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3:03 PM
we're already in this chat room, why would we need another platform?
 
The Darwin Lagrangian (named after Charles Galton Darwin, grandson of the naturalist) describes the interaction to order v 2 c 2 {\displaystyle {\frac {v^{2}}{c^{2}}}} between two charged particles in a vacuum and is given by L = L f +...
I'm guessing it's to do with this
 
@ACuriousMind fun... Choose your avatar and quick picture there... (of your halloween avatar, you chose)
 
you and I have very different ideas of what 'fun' is :P
15 messages moved to Trash
3 messages moved to Trash
@bolbteppa I'm disappointed that Lagrangian not about some physicist trying to model evolution as a classical mechanics system :P
 
@bolbteppa : maybe. But maybe not. This doesn't sound right: "The Darwin interaction term is due to one particle reacting to the magnetic field generated by the other particle". The electron isn't a particle that generates a magnetic field. It "is" electromagnetic field. I like this paper by Art Hobson: arxiv.org/abs/1204.4616
 
Yeah I don't think it's the same thing
 
3:12 PM
@Azmuth it is cool I will try
 
@JackRod ok, then tell me to join :)
 
But can we do it sole time later?
 
@JackRod yep
Phone overheated :P
 
No
I am online for some work
 
cool
 
3:14 PM
@JohnDuffield the photon is 'the electromagnetic field', the electron 'is' (part of) the electron-positron field
 
I'm out
 
Art Hobson explains the double slit experiment in section IV. THE 2-SLIT EXPERIMENT. I think it's more or less right.
 
Art says in the abstract "The field for an electron is the electron"
On page 9 he explains the electromagnetic potential (eq. 3 on page 8) creates and destroys photons
 
> Drinking moderate amounts of alcohol before writing can actually enhance your creativity.
๐ŸŽƒ๐Ÿ‘ป
apparently emojis work here
 
When you say X is a wave, you're just trying to pretend that the wave is some kind of sheet (string theory lol) that's not made up of constituent particles, it makes zero sense, this is the whole reason why people say even gravity is made up of gravitons, the whole of physics assumes a point-particle model, the point is the laws determining where those particles end up being measured is related to the laws waves satisfy
 
3:21 PM
@bolbteppa : sure. But in electron-positron annihilation to gamma photons, I don't buy the assertion that two quanta pop out of existence like magic whilst two other quanta pop into existence like magic.
 
Could anyone help interpret the phrase (from P+S) "the $\gamma$ matrices are invariant under simultaneous rotations of their vector and spinor indices." This is just below the derivation of $\Lambda_{1/2}^{-1}\gamma^\mu \Lambda_{1/2}={\Lambda^\mu}_\nu \gamma^\nu$. Is the implication that the $\gamma$ matrices exist within both $\Bbb R^{1,3}$ and $\Bbb C^4$? It's kind of odd to me that we can equate the action of the spinor representation and the 4-vector representation of $\mathfrak{so}(1,3)$
 
-2
Q: Seperation b/w 2 bars connected with spring and block in the middle

Fahamhow to attempt this ques...guys may you please give a hint as to how to start...I want to first attempt with a hint rather than seeing the solution. I tried to integrate the net spring force and the got the accleration....but the equation of force thats increasing is not given anywhere in the que...

 
wow 10th time lucky on the typos
 
@JohnDuffield how do you explain beta radiation then, a neutron magically turning into a proton, an electron and an anti-neutrino
 
@bolbteppa : the whole of physics does not assume a point particle model. It's the wave nature of matter, as vindicated by experiment (en.wikipedia.org/w/…). It's quantum field theory. Not quantum point-particle theory.
 
3:25 PM
@Charlie the phrase is just the verbalization of the equation you wrote. The $\Lambda_{1/2}$ act on the spinor indices, the $\Lambda$ on the r.h.s. acts on the vector indices. If you multiply the whole thing from the left with the inverse of the $\Lambda$ on the r.h.s., then you get a "simultaneous rotation" of both indices on the l.h.s. that leaves the $\gamma$ invariant since the r.h.s. is then just $\gamma^\mu$.
 
and the fields create the point particles, what is a field if it's not made up of point particles
 
@bolbteppa I feel you're getting into realms of ontology that physics really doesn't say anything about :P
A field is just an assignment of numbers or operators to positions in spacetime. The theory really doesn't care what we humans think it's "made of"
 
@bolbteppa : I don't have to explain beta decay. It ought to be enough for me to point you at neutron diffraction and electron diffraction. But I could start with this: electron capture does what it says on the tin. And see this: pbs.org/wgbh/aso/databank/entries/dp32ne.html.
 
We all know the game is to pretend everything is like a big sheet or a water/air 'field' as if it's not made up of constituent particles, it just makes zero sense
 
@bolbteppa : it makes sense when you appreciate that a photon is a wave, and that you can make electrons and positrons out of photons in gamma-gamma pair production, and then you can diffract them. Because of the wave nature of matter.
 
3:32 PM
@JohnDuffield "I don't have to explain beta decay" surrender if I've ever seen it ;) An anti-neutrino particle pops into existence from thin air and you don't buy it but can't explain it
 
I'm having deja vu
Mar 25 '18 at 22:26, by ACuriousMind
@bolbteppa @JohnDuffield This discussion is clearly not productive. Why continue it?
 
@bolbteppa : I didn't say I can't explain it.
 
@ACuriousMind haha
 
@ACuriousMind It's 2020. Chill
 
@Azmuth what do you mean?
 
3:35 PM
@bolbteppa : see this rain.org/~karpeles/einstein.php where Einstein said a field is a state of space.
 
@ACuriousMind 2020 refers to year...
 
(Pair production is another example of particles popping into existence out of thin air)
 
yes, I've seen the video
 
@Azmuth ...and what does the current year have to do with this conversation being essentially a repeat of what they already did over 2 years ago?
 
then one particle goes into the black hole and hawkin radiation comes out
nothing. Leave it...
 
3:37 PM
@bolbteppa : no it isn't. Pair production (in its simplest form) is where two gamma photons interact with on another to form an electron and a positron, each of which has a wave nature, and each of which has spin. If these get too close the reverse process occurs. And I for one do not believe in magic.
 
@JohnDuffield You've literally linked the same thing there as you did 2 years ago. If you're just here to repeat the patterns of not listening and moving the goalposts that in part earned you your last chat suspension, you're not welcome here.
 
Simultaneous rotation of the vector and spinor indices to me sounds like $$\Lambda_{\frac{1}{2}}^{-1} \gamma'^{\mu} \Lambda_{\frac{1}{2}} = \Lambda^{\mu}_{\,\,\,\nu} \Lambda_{\frac{1}{2}}^{-1} \gamma^{\nu} \Lambda_{\frac{1}{2}} = \Lambda^{\mu}_{\,\,\,\nu} \Lambda^{\nu}_{\,\,\,\rho} \gamma^{\rho} = \Lambda^{\mu}_{\,\,\,\rho} \gamma^{\rho}$$
no?
 
@JohnDuffield I'be an idea, make another chat room, it's easy and invite real physicists there.
 
@bolbteppa why did the two $\Lambda$ fuse into one in the last step? :P
the indices should work out such that they cancel each other (but maybe the $\Lambda_{1/2}$s need to be the other way around, too).
 
$K_\nu^\mu K^r_\mu = K^r_\nu$
It's magic, it's magic!
 
3:41 PM
what you've written there is valid though not magic :P
 
:P XD :)
 
Ah I just meant to say 'group property' but the indices don't match doing it blindly
 
I have been reading the responses, I'm just doing something atm i'll answer in a second
 
Okay... roger that.
 
Oh wait...
 
3:43 PM
what?
 
@ACuriousMind : I'm talking physics and giving references to bona fide physics papers. Like Schrödinger's quantization as a problem of proper values, part II. On page 26 he said โ€œlet us think of a wave group of the nature described above, which in some way gets into a small closed โ€˜pathโ€™โ€.
 
Indices up, transforms under the inverse technically right
 
@JohnDuffield I see you haven't changed, then. See you in another year.
 
@JohnDuffield Your papers are the best ones! I appreciate your collection.
@ACuriousMind Suggest some Podcasts from your favorites
Mine - Codeblocks, Mr. Robot, Genius, Serial, the Python Magician, 7 min daily
 
3:46 PM
@Azmuth And you stop pinging me with your random personal questions. Since this is the second time I have to tell you today, have some time off, too.
 
That sentence has thrown me off, yeah you want
$$\Lambda_{\frac{1}{2}}^{-1} \gamma'^{\mu} \Lambda_{\frac{1}{2}} = (\Lambda^{-1})^{\mu}_{\,\,\,\nu} \Lambda_{\frac{1}{2}}^{-1} \gamma^{\nu} \Lambda_{\frac{1}{2}} = (\Lambda^{-1})^{\mu}_{\,\,\,\nu} \Lambda^{\nu}_{\,\,\,\rho} \gamma^{\rho} = \gamma^{\mu}$$
em is it to do with $\psi(x) \to \Lambda_{\frac{1}{2}} \psi(\Lambda^{-1}x)$
 
Ok I'm back, just regarding the $\gamma$ matrix stuff earlier, what's being done implies that the $\gamma$ matrices exist in both $\Bbb C^4$ (so we can act on them with the spinor representation of the Lorentz algebra) and the 4-vector space (so we can act on them with the 4-vector representation of the algebra)
@bolbteppa isn't this just the statement that the $\gamma^\mu$'s are invariant under a simultaneous forward and inverse spinor/4-vector transformation?
I don't think I follow
 
@Charlie Yes. Each $\gamma^\mu$ (fixed $\mu$) is a 4-by-4 matrix, and the four of them form a 4-vector (of matrices).
if you write out the $\Lambda_{1/2}$ matrices in index notation, too, you have that the $\gamma^\mu_{a\dot{a}}$ have both vector and spinor indices. (If you haven't come across the dotted notation yet, replace it by however you label spinor indices :P)
the normal $\Lambda$ acts on the $\mu$-part, the $\Lambda_{1/2}$ acts on the spinorial $a\dot{a}$-part
 
Ok, I guess they are just specific $4\times 4$ matrices, which I guess makes me wonder why it's necessary to relate them to the clifford algebra
oh they have two spinor indexes o_o
 
@Charlie nooo, it's just the dotted notation, you might write this as a single index in another notation
 
3:53 PM
ah ok
 
I'm trying to remember what the technical name for this notation is but I'm drawing a blank
@Charlie I don't understand what you mean by "relate them to the clifford algebra"
the abstract $\gamma^\mu$ generate the Clifford algebra
there is a unique irrep of the Clifford algebra in which they are 4-by-4 matrices
 
The sentence in the book says
$$\Lambda_{\frac{1}{2}}^{-1} \gamma^{\mu} \Lambda_{\frac{1}{2}} = \Lambda^{\mu}_{\,\,\,\nu} \gamma^{\nu}$$
means "that the $\gamma$ matrices are invariant under simultaneous rotations of their vector and spinor indices (just like the $\sigma^i$ under spatial rotations). In other words, we can "take the vector index $\mu$ on $\gamma^{\mu}$ seriously""
Is this saying
$$\Lambda_{\frac{1}{2}}^{-1} \gamma'^{\mu} \Lambda_{\frac{1}{2}} = (\Lambda^{-1})^{\mu}_{\,\,\,\nu} \Lambda_{\frac{1}{2}}^{-1} \gamma^{\nu} \Lambda_{\frac{1}{2}} = (\Lambda^{-1})^{\mu}_{\,\,\,\nu} \L