The h Bar

John Rennie

4:32 AM

@Stupidquestioninc it's a Young's slits diffraction pattern.

John Rennie

4:55 AM

The image is from Wikipedia

Superfast Jellyfish

@FakeMod you should look at double pendulums next! And their phase portraits.

Faded Giant

5:45 AM

@ACuriousMind no, clearly Knight Rider ;-)

skullpatrol

6:04 AM

FakeMod

@SuperfastJellyfish That would be interesting to implement :-))

Stupid question inc

7:56 AM

so this was the experiment conducted to prove light is a wave :-(

John Rennie

8:24 AM

@Stupidquestioninc this is one of many experiments that shows light has wavelike properties.

Light is neither a wave nor a particle. It is a quantum field. But it can behave like a wave in some circumstance and behave like a particle in other circumstances. When we diffract light it is behaving as a wave.

Stupid question inc

wow sounds interesting

John Rennie

@Stupidquestioninc I wrote an answer on this on the main site. Give me a moment and I'll try and find it.

Stupid question inc

@JohnRennie sure take your time

John Rennie

85

A: What is the relation between electromagnetic wave and photon?

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field. QED is a complicated theory, so while it is possible to do c...

@Stupidquestioninc that's the answer I was thinking of.

Superfast Jellyfish

@FakeMod in fact the behaviour becomes very curious for certain initial conditions

Stupid question inc

8:36 AM

looks like I don't have enough knowledge to understand these

sorry I am new to physics 😅

John Rennie

@Stupidquestioninc you don't need to worry about the details when you are first learning physics.

You'll learn all about quantum fields if you decide to do a physics degree.

Stupid question inc

oh but I am going to do math degree so is there steps I need to follow in order to reach there?

John Rennie

I wouldn't worry about it.

Stupid question inc

surely physics has something beautiful and fun deep down there

John Rennie

Some professional mathematicians do study the maths of quantum field theory. This area is known as mathematical physics.

Mathematical physics

Mathematical physics refers to the development of mathematical methods for application to problems in physics. The Journal of Mathematical Physics defines the field as "the application of mathematics to problems in physics and the development of mathematical methods suitable for such applications and for the formulation of physical theories". == Scope == There are several distinct branches of mathematical physics, and these roughly correspond to particular historical periods. === Classical mechanics === The rigorous, abstract and advanced reformulation of Newtonian mechanics adopting t...

Stupid question inc

8:43 AM

but I am going to do a pure math this sounds applied one

found one course

goodtheorist.science

John Rennie

Yes, if you do pure maths you're unlikely to find yourself working in mathematical physics.

@Stupidquestioninc note that Gerard 't Hooft is a physicist not a mathematician.

Stupid question inc

@JohnRennie yes but I have interest in both field

Ankit

@John can you tell me whether a spring balance measure force if it's other end is not fixed ?

Stupid question inc

but there is some constraint so I choose pure math which is my primary interest

Ankit

This question is for everyone present here.

John Rennie

8:48 AM

Maths degrees don't force you to choose an area of maths when you start, so you can wait until your last year to decide. At that point you can choose what you find most interesting.

@Ankit if the other end of the spring isn't attached to anything the tension in the spring will always be zero i.e. the spring will be at its natural length.

Ankit

@JohnRennie so it will show zero deflection ?

John Rennie

Yes

Ankit

@JohnRennie I have posted a question on the main site. Can you help me with that ?

I will share you the link if you agree.

John Rennie

2

Q: Did we just misunderstand gravity?

Einstein told us that the balance under free fall shows zero deflection because gravity is a curvature in the fabric of spacetime. Now let's assume we have a spring balance in gravity free space (though there is gravitational force between me and the balance) as shown below. If we somehow apply a...

This one?

Ankit

@JohnRennie Yes this is the one .

John Rennie

8:53 AM

The balance will show a deflection, and the reason it shows a deflection is that the balance has a non-zero mass.

Suppose the mass of the balance is m, then if you are accelerating the balance at some acceleration a you need to apply a force to it of F = ma. Yes?

Ankit

@JohnRennie but you said no deflection earlier.

Is it only for nonzero mass ?

John Rennie

@Ankit if the spring is massless, or approximately massless i.e. very small mass, then the force you have to apply to accelerate it is approximately zero. Yes?

Ankit

@JohnRennie yes.

John Rennie

So in that case you can accelerate the spring as much as you want and it will read (approximately) zero.

But if the other end of the spring is attached to something that has a non-negligible mass (i.e. the rest of the balance) then the force won't be (approximately) zero.

Ankit

@JohnRennie But the force was applied on that system of spring and whatever inside the balance. So that total system will accelerate as a whole . Then why should the spring stretch ? I couldn't grasp it.

John Rennie

8:58 AM

Consider a simplified model of the balance. We have a point mass m attached to one end of the massless spring, and we pull on the other end of the spring to make the whole system (point mass and spring) accelerate with an acceleration a.

OK so far?

Ankit

Yes

John Rennie

The mass is accelerating at an acceleration a. Yes?

Ankit

Okay.

John Rennie

So there must be a force F = ma acting on the mass to make it accelerate. Yes?

Ankit

Oo . So the force on that body is by the spring and thus the spring also experiences the same force ?

In opposite directions ?

That's why stretching ?

John Rennie

9:01 AM

Yes. The only thing connected to the mass is the spring, so it must be the spring exerting the force on the mass. That means there is a tension in the spring equal to the force F = ma.

Ankit

Okay great 😊.

John Rennie

And that means the spring must be stretched by a distance x where kx = ma.

(where k is the spring constant)

Ankit

But what if we consider the ideal situation ?

John Rennie

You mean everything is massless?

(or of negligible mass)

Ankit

@John Suppose we have just a spring and nothing at its ends . Now if we apply a force will it stretch ?

John Rennie

9:04 AM

Massless spring or massive spring?

i.e. does the spring itself have a mass?

Ankit

Very negligible mass

John Rennie

OK, so what force do you need to apply to accelerate a zero mass at some acceleration a?

Ankit

Don't take zero . Its a mysterious number 😂. I said of extremely low mass.

John Rennie

OK, so the force would be ε, where ε is some exceedingly small number given by ε = ma.

Yes?

Ankit

Yes

John Rennie

9:07 AM

Then the extension of the spring would be x = ε/k, where k is the spring constant.

And since ε is an exceedingly small number that means x is an exceedingly small number as well.

In the limit of m -> 0 we get x -> 0

Ankit

So it will negligible as well.

@JohnRennie my general doubt was that when we fall at g we don't feel that force . Then why isn't it applicable when we accelerate with some value a due to a force f = ma ?

John Rennie

Yes

Ankit

I think we will not feel that force too.

John Rennie

Go back to the simplified case of the point mass and the spring. If you accelerate the system by pulling on the end of the spring then:
1. you apply a force to the end of the spring
2. the other end of the spring applies a force to the mass

But now suppose you and the spring and the mass are falling freely in a gravitational field. This time:
1. gravity applies a force to you
2. gravity applies a force to the mass

So now the spring is not applying any force. The only force is being applied by gravity, and gravity applies the force to both you and the mass. The spring doesn't hav to apply any force, so it isn't stretched.

Ankit

Hmm. Thanks for your answer

John Rennie

9:17 AM

:-)

Ankit

@JohnRennie is it possible that work done by a force is negative while power is positive ?

John Rennie

Power is rate of work: P = dW/dt. Yes?

Ankit

Yes

John Rennie

So power and work will have the same sign.

Ankit

Okay.

If work done is decreasing with time . Doesn't this mean that it's rate should be negative.

?

John Rennie

9:21 AM

It means dP/dt would be negative

Ankit

Like if velocity is decreasing doesn't this mean that acceleration is negative of velocity ?

Emilio Pisanty

@DanielSank Update: TeX Live 2020 Release October 1, 2020

(in case it's useful)

John Rennie

@Ankit Suppose in the first second you do 10 joules of work, in the second second you do 9 joules of work and so on.

Is this what you mean by the amount of work is decreasing?

Ankit

9:39 AM

@JohnRennie yes .

John Rennie

So the average power in the first second is +10 J/s, the average power in the second second is +9 J/s, and so on.

Charlie

11:32 AM

'elo

Jack Rod

12:50 PM

I am folating with idea that 80 percent of the universe I meant the matter is doing nothing

Charlie

huh?

Jack Rod

Seriously we talk about dark matter dark energy and even in some of the blogs I have seen the people claiming that 50 percent universe is missing

But yes most of the matter in our surroundings is useless

Charlie

depends how you define "useless" in this context

John Rennie

I was going to enjoy a long cycle ride this afternoon but it's started raining and cycling in the rain is no fun.

Raining in England - who would have thought it?

Azmuth

lol

I was going to sleep, but you know I sleep everyday so it's becoming boring to sleep each day.

Charlie

1:18 PM

Just had a thought, since both the momentum and position operators are unbounded they have continuous parts of their spectrum, why doesn't this imply that all other observables $\hat O(\hat x,\hat p)$ are unbounded? Does this imply we can construct bounded operators as functions of unbounded ones? That seems odd to me for some reason

Azmuth

Position and Momentum operators are unbounded sure?

John Rennie

@Charlie can you give an example?

Charlie

@Azmuth are you saying they aren't?

Azmuth

@Charlie There's no reason for Position to be unbounded...

(I guess)^

Charlie

@JohnRennie I guess for instance the quantum harmonic oscillator has a purely discrete Hamiltonian, but this is constructed as a function of the position and momentum operators whose eigenvalues are continuous

Azmuth

1:22 PM

$\int x\psi^* \psi dx $

^that's bounded...

Charlie

hmm

Azmuth

@Charlie There are cases (such Bloch Crystal) where Momentum and Position are discrete too.

Jack Rod

@JohnRennie hi

Had a wonderful trip

John Rennie

@JackRod hi :-)

Jack Rod

Most scary moment

When my eye was watching

The ground

@JohnRennie

John Rennie

1:30 PM

The bungie jumping?

Jack Rod

Yes sir

John Rennie

It must be very weird to see the ground rushing up towards you.

In most cases that would be the last thing you would see :-)

Jack Rod

You know friend almost fainted after the jump

John Rennie

Wow!

It must have been pretty scary then! :-)

I have never done a bungie jump, and I fear I'm getting a little old for it now.

ACuriousMind

1:55 PM

@Charlie Functions of unbounded variables are not necessarily unbounded! (e.g. $x\in\mathbb{R}$ is unbounded but $\sin(x)$ is bounded)

since functions of operators are essentially just functions on the eigenvalues, the same logic applies to operators

also, not all operators are functions of $x$ and $p$, see: spin

Charlie

ah ok ty that helps

ACuriousMind

but also not all unbounded operators have continuous parts in their spectrum - the squared angular momentum operator has a fully discrete spectrum but is unbounded.

Slereah

2:48 PM

Or the particle number operator!

ACuriousMind

yeah, that one's neither a function of position nor with continuous spectrum

shelton Benjamin

3:11 PM

The coefficient of restitution is i dependent of frame is it the reason this is only proper defined term for determining nature of materials

Ankit

4:26 PM

Can someone tell me how a photon can be an energy packet ? Like how can we describe energy in the form of blocks. How can energy be confined to a particular point of space ?

Why don't they just spread out in space all around ?

@JohnRennie. @ACuriousMind it will be helpful if you explain my questions .

Has this question been asked earlier on this site ? If yes then please share the link. If not , should I ask it there ?

Charlie

5:40 PM

If we have a Lie group that isn't connected like the Lorentz group, is it meaningful to talk about a representation of each disconnected component separately or must we talk about a representation of the group as a whole full stop?

I want to say it might depend on whether each distinct disconnected component forms a subgroup, for which I think the restricted Lorentz group is but as for the other bits I'm not sure

I guess more generally the question is do topologically disconnected subspaces of Lie groups separately satisfy the axioms of a group

ACuriousMind

@Charlie No component except the identity component is a subgroup because none of them contains the identity!

Charlie

AH

of course

duh

ACuriousMind

Further, all the components of a Lie group are just "copies" of the identity component, i.e. there is some element $g_i\in G$ such that $G_i = g_i G_0$ for $G_0$ the identity component and $G_i$ some other component.

Charlie

wait is each disconnected component isomorphic to the component connected to the identity?

ACuriousMind

as a manifold they're isomorphic, yes

Charlie

5:49 PM

:o

ACuriousMind

(not as a group because the components are not groups, as we just established :P)

Charlie

was the point you were making above that each component contains a "psuedo"identity element

oh wait no

I think i see

we can generate another component as a coset of some element with the identity component

Slereah

every connected component of a lie group is homeomorphic to the identity component

is the point

Ankit

@ACuriousMind can you plese respond to my question ?

Or someone else.

It would be a great help.

Charlie

I see that, but what's written above refers to the group structure, that $\exists g_i\in G:G_i=g_iG_0$

which is pretty neat

ACuriousMind

5:53 PM

@Ankit Please don't ping people to answer your questions unless you have reason to believe they might specifically be interested in it.

What you're looking for is a crash course in quantum mechanics, not some short answer.

Ankit

@ACuriousMind Oo sorry for this.

I thought someone can provide a better explanation here

ACuriousMind

@Charlie my point was that there isn't really any additional structure hidden in the non-identity components, it's just a discrete group $G/G_0$

Charlie

@Ankit Maybe try reading this, this, or this?

ACuriousMind

many Lie groups (but not all) can be written as the semi-direct product of $G/G_0$ and $G_0$ and hence the question of their representations reduces to finding representations of the discrete group and the identity component

Sir Cumference

the bad thing about strictly reading textbooks is i found out way later that it's pronounced "lee groups"

ACuriousMind

5:57 PM

this is why people usually talk about $P$ and $T$ (parity and time inversion) separately from the continuous Lorentz transformations - the Lorentz group is the semi-direct product of its identity component and the finite group of two elements of order 2 ($\mathbb{Z}_2\times \mathbb{Z}_2$)

Charlie

ok ty that's interesting

Ankit

6:15 PM

@Charlie I have checked out those links but do you think they answer my questions ?

JingleBells

In Mendel genetics, how do we know which gene is recessive and which dominant? I mean, in the square thingy we write the dominant with a capital letter and the recessive with a lower letter but in real life, how does it work?

ACuriousMind

@JingleBells It's complicated :P

JingleBells

So in examples in school, we're told which one is dominant and which is recessive but in real life it's hard to figure out?

ACuriousMind

@JingleBells oh, I thought you were asking about what the genetic mechanism behind it is. If you just want to know whether an allele is dominant or recessive, do what Mendel did and count its presence in offspring

JingleBells

If a father has white skin and the mother has black skin and they reproduce, how do I represent the color of the skin in the Mendel matrix?

ACuriousMind

6:28 PM

you don't because skin color isn't a monogenic phenotype nor does it follow a dominant/recessive pattern :P

(and the "Mendel matrix" is a called a Punnett square, btw)

JingleBells

Is there some global universal database of which genes are recessive and which dominant? For example, in terms of hair color, black is dominant, and blonde is recessive? I know I'm wrong but I'm struggling to understand how it is determined that when we do the Punnett square, the dominant is a always on the left...

ACuriousMind

@JingleBells such a database wouldn't make sense because "dominant" or "recessive" isn't a universal property of an allele, it's always just relative to the other allele present (i.e. it might be that in a pair A/B A is dominant, but in a pair A/C C is)

JingleBells

^ yes, that's what I'm struggling to understand. What determines which is dominant and which recessive?

ACuriousMind

It depends on how the alleles actually produce the phenotype. E.g. it might be that B codes for nothing, while A and C code for different proteins that when present change the phenotype (e.g. these proteins are a specific color). Then B is always recessive in contrast to A and C, but what happens in the pair A/C depends on how the two proteins interact or how the colors mix

I.e. it's much easier to observe which is dominant/recessive than figure it out from first principles (that's almost impossible)

JingleBells

So if 100% of the kids are white-skinned even though dad is white and mom is black, it means black is recessive, and vice-versa? You just have to work through all the possible Punnett squares (agh, Mendel matrix sounds cooler) and compare the possible outcomes to the real offsprings to figure out which is dominant and which is recessive?

ACuriousMind

6:44 PM

yes (but note that once you get to polygenic traits it becomes substantially more difficult since you first need to figure out how many genes there are even involved to compute the "possible outcomes")

I'm not quite sure how that works except through a lot of guesswork, actually

JingleBells

Hmm, if for one organism the black gene is dominant but for the other organism, it's white. How does that work? If organisms don't agree on which is dominant and which is recessive?

ACuriousMind

well as long as they're the same species they should agree or you haven't found all the genes (or other factors) influencing the phenotype :P

JingleBells

So it's not possible that dad is Wb and mom us Bw?

ACuriousMind

no, that doesn't make any sense

JingleBells

So there is a universal database of which gene is dominant compared to another? (say, for humans)

ACuriousMind

6:49 PM

of course if there are multiple genes it can happen that for one gene the allele for one color is dominant and for the other gene the allele for the other color is dominant

Charlie

@Ankit I'm not entirely sure what your question is, but they have similar phrases

ACuriousMind

@JingleBells I don't think there's a database but you probably can find studies/articles for that for all the genes that have been studied

JingleBells

Cool, so we can conclude that in humans, for example (an unrealistic one), the black skin gene is dominant when paired with white?

I guess it's the molecular structure of the alleles that determine which dominates over which. (e.g black over white in humans)

Charlie

When the word "pathological" is used in physics (eg. "a pathological case") what do we mean?

Is it along the lines of "it's there in the mathematics but it's non-physical"?

ACuriousMind

@JingleBells I don't think you can say that one color is dominant over the other in humans - as I said there are multiple genes involved and heterozygotic carriers usually end up with a color that's neither of the parents' color, i.e. there's "incomplete dominance" here where the phenotypes mix.

@Charlie That is a common meaning, yes

Charlie

6:56 PM

ok thanks

JingleBells

@ACuriousMind Yes, I get that but suppose it's a monogenic phenotype. Then it makes some sense?

ACuriousMind

I'm not sure why you insist on taking skin color as an example if you're going to ignore completely how it works in reality :P

JingleBells

Cuz I can't think of any monogenic phenotype

ACuriousMind

what's wrong with Mendel's peas and the colors of roses? :P

JingleBells

Rose color is not continuous (like skin color)?

I'm looking for a binary phenotype but I can't think of anything

ACuriousMind

7:00 PM

not sure about roses actually, but the peas' flower colors definitely are

JingleBells

7:12 PM

Hmm, so when the whole system runs and ribosomes build proteins..., something determines which one of the two alleles from the two chromosomes (of mom and dad) to take the info from, right? One is dominant, the other recessive. How does the ribosome know which one to build a protein for?

ACuriousMind

37 mins ago, by ACuriousMind

It depends on how the alleles actually produce the phenotype. E.g. it might be that B codes for nothing, while A and C code for different proteins that when present change the phenotype (e.g. these proteins are a specific color). Then B is always recessive in contrast to A and C, but what happens in the pair A/C depends on how the two proteins interact or how the colors mix

it doesn't decide, both strands are read, it depends on how they interact (and of course there's many more complicated situations than just the one above)

JingleBells

So the ribosome builds both proteins (one for the dominant allele and one for the recessive) and then the proteins fight? :D

ACuriousMind

uhhhh...sort of? At this point you'd probably better ask a biologist :P

JingleBells

Alright, thanks!

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