the field of laser physics and would have received it even if the video in question had come to light at an earlier stage."

"The Royal Swedish Academy of Sciences awards the Nobel Prize to those who have made the most important discoveries or inventions in the fields of physics and chemistry. N…

And here I thought this years physics laureates represented a small step forwards... apparently balanced by a step back.

Anonymous

@KyleOman I couldn't quite understand the uproar over the video. It might be a bit silly, but I didn't find anything offensive in particular.

Anonymous

I think I concur with the spokesperson in this case.

Anonymous

> The video “echoed the attitudes that the Royal Swedish Academy of Sciences does not share,” its general secretary, Göran K. Hansson, told AFP, stressing that the filming took place long ago.

Anonymous

Hmm

Al Nejati

08:48

of course the swedish academy would say that

they have to uphold the image of the prize as sacrosanct, timeless, and eternally justified :)

which, to be fair, it almost always is (the physics nobel, at least)

Anonymous

@AlNejati Lol

Anonymous

@AlNejati Yes, the physics Nobel more or less seems to have gone to deserving people. At least in the last 2-3 decades.

Anonymous

On the other hand, the Peace Nobel has always been far more controversial.

Kyle Oman

09:12

@Blue we live in a time where gender imbalance is a recognized issue (including in 2011, when the video was made). I'd invite you to think about whether changing the gender of each person depicted in the video changes the message, and how. It may have been born out of an attempt at humour, but that doesn't necessarily make it right.

Yes it was filmed "long ago", but that seems a bit flimsy as an argument when the awarded research was also conducted "long ago". I'm not trying to say that the physics is undeserving of recognition, just that I think it's important to have conversations about who we hold up as exemplary individuals in our field, and why, and why not.

Al Nejati

as I've said before, I maintain that us physicists/mathematicians/engineers/etc. shouldn't really give our opinions on these things

we lack the historical context and social/psychological knowledge to make any meaningful contribution and we usually wind up making fools of ourselves

Anonymous

@KyleOman I agree gender imbalance is indeed an issue to worry about. But I wouldn't really read too much into that video as that often leads to false conclusions. In any case, the video isn't a strong evidence of 1. gender bias on Mourou's end 2. misuse of funds. There are so many other dire "issues" in the world to worry about, that I wouldn't put gender imbalance even in the top 20, but that's just me.

Anonymous

@AlNejati Applies to almost everyone really. Anyone at this point of time putting forward arguments (irrespective of their validity) which go against feminism is prone to attacks. So, yeah, indeed. It might be just better to be silent on these matters, and especially so when you're a top scientist. (Not that I'm supporting anyone here)

Kyle Oman

My attitude toward the equality issue is (1) to try not to be part of the problem myself and (2) to say something when I see things that I think are contributing to the problem. One could argue that by silently standing by, one becomes part of the problem; I try not to do that.

Al Nejati

I don't view it as a binary decision between either staying silent or taking (possibly misguided) action.

Anonymous

09:25

@AlNejati That

Al Nejati

Instead one can defer to the people who are better situated to react to these issues.

Imagine if it were reversed and a psychologist decided to give a talk on string theory.

best case scenario, they might give a 'correct' view on the theory, but it certainly wouldn't be a contribution to the field and no one would come out any smarter :)

Emilio Pisanty

13:08

@JohnRennie Here I was feeling proud about my straight week on the rep cap

but I bow to your straight ten days

data.stackexchange.com/physics/query/908629/rep-cap-streaks

and to Luboš's completely, absolutely ridiculous 21 days on the rep cap

... which he achieved immediately after joining the site

he basically just up and got 5.5k on his first three weeks, comprising some 230 posts

Kelthar

13:24

any idea on how to check the independence of a set of summations?

Emilio Pisanty

@Kelthar that's not really enough information to tell what you're asking

Kelthar

I don't know if it's very visible, but I have those expressions inside the brackets

the goal is to check that c_theta is surjective

c_theta is a vector whose component j is given by one of those expressions

Emilio Pisanty

@Kelthar $c_\theta$, absent additional information, is just an object, not a function

it doesn't make sense to ask whether it's surjective or not

Kelthar

13:48

yeah, sorry. $c_\theta$(w, v). $c_\theta \in \R^{D(D+1)/2}$, $w \in \R^{k\times D}$ $v \in \R^{k}$

I guess I'm a noob in this formalism

Ultradark

What is cimplenenyarity and why is it so important?

*complementarity

Emilio Pisanty

@Ultradark in what context?

Ultradark

In terms of light being a particle and a wave. Apparently complementarity says that these two descriptions of light are mutually exclusive features

And I agree with that im just trying to wrap my head around the concept

Emilio Pisanty

@Ultradark that's... a sub-optimal description, in my opinion

Ultradark

I know sorry

But how am I supposed to describe the concept if I don't understand it

Emilio Pisanty

14:04

@Ultradark I'm not bashing you. I'm bashing whatever source you're reading that gave you that terminology.

"complementarity" is a loaded term, and it generally refers to complementary pairs of variables like position and momentum, energy and time, etc.

it's a bad idea to use it for other things

Ultradark

And BOTH of these variables must be taken into account in order to fully describe what's going on?

Emilio Pisanty

@Ultradark no. On the contrary, knowledge of one of those variables actively reduces how much you can say about the other member of the pair.

Ultradark

Even energy and time?

Emilio Pisanty

It's much easier if you think of them in terms of the pair (position of a wavepacket, wavelength of a wavepacket)

If you want to know the wavelength of a wavepacket very precisely, then you need many periods of the wave to measure the wavelength. But if you're able to do that, then you have a very broad wavepacket and you won't be able to say anything very meaningful about its position.

@Ultradark Yes, even energy and time. If you want to measure energy very precisely, then you have a fundamental requirement that your measurement needs to take very long. (More precisely, if you want a measurement to within an uncertainty $\Delta E$, then your experiment needs to last at least a time of order $T = \hbar/\Delta E$, where $\hbar$ is the reduced Planck constant.)

Uncertainty principle

In quantum mechanics, the uncertainty principle (also known as Heisenberg's uncertainty principle) is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which certain pairs of physical properties of a particle, known as complementary variables, such as position x and momentum p, can be known. Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the standard deviation...

that's the Heisenberg Uncertainty Principle at work.

Abcd

14:47

A polariser absorbs electric field components of one plane.

How are polariod cameras polariods?

I dont even see the difference between ordinary cameras and polaroids, just that there's a little writing space under the picture of Polaroid cameras..

Anonymous

@Abcd What do you mean by polaroid cameras?

Abcd

@Blue please google polariod cameras

google.co.in/…

Anonymous

That's not the response I was looking for.

Anonymous

If you're referring to instant cameras manufactured by the Polaroid Corporation, that's an entirely different thing.

Abcd

15:05

@Blue polariod cameras arent famous in India

I am talking about such cameras^

Do they use the concept of polarisation of light? If yes, then how?

Anonymous

@Abcd Those are instant cameras which are colloquially referred to as polaroid cameras. The "polaroid" in that name has nothing to do with polaroids which you learn in physics. It comes from the "Polaroid" Corporation.

Abcd

@Blue Oh m shocked

Anonymous

And that is exactly why I had asked you to clarify your question in the first place.

Abcd

how can I clarify when I dont know anything about it.

I tried to calrify by posting the image of the camera

@Blue Do you know any common use of physics polariods??

Anonymous

@Abcd Wikipedia states some common applications.

Abcd

15:14

Oh only horizontal components allowed through sun glasses

Anonymous

Polarizing filter (photography)

A polarizing filter or polarising filter (see spelling differences) is often placed in front of the camera lens in photography in order to darken skies, manage reflections, or suppress glare from the surface of lakes or the sea. Since reflections (and sky-light) tend to be at least partially linearly-polarized, a linear polarizer can be used to change the balance of the light in the photograph. The rotational orientation of the filter is adjusted for the preferred artistic effect. For modern cameras, a circular polarizer (product labeling abbreviation: CPL) is typically used; this comprises firstly...

Anonymous

@Abcd That page explains it well ^

Abcd

@Blue hmm

Anonymous

The images are rather nice on that page.

Abcd

yup

Emilio Pisanty

15:30

yo @DanielSank was this visible from your location?

SpaceX's Spectacular Twilight Launch of SAOCOM

►

::super jealous::

Swapnil Das

16:12

I'm really not liking this excessive feminism in the name of the Nobel Prize. Enough is enough.

“We are disappointed looking at the larger perspective that more women have not been awarded,” Göran Hansson, the vice-chair of the board of directors of the Nobel Foundation, said at the time. “I suspect there are many more women who are deserving to be considered for the prize.”

What nonsense?

DanielSank

@EmilioPisanty Yes, it was very neat to watch.

My windows started vibrating and I thought maybe there was an earthquake. Then I got a message from my friend saying that there was a launch.

Went out on the balcony and saw two things: the orange streak of the rocket coming down, and the white glow of the satellite up in the sky.

John Rennie

@EmilioPisanty I suspect that was when I first began to realise that I could get the legendary badge. I went a bit mad and answered everything in sight. Most answers will get one or two upvotes so answer 20 questions in a day and you're pretty much guaranteed a rep cap.

Abcd

Hi @JohnRennie ?

John Rennie

@Abcd hi

Abcd

@JohnRennie can we add electric fields using complex numbers

John Rennie

16:27

Yes. A phasor is just a way of combining the amplitude and the phase and when you add two phasors you are in effect just doing a vector sum.

Abcd

@JohnRennie sorry I meant complex numbers not phasors

John Rennie

Phasors are complex numbers ...

That is a phasor is a way of using a complex number to represent an electric field.

Abcd

@JohnRennie Please tell how to add $E_o \sin (\omega t)$ and $E_o\sin (\omega t + \phi)$ using complex numbers

Emilio Pisanty

@DanielSank ::jealous::

$E_o \sin (\omega t)$ plus $E_o\sin (\omega t + \phi)$ equals $E_o \sin (\omega t)+E_o\sin (\omega t + \phi)$

done

John Rennie

Just write them as $E_0e^{i\omega t}$ and $E_0e^{i(\omega t + \phi)}$

Emilio Pisanty

16:32

@JohnRennie that's... not an approach I would condone at all

Abcd

:47084531 is it a lengthy method?? Why does she disapprove of it?

John Rennie

@EmilioPisanty oops, I thought you were replying to my strategy for hitting the rep cap ...

Emilio Pisanty

15

A: What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$?

It doesn't really play a role (in a way), or at least not as far as physical results go. Whenever someone says we consider a plane wave of the form $f(x) = Ae^{i(kx-\omega t)}$, what they are really saying is something like we consider an oscillatory function of the form $f_\mathrm{re}(...

John Rennie

@Abcd he not she :-)

Emilio Pisanty

particularly the bit with

> many authors are pretty lazy and they are not as careful with those distinctions as they might.

@JohnRennie it doesn't serve anyone to conflate the real-valued physical fields and the complex-valued amplitudes used to represent them

Abcd

16:34

@JohnRennie Oh, all this while (like for 1.5 years) I thought Emilio was a female name.

Emilio Pisanty

@Abcd I can confirm it is a male name.

John Rennie

I guess Emiliana would be the equivalent female name

Emilio Pisanty

@JohnRennie in Spanish-speaking countries, you'd just use Emilia

Abcd

...So there's no active female in the h bar.

Anonymous

@JohnRennie Just Emilia would suffice :P

John Rennie

16:35

In the Romance languages the "o" ending tends to be maculaine and the "a" ending feminine

danielunderwood

Huh I never really thought of that being true with proper nouns for some reason

Emilio Pisanty

@Abcd if you really want to use a complex-numbers formalism to add $E_1(t) = E_0 \sin(\omega t)$ and $E_2(t) = E_0 \sin(\omega t+\phi)$, you start off by writing them as the real part of a complex exponential

Anonymous

"a" endings are quite common for female names in several languages. Even in Sanskrit, for example.

Abcd

@JohnRennie But there's no cosine

John Rennie

@Abcd no, the ladies are even more underrepresented in the PSE chat than they are in the Nobel prizes.

Emilio Pisanty

16:36

@Abcd that's why John's method is reprehensible

Abcd

@EmilioPisanty Oh, I see.

Anonymous

@Abcd Just put an $\text{Im()}$ around JR's expressions

Abcd

@Blue Ikr

Anonymous

It's fine mathematically

John Rennie

Actually, now I stop to think about it, I'm not sure phasors are that useful for adding electric fields. We use them a lot in circuits but I don't see them used as much in interferometry.

Emilio Pisanty

16:38

instead, you write $$E_1(t) = E_0 \sin(\omega t) = \mathrm{Re}(i E_0 e^{-i\omega t})$$ and $$E_2(t) = E_0 \sin(\omega t) = \mathrm{Re}(i E_0 e^{-i(\omega t+\phi)})$$

danielunderwood

@JohnRennie I mean some of them could be women and just not saying it. I wouldn't be surprised if there really weren't any based on the gender distribution of people I've met in physics though

Emilio Pisanty

and then you add them up

the real part is linear so you can open up the sum

Abcd

@EmilioPisanty I think writing as Im(JR's expressions) is better

Emilio Pisanty

$$E_1(t)+E_2(t) = \mathrm{Re}(i E_0 e^{-i\omega t}) + \mathrm{Re}(i E_0 e^{-i(\omega t+\phi)}) = \mathrm{Re}(i E_0(1+e^{-i\phi}) e^{-i\omega t})$$

and you're done.

@Abcd whatever rocks your boat. They're equivalent so you can use whichever you find most convenient.

Abcd

@EmilioPisanty ikr. I don't like the Re() thing coz of the annoying negatives and extra i-s.

Emilio Pisanty

16:40

@JohnRennie that's 'cause you're not seeing much (or the right kinds of) interferometry. They're plenty useful.

@Abcd well, that's because you chose sines instead of cosines in the first place.

Folks generally don't take very kindly to having Im instead of Re in those kinds of expressions.

CooperCape

I've always preferred Re() over Im()

Emilio Pisanty

it makes it more likely that you'll slip up with a sign somewhere.

CooperCape

Just to add my opinion for no particular reason

Emilio Pisanty

20 secs ago, by CooperCape

I've always preferred Re() over Im()

case in point

Abcd

So we have $\text{Im}(E_o(\cos \omega t+ i \sin \omega t)(1+\cos \phi + \sin \phi))$

Then?

Emilio Pisanty

16:42

@Abcd oh god, no

Abcd

??

Anonymous

@CooperCape Forget all that use the full blown $\sin(x) = (e^{ix}-e^{-ix})/2i$ ;)

Emilio Pisanty

if you already have things as an imaginary part of a complex expression, it makes no sense to split up the $e^{-i\omega t}$ into anything else

also, mind the signs

Abcd

Then what should be done?

CooperCape

@Blue wait how else would you calculate sin(x).

I don't own a calculator on principle

/s

danielunderwood

16:44

$\sinh(ix)$ of course

Emilio Pisanty

generally speaking, if you're using $i=\sqrt{-1}$ for your imaginary unit, there's a strong bias towards representing your complex oscillations as $e^{-i\omega t}$

Anonymous

@CooperCape I meant that as an alternative to writing Re() or Im() at all

CooperCape

@Blue I know it was my crap joke lol

Emilio Pisanty

if you want a positive sign, use $j=\sqrt{-1}$ and $e^{j\omega t}$ like the engineers do

there's nothing wrong with either convention

CooperCape

@danielunderwood and then you can rewrite $e^{ix}-\cosh(ix)$

I like this

Emilio Pisanty

16:45

but you shouldn't mix-and-match between the two unless you really know what you're doing

using $e^{i\omega t}$ is just a recipe to help ensure that a high fraction of your readers will make a sign error when interpreting your text.

@Abcd what I did above

6 mins ago, by Emilio Pisanty

$$E_1(t)+E_2(t) = \mathrm{Re}(i E_0 e^{-i\omega t}) + \mathrm{Re}(i E_0 e^{-i(\omega t+\phi)}) = \mathrm{Re}(i E_0(1+e^{-i\phi}) e^{-i\omega t})$$

that's really it

you won't get it any simpler than that

Anonymous

@CooperCape Anyway, this conversation reminds of the marks I lost in my first circuit theory test for writing $i$ instead of $j$ everywhere - for not following the damned EE convention! :P

danielunderwood

@Abcd it may be worth watching the first couple lectures of a basic complex analysis course. I think a lot of people in physics never really take one and get introduced to complex numbers in an odd way

Emilio Pisanty

@Blue it does feel pretty icky to write $j$ for $\sqrt{-1}$, no?

CooperCape

@Blue God damn house rules

danielunderwood

Wait are both $i$ and $j$ $\sqrt{-1}$ or should we have $j = - \sqrt{-1}$?

Emilio Pisanty

16:49

@JohnRennie oh, I think your rep-cap strategy is perfectly good

enumaris

$4+2j$

danielunderwood

Don't EEs sometimes like to use $i$ for current or something?

enumaris

shudders

Emilio Pisanty

@danielunderwood The best way to think about it is that $j=-i$ and $i^2=j^2=-1$

the symbol $\sqrt{-1}$ isn't particularly useful other than a nominative statement, due to the branch problems with $\sqrt{z}$

Anonymous

@EmilioPisanty Not anymore. After a lot of struggle, they finally managed to convert me into an engineer XD (It was hell initially seeing $j$ everywhere)

danielunderwood

16:50

Ahh I never really knew there was a difference in definition. I thought EEs were just odd

Emilio Pisanty

@danielunderwood it's not so much a difference in definition. It doesn't even make sense to treat it like that.

it's just that all of the phasor formalism is untouched if you flip the signs on all the $i$s

morally, when you're extending the reals to the complex numbers you "don't know which way" the positive imaginary unit should go, so there are two equivalent ways to extend the formalism.

EEs tend to prefer positive coefficients in their $e^{j\omega t}$s because they work primarily with circuits

physicists put a premium on space over time when writing things like $e^{i(\mathbf k \cdot \mathbf r - \omega t)}$ so we like negative signs on our $e^{-i\omega t}$s

EEs like to use $j$ for the imaginary unit because that frees up $i$ for current

physicists like to use $i$ for the imaginary unit because we're sensible people

the upshot is that there's a huge fraction of the literature over both physics and EE that can be reconciled by pretending that $j=-i$.

@Blue I guess if you're around ick for long enough, you get desensitized

(cf. supra re: "sensible people".)

danielunderwood

I guess I've just always seen kind of a "right hand rule" with how to represent the positive real and imaginary axes. Sounds like that isn't necessarily the case though

Now EEs just need to come up with something that needs quaternions

Related, are people that use $\mathbf{i}, \mathbf{j}, \mathbf{k}$ as Cartesian unit vectors actually using the non-real parts of a quaternion for some reason?

Anonymous

17:11

@danielunderwood I don't think they're related

Anonymous

Wait, hmm

Anonymous

There may be some relation in the historical context but I doubt there's any equivalence

Anonymous

> Hamilton knew that the complex numbers could be interpreted as points in a plane, and he was looking for a way to do the same for points in three-dimensional space. Points in space can be represented by their coordinates, which are triples of numbers, and for many years he had known how to add and subtract triples of numbers.

Anonymous

> However, Hamilton had been stuck on the problem of multiplication and division for a long time. He could not figure out how to calculate the quotient of the coordinates of two points in space.

Anonymous

> From the mid-1880s, quaternions began to be displaced by vector analysis, which had been developed by Josiah Willard Gibbs, Oliver Heaviside, and Hermann von Helmholtz. Vector analysis described the same phenomena as quaternions, so it borrowed some ideas and terminology liberally from the literature of quaternions. However, vector analysis was conceptually simpler and notationally cleaner, and eventually quaternions were relegated to a minor role in mathematics and physics.

Anonymous

17:18

So basically the $\{i,j,k\}$ Cartesian vector space is sort of a subspace of the vector space formed by the quaternion basis vectors.

Anonymous

@danielunderwood $1,i,j,k$ are all reals. It doesn't make sense to call $\mathbf{i}, \mathbf{j}, \mathbf{k}$ the "non-real" parts

Anonymous

The 4-dimensional quaternion vector field is defined over the field of real numbers

Anonymous

But yes, it seems like you can derive the properties of Cartesian vectors from quaternions

Anonymous

> Vector calculus was developed from quaternion analysis by J. Willard Gibbs and Oliver Heaviside near the end of the 19th century.

Anonymous

Nah, I guess I was wrong

danielunderwood

17:25

Does it make sense to call $i, j, k$ reals since they have the properties $i^2 = j^2 = k^2 = ijk = -1$ that seems to define the quaternions? Certainly their coefficients are kind of reals in the sense that you can have a quaternion coefficient, but that would be equivalent to a quaternion with all real coefficients

Anonymous

@danielunderwood I was wrong indeed

danielunderwood

The use of them as unit vectors does make sense if vector calculus came out of quaternions though. And the pair products of bases are quite like cross products

Anonymous

@danielunderwood I was learning the properties of quaternions on the go. I have never studied it before. :P Anyhow, it does seem you can derive coordinate geometry from restricting it to just $\{i,j,k\}$

danielunderwood

I'm not familiar with them as I could be. I can do some computations with them, but don't feel that I truly understand them

Anonymous

So basically what are the properties we expect our ordinary vectors to follow?

danielunderwood

17:29

I was just always troubled by the convention of using i, j, k for Cartesian bases and thought there may be a connection once I saw them as quaternion units

More Anonymous

@Blue any thoughts on my question physics.stackexchange.com/questions/433329/… ?

Anonymous

Addition, Subtraction are simple to derive

Anonymous

Even quaternions follow them

Anonymous

Now dot product, cross product and scalar triple product and vector triple product

danielunderwood

@Blue Well they certainly have addition and scalar multiplication like we'd expect. We expect $x \times y = z$, which works with $ij = k$. We'd expect $x \times y \times z = 0$, but $ijk = -1$, though I'm not sure how to interpret that since -1 is the real part of a quaternion. I have no idea about dot product

Anonymous

17:33

@danielunderwood But $ijk$ is not how vector triple product (or double cross product) is defined for quaternions I guess

Anonymous

Read the section: Quaternions and the geometry of R3

Anonymous

> A quaternion of the form a + 0i + 0j + 0k, where a is a real number, is called scalar, and a quaternion of the form 0 + bi + cj + dk, where b, c, and d are real numbers, and at least one of b, c or d is nonzero, is called a vector quaternion. If a + bi + cj + dk is any quaternion, then a is called its scalar part and bi + cj + dk is called its vector part

Anonymous

So essentially all our Cartesian vectors are vector quaternions

Anonymous

Modulo the notions of multiplication and division

Anonymous

17:36

So it pretty much it seems that what you initially expected is correct: Cartesian geometry is a strict subset of quaternion geomtry

Anonymous

The thing to keep in mind is: Cross product and Dot product are not equivalent to "multiplication"

Anonymous

We don't have any notion of $\vec{\mathbf{i}}\vec{\mathbf{j}}$ in coordinate geomtry!

Anonymous

We only have notions of $\vec{\mathbf{i}}.\vec{\mathbf{j}}$ and $\vec{\mathbf{i}}\times \vec{\mathbf{j}}$

Anonymous

(which also exists for quaternions)

Anonymous

(I'm sorry for having posted complete nonsense initially :P. I guess I'm making sense now.)

More Anonymous

17:45

Thoughts any one?

1

Q: Using a time-like boundary as a computer?

Question and Summary Using classical calculations and the Robin boundary condition I show that one calculate the anti-derivative of a function within time $2X$ I can compute an integral $$\frac{2 \alpha}{\beta} e^{ \frac{\alpha}{\beta}X} \int_0^{X} e^{- \frac{\alpha}{\beta}\tau} \tilde f_0...

danielunderwood

Well multiplication is only a matter of syntax. If we use $i, j, k$ for "quaternion units" and $\mathbf{x}, \mathbf{y}, \mathbf{z}$ as the Cartesian units, I'd say that the "quaternion product" $ij$ is equivalent to $\mathbf{x} \times \mathbf{y}$ and similar if you make the associations $i \leftrightarrow \mathbf{x}, j \leftrightarrow \mathbf{y}, k \leftrightarrow \mathbf{z}$ with the exception of the triple product

Anonymous

@danielunderwood How would you show the equivalence of quaternion multiplication and vector cross product?

Anonymous

It doesn't make sense to me...

Anonymous

Cross products are already defined for quaternions in a different manner

Anonymous

(which is equivalent to the vector cross product)

Anonymous

17:49

danielunderwood

Actually there is an issue in the fact that you'll end up with real terms. I think they're equivalent if you throw away the real part and make the diagonals 0...but the diagonal is one of the defining parts of quaternions

What's written on wikipedia is carrying out the multiplication $(ai + bj + ck)(di + ej + fk)$ and throwing out the real part I believe

Anonymous

@danielunderwood Yep, cross products are defined by putting the real term to 0

Anonymous

Essentially that's what they're calling the "vector part"

danielunderwood

Ahh I see what you're saying about Cartesian geometry being a subset of quaternion geometry

Anonymous

What are quaternions, and how do you visualize them? A story of four dimensions.

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Anonymous

17:54

Damn, I need to watch this!

Anonymous

> Oliver Heaviside @1:57 is freaking Wolverine / Hugh Jackman

Anonymous

LOL

danielunderwood

I think I watched that when it came out, but I don't remember a ton out of it

Also since you seem to like online lectures, you may want to check this guy out. His lectures may be my favorite from the ones I've seen online, though they're usually sufficiently mathy for me to get lost at some point youtube.com/channel/UCpHjg_Qmzxm3xaAWRrwQPCA

Anonymous

@danielunderwood Osborne, hehe. His ones are a bit high level

Anonymous

Maybe during the vacations

danielunderwood

17:59

Yeah I wish he had slightly lower-level ones, but I've enjoyed the ones I've watched. I think a large part of it may be the blackboard washing that gives a few moments to digest what he just said

Anonymous

Lol

Anonymous

Yeah

vzn

> The video was anonymously uploaded to YouTube in 2013, but began circulating much more widely after reporter Leonid Schneider discussed it on his website, For Better Science, the day that Mourou won his Nobel. According to Le Monde, before this week, “Have You Seen ELI” had only garnered a few hundred views. It now has more than 67,000. motherboard.vice.com/en_us/article/9k7zm5/…

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