12:00 AM
and inductive reactance only positive imaginary values?
so magnetic impedance should only have imaginary values which are positive?

reactance is part of electrical impedance as is susceptance
and electrical impedance has a real and imaginary component too\
but magnetic impedance does not have a reactive component, its imaginary component is called permeance

fascinating

@antimony the real part of magnetic impedance is called magnetic resistance which is not the same as normal resistance (electrical resistance) but does server a similar function (dissipates energy as heat)
an
err sorry getting used tot he interfae

me too

@antimony electrical impedance, the kind your used to, describes electrical current and voltage as it moves around a circuit through wires made of copper and components like inductors and capacitors as your used to
@antimony however mangentic impedance is what you'd use to ddescribe similar things (but they are physically very different) in a magnetic circuit
@antimony in a magnetic circuit the wires are replaced with iron which conducts magnetic fields rather than electrical fields through it and what acts like a capacitor or an inductor in such a circuit looks a bit different than what those things look like in an electric circuit but behave similar within the analogy

12:05 AM
incredible
This whole time I thought I'd been accounting for both electric field (capacitor) and magnetic field (inductor)
well, i mean i know any discrete circuit element has an inherent RLC component

@antimony well you do normally account for both fields
@antimony this isnt accounting for anything new it is just accounting for things in a different way

yes i see

@antimony this is a different model for similar ideas

i suppose certain applications it is preferred to use the magnetic model?

@antimony its not really used all that often honest, but there is some sort of analysis where it is useful, I see it used in radar analysis a lot where they want to calculate how a material will absorb the field
@antimony you know how if you have an inductor and it uses a ferrite core, particularly at high frequencies beyond the capabilities of the core, the core will heat up and get hot?

12:13 AM
ahhh yes i see
eddy currents or some such? (always glossed over a bit in my current level of studies)

@antimony well in the electrical world what would would happen is we call that ESR, effective series resistance.. what that means is if the inductor is heating up we would see that as if the inductor had resistor in series with it at that particular frequency, if we measured it with a current meter it would look exactly like a resistor despite there being no resistor
@antimony however if we viewed the core as a magnetic circuit, then we would describe it as the core material itself have a complex impedanceto the magnetic field propagating through it, which likewise would have a magetic resistance

it is like the frequency where the inductor self resonantes?

@antimony this would happen even without it self resonanting

wow, incredible
oh i see

@antimony think of an induction furnace, it uses the magnetic resistance of a material to heat it up

12:18 AM
ohhh

@antimony basically because gold or whatever is being smelted has a magnetic resistance it will convert the energy int he changing magnetic field into heat and melt
@nt

this is soley dependent on the material permeability?

@antimony the equivelant of that with an electric field would be if you just hooked a car battery up to a piece of metal and used its electrical resistance to heat it up and melt it by putting regular current through it

hahahah

@antimony well using the term "solely" in physics never goes too well :)

12:19 AM
true :)

@antimony complex permiability and electrical resistance are often the two major components at play, but there are others that are rather nuanced and complex

i see

@antimony the three types of loss is "static hysteresis loss", which is loss that occurs if the material becomes magnetized across each cycle, this loss only applies to ferromagnetic material like iron. then there is also some sort of wall current and spin loss that i dont really understand which is called residual loss..
but the main component of loss in most materials is a combination of complex permiability and electrical resistance. That is the eddy current stuff you were talking about but best not to think in terms of eddy current if your talking magnetic circuits in my opinion

ahhh i see

@antimony if you think of things in terms of magnetic circuits think of the magnetic field flowing through the material as being the same idea as a current flowing through an electrical circuit

12:24 AM
ok nice

Anti
@antimony reason im in here asking questions is im writing a blog aritcle explaining all this and wanted to make sure i had the right equations
@antimony the article is intended for people like you to understand and I dscribe all sorts of "electrical circuit duals".. and this is the magnetic circuit dual
@antimony one really interesting thing about magnetic circuit duals is that the main units are flipped.. the "current" in a magnetic circuit is measured in volts and the "potential" is measured in ampere

very nice!
i see
i suppose it kind of makes sense seeing the reciprocal of units in the analogs
This is lovely to me since I always was used to alot of hand waving occuring around magnetic phenomena

@antimony yea all the equations look very similar but they are all turned on their head :)
@antimony me too which is why i spent the last week studying this, it is new to me too as of a week ago
@antimony you will probably like the blog post once I'm done with it.. i have a half-finished draft up on my beta site right now but it isnt complete so may not be great for you and may have some errors until im done with it

nice
pls feel free to tell me the blog url

@antimony here is the beta site for it with the draft blog entry: beta.jeffreyfreeman.me/…
@antimony the normal blog you drop the beta in the url which is where you normally want to go as i dont always update the beta site at all, it only changes when im renovating

12:38 AM
very nice
thanks :)

enjoy :)
@antimony but seriously check back in a few days when i actually finish the part on magnetic circuits, ill be adding a lot more content on magnetic circuits before publishing.

copy that
and thanks for putting me onto this wonderful alternative model!

@antimony no problem.. when i first realized it was a thing i didnt know that actual physicists had proved out the ideas.. when i told my EE friends they thought i was wrong and nuts.. then i realized it was a real thing and my ideas were proven out long before i had them in more detail than my original thoughts, that was a week ago and ive beensstudying and writing about it since then
@antimony here is the wikipedia page that talks about it, though i think it does a really poor job explaining it: en.wikipedia.org/wiki/…

ahh thanks
ok i see
i had previously *heard of the gyrator-capacitor model, but never knew much about it
nor the extent of it's scope

@antimony its kinda cool when you realize what a capacitor and an inductor (or rather their analogous counterpart) is in the model
@antimony they are sort of their opposites of the electrical one.. in an electrical capacitor you have two thin wires that attach to two plates made of the same material as the wires, and then with a resistor in the sapce between them
@antimony in the magnetic circuit its similar in that it is two thin magnetic wires (iron) coming to two big plates (also made of iron) but instead of the space between them being a high resistance (electrical or magnetic) it is low resistance, so the space between is filled with iron too.. basically the opposite of an electrical capacitor
@antimony a magnetic capacitor stores magnetic flux jsut like an electric capacitor stores energy in electric field
@antimony now whats really interesting to me though is the magnetic inductor.. just like an electric inductor stores energy in magnetic flux a magnetic inductor stores energy in an electric field.. to make one of those you would take one of the iron wired and wrap some copper wire around it to make a coil and then on the two ends of the coil you would put a regular electrical capacitor..
so a magnetic inductor is still an inductor but in the reverse orientation where the magnetic wire is where the ferrite core and the leads of the inductor form the capacitor part to store the electric field

1:05 AM
thats pretty awesome
i'll have to take a closer look when i get a chance

@antimony hopefully my paper will help when im done writing it. Honestly the material that exists on this is very hard to follow and it really doesnt explain things well the stuff I could find.. thuis why i invested some much time in this article
@antimony so sadly unless you find a better source (and if you do let me know) you may need to rely on my blog post when its done

yeah i noticed its quite messy on wikipedia
if i see something better i'll let you know
though i suspect you will have finished the article before i have time to delve too deeply

@antimony probably

i love electrical equivalent models
for example did you know there is one for thermal
and acoustics
(i'm sure you did, but just in case)

1:24 AM
@antimony im not sure id call that an electronic model exactly but yes.. in fact what we are really talking about is wave theory
@antimony waves are (almost) the same in every medium and thus the models to describe them are all very similar if not the same
@antimony have you seen the bell labs video where they describe electrical waves using mechanical ones?

1:47 AM
sounds familiar, can't remember if i've seen that one
would the thermal model really contain the wave information though?

2:39 AM
the acoustic one, definitely though :_
:)

2 hours later…
4:53 AM
What do they mean by this question?
"Calculate the exact spot between the Sun and the Earth where the forces cancel each other."
Like does (m_1)(m_2)(G)/r^2 have to equal 1?

@NorthLæraðr suppose you are between the Sun and the Earth, with the Sun on your left and the Earth on your right.

Okay
Oh, I see. Where should I be so that the force of gravity is equal on both sides

Yes :-)

Hm... how would I go about doing that?

Call the Earth to Sun distance d, and suppose you are a distance r from the Sun, so your distance from the Earth is (d-r). OK so far?

5:02 AM
Yup

And call the Sun's mass M and the Earth's mass m.

Okay

Then the gravitational acceleration due to the Sun is a(Sun) = GM/r²

Right

And the gravitational acceleration due to the Earth is a(Earth) = Gm/(d-r)²

5:03 AM
Okay

And set the two equal ...

Hold on...

Yes ... ?

No, I'm just moving the variables around :P

:-)

5:06 AM
Hmm, okay do I factor the (d-r)^2?

When you set a(Sun) = a(Earth) this is going to give you:
GM/r² = Gm/(d-r)²

Mhm

What I would do is rearrange this to:
r²/(d-r)² = M/m
Then you can square root both sides:

ah...

r/(d-r) = √M/m

5:10 AM
and how do we move the d from (d-r)?
Oh, is it 2r=(√M/m)(d)?
@JohnRennie ?

You multiply both sides by (d-r) to get r = d√M/m - r√M/m
OK so far?

Yes

Then add r√M/m to both sides to get r(1 + √M/m) = d√M/m
And finally divide both side by (1 + √M/m) to get r = d√M/m/(1 + √M/m)

Ohhhh, I see.
I was like working out how you got (1 + √M/m), but I figured it out
That seems... much more complicated than what I though my teacher would've wanted
But I don't know for sure :P
Thank you so much!

:-)

5:27 AM
I'm trying to understand how h-INDEX works...
still no clue

Wait... hold on, I think I just got pranked by my Physics teacher
After all that algebraic manipulation
The answer (with correct significant figures) is 1.50*10^11m, which is literally the same thing as the distance from the Earth to the Sun
So the attraction between the Earth and the Sun would be when I'm... on the Sun or on Earth?
More precisely, at the center of the Earth and the Sun?

@NorthLæraðr It's probably more useful to calculate the distance from the centre of the Earth i.e.
r(earth) = d(1 - √M/m/(1 + √M/m))
I get r(earth) = 2.6 x 10^8 metres

5:43 AM
ohh...

The radius of the Earth is 6371km, so that distance is about 41 times the radius of the Earth.

wait what is r(earth)? Radius of earth?

It's just that the Earth-Sun distance is so large that 41 times the radius of the Earth looks really small by comparison.
@NorthLæraðr no, r(earth) means the distance of the equilibrium point from the centre of the Earth.

My brain is like not functioning, it's 11 where I'm at..
I'm so confused now

I'll draw a diagram ...

5:49 AM
How did you get from r = d√M/m/(1 + √M/m) to r(earth) = d(1 - √M/m/(1 + √M/m))

@JohnRennie Sir, I have a small doubt. Can a fusion reaction occur with $\beta$ and $\gamma$ decay?

@JohnRennie Love the diagram, btw :P

:-)
The result I got was r(sun) = d√M/m/(1 + √M/m)
Yes?

Right

5:54 AM
And r(sun) + r(earth) = d, so r(sun) = d - r(earth)
So if we use this to substitute for r(sun) in my equation we get:
d - r(earth) = d√M/m/(1 + √M/m)
OK so far?

Yeah

Rearranging this gives:
r(earth) = d - d√M/m/(1 + √M/m)

@Azmuth say the h index is 5. This tells you that the person has 5 papers with citations of at least 5. And all the rest of the papers have citations less than 5

And taking out the factor of d on the right side gives:
r(earth) = d(1 - √M/m/(1 + √M/m))

Okay, so my friend just texted and asked if it's just half the distance from the Earth to the sun, since the sun and the Earth both pull with the same force

5:57 AM
@ArnavMahajan hi :-) Can you clarify what you are asking?
Beta and gamma decay are not obviously related to fusion.
@NorthLæraðr so now you can tell him the answer is "no" :-)

@JohnRennie Hm.
Right

It would only be halfway if the Sun and the Earth had equal masses.

@JohnRennie So the question was: $4H_1^1 => \alpha + 2e^+ + energy$. It was asked that this reaction represents? (1) $\beta$ decay (2) $\gamma$ decay (3) Fusion (4) Fission

This is what happens in the Sun.

This question has multiple options correct

6:05 AM
Th way it works is a proton undergoes reverse beta decay i.e. it emits a positron and turns into a neutron p -> n + e+ + neutrino
Then a proton combines with this neutron to form a deuterium nucleus.
This is the full reaction:

@JohnRennie I don't doubt that you are wrong, it's just this problem is waaay more complex than any of the other questions in the worksheet.

@ArnavMahajan option (3) is correct because it's a fusion reaction, and option (4) is wrong because no fission is involved. Yes?
@NorthLæraðr welcome to physics :-)

well... it looks out of place considering every single other question is very straightforward. Idk, I've never had to do this much algebraic equation in one worksheet, but I guess first time for everything?

Perhaps your teacher has an evil sense of humour :-)

Maybe. Or maybe he's just really bad at phrasing them
I don't think it was meant to be this convoluted, but I think how it's phrased might call for those steps above

6:14 AM
Imho, set it aside and move on :-)
aka time management

@skullpatrol Probably the best advice. My brain is not functional right now. I thought it was due today, but it's due on Saturday

Doing homework at 11 p.m. the night before it's due. It's good to see that nothing has changed in the 50 years since I was at school :-)

I had marching band practice right after school until 7. This rarely ever happens :P

I used to make excuses like that as well.

Alright, sure

6:18 AM
if it wasn't for the last minute, nobody would get anything done

:-)

Geez guys, I actually didn't procrastinate on this... but who'd believe me, I'm just a random internet user
@skullpatrol Truth

@NorthLæraðr no offence intended.

Lol, it's fine

For the record I still leave programming jobs to the last minute then pull all nighters to finish them.

6:21 AM
Now this choir recording on the otherhand that's due in 40 minutes....

6:31 AM
Alright, thank you everybody. Good night!

cya, pal

1 hour later…
8:01 AM
how are we already almost in october of 2020
it feels like just recently everyone was talking about the new decade beginning

"october"...I think I faintly remember what months were

i swear the years are getting shorter

Would you like even more of 2020? :P

well nooo
it's a weird paradox, time speeding up is freaky but i'd never want to go back

8:18 AM
time? oh yes, we just had the autumnal equinox

I dun get little dennis now a days in Newspapers

The Calvin and Hobbes series finished years ago. The whole collection is available as an ebook.

1 hour later…
9:53 AM
@B.Brekke Thanks a lot B. Brekke for your response.

5 hours later…
3:02 PM
anyone else getting scam calls at least once per day?

I get them very occasionally on my mobile

3:25 PM
I'm very confused about Ampere's Law
I've traditionally written it as $\nabla \times B=\mu_{0}\mathcal{E}_{0}\frac{\partial E}{\partial t}+\mu_{0}{j}_{c}$
But sometimes I also see it written as $\nabla \times H=\frac{\partial D}{\partial t}+J$
How? Where does the ${\mu}_{0}$ go?
Where does ${\epsilon}_{0}$ go?
If anyone can help, that would be great.

@DarkRunner That should probably be a $D$ instead of the $E$

@SuperfastJellyfish Fixed, thank you

Any chance that's just a different choice of units?

Hm, ... maybe one in which $c=1$?!

$B/\mu_0=H$ and $\epsilon_0 E=D$

3:29 PM
Because I know that $\mu\epsilon=\frac{1}{c^2}$

ah ^

Ah yes, yes. Thanks a lot! @SuperfastJellyfish

These are generally used to describe electromagnetic fields inside materials. As the fields inside depend on how the material responds to the external field.
:)

2 hours later…
5:21 PM
@SirCumference same for me. And I'm still 18.

5:50 PM
@JohanLiebert it's gonna get worse
i feel like 50% of college for me was my first year
the rest flew by

I've been out of school 6 years and it feels like nowhere near that long.

It might be because once you're out of school there isn't a huge change up as you go into the next year, at least not if you're working, so the years feel less distinct

2 hours later…
7:43 PM
@JohnRennie Wow, you were right. My physics teacher is evil
He literally gave us an ultra-algebraic heavy qs

8:14 PM
Quick question, when we quantise a vector or spinor field (or more generally anything with more than one component) do we produce a quantum field operator for each component or does the entire object become one operator? I'm about 90% sure it's the former, which would immediately solve another problem I've been wondering about
ie. $$V^{\mu}(x) \rightarrow \hat{V}^\mu(x),\quad \psi^\alpha(x)\rightarrow \hat{\psi}^\alpha(x).$$

@Charlie it's the former

ok great, thanks

3 hours later…
10:53 PM
@SirCumference lol! True. Felt the same for my high school.

11:12 PM
@bolbteppa I think i'll decide to remember 2020 as the year we found this
maybe i'm optimistic but it sounds super promising. who knows, perhaps gas planets like jupiter could have bacteria in their atmospheres too

11:41 PM
Probably