@qwerty when you have nothing to choose basis tangent vectors with, by definition, you would have no natural choices to make. However, when you have coördinates, their partial derivatives privileges them as basis tangent vectors because it makes your computation easier to work with; you would just have one consistent set of variables to keep track of; much less than when you have arbitrary basis tangent vectors at every point in spacetime
@ACuriousMind The better and much more precise answer should be that the $S$ function is a function of initial (time,x) and final (time,x) where either x can be independently chosen to be position or momenta but not both.
@Loong Any ideas on a possible source of 477 keV gamma? We just got a question posted as an answer on a 10 year old question: physics.stackexchange.com/a/842542 I suppose I should flag it as Not An Answer...
@RyderRude I've never been completely comfortable with the arguments around Searle's Chinese Room, either for or against. IMHO, it's dangerous to speculate about such a thing when we've never actually constructed a functional one. And even if we did make something that kinda behaved in the desired way, there's the danger that some people would perceive it as being more intelligent or conscious than it actually is, just like they currently do with ChatGPT etc.
@RyderRude I still haven't finished reading that Chalmers paper. Even he acknowledges that IIT is flawed, and he cites an article by Scott Aaronson that elucidates some of its failings, principally that it doesn't really give us a good numerical measure of the intelligence potential of a system. But I guess Chalmers is mostly using it as a proxy of a more valid measure.
@RyderRude Hofstadter coined "heterological" for Gödel, Escher, Bach.
@PM2Ring yes. i think it is like a dilemma. if something behaves identical to humans, i don't think i would be able to treat it as a machine even if my theory "said" that it was a machine
society would (or maybe should) treat such machines as humans and it would be taboo to talk about whether or not they r consciousness
@imbAF It's not easy to detect neutrinos, especially if they don't have a lot of KE. And even in our best detectors, we only detect a tiny percentage of the neutrinos travelling through the detector. Getting accurate energy measurements from them is difficult...
@PM2Ring yes. i take IIT as some starting point. maybe it will transition to other programs. or maybe it will just fail and results will come out of its failure (like Hilbert's program)
> Neutrinos are notoriously difficult to detect. Our best neutrino detectors, using the Ga → Ge → Ga reaction sequence (aka, the "Alsace-Lorraine" technique), have a threshold of 233 keV. And even then only a tiny fraction of the neutrinos passing through the detector actually trigger the reaction sequence. A well-known heuristic is that a light-year of lead would only absorb ~50% of the neutrinos passing through it (assuming the neutrinos have energies typical of those we can detect).
> So neutrinos are effectively invisible unless their kinetic energy is at least on the order of a million times greater than their rest mass.
@RyderRude I don't think we're obliged to treat any machines as humans. OTOH, if a machine can really pass the Turing Test, I'm happy to treat it as an intelligent entity (unless & until it demonstrates that it isn't). OTOH, the Turing Test was never intended to be some ultimate test of intelligence, it was merely a starting point, created when there was nothing else.
@RyderRude Yes. I think we need something like Kolmogorov complexity, but that is notoriously hard to calculate for anything beyond the simplest algorithms.
@PM2Ring but don't u think it would be immoral to treat it badly if it behaves like a human? like, say, someone meets an AI person in the future. and later they find out that it is AI (but not because it behaved like AI, they found it out some other way)
@RyderRude Yes. It's not too hard to see the equivalences. The version I find easiest to understand is Turing's Halting Problem. But maybe that's because I'm a programmer. :)
@RyderRude Huh? Why does that require me to treat an AI as human? I can treat cats nicely and morally, as cats. That doesn't mean I consider them to be funny shaped humans with limited speech capacity. ;)
@RyderRude That's pretty disturbing. And also weird. Anyone who's spent time interacting with dogs has surely noticed that dogs have some kind of consciousness.
@naturallyInconsistent I'm not sure how the principal function is supposed to be a function of momenta. The principal function $S(q_f,t_f; q_i,t_i)$ is the action for the solution to the equations of motion $q(t)$ with $q(t_i) = q_i, q(t_f) = q_f$. How can you bring in momentum?
@RyderRude Once again I ask you to research things more carefully before just asserting stuff. This story is likely apocryphal and vivisections were rather common in Descartes' day - it was not a unique belief of Descartes due to his philosophy, all the scientists were dissecting animals, essentially. He also had a dog called Monsieur Grat whom he seems to have liked a lot.
Descartes was a pretty complex guy. He did great maths, physics, and philosophy. But he was also ensnared by religion, like most people for most of history. He was also a professional soldier, and fought as a mercenary.
@RyderRude Yes. That phi function is the numerical measure of the potential intelligence capacity of a system. But Aaronson shows that it doesn't have the properties that it needs to have.
But I'll shut up for now. I want to read this discussion about Fourier...
@ACuriousMind Im not saying anything nonstandard to modern physics. You can check some of the textbooks that cover this weird thing and realise that the Legendre transform converting the action integral from position to momentum variables can also be seen as a Fourier transform. This is not surprising because, as you know, the semiclassical limit of the underlying quantum path integral can be done either way.
@naturallyInconsistent I think we must be talking past each other - the standard Legendre transform from Lagrangian to Hamiltonian mechanics turns velocity into momentum, not position!
@naturallyInconsistent That's more or less part of the Legendre transform. What does that have to do with Hamilton's principal function being defined as $S(q_f,t_f;q_i,t_i) = \int L(q(t),\dot{q}(t),t)\mathrm{d}t$ for $q(t)$ a solution of the equation of motion with the given initial and final positions/times?
Of course you can also rewrite the definition in terms of the Hamiltonian, but that doesn't change what the $S$ is a function of
@PM2Ring while you are saying you are interested in myow Fourier discussion, I'm actually interested in your comment reply here. I'd have expected that we won't be computing any speed for neutrinos because the measurement uncertainties imply that we can only obtain null geodesics, i.e. $E=|\vec p|$. That is, any speed estimates would come from assuming a value for $m$ of neutrinos and then working out the $\gamma$ or $|\vec v|$ from there.
@ACuriousMind As I said, you can change the S to be a function of momenta too. It doesnt just have to be a function of position. You can work purely in the momentum space, though in the free particle case that is trivial.
Also you seem to be again confusing position and velocity - in classical mechanics, it's the velocity you get rid of, not the position, when you "go to momentum space"
@ACuriousMind I don't see the point in considering them separately, but even if we do strictly separate them, you can still obtain the relations we are talking about by taking the semiclassical limit of path integration, as I said earlier. Also, a source I'm reading is defining Hamilton's principal function as $S(q,P,t)$, i.e. as a $F_2$ type generating function. I'm not sure what it is you are fixating about
@RyderRude no, you are just wrong, and you have no idea what is being talked about, so please just go away
@naturallyInconsistent Now we're at least getting closer to the source of the confusion - you started by saying $S$ "is a function of position or momenta but not both" and now you've written down $S(q,P,t)$ which is clearly a function of both. How does that fit together?
@ACuriousMind Again, I made it clear in the first chat comment I had here: For either of the initial time or the final time, you may choose independently to have either the position or the momentum, but not both position and momentum at any one single time slice.
@naturallyInconsistent Well, we can't see any neutrinos with gamma under a million or so, and of those we detect on the order of one per billion passing through our detector. You can't get much precision from that. And yes, their worldlines are virtually indistinguishable from null geodesics.
We have detected some really high energy neutrinos: the SN 1987A supernova in the LMC emitted a bunch of neutrinos (and antineutrinos), but we only managed to detect about 30 of them in total.
That is, you may have $S(t_f,p_f,t_i,q_i)$ if you really wish to consider it. This would just be very much connected to the time-slicing unitary time evolution in phase space of quantum theory, where you bounce between position and momentum variables every time slice.
@naturallyInconsistent Yeah, I don't really understand what you mean by that. Can you explicitly write down (or link some place where this is done) Hamilton's principal function in terms of initial/final momenta?
The generating function approach in its usual form (e.g. Wiki) treats your "momenta" as integration constants and the coordinates $q$ as the variables to arrive at the HJ equation
@PM2Ring Again, my point is that you can only deduce gamma from assuming some specific value for the mass; the only thing we kinda did measure is the mass differences of the neutrinos responsible for the neutrino oscillations. I don't see why we should be working with gamma and v when it is this level of tenuous. Gamma being greater than a million should be a good reason to stop arguing using gamma at all...
I don't really understand how your $S(p_f)$ is supposed to lead to the HJ equation, since it's the $q_f$ with respect to which the "$q$-differentiation" in the HJ equation $\partial_t S = H(q,\partial_q S, t)$ happens
@naturallyInconsistent Oh, sure. That's just a rough estimate on the gamma, because there's still at least an order of magnitude of uncertainty on the neutrino rest mass, for the 3 mass states. I forget the details, but I think ProfRob has the latest values, either on Physics.SE or Astronomy.SE.
I would accept saying that the principal function is a function of both position and momenta if we take the generating function approach literally, but I'm really confused by the claim that it can be a function of momentum only
@ACuriousMind Well, for starters, it would be modified, not least because differentiating w.r.t. $p_f$ will instead give you $q$ so that it should be $\partial_t S=-H(\partial_p S,p,t)$ kinda thing, minus signs not sure
@PM2Ring Yeah; that little nitpick is what I'm going after. Like, when we are so uncertain about the appropriate rest masses to put in, maybe we can just work with the 4-momenta that we know much better, than discuss the gamma factor that we are so unsure of.
Astrophysicists still mostly ignore the neutrino rest mass, and treat them as a form of radiation. Unless they're specifically interested in neutrino flavour or mass states for some reason.
@naturallyInconsistent Okay, but then you've chosen a generating function that makes the positions constants of motion instead of the momenta. I agree that's probably equally possible, but that's not what the standard HJ equation is about. So I'm confused why you said very explicitly it would be more correct to say Hamilton's principal function can be a function of momentum.
Yes, you can do something similar with the roles of position and momentum exchange, but that's not normally what people mean when they say "Hamilton's principal function" or "Hamilton-Jacobi equation"!
i think there is a symmetry between position and momentum space. the only thing that could spoil the symmetry is that the momentum space is a cotangent space
but I still think there is a version of PLA on the momentum space
@naturallyInconsistent I only mention that 1 million gamma thing to give people a rough idea of how close to c detectable neutrinos are moving, and how miniscule their rest mass is compared to their KE. Of course, most neutrinos are the undetectable neutrinos of the cosmic background. ;)
@ACuriousMind Again, I think you are being excessively strictly separating Hamilton's principal function from the action; physicists call them both $S$ for good reason and they are provably the same entity if you actually compute them for the few cases that we can get them analytically.
if I take the phase space action : it is a function of (q,p,v). interestingly, dp/dt doesnt appear in it which points to an asymmetry between position and momentum space
When you consider the action in general, you see that the semiclassical limit coming from path integration necessarily require that you can have an analogue of HJE for purely momentum space i.e. $S(t_f,p_f,t_i,p_i)$ and for mixed e.g. $S(t_f,p_f,t_i,q_i)$ cases
@naturallyInconsistent Yes, I'm very strict about what things in classical mechanics are functions of what and clearly differentiating them. It's kind of my thing ;P The usual action is a functional on paths (either on configuration space or phase space), Hamilton's principal function is a function of the initial and final positions of a path in configuration space.
These are mathematically two different functions, and a function that is a function of the initial and final momenta of a path in phase space is a third different function still.
I'm extremely nitpicky about this because all the usual confusions like "how are position and velocity independent in the Lagrangian" etc. arise precisely because people are not careful about what is a function of what
@ACuriousMind I'm not sure of the utility of this strict separation. But I think you already fully understand the point I'm making and where I'm going with that; it would be silly to link you to stuff that you clearly are aware of.
@ACuriousMind Yes, when miao miao is teaching, miao miao would also be absolutely anal about this too. The pedantry is very kind to students.
actually Qmechanic is even stricter and separates the initial/final "Dirichlet action" $S(q_f,t_f;q_i,t_i)$ from the Hamilton principal function $S(q,\alpha,t)$, which I have to agree with in principle, too :P
suppose we take the phase space Lagrangian which is, $p\dot x - \frac{p^2}{2m} - V(x)$. on shell we have, dp/dt = -V'(x) and dx/dt=p/m. if the first equation is invertible, then we can eliminate both x and v in favor of p and dot p
@RyderRude You should have just searched very deeply for the answer. The momentum space Coulomb problem is solvable because the configuration space Coulomb problem has been solved. It is, however, more difficult in momentum space because the Coulomb potential is $1/r$ in position space and yet $1/p^2$ in momentum space, causing the Coulomb potential to look more singular in momentum space. Not to mention that even defining the Coulomb potential tended to be taking the massless limit of Yukawa.
And path integration of Coulomb problem, even in position space, is horrible. The space has torsion.
You'd know that if you actually studied the problem as an academic should.
@naturallyInconsistent You're misunderstanding the question, he's just asking what happens if you literally exchange position and momentum in the standard Hamiltonian, not about the actual momentum space version of the Coulomb problem
The conversion of the Coulomb potential into momentum space version is an unavoidable milestone in the standard physics curriculum. If you had studied properly, you would have known what it is we were talking about.
@qwerty I mean, as far as I can tell, they did indeed settle an old conjecture in theoretical computer science. That's a breakthrough for computer science, not necessarily for the rest of us, not quite sure what your concern is?
and in the context we were discussing, Fourier transforms aren't involved. we have to do a sort of Legendre transform to switch to the momentum space Lagrangian
like, start with $H=\frac{p^2}{2m}+V(x)$ and then do $L(p\dot p)= q\dot p- H$, and eliminate $q$
@ACuriousMind oh it just that oftentimes the marketing overhypes the reality of a result. i guess they didn't directly mention any practical consequences for "the rest of us" but i wondered if there were
@qwerty Real-world hash tables operate often under different optimization constraints than just this theoretical setting, so no, it's not obvious that it has any direct impact on anything practical. Doesn't mean it's not a breakthrough (settling math conjectures in algebraic geometry or whatever also rarely has direct real-world impact :P)
i think we have different $1/r$ in the field theory context and in the QM context. In the QM context, the operator is $1/r \delta (r-r')$, while in the Field theory context, the operator is $\frac{1}{|r-r'|}$
this is why their Fourier transforms are different
I find this entire discussion about the Fourier transforms of operators quite bizarre :P The position operator $r$ is a derivative in momentum space. So its inverse in momentum space is, formally, the inverse of the derivative operator, which, if you want to write it down in analytic terms, will become an integral operator
- unsurprisingly, it's an integral involving the fundamental solution/Green's function, because that's what Green's functions are - the integral kernels of the inverse of their differential operator.
Again, if one simply consults the standard curriculum, it is unavoidable to see what to do with inverses of differential operators. The whole $\frac1{\nabla^2}$ operator
@ACuriousMind the inverse operator we have here is $\frac{1}{\sqrt{dp_x^2+dp_y^2+dp_z^2}}$. i think this becomes the Green's function of the operator $\sqrt {dp_x^2+dp_y^2+ dp_z^2}$
by dp, i mean partial derivatives @ACuriousMind
i am just saying that the Fourier transform of the Coulomb potential can mean different operators in different contexts
in QM, we mean to transform $1/r \delta (r-r')$. in field theory, we do $1/|r-r'|$
So if the reflection at far-field is due to diffraction then the surface of the mirror no longer form a consistent image (in the sense of faithful reproduction of a source as viewed from the far-field observer)?
@RyderRude Your notation is terrible and no, it's not "the Green's function", it's an integral operator that is the convolution with the Green's function (and it's actually even worse here because the square-root of the differential operator itself is not a "pure" differential operator anyway). Again, this is not the Fourier transform of the function $f(r) = \frac{1}{r}$ in any sense.
@ACuriousMind i meant to say that the operator's $(p,p')$ basis components is the Green's function. the operator itself is the convolution ofc
@ACuriousMind that's what I said. the Fourier transform of coulomb potential in the QM context is different from the field theory context. in particular, it is different from the FT of the function 1/r
in the field theory context, we r transforming the operator $1/|r-r'|$. in the QM context, we r transforming $1/r \delta (r-r')$
by operator, I mean (r,r') components of the operator
yes. i agreed that the last formula is correct. it is just that the result of that is some crazy operator. this crazy operator is why we never do the QM coulomb problem in Fourier space
@qwerty That does look interesting. But as ACM said, there are lots of ways to build a hash table, eg en.wikipedia.org/wiki/Cuckoo_hashing And theoretical performance isn't necessarily a strong indicator of how your hash table will perform with your actual real-world data. Your linked article doesn't give much detail, but it reminded me of en.wikipedia.org/wiki/Skip_list
For this specific case, the integral operator corresponding to the inverse radius is known quite generally and explicitly - it is convolution with the Riesz potential at $\alpha = 1$.
I don't know what you mean with "crazy operator" It shows you that $(V\psi)(p)=\int \mathrm dp^\prime\, V(p,p^\prime)\psi(p^\prime)$ with the usual physics notation. The matrix elements $V(p,p^\prime)$ are just the FT, as you can explicitly compute; of course the operator is "non-diagonal" in the physics sense, i.e an integral operator. Perhaps I misunderstand what you mean here.
If you had opened the Wiki article about the Riesz potential, it explicitly talks about how it's the inverse of powers of the Laplacian (and $r$ is the square root of the Laplacian in momentum space). You don't need to think, you just need to do the math.
@ACuriousMind ok so the Fourier coulomb potential operator from field theory is $\frac{1}{p^2}\delta (\vec{p}-\vec{p'})$, while the operator u wrote is a different convolution
so they're still different. it is just that the QM one doesn't look crazy in the end @ACuriousMind
i didn't expect that it would come out not looking crazy. but it seems like Reisz has worked out the general formula for Laplace powers. it is really pretty
@James I haven't really heard anyone use "near-field" and "far-field" in the context of visible light. Since it's not produced by antenna-like processes, I'm not even sure what that means in this context. Where does this question come from?
@James the interference for the reflection happens in the material, not in some "region" outside the mirror. See e.g. physics.stackexchange.com/a/83118/50583
@James Ah, that explains why your question makes no sense - the LLM has just generated nonsense you simply repeated. There is no such thing as a "specular region" in reflection - if you search for that exact term, the only thing that comes up is image classifiers that try to distinguish specular and non-specular regions in a picture, i.e. something entirely different.
@TobiasFünke i think i misunderstood. in ur initial comment, by $1/p^2$, u meant the operator $\int \frac{1}{|p-p'|^2}\psi (p')dp$, while i thought u meant the "multiplication by $\frac{1}{p^2}$ operator"
also, i wasn't familiar that $\frac{1}{\sqrt{\sum \frac{\partial ^2}{\partial p_i^2}}}$ could be simplified to sort of look like a $1/p^2$ operator. i hadn't encountered this in QM. i had only encountered the a "multiplication by 1/p^2" operator in field theroy. so that's what I thought u meant
RR, sorry I honestly could not follow the discussion (not having mathjax does not help either hehe). But anyway, if your could resolve some of your issues a bit that's nice.
@ACuriousMind no, but this takes a lot of writing just to explain. We have mathematical systems in which quite many vectors (not least the non-null non-zero ones) have inverses, and the results of that analysis turns out to produce the (vectorial) spherical harmonics = multipole expansion that is standard to classical electrodynamics; those formalisms then show how to extend it to arbitrary dimensions and arbitrary (spinor) tensors
@TobiasFünke needless to say, the keyword is geometric algebra. The argument is made in full and in full generality in Doran & Lasenby.
when is it valid to substitute EoM into Lagrangian? e.g. if i replace p with $m\dot x$ (using the EOM $\dot x=p/m)$ in $L=p\dot x -H$, i get back a valid Lagrangian
@RyderRude You need to ask a much more specific question to get any kind of useful answer. It's not clear what you mean by a "valid" Lagrangian. The equation $L=p\dot{x} - H$ doesn't hold for arbitrarily tuples $q,\dot{q},p$ and the Lagrangian is not a function of $p$ at all. See also this answer of mine.
@ACuriousMind thanks. what i meant is that (forgetting the machinery of tangent and cotangent spaces. just take $p$ and $q$ as arbitrary dynamical variables of some theory). we have a Lagrangian $L=p\dot {q} -\frac{p^2}{2m}+V(x)$ whose EL eqns are Hamilton's equations. now, if i replace $p$ with $m \dot {x}$ (using the EOM $\dot x =p/m$), i get $L=\frac{1}{2}mv^2-V(x)$. Then I take EL eqns of this. And new EL eqns are equivalent to the previous EL eqns
as in, the new EL eqns are $md^2x/dt^2=-V'(x)$. one can re-introduce $p$ now and recover the previous EL eqns
now, if i had taken the original $L(p,\dot q, q)$ and replaced $\dot q$ with $p/m$, the new $L$ i would get would give non-sense EL eqns
so it seems like sometimes one can substitute EoM into Lagrangian, and other times, one cannot
1. You can't just "forget" that $p$ and $\dot{q}$ are different variables. The Lagrangian is a function of $q$ and $\dot{q}$, not of $p$. 2. Again, you need to be much more careful. You switched your notation between $x$, $q$, $\dot{q}$ and $v$ between every equation you wrote down. The E-L equations are not Hamilton's equations (you can derive Hamilton's equations from the E-L equations, but they are not the same).
@ACuriousMind i want to approach this at the level of variational problems (not at the physics level). just take the pure math variational problems defined by $L=p\dot {q} -\frac{p^2}{2m}+V(x)$. $p$ and $q$ are meaningless real-valued variables. i am not giving them the meaning of position and momentum (contd...)
(contd).. I understand the EL eqns r not Hamilton's eqns. i meant that they look like them. and finally, my question is that if i replace p with dot q, i get a new variational problems that is compatible with the previous one. but if i replace dot q with p, I get non sense
You get Hamilton's equations as E-L when you consider the action principle for the Hamiltonian action $S[q((t),p(t))] = \int p(t)\mathrm{d}q - H(q(t),p(t)) \mathrm{d}t = \int (p(t)\dot{q}(t) - H(q(t),p(t))\mathrm{d}t$. The integrand is not in terms of a function $L(p,\dot{q},q)$, and your statements about "replacing" $p$ by $\dot{q}$ and vice versa make no sense to me.
You can only come up with this idea if you make exactly the mistakes I complain about in the answer I linked, which is not respecting what is actually a function of what.
sorry... what u wrote seems to be $\int L(p(t), q(t), \dot q (t)) dt$, where $L=p\dot q - H(p,q)$. what am i missing?
i am taking the configuration space of the Lagrangian to be $(p,q)\in R^2$. By replacing p with dot q, i mean taking the EL eqn dq/dt=p/m, and using this to replace p with dot q in the Lagrangian
Again, in the phase space/Hamiltonian formulation there is no $\dot{q}$ variable. There is $\dot{q}(t)$, i.e. the time derivative of the "position part" of a path $q(t)$, but there is no $\dot{q}$.
@ACuriousMind i just think it is useful. but anyway, pls forget phase space here. $(p,q)$ is defined to be my "configuration space" here. i will re-named them $(a,b)$ to avoid confusion. so my Lagrangian is $L=a\dot b -\frac{a^2}{2m}+V(b)$. i take the EL eqns, and then get $\dot b=a/m$ as one of the eqns. i then plug this back into my Lagrangian. and take variation of my new Lagrangian, and what I get is an EL eqn that is compatible with the previous two EL eqns
so it is like plugging the EL eqn into the Lagrangian did not screw up the variational problem like it usually does
@TobiasFünke yes, that is quite necessary; the geometric product of two vectors gives a scalar part and a bivector wedged part, and that then beautifully extends to higher dimensions and beyond bivectors. The resultant theory happens to have enough structure to define inverses for vectors unambiguously, amongst other niceness
@RyderRude That's because this Lagrangian formulation has no kinetic term for $a$, so $a$ is just a Lagrange multiplier for the constraint $\dot{b} - \frac{a}{2m} = 0$. That you can plug in ("solve") constraints into the action, eliminating $a$, without destroying the variational principle is perfectly normal.
@ACuriousMind thanks. i will have to read more about constraints... ___
@ACuriousMind is it intuitively becuz the constraint defines some sort of a constrained sub manifold and then we apply the variational principle on the submanifold?
well, it's not exactly a Lagrange multiplier but it's the absense of the kinetic term that ultimately leads to this being a constraint instead of a "proper" equation of motion
@RyderRude Yes - the equations of motion here do not have solutions for arbitrary tuples $(a,\dot{a},b,\dot{b})$, only for those where $\dot{b} = a$ and you cannot freely choose $\dot{a}$ as it will be determined by the solution via $\ddot{b}(0)$, meaning there is two d.o.f. less than you would think.
plugging in the constraint just eliminates the superfluous d.o.f.
@RyderRude That book is about the Hamiltonian formulation, but sure: In Hamiltonian terms, The Legendre transform is singular - $p_a = \frac{\partial L}{\partial \dot{a}} = 0, p_b = a$ - leading to the second-class constraints $p_a = 0, p_b = a$. This just eliminates $a$ from the start and leaves you only with $(b,p_b)$ as the Hamiltonian dynamical variables.
@SignorFeynman I mostly refuse to use emojis - long live ASCII
Is there a connection between a Lie derivative and a Material derivative? The second one being: ${\displaystyle {\frac {\mathrm {D} y}{\mathrm {D} t}}\equiv {\frac {\partial y}{\partial t}}+\mathbf {u} \cdot \nabla y,}$ ?
@User198 for scalar functions, the directional covariant derivative and the Lie derivative co incide
we have a (tensor function) $f(y,t)$. we can compute $v^{\mu} \nabla _{\mu} f(y,t)$. the $\nabla _{\mu}$ refers to the Covariant derivative. it equals the partial derivatives for scalar functions
I have a question about collisions. If we consider the reference frame where one of the two particles is at rest, then I presume that one cannot talk about com energy, right?
@User198 The second term is clearly the Lie derivative of the function $y$ w.r.t. the vector field $\mathbf{u}$, no? I'm not really sure what the question is.
