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
+...
@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
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
@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 $\mathfrak{so}(1,3)$
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...
@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.
@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 $\gamma^\mu$.
@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.
@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
@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?
@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’”.
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?
@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
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…
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
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?
@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 $\gamma^\mu$.
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
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.
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
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?
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.
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?
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?
@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.
@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.
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.
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.
@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.
@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.
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.
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
