i see

i am on a collision course with relativity :)

and probably always was at this point

if you ask too many questions, you end up at relativity?

yes, pretty much

haha

lovely

and then you can start to wonder how that meshes with normal Galilean relativity - why do these equations predict a constant speed of light for observers moving at different speeds?

and boom, there's special relativity

ahhh i see

lovely

thats pretty awesome

i have a notion relativity commences when we think of particles moving at high velocity?

so if i am concerned with EM waves in vacuum and materials & the photon/particle relation is ignored, since its classical

the wave in the material as both the EM field and the displacement of oscillating charge, does some discrete element of that become particle-like in classical EM? and thus relativity

no, it doesn't have to be about particles

ohh

I mean, a wave - the EM wave in this case - can also have a velocity

right

and if you're looking at a wave e.g. on a surface of a lake, the speed of that wave relative to you changes when you start moving relative to the lake

ohhhh

good point

the Michelson-Morley experiment is exactly about finding out whether light/EM waves behaves like that or not

the answer is...not, and that's where relativity starts

ahh i got you

fascinating

so the frequency will change, but not the velocity of the wave?

yes, that's called redshift/blueshift for light

oh lovely

so the Michelson-Morley experiment tell us the velocity of the observer does not effect the (phase?) velocity of EM in vacuum?

well, what exactly it told us was a considerable matter of debate at the time :P

haha :)

but yes, today we read it as telling us that the speed of light is not relative - all observers see the same speed of light in all directions

the rest of special relativity is essentially just working out the math how to make this strange statement into a consistent physical theory

gotcha

lovely

and when we say speed of light we mean the phase velocity?

(from which we then derive the other velocities)

::googles "phase velocity" because he can never remember which is group and which is phase velocity::

yes, phase velocity

lovely :)

and the object to fix it into a consistent physical theory is basically spacetime?

yeah - you get length contraction and time dilation etc. basically just thinking very carefully about light bouncing between mirrors :P

haha right

so i have x,y,z,t and we need some way to interrelate x,y,z and t with eachother, in addition to the usual methods of time-evolution equations etc (eg. what my field is doing over time within space), but also another relationship which includes the velocity of one or more observers?

I mean at this point you're just asking the questions whose answer is "learn special relativity" ;)

hahah ;)

thanks for your help mate, really appreciate it

you've helped me alot!

now i need to evaluate where to fit relativity into my schedule haha

it's really not all that complicated (but there's like a million ways to teach it and everyone seems to have a different approach they find the simplest)

ah i see

the more i think about it, the more it seems like a problem if the speed of observers can change but the speed of my EM waves is not only fixed relative to eachother, but therefore always constant

that's what I mean by there being some stuff to work out to make this into a consistent physical theory ;)

:)

Hi All...

Hello @JohnRennie Sir

Hi :-)

@JohnRennie Sir we know newton's 3rd law is behavior of two different objects. Which automatically results law of conservation of momentum for two different objects. What is the law for two different objects at which we apply and get law of conservation of energy?

Conservation of energy is complicated because there are different forms of energy e.g. KE, PE, heat, etc. You cannot derive conservation of energy from simple mechanical concepts.

It is actually derived from some complicated maths called Noether's theorem.

@JohnRennie If we take simple system of two different objects and no external force. What happened in that situation?

But suppose the two objects are balls of clay. When they collide they will go "splat" and stick to each other i.e. it will be an inelastic collision and the KE will be converted to heat. So KE is not conserved in this process.

@JohnRennie Also pls take perfectly elastic collision between two different objects. Ideal situation need to be considered.

But a perfectly elastic collision is defined as one that conserves KE. So you are taking conservation of KE as your starting point then trying to derive conservation of KE. So it would be a circular argument.

@JohnRennie Aye Sir

@JohnRennie My point is that. We know law of conservation of momentum is derived from newton's 3rd law. I can understand that, there is systematic way to approach law of conservation of momentum. But i don't find any law or criteria at which we can say anything about law of conservation of energy.

@JohnRennie just a general question. I've seen many definitions that describes momentum in their own way. And then the conclusion to which I can end up with is that the momentum is just a product of mass and velocity that's all :-( I'm not getting a proper explanation for this term.

I've done so many questions based on that but couldn't find the proper meaning of this term. Everytime I see the definition for this I only get is 'it's a product of mass and velocity'. So can you please explain what is the physical sense of this term (without telling it's just mass times velocity).

Conservation of momentum is actually not derived from the third law. It's actually the other way round. Conservation of momentum is also derived from Noether's theorem. Then from conservation of momentum we can derive the third law.

@JohnRennie I do have got some different definition for this other than mass times velocity but still because of different explanations for this term, I get confused

@JohnRennie Okay sir. How elementary students can find systematic way or argument for law of conservation of energy?

@123 No. Indeed before Emmy Noether proved her theorem all physicists just had to assume energy was conserved without being able to prove it.

There is no simple proof of it.

@TejasDahake Different definitions for momentum?

@JohnRennie Aaaaahh.. That's what i wanted to know what tool people have before noether's theorem at which they believe energy conserved. You answered it.

@JohnRennie yes

What other definitions for momentum are there?

That's not a definition. A definition is something that allows us to calculate actual values. So p = mv is a definition because we can put in values for m and v and get the numeric value of p.

@JohnRennie somebody tells it as it's just a quantity which equals to mass times velocity. somebody calls it as it is a quantity which keeps the object in motion etc.

@JohnRennie Sir how can we understand or explain law of conservation of energy and momentum in simple words from noether's theorem?

@TejasDahake a quantity which keeps the object in motion is just a vague description not a definition.

@123 I'm afraid not because it is complicated. Fir now you will just have to accept it.

@JohnRennie I have learnt something about noether's theorem time invariant/symmetry, translational symmetry, rotational symmetry.

@JohnRennie then what is the exact meaning of this term?

Why momentum was defined? Why only mass times velocity

I want to know what the physical sense does it make

@TejasDahake There isn't a "meaning". In physics momentum is a conserved quantity related to a symmetry called shift symmetry.

I'd guess you're trying to get an intuitive feel for what momentum is, but intuition comes only with experience.

@JohnRennie shift symmetry?

@JohnRennie yes that's what I'm asking for.

@TejasDahake This quite a nice introductory article

@JohnRennie that was a nice article

That was all about conservation of momentum

@TejasDahake It is a fascinating subject, but it's pretty advanced. Students wouldn't normally study this until the second or third year of a physics degree.

@JohnRennie yes, there were some of the points mentioned which, to me, were somehow new

@JohnRennie So for now, will I have to just stop thinking about the exact feel of momentum?

And to just apply it blindly where it's needed?

Basically yes, thouhg I think saying you are applying it "blindly" is an exaggeration. For example you know momentum is only conserved when no external force is acting. Yes?

@JohnRennie Yes ofcourse I know it, basically it's only the thing which is taught at the very beginning of the lectures while studying momentum

Newton said that momentum is the "quantity of motion", but that doesn't actually tell us what momentum really is. ;) However, we can just treat momentum as a fundamental property.

In Newtonian mechanics, momentum is proportional to mass and to velocity. That defines a relationship between those 3 quantities. But it doesn't explain what either momentum or mass "really" are.

@PM2Ring that is really weird.

@JohnRennie I visited your profile. And I saw that you are a chemist, but interestingly you are the most recognised person on Physics Stackexchange which is very nice. But i'm very curious to know what's the story behind.

I went to Cambridge intending to do physics, but along the way I got really interested in quantum chemistry, i.e. calculating the properties of molecules using quantum mechanics, so in my final year I did mainly quantum chemistry.

Then for my PhD I investigated a solid state photoreaction between metallic silver and germanium selenide.

This would be classified as physical chemistry, though I would have said it was more on the physics side than the chemistry side.

Then when I left university I worked in colloid science, which again is physical chemistry, and again I was more on the physics side than the chemistry side.

So I'm officially a (physical) chemist, but I was always a physicist at heart :-)

And now I'm retired I can mess around and do whatever I want!

@JohnRennie that is so much of hardwork invested therein :-) . You deserve a lot. The science is able to survive because of people like you are here in this world. Or there is nothing other than things like engineering and all which is just an applied science. But the people who are scientists gives birth to new concepts. Which i think is more important than just applying the existing knowledge.

The reason scientists do science is selfish really. We do it because it's fun!

How to derive Yang-Mills theory from the massless representation of the Poincare group?

Sorry, I meant to derive the Yang-Mills equation

I was reading induced representation of the Poincare group. One can derive Maxwell equations from the massless representation. I've been wondering how to generalize it to Yang-Mills equations.

@JohnRennie I love to read that, really

I often have arguments with other students about applications being just a consequence :P

No, no, we have the best interests of humanity at heart and we strive endlessly for that cause!!

The first time I said "I've no interest in helping people" (with science I mean) I got some people laughing

It came out more brutal than I expected

But such discussions pop up periodically and I end up saying that in my opinion science at its heart is libido sciendi

I'm striving to find a book named "Constrained Hamiltonian systems" printed in 1976 but it seems it cannot be found D:

Yo @JohnRennie

at a 'spoons in Canary Wharf

Thought you'd find it cool

Inducted into the Scottish engineering hall of fame in 2014.

Dummy question, how do you replace the gothic symbols for Lie algebras in handwritten notes? Just lowercase letters?

why do we have to remember chemical reactions

@Feynman_00 just learn to hand-write fraktur :P

@goodplayer1 because organic chemistry is a memory based subject.

its hard to remember things

I don't have to remember any chemical reactions, I don't know what you're talking about!

@ACuriousMind it's not hard; you just need a suitable dip pen :-)

@Mad it's just the usual ambiguity between a derivative and a difference quotient

why is chemistry so random

nothing makes sense

also, for Lie algebras you really only need to learn g,h,j,k,l and perhaps m and n

@goodplayer1 it's a skill that you must practice and it gets a lot harder in biochemistry.

@goodplayer1 it's not random it's just complicated

hmm you are right

its complicated

It only feels random at the beginning because you're not used to it yet

@ACuriousMind Not only can you write a proper $\xi$ but you can even write fraktur

i wish i had good teachers in my school

i really want to understand things

its nice to be around here

@Feynman_00 I had an algebra prof who insisted on spending the better part of an entire lecture showing us how to draw the fraktur letters he used for ideals and algebras

and I can write a $\xi$ because I actually learned Ancient Greek :P

i get some help here

gorgeous!

I can't even write a proper $\Sigma$ after 3 years lol

@ACuriousMind s, u, o ?

It looks like some people use $\mathfrak{g}$ for a generic Lie Algebra but regular lower case letters for Lie Algebras of Lie Groups

This is what the book I'm using does

Unless the author was just lazy :P

that's a weird convention because every Lie algebra is the Lie algebra of a Lie group :P

Oh, I never thought of it like that

I've introduced the concept of Lie Algebras as a vector space with a Lie bracket and then defined the Lie algebra of a group as a special set that turns out to be a Lie algebra in the first sense

But thinking in terms of differential geometry makes it reasonable although the association is not one to one

Any holomorphic function is locally a Conformal Killing vector on a sphere. But globally defined CKV is $a+bz+cz^2$...Why is this so?

@ManasDogra see physics.stackexchange.com/q/108472/50583, physics.stackexchange.com/q/391624/50583

In the degenerate time independent pertubation theory

the matrix of the disturbance in an eigenspace of a degenerate eigenvalue is not diagonal right?

Is it possible to know how much Gray units Madam Curie got exposed to on average? while working

How come I don't have to change limits of integration for $\int_{e^y}^{y}y\frac{\ln x}{x}\text{d}x , y>0$

I did the substitution and I got the same answer as the textbook but isn't it wrong because e^y > y for all y>0?

@Mad maybe about 10 mSv per week. But I think no official attempt has been made to estimate her total dose.

@Obliv it doesn't really matter either way if you keep track of what you're doing. For $F(x) = \int f(x) dx$, $\int_a^b = F(b) - F(a)$ but all swapping the bounds does it $\int_b^a = -\int_a^b = F(a) - F(b)$.

i thought it mattered, it's been a while.

Thank you :) @DanielUnderwood

I mean you will get a negative if you flip the bounds, but at least to me that just falls into the bucket of keeping track of exactly what you're calculating

yeah, it would matter on a test as well, so at least I have confirmation that the textbook missed that.

@Loong what was the main source of radiation?

That is the tricky part since there were so many sources.

external radiation, mostly from Bi-214 and Pb-214

incorporation of Ra-226 (via ingestion and skin)

inhalation of Rn-222 progeny, mostly from Po-218 and Po-214

probably also incorporation of Pb/Bi/Po-210 when she isolated polonium

@imbAF yes, not necessarily diagonal

@NiharKarve One more thing. In the non degenerate case we can write a perturbed eigenstate as a summation of different correction terms right?

I only got 2 µSv last week from looking at a few waste drums. Radiation protection has changed a bit in the last 100 years. ;-)

$|n\rangle=|n\rangle_0+\lambda |n\rangle_1+...$

yep

But that;s not the case when degeneracy is involved right?

What I mean by that

I don't get what you mean, you can still write $|n\alpha\rangle = |n\alpha\rangle_0 + \lambda|n\alpha\rangle_1 + ...$

But isn't it different

the eigenstates we find in the eigenspace to a degenerate eigenvalue

are the new eigenstates to the perturbed system

but if you can write an eigenstate of the perturbated system like how you say you can, then what is the expression for $|n\rangle_1$? Is it like the expression for the un perturbed system, but here we use the eigenvalues that are expressed as a lin. combination of the degenerated eigenvalues?

@imbAF All I'm saying is that there exists a perturbation series for the exact eigenket, whose leading-order term is the unperturbed eigenket - nothing mysterious about that

if you check this video around the 4:40 min mark , you'll see that the person here simply says that these new eigenvectors are expressed as a linear combination of the eigenvectors belonging to the degenerate eigenvalue. At no point it is said that what we get, are the first order correction terms of the eigenstate youtube.com/…

the explicit expression for $|n\alpha\rangle_1$ has a part coming from contributions from the degenerate eigenspace, and another from contributions from the orthogonal complement

you are describing this part here : wikimedia.org/api/rest_v1/media/math/render/svg/… ?

yep

the first term is the normal expression coming from nondegenerate pertubation theory, just restricted to states outside the degenerate subspace

I understand that part

the 2nd part is the weird thing

you need section 1.2.2 in ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/…

Ok thx for this

One more thing

Is it correct to say that while in the unperturbed degenerate system, the Hamiltonian is a diagonal matrix in the basis spanned by it's eigenstates, in the perturbated degenerate system the new hamiltonian is not diagonal in the basis spanned by the eigenvectors of the unperturbed system, unless we substitute the degenerate eigenvectors (of the unperturbed system) with those that are expressed as a linear combination of them,that we find when we study the disturbance operator in the sub space>

Kinda complicated to describe it

basically when, the new perturbated Hamiltonian, is a diagonal matrix? Which vectors are necessary for this to happen

@imbAF I think this is right

So, we use the unperturbed eigenstates, even for the perturbated hamiltonian, when trying to express it as a matrix?

that's known as finding a "good basis"

@imbAF when trying to express what as a matrix?

the Hamiltonian

when you try to express the non disturbed one, you can use it's eigenstates and it will be diagonal

if degeneracy doesn't come into play

is it correct to say that even the perturbated hamiltonian, in the basis of eigenstates of the unperturbed hamiltonian, is still diagonal

or is it that for the perturbated hamiltonian, we need to use the perturbated eigenvectors?

But I don't think it's the case

@imbAF not necessarily

only when you choose this "good basis"

Ok

thanks

np

a common example of this is in the Zeeman effect

for instance, suppose your unperturbed H is the hydrogen Hamiltonian and the perturbation is $\alpha\mathbf B\cdot(\mathbf L + 2\mathbf S)$

can we take smth simpler

like the 2d QHO?

@NiharKarve you could either use $\{|n\ell m_\ell m_s\rangle\}$ as your unperturbed basis, or you could use $\{|n\ell j m_j\rangle\}$

Only in the latter basis is the perturbation diagonal

@imbAF the details aren't important, I just wanted to give a concrete example

aha

by pertubation you mean $H=H_0+\lambda W$ right?

well by perturbation I mean $\alpha B\cdot(L + 2S)$ (but of course $H = H_0 + \alpha B\cdot(L + 2S)$ will also be diagonal)