The h Bar

6:05 AM

@JohnRennie - I saw your answer here: physics.stackexchange.com/users/1325/john-rennie

And I have a question: what if I replace that moving rod with a slanting moving rod? How about a semicircle instead of that moving rod?

@McSuperbX1 hi

Hello!

I will continue with my question: So should I consider the effective vertical length OR the actual length of that new wire I'm using now and why?

The link you give just links to my profile. It doesn't link to an answer. What question are you asking about?

Q: What is meant by the word "length" in definition of surface tension?

Oh sorry..

1

Surface tension is defined as the force applied per unit length. What is that "length" belonging to? I can imagine force being applied per area but not length. "It would take a force of 72 dynes to break a surface film of water 1 cm long." Film is two dimensional. Isn't it? My textbook didn't ...

Here you go =]

The simplest way of looking at this is just to look at the area that film changes by as you move the rod. The surface tension is also the surface energy density i.e. the surface tension is the energy per unit area of the film.

Suppose you move your rod (slated, semicircle, or whatever) some distance $x$ and this increases the surface area of the film by $A$. Then the enrgy of the film has increased by $E = \gamma A$ so the work done by the moving rod has to be equal to $\gamma A$.

6:12 AM

But what about the force on that rod?

W = F * x

The work done by the rod is just $W = Fx$, so there is a total force of $F = \gamma A/x$

so i just divide the work done by the distance travelled horizontally?

Ahh

That is smart!

Note this is the total force i.e. the vector sum of the forces on all the infinitesimal parts of the rod.

Yes absolutely.

The force is always normal to the rod so if the rod is curved the forces along it will be in different directions.

6:14 AM

Which means I can also consider the net vertical length of the rod

to find the force?

But you can resolve the forces into components parallel to and normal to the direction of motion. Only the components parallel to the direction of motion do any work.

So what I just said is correct then?

Yes, or at least yes for the example I gave of the rod moving along a U shaped frame.

Alright! Sounds great. Thank you.

Also, I hope you would not mind if I ping you often asking more questions.

(preparing for the JEE. again..)

No problem. I'm always happy to chat about physics :-)

6:17 AM

Fantastic. Thank you. Have a nice day.

If you want to ask about specific JEE questions ask in the problem solving room.

Understood.

:-)

PM 2Ring

7:07 AM

I'm not happy with this sentence: "Photons always travel in vacuum, even in the medium, between the atoms." From physics.stackexchange.com/a/484184/123208

7:20 AM

it looks ok, in context

PM 2Ring

7:32 AM

It's not exactly wrong, but I'm still not happy. ;)

8:09 AM

Hello folks

8:32 AM

ello

"It is easy to show that the dimension of a differentiable manifold is uniquely determined. For a general, not differentiable manifold, this is much harder."

oh no

Q: Can you help me to make my question more specific?

8:47 AM

@RyanUnger Riddle me this, Batman

Nambu-Goto action is $\approx \int_\Sigma d\mu[g}$

But obviously this will likely be divergent in general

It should probably be done on some subsets, but then are boundary terms necessary?

Physics Meta

12:35 PM

0

I did this question but they say it's too broad, I modified but I think they consider it still to broad. I don't understand why, can you help me? What is the approach to the mathematical modeling?

discussion specific-question

Ryan Unger

12:52 PM

@Slereah do you want me to tell you how it’s done

@Slereah you always do them on subsets

@JohnRennie 👍🏻

@RyanUnger he has always said that

1:26 PM

blah

Trying to come up with singularities for that question on static spherically symmetric spacetimes

Basic idea isn't too hard but then it's a lot of Riemann tensor shit

Ryan Unger

what is the question

physics.stackexchange.com/questions/484228/…

Pretty sure I can just take some spherically symmetric thing that has some Badness of the form $\approx \sin(1/(r-a))$

But showing it will take some computation

Ryan Unger

@Slereah doesn't RN work

aren't there naked sings for large charge

True

Simplest one is just Minkowski minus the origin

but I suppose that is cheating

I also tried some quasi-regular singularity but it's tricky to find a conical singularity that's spherically symmetric

Easier in $2+1$ dimensions

1:47 PM

arxiv.org/pdf/1512.06497.pdf

what is this

but yeah RN is probably, if not simpler, at least easier to find

Still should probably do that $\sin(1/r)$ example

bc I think it is neat

2:18 PM

Hm

That example will be tricky tho

bc I need the Riemann tensor to be bounded but fucky

but the derivatives of $\sin(1/r)$ are unbounded

so the metric components should be antiderivatives of it

enumaris

omg

had to come to the office at 7:20

sleepy af

And the antiderivative uses the cosine integral function

not the most pleasant

Seems tricky to keep things all bounded

Though I think if I just say that $g_{rr,rr} = \sin(1/(r-a))$

I should be golden

or something similar to keep all things nice

enumaris

"bounded but fucky" - a very technical description

Bounded but not $C^0$

if you prefer

enumaris

2:56 PM

no no

I very much prefer the former

ChoMedit

Could someone check Goldstein classical mechanics 3e p.353? I think in formula (8.61), $H_c$ must be $H'_c$. Am I missing something?

Hello !

If a= 2 icap + 2x jcap then how to find magnitude of velocity at t=3 seconds ,given at t=0,x=0,v=0

@JohnRennie are you available ?

3:17 PM

@Jasmine just to check, do you mean $a_x = 2$ and $a_y = 2x$ ?

@JohnRennie yes that's what is given

I'm guessing that when you write icap you mean i with a hat i.e. the unit vector $\hat i$.

If $a_x = 2$ that means:

Ok I am phone so

$$ \frac{d^2x}{dt^2} = 2 $$

And given your boundary conditions $x = v_x = 0$ at $t = 0$ that immediately gives us:

$$ x(t) = t^2 $$

If $a_y = 2x$ then we can substitute for $x$ to get:

$$ \frac{d^2y}{dt^2} = 2t^2 $$

Integrating gives:

$$ \frac{dy}{dt} = 2t^3/3 $$

Then:

$$ y = t^4/6 $$

Ok

3:24 PM

Oh, hang on, you want velocity not position ...

So $v_x = 2t $ and $v_y = 2t^3/3$

I am getting 6root10

That seems correct

I was doing weird stuff that's why not getting answer then

Thank you sir !

:-)

I hope I can ask such questions in PSS room that was little off for a while so asked here

Hii @JohnRennie

@Jasmine yes, just ping me in the PSS room and if I'm online I'll answer.

@user8718165 hi

3:29 PM

@JohnRennie I have a doubt...are you eating now?:-)

@JohnRennie Ok :)

@user8718165 no, you can ask now.

They are different names for the same thing.

It's the frame in which the total momentum is zero, and the total momentum is the centre of mass velocity times the total mass.

@JohnRennie just deleted it 'coz couldn't format it properly.

So if the total momentum is zero that means the velocity of the centre of mass must be zero.

@JohnRennie Is it something like...when you are sitting at the COM and looking at other particles of the system...in pure translation nothing is moving relative to you so p is zero

3:34 PM

I wouldn't say nothing is moving relative to you.

For example suppose you have two identical masses attached to the ends of a spring and the masses are oscillating.

@JohnRennie okay

The centre of mass/momentum frame is the frame in which the middle of the spring is stationary i.e. the masses are equidistant and their velocities are equal and opposite.

So if you're sat at the middle of the spring looking at the masses then they are moving, but in such a way that the total momentum you measure is always zero.

@JohnRennie if the spring system was performing translation along with oscillation then also p is zero(from COM frame)?since no force is acting on the COM?

@user8718165 the two masses have equal and opposite velocities, so the two momenta are $+mv$ and $-mv$. The total momentum is $+mx + -mv = 0$ i.e. the total momentum is zero.

@JohnRennie okay sir...got it.what if some external force acts? then it won't be zero..?

3:43 PM

Yes, when you calculate the COM frame this normally assumes no external force is acting.

@JohnRennie thank you so much! got it :-)

Q: Is there any hard scientific evidence that the alpha particle is tetrahedral?

:-)

Secret

3:55 PM

36

I'm writing a piece on the nuclear force, and I'm struggling with something. I always thought of the alpha particle as something with a tetrahedral disposition. If you search the internet on this there's plenty of hits. Ditto if you search for images: The alpha particle is usually depicted as ...

nuclear-physics neutrons strong-force

it's actually an interesting question to ask about those rough illustrations we often found

Meanwhile using boteppa's advice, I think there really isn't any known picture for QFT entities

if you put too much emphasis on fields, then as per boteppa's warning, you will over interpret the maths when the physics is mostly states at infinity

If you put too much emphasis on Fock states, then the whole thing breaks down in the interaction regime

Does QFT have an ontology at all, sometimes I wonder

4:33 PM

@JohnRennie

@Akash.B hi

@JohnRennie I have a question

@Akash.B yes?

Do photons loose its energy?

Gotta check whether my theory is right

@JohnRennie

Can you explain what you mean by that? By photon we normally mean a quantum with a certain frequency $\nu$ and an energy $h\nu$.

The photon can transfer its energy into something else, e.g. by exciting an electron in an atom, but it can't change its frequency or energy without interacting with something else.

4:38 PM

@JohnRennie I recently came across an article

It says that the wavelength of light varies with the expansion of space

Yes, that's the cosmological redshift

How's its related?

I'm not sure how you describe the evolution of a single photon when it experiences a cosmological red shift. The same difficulty would apply to a gravitational red shift.

Okay

Ultimately photons are described by quantum field theory, and I believe QFT gets complicated when you have a curved spacetime but I'm not sure exactly what happens and how it explains cosmological and gravitational red shifts.

4:42 PM

@JohnRennie can i introduce my new theory?

@Akash.B ok

Its name is spring theory

According to me photons behaves like a spring @JohnRennie

The problem is that we already have a theory that describes what photons are and how they behave. That theory is called quantum electrodynamics. It's been experimentally tested to high precision so we know it works.

Oh I see

Then I am withdrawing

That's always the problem with trying to build new theories. The new theory has to be compatible with the old, experimentally tested theory.

4:50 PM

There also has to be a need for a new theory. If you do not have a phenomenon your theory explains that the old one does not, it is of very little interest to anyone.

Isomorphic, to within the limits of experimental evidence. If it predicts new behavior that haven't been tested before, it may still be worth testing.

@JohnRennie well the theory i proposed seems to solve my question

But it's still fun thinking about new theories. I used to do it all the time as a teenager, but sadly when I got to university I realised none of my ideas were of any use.

@Akash.B "photons behave like springs" is not a theory

Oh well :-)

4:52 PM

It's hard to develop new theories if you don't understand the specifics of what the old theory is saying. Or to use a similar, very well worded sentiment "In order to claim that "physicists" misunderstand something, one should first demonstrate that one in fact knows how physicists understand something."

@ACuriousMind hmm, so a photon is described by a linear object that has a tension and can oscillate. Where have I heard that before? :-)

@ACuriousMind just an imagination of a 16 year old boy

A theory is a consistent net of ideas, both qualitative and quantitative. Unless what you have produces predictions for e.g. the angle of electron and photon in Compton scattering, it does not play in the same realm as theories like QED.

Generally new ideas are of the most interest if they deal with behavior that is already observed but not tested, but if it is consistent with existing evidence, it can still be worthwhile to see if you can come up with a test that differentiates which of the two better describes reality.

@Akash.B it's good fun thinking about this stuff. Don't let us put you off.

4:52 PM

@JohnRennie In bad popularizations of string theory? :P

Well it remains me of Einstein's quote

@ACuriousMind I understand string theory! I've read The Elegant Universe! :-)

::eye twitches::

(Twice actually. And I thoroughly enjoyed it.)

Haha I think "the elegant universe" is my entire understanding of string theory :(

4:54 PM

Why twice?

It's a well written book and I think it does as well as any popular text could be expected to.

"You have to first learn the rules of the game ,then play better than others"

Did he really say that?

Yes he did

Although ideas that raise new questions are also worthwhile, at least in my opinion. Possibly more so, as I think good questions are harder than good answers. (Not that answers are easy, mind, but we mostly seem to stumble across the good questions by accident.)

4:56 PM

@skullpatrol back in 1999 we'd only just invented electricity. There wasn't the huge amount of stuff to read on the Internet that there is now. When I found an enjoyable science book I'd frequently read it twice or more.

@JohnRennie I'm not saying it's possible to do better! It's just unfortunate that so many people think that one can attempt to make technical arguments about physics based on such popularizations (string theory and QFT being the most frequent, but far from the only victims)

@JohnRennie why space is dark even though it has many stars?

Hello Mr Olbers.

2

Strange things are happening in universe

@Akash.B not all the universe's stars are visible yet. See observable universe

4:57 PM

Space is bright. You're just indexing your idea of dark against a planet irradiated by a very close star.

@ACuriousMind I suspect it's human nature to overestimate the depth of one's knowledge. I bet it was happening back in Ancient Egypt.

Also other reasons

Olbers' paradox

In astrophysics and physical cosmology, Olbers' paradox, named after the German astronomer Heinrich Wilhelm Olbers (1758–1840), also known as the "dark night sky paradox", is the argument that the darkness of the night sky conflicts with the assumption of an infinite and eternal static universe. The darkness of the night sky is one of the pieces of evidence for a dynamic universe, such as the Big Bang model. In the hypothetical case that the universe is static, homogeneous at a large scale, and populated by an infinite number of stars, then any line of sight from Earth must end at the (very bright...

@Akash.B citation needed :-)

@Akash.B You're 3 pounds of jelly piloting a sack of meat and you think the rest of the universe is strange?

@SirCumference I already linked that, just without the one-box :P

@skullpatrol I suspect Strange things are happening in universe is one of the very statements physicists can make with complete confidence :-)

5:00 PM

@ACuriousMind Yeah I'm blind :P

Everything is as usual, then :P

The laws just change when the size of everything decreases

Feynman said that

@Akash.B what

Oh you mean different scales

yes

5:03 PM

How can a universe be created with such complexities?

@Akash.B Ever looked at a fractal?

@Adirian fractal?

Google "Mandlebrot". Simple rules + iteration can give rise to enormous complexity.

Scale invariance

In physics, mathematics and statistics, scale invariance is a feature of objects or laws that do not change if scales of length, energy, or other variables, are multiplied by a common factor, thus represent a universality. The technical term for this transformation is a dilatation (also known as dilation), and the dilatations can also form part of a larger conformal symmetry. In mathematics, scale invariance usually refers to an invariance of individual functions or curves. A closely related concept is self-similarity, where a function or curve is invariant under a discrete subset of the dilatations...

@Akash.B A shape that repeats itself and got the attention of metaphysicists

5:08 PM

@SirCumference And people on psychedelics

@JMac Turn the gain up on your pattern recognition high enough and you'll start seeing fractals everywhere. Whether or not they're real, of course, is a question best determined after the gain is dialed back down.

Really got the attention of mathematicians, who also happened to be frequently on said psychedelics at the time.

Feynman said he experimented with LSD.

@Adirian I've definitely done a couple experiments in university that looked at the effects of that. All extra-curricular, of course.

@skullpatrol everyone experimented with LSD in the 60s :P

yup, flower power

5:18 PM

@JMac It provides an interesting shift in perspective. Don't quite get why some people's brains fall out, though. It just erases all your mental filters.

today it's heroin and fentanyl

@skullpatrol The people doing those hard drugs are rarely the ones doing psychedelics in my experience.

@skullpatrol I fail to observe a large counterculture notorious for their use of heroin, so: What are you saying?

i meant in terms of popularity

nothing else

???

5:26 PM

pretty different reasons for the popularity. you can trace a lot of the heroin/fentanyl issues to prescription opiods and subsequent addiction.

LSD was popular in the 60s

Yes; but not because LSD-like products were routinely prescribed to treat pain.

true

@skullpatrol As were and still are a lot of other drugs like caffeine, nicotine or THC. What's your point in bringing up heroin of all things?

Fentanyl is popular today

5:29 PM

So is alcohol. Why are we enumerating "popular" drugs?

no idea :P

You started it :-)

15 mins ago, by ACuriousMind

@skullpatrol everyone experimented with LSD in the 60s :P

@skullpatrol No, you did by randomly saying heroin and fentanyl are popular and then not explaining what that was supposed to have to do with the ongoing conversation!

ok, ok I'm guilty as charged

I thought you emphasized the word "everyone"

which implies "popularity," no?

How did we get here from fractals

@SirCumference I made a joke about psychedelics and things quickly spiraled downhill

2