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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.
🎃👻
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

how 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 “let us think of a wave group of the nature described above, which in some way gets into a small closed ‘path’”.

Indices up, transforms under the inverse technically right

@JohnDuffield I see you haven't changed, then. See you in another year.

@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

I'll simply refer to them as the Spinor Index Notation Of Representations or spinor notation for short

Oh wait are we talking about a representation of the Clifford algebra on $\Bbb C^4$ and $\Bbb R^{1,3}$
I can't tell if I'm being completely duh
In P+S the gamma matrices are introduces as a $4\times 4$ representation of the Clifford algebra, it didn't occur to me that those representations might be on the 4-vector space and spinor space we're actually using lol

4:00 PM
yes, the Clifford algebra is represented on the spinor space - that's how we get the spin representation, since the $[\gamma^\mu, \gamma^\nu]$ are the generators of the spin algebra, rememeber?

lol ok I get it now, that was not smart of me, ty

don't worry
most things seem silly after one has understood them :P

P+S should come with a manual so people like me can have the obvious information spoon fed to them as we read it :P

if you find the land of sanity please send me a route description
haven't been there in ages

just use the ol' method
where spinors are square roots of vectors

4:03 PM
::twitches::

$\psi^2 = V$

If $\psi^2 = \psi \otimes \psi$ and $V = V_{\mu} \sigma^{\mu}$ you've got a winner

Is "that the $\gamma$ matrices are invariant under simultaneous rotations of their vector and spinor indices" not a no-no use of the word "invariant"?
We haven't shown invariance, just that the spinor and vector transformations are equivalent
Or rather, as equivalent as they can be in two different vector spaces

45 mins ago, by ACuriousMind
@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 $\gamma^\mu$.

ah :P
got it

4:11 PM
it's "invariant" under the "simultaneous rotation" of the transformation I described there

ahh

but this is an excellent example of why formulae are sometimes much clearer than words :P

4:22 PM
Reading it, it really doesn't sound like the thing you just said of multiplying with the inverse but it can't really mean anything else right, unless that inverse thing I mentioned above which I don't think it can be...
These notes closely follow P&S and below (3.72) they don't say this they just say it means you can treat $\mu$ as a vector index

4:38 PM
Does anyone know where the $\omega / \phi$ factor comes in the kinetic term for the scalar field in the Brans-Dicke Lagrangian

What justification apart from "it produces interesting/correct e.o.m." are you looking for?

In particular, why the 1/phi? It doesn't pop up for a general scalar field

How do we calculate radial and tangential component of gravity in projectile motion?

What do you mean? @123 What is the "tangential" component of gravity?

yeah you need to be a bit more specific there

4:49 PM
Hi @Charlie . We use rectangular components in projectile motion.

Sure, but why would there be a component of gravity in the direction tangential to the motion of the projectile?
If you draw a force diagram for a projectile flying through the air the force of gravity always points down

@Charlie sure but you can certainly write it the basis of the direction of motion in an instant ("tangential") and one orthogonal to it ("radial")

if we rotate rectangular axis then we can always find components along surface and vertical
components always depend on the axes we chosen.

That's what I was going to assume, it just seems like an odd thing to want to calculate
What part of the calculation is confusing @123?

because in the morning i discussed about projectile motion.

4:53 PM
Take the dot product of $\vec F_g$ with the tangential vector $\vec v_t$

my concern was that about projectile.

Oh wait I see what you meant

@123 as a function of time?

In uniform circular motion we find radial component of acceleration at which only direction changes but velocity is constant $a_c = \frac{v^2}{r}$ and tangential velocity at which velocity change.
The same analogy i wanted to use for projectile. as in uniform circular direction changes can be seen by radial component.

Wait are you talking about a projectile orbiting a central massive object or a projectile being thrown through the air on the surface of the Earth?

4:55 PM
I see the direction changes also in projectile but it is not encounter in projectile motion why?

well, for one thing a projectile doesn't really have a "radial" component
it's not moving in a circle, it's moving in a parabola
there aren't nice coordinates for that like there are for the circular case, so you don't really profit from the decomposition here in that it would make computations easier or anything
it's much easier to just consider the projectile motion in the usual Cartesian coordinates where gravity is aligned with one axis

@ACuriousMind Ookay. but in projectile we see direction changes why we don't encounter this phenomenon?

if we "see" it we "encounter" it, no?
I'm not sure what "phenomenon" you mean, either

There are certainly parabolic coordinates

but they aren't nice :P

4:59 PM
And if you cared to use them for projectiles, you would also get centrifugal forces
I'm sure it's available in the Big Book of Coordinates
What is it called again
possibly this

pls explain what happen to the force/acceleration when direction changes in any motion. e.g: curvilinear , circular etc.. what is same in all these
What about the force if it changes only direction not change it velocity?

I'm really not sure what you're trying to ask

What i thought as in uniform circular motion two parts of acceleration one is responsible for only direction changes (radial) and another is velocity changes (tangential). It is same in all types of motion. Is this true or not?

you can view all motion like that, yes

In projectile i can see the direction is changes in projectile motion it doesn't matter motion can be parabolic or any general curve.
Where is radial component in projectile motion which is responsible of changes direction of projectile?

5:08 PM
I wouldn't necessarily call it "radial" but it's just the direction orthogonal to the tangent

@ACuriousMind Yes you are right. where it is in projectile motion?

I don't know what you mean by "where is it" if "it's the direction orthogonal to the tangent" doesn't answer it

And if you are asking about the force responsible, it is just the component of gravity in that direction

when i derive equations of projectile motion it does not have any component which is orthogonal to tangent. at which i can calculate the direction change of projectile.
@NiharKarve I know all formal definition and ideas frnd. But i need to analyze the motion of projectile in term of radial and tangential way.
because radial component does not change velocity only direction. i want to see this component in projectile. which shows path/trajectory of projectile is parabola.
I know in formal equations gravity changes velocity in y-component by which changes displacement occurred. but which can be seen as change in direction.
If this is true so the idea of radial and tangential component is not true. because in projectile velocity changes can also change direction not required radial component.

@ACuriousMind Come to think of it, in the Brans-Dicke Lagrangian, how do $\phi R$ and $\frac\omega{R} (\partial\phi)^2$ even have the same dimensions?

5:22 PM
There are two components of force one is radial and another is tangential. radial components of force spent in direction change and tangential component spent in accelerate/decelerate the object.
Where radial component of force which is responsible of projectile having parabolic motion.

@NiharKarve Don't you mean $\frac{\omega}{\phi}(\partial\phi)^2$ and aren't both $\omega$ and $\phi$ dimensionless so both terms just have dimensions of a double spacetime derivative?

Hold on, $\phi$ is dimensionless?

ah, no, $\phi$ would usually have mass dimension 1 wouldn't it?
but then both terms have mass dimension 3 (derivatives have mass dimension 1) and it still works out
it doesn't actually matter what dimension $\phi$ has - both terms have 1 power of $\phi$ effectively and 2 spacetime derivatives and $\omega$ is dimensionless, so they have the same dimension

Whoops, looks like my confusion stemmed from a typo, someone wrote that the Ricci tensor has dimension M^-2 when they meant m^-2 (as in metres). So can I postulate the inclusion of 1/phi from a dimensional standpoint too?

5:37 PM
now that you mention it, I guess you can ;)

Awesome, thanks!

5:50 PM
@123 For circular motion, the radial and tangential components are usually from the point of view of the moving object. If you do that with parabolic motion, you can get tangential and radial components; but since the path isn't circular or constant velocity along the path, both components change over time and it's usually more complicated than it's worth to solve like that.

3 hours later…
8:27 PM
@JMac Thanks .. can we get radial component for which we can see the direction changes as parabolic in projectile motion?
Because in rectangular coordinate y-components gravity completely consumes in changing velocity. not direction.

@123 I would say in parabolic motion, the gravity changes the velocity and direction in Cartesian (rectangular) coordinates. If the gravity wasn't there, the object would keep moving in a straight line with a uniform velocity.

how can we say it changes direction. What component of gravity does that?

The vertical component. Without that, the object moves in a single direction (the one it was initially thrown). If without gravity, the path of the object does not change direction, I think it's fair to say that the gravity is changing the direction of the object.

What i see gravity has only y-component. So the acceleration due to gravity changes velocity continuously and if velocity changes its position changes along y-components which seems to be direction change.
@JMac Yes you are right. Pls clear me one thing.
In this form $a = a_{radial} + a_{tagential}$ in this formula radial only changes direction but velocity is kept constant example is centripetal acceleration $a_c = \frac{v^2}}{r}$. right or wrong?
and tangential responsible of changing velocity.
But in the case of projectile motion velocity changes does direction change here velocity is not constant.

In the radial frame of reference, yes. But in that frame of reference, you follow the object relative to a circular curve. So for parabolic motion, the frame of reference is always rotating relative to the rectangular reference frame, and the tangential velocity is always changing along with the radius of curvature.

8:42 PM
In example of uniform circular motion velocity does not change when changes direction.
@JMac You are right. But it is true radial, tangential frame always rotating at every point in every curvilinear motion. It could be projectile , circular , or any any general curve.
What is the special name of radial, tangential frame of reference. ICOR or any else?

@123 Yes but when the rotation is circular the radius of curvature is constant, so for example the $r$ in $\frac {v^2}{r}$ stays the same. In parabolic motion, that $r$ is always changing because the curve doesn't have a constant radius of curvature.

@JMac Yes you are right. Due to radius of curvature is same. It means centripetal acceleration is changing in projectile hue to r is changing?

And because tangential velocity is changing too. Both $v$ and $r$ are changing in parabolic motion, so radial coordinates don't make as much sense.

Why $\bf{v}$ is changing of radial component?
radius of curvature change means only r change not v.

@123 Because it doesn't keep moving in the same velocity in that direction either. The tangential direction is between the x and y directions, and since velocity in the y direction is always changing, so is velocity in the tangential direction.

8:58 PM
Ookay.. So it is fine if radial component changes velocity?

Well in the radial coordinates, the change in velocity basically determines how quickly the reference frame rotates relative to how fast it's going.

Pls tell me what is the name of coordinate radial and tangential components?
@JMac Hmm.. That's new information for me.
Why radial is constant in uniform circular motion, because it is changing the same way due curvature is same?

@123 Yeah uniform circular motion means tangential velocity is constant, and the radius is constant (since it's a circle), so $\frac {v^2}{r}$ is also constant.

Hmm... Thanks.. For the help.

It might help to think of how changing variables changes things. For example, if you want to make the circle start to spiral inwards, you can increase radial force while keeping tangential force the same. That means tangential velocity should stay the same, but since $a_c$ increases, $r$ must decrease until $a_c = \frac {v^2}{r}$ again; which makes sense since you're pulling inwards harder on the object than you need to for it to be circular.

9:13 PM
Very good example. can you give the same example for projectile.

Projectile motion is a mess in those coordinates. The tangential velocity is constantly decreasing until the apex of the parabola, then constantly increasing again; and the radius decreases until the apex and the increases again; but I don't know how much each variable changes relative to each other; which I think you would need to know.

Ookay.. no i don't need calculation just first need explanation.
which you given to me. i was concerned about projectile motion in radial, tangential coordinate to see the direction changes effect. because rectangular coordinate doesn't show direction change in projectile.

I find it does. As long as you just think of it as a constant velocity in the horizontal direction, and a changing velocity in the vertical direction, you get a good idea of the how the object is changing direction. The direction of a parabola actually changes at a constant rate in rectangular coordinates

@JMac Yes you are right. rectangular component does not separate direction changes component of force.

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