I stumbled upon "a symplectic structure on a manifold can be thought as an extra structure which allows one to do Lagrangian mechanics on it", so I was too curious and googled what Lagrangian mechanics is.
@yuggib You're on a point on the trajectory. To move onto another point in the trajectory which is infinitesimally close to the original, you need to use $T - V$ amount of energy, right?
@BalarkaSen you may say that moving from one point to another on the path "transfers" a small bit of kinetic energy to potential energy (or vice versa)
Possibly, I don't know much about the Legendre transform to confirm or deny :P Thanks, though, I'll try reading up on them at some point of time. I'll write down the keywords.
I need to get back to math for now. Thank you very much!
In fact, in Hamiltonian mechanics you should write the evolution equations (also for functions of position and momenta) using the Poisson brackets, and they should essentially be the symplectic form
(I am not so confident with the language either, so take what I say with the due care for possible mistakes)
@BalarkaSen It is always time to get back to math ;-)
Hey guys, i have a problem with inertial-non inertial frame of references. Relativity of motion tells us that we cannot tell if i am moving and you are still, if i am still and you are moving or we are both moving. The same applies to acceleration! The observer that is accelerating it a non inertial observer, right? So, is there an objectively non inertial observer? If so how do we know who is one?
@BalarkaSen one interpretation of the Lagrangian is that is minimizing, at every point along a trajectory, the excess of kinetic energy $T$ over potential energy $V$, so if you view this from a purely kinematical perspective you see that $T$ describes the motion of a free particle but then adding $V$ is basically a constraint on the path the motion takes so that minimizing the Lagrangian is minimizing constraints on the free motion as much as possible.
Another way to think of it is that it is just a global representation of the local Newton's law $F = ma$, and you can obtain the Lagrangian explicitly by integrating $F = ma$ against potential (i.e. virtual) paths (it's worked out in Lanczos' book).
So you can view Lagrangian mechanics as a global form of $F(t,x,v) = ma$, and then Hamiltonian mechanics is just a generalization of $F(t,x,p) = dp/dt$ i.e. using momentum $p = mv$ instead of velocity $v$, 8.1 of Neuenschwander's Noether book works this out nicely.
@BalarkaSen let me clarify just in case: The classical motivation for looking at a Lagrangian comes from wanting to analyzing $F(t,x,v) = ma$ along the whole path. Integrating this against paths results in the action integral of the Lagrangian. From this expression we see that the Lagrangian is just the excess of kinetic over potential energy at a point, integrating it over a path (Lanczos Variational Principles Ch. 4) gives this along the action, and then finding the equations of motions (i.e. minimizing the action)
Looking directly at Newton's equations (Neuenschwander Ch. 8) we can motivate Hamiltonian mechanics by re-expressing $F = ma$ as $F = ma = m \partial_t v = \partial_t (mv) = \partial_t p$ we have $F = \frac{dp}{dt}$ and using $F = - dU/dx$ http://physics.stackexchange.com/a/54922/25851 this can be re-expressed as $\frac{dp}{dt} = - \frac{dU}{dx}$, which is one of Hamilton's equations. Since it's first order while Newton was 2nd order we need a second equation for $x$, so using $F(t,x,p) = dp/dt = \frac{dp}{dx}\frac{dx}{dt}=\frac{dp}{dx}\frac{p}{m}$ this reduces to $\frac{p^2}{2m}+U=E = H$ g…
Hamiltonian mechanics can just be interpreted as finding the path along which the total energy of the particle is minimized all over the path, right? I mean, $T + V$ has a concrete meaning as the total energy of that particle at that point.
Another way to motivate what we just did from Lagrangian mechanics is the desire to re-write Lagrange's 2nd order equation as a system of first order equations (Gel'fand CoV), this is where the symplectic structure comes from, analyzing a system of ode's simultaneously. If we integrated $F(t,x,p) = dp/dt$ against paths we would get the Hamiltonian formulation of Lagrangian mechanics, i.e. the Hamiltonian form of the Action/Lagrangian.
However since the momentum may also be expressed as a derivative of the Lagrangian, $p = \frac{\partial \mathcal{L}}{\partial v}$, we can use a classical trick to find the Hamiltonian directly from the Lagrangian, the method of Legendre transforms. The Legendre transform is thus just an old geometric trick to take a surface expressed via points, say $(x,y)$, and re-generate the surface using the points $x$ and tangent lines to the $y$ points (Goursat, Vol. 1) all along the surface. What Landau (Vol. 1) does is to skip the classical $F = ma$ motivation and use symmetry principles along with …
Yeah you can think of the Hamiltonian formulation that way on a basic level
So it's like Lagrangian is about minimizing interconversion of different forms of energy while Hamiltonian is about minimizing total energy
Yeah, I just gave tangent lines for my $(x,y)$ example, and there is a great pdf showing visuals of the Legendre transform online for a curve $y = f(x)$ arxiv.org/pdf/0806.1147.pdf to help get used to it
I deal with my learning using the following thinking process:
Geometry + Rule
For example, in the case of energy levels in molecular states, if you have a product rotational energy level that is of similar energy to an electronic level, then you expect mixing to happen
The geometry here is that in an energy level diagram, the phrase "for similar energy" means they are nearby in a "space" where positions are notated by the physical quality energy
The rule is that: Whenever you have levels that are close together (based on the geometry mentioned above), mixing can occur
What I need more to do physics problems, however, is the following way of thinking:
we know that sqrt(2h/g) is time for something to fell on the floor under uniform acceleration in newtonian mechanics $$v_0>\sqrt{\frac{(d^2+h^2)g}{2h}}$$ The messy sqrt contains the inverse of this multiplied by some distance, suggesting it is some kind of velocity (because the dimenions you get are [L][T]^-1 which is the same as velocity This suggest v0, the bullet velocity has to be greater than some kind of critical value
Because from what you read up there, the way I think is like C programming, in that given some stuff obeying some conditions, I output and deduce results
But I don't know how to break free of this kind of thinking because I cannot really grasp how most people think
In particular, how do people make guess of something that cannot be deduced from formulae, such as when doing some electrodynamics problems, how do they know that they need to e.g. assume the magnetic field is uniform, in order to continue on the calculation?
leap of faith is also something that is kinda beyond my understanding
I know that by doing more problems, I can obtain this kind of thinking (which has proved itself given I suck at electromagnetism back in year 2, and getting better at year 3)
But how does this work, e.g. what are people thinking when they slowly acquire these ways of thinking?
or more accurately, what is the mechanism that is happening in terms of thinking process that ultimately lead to thinking in a non algorithmic way when one do problems?
This is not a diary, if you can deduce what it truly means
*He has entertainment, while I have none* *Once the conspiracy, and now it is none* *The truth is clear, but no one cares* *While questionable, he fares well then me* *As for those who disgree, he still amazes* *This is obvious, tachyon, relativity and the past.* *Because it is your choice* *to choose the path of stillness*
*It will still happen, until the end of time* *The purpose is simple, because they don't respect the gods* *In response to the crisis, there's one final option* *Operation (encrypted)* *It does not matter* *For no one cares* *Until it is too late*
Hi guys, how is the quantum mechanical time reversal operator defined exactly? Zee writes that if the time reversal operator is defined as $\Psi(t) \mapsto \Psi'(t') = T\Psi(t)$ whereas $t' = -t$, then one must conclude that $T^{-1}(-i)T=i$. How is that even defined? $T$ is an operator that acts on the Hilbert space $\mathcal{H}$ of states. $i$ is a scalar, it is NOT an element of $\mathcal{H}$, so how can $T$ even act on $i$ and $-i$?
is it meant like $T^{-1}(-i T(\Psi)) = i\Psi$?
seems the only thing that makes some sense, treating the scalar $i$ as some sort of operator $\Psi\mapsto i\Psi$
@dmckee I have a special relativity conceptual question, I am not sure if this way to understand it is considered too misleading to be used as a guiding principle to more complicated problems
@ACuriousMind I was talking to my LA TA and he said homomorphisms only preserve compactness if they are continuous (this is a sufficient condition). Unfortunately, I don't see how the $f(a)$ we determined yesterday is continuous.
@tpg2114 My faculty advisor in the math department works on spectral theory in algebraic topology or something like that...
@0celo7 Oh boy. Easy way: Everything's smooth with Lie groups.
Hard way: Since $\omega$ is left invariant, $R_g^\ast\omega = C^\ast_{g^{-1}}\omega$ where $C_{g^{-1}}$ is conjugation by $g^{-1}$. Then by the property of a volume form, $C_{g^{-1}}^\ast\omega = \det(\mathrm{Ad}_g^{-1})\omega$, where $\mathrm{Ad}_{g^{-1}}$ is the linear map on $\mathfrak{g}$ induced by conjugation by $g^{-1}$. Since $\mathrm{Ad}$ is smooth in $g$, inversion is smooth, and $\det$ is smooth, this is smooth as the composition of smooth functions.
@Bass Yes. Formally, you could define that by defining the operator belonging to the complex structure that underlies mulitplication by $\mathrm{i}$ and state that $T$ act on it by inverting it's sign, what you wrote there is the less convoluted way.
@0celo7 Conjugation by a group element $g$ is the map $G\to G, h\to ghg^{-1}$.
@dmckee We know that photons travel along null curves in spacetime and that it does not have a well defined 4-velocity because the proper time is 0
We know that special relativity is based on the two postulates that laws of physics does not change with inertial frames and the speed of light is the same for all observers
We also know that in O's frame, O can measure the time some events took place and where it is by using light signals. Therefore one can easily work out time dilation and length contraction when the event is in some relative motion to the observer
Sorry for this long question, I have tried to simplify it...
@dmckee Correction: As drawn in the spacetime diagram, I have plotted the light cones that can be radiated from the light source in all direction as it travels towards the right for each lightsecond passed in O's frame.
@0celo7 Yeah, but sometimes it's really useful to remember that thing is really mostly a derivative/Jacobian, the ${}_\ast$ somewhat obscures that, I feel
Reading Ref. 1 I admit I am a little lost in some places. I was hoping that someone in this area could explain the basic premise of integrating the Lie bracket and further is it connected to nlab's Quantomorphism group? If this is so, can we 'derive' a quantum system by Lie integration of a Poi...
Temporarily locked question to avoid migration cf. meta discussion about on-topicness of math questions. Input to how we proceed in this question is welcomed. ...[June 15th, 2017: Reopened.]
@Qmechanic I'd prefer that the close voters explain why they feel that question is off-topic. Unfortunately, locking the question makes it a bit hard to see where such comments for this specific question should be left.
Let be $(M^{n+1},g)$ a spacetime (Lorentz manifold, connexe and time-oriented), $n\ge 2$, and $S\subset M$ a null hypersurface (codim $S=1$ and the restriction of $g$ to each tangent space $T_p S$ is degenerate).
If $K$ is a null vector field of $S$, show that the integral curves of $K$ are nul...
"A complete orthonormal set of eigenfunctions of the wave operator is found in the special case of a spacetime in which the total deficit angle is $2\pi$"
Wait what
2 pi sounds a bit much
Isn't that the whole spacetime
Oh wait, Gott spacetime is two cosmic strings I think
@Qmechanic Thinking about this, how is anyone supposed to stumble upon this? The close voters already voted and won't usually return to the question, and the lock took this out of the review queue.
@Qmechanic @ACuriousMind Thank you both for your help with regard to the close voters. I was quite surprised to be honest! I expected perhaps one or two down votes due to apparent lack of prior research maybe (despite reading quite extensively in this area for a while).. but not off topic!
Why is area/determinant anti-symmetric? I know how to prove it from the fact that area is alternating $f(u+v,u+v)=0\rightarrow f(u,v)=-f(v,u)$ and I can obviously visualize this alternating property, but I can't visualize or see why area is, a-priori, anti-symmetric. Thoughts?
In other words, if you were motivating the determinant, how would you motivate anti-symmetry without going through this alternating argument?
I haven't been on here very long but it seems like people are a bit harsh on naiive questions. For example Gravity between two masses was getting downvoted last I checked. While these questions are very basic, they are not disingenuous or abusive, and indeed represent a very common line of reason...
Hmm... I don't see why that is or why it has to be, also that idea reduces to a 2-d argument where you say $f(a,b)$ is the area starting from $a$ and then going counter-clockwise to $b$, so that $f(b,a)$ is obviously it's negative. I don't believe that argument or understand why we want to attach orientations, right now I can only see the alternating $f(a,a) = 0$ idea justifying an area/anti-symmetry $f(a,b)=-f(b,a)$ interpretation. There's just a tiny step but it's not clear to me.
@JohnDuffield any thoughts on why area/determinants is/are anti-symmetric?
@bolbteppa You can't have unoriented area as linear function in each argument because it has to be alternating, and alternating forces antisymmetry, which is then interpreted as assigning orientation. There is no a priori reason for anti-symmetry - the a priori thing we want is alternating, and that's mathematically equivalent to antisymmetry (except in characteristic two).
You might also give up linearity instead - just take the absolute value of $f$.
I'd say there's nothing really forcing you to do one or the other, but choosing it to be linear is just nicer to work with
Lots to like in that graphic. The overcomplicated slide. The 'audience' doing everything but paying attention. The nice linear representation of the living and dead kittens. Nice.
@GeorgeSmyridis Non-inertial <-> experiences forces. There is a long history of this, going all the way back to Newton. A thought experimental way to tell would be to place some test masses around and make sure their velocity stays constant relative to you.
@JamalS Jost's Mathematic Concepts gives a nice intro to moduli spaces as merely analyzing the space of all relations instea of just one (compare to books on foundations/analysis/discrete-math where you usually only discuss one relation): "When, instead of looking at a single relational structure, we would like to understand all possible such structures, we are led to so-called moduli spaces (the name being derived from Riemann’s investigation of the various conformal structures (Riemann surfaces) that a given topological surface can be equipped with)."
user54412
@GeorgeSmyridis For rotation, you can imagine connecting two masses by a taught string and then cutting the string. If the masses fly apart, they must have been truly rotating. See also Newton's bucket.
@TanMath hey check out my latest on QM man, some ref to QM biology with you )( in back of mind :) ... are you taking any physics classes? any laboratories? have been feeling like it would be nice to hook up with experimentalists somewhere, even undergrads, but they seem so rare in cyberspace at least... :(
@ACuriousMind I'm curious how much you think that meta discussion applies to the current question. For instance, I wouldn't have voted to close the current question (as much as I think "quantomorphism" is a made-up word ;), but I'm still adamantly opposed to "what is a delta function" being considered physics.
@ChrisWhite It's not really a fit because that "Dirac delta" question is about a pure mathematical object that is used in physical contexts, while Poisson brackets and Hamiltonian systems pretty much are physics that happens to be mathematical. I think the current question is much more clearly on topic than the subject of that meta debate
And I would be with you in closing the question about the Dirac delta.
Which is why it is puzzling that the question got three close votes.
For the record, here's the review history of that one.
Perhaps @Qmechanic could superping the close voters from here?
For example, I want to write an answer to the topic:
Be nice with the comments
that has the address:
"meta.physics.stackexchange.com/questions/7243/be-nice-with-the-comments"
If I press "log in" the effect is zero. Why?
I am redirected to the same link:
without any warning like "you are no...
@ACuriousMind : Ah, found the review. @JohnRennie @@Gert @@user36790 : Since you reviewed this question, you might be interested in venting/defending an opinion about whether this question should be migrated or not. See also here.
In the context of Dirac spinors, when one writes $\bar\psi=\psi^\dagger\gamma^0$, is $\psi^\dagger$ just the complex conjugate of $\psi$ (plus maybe vector transpose), or is there more to it?
I'm studying QM & Electro for my undergraduate degree, and one of the books begins with polarization, and then quantum states, why would they do that? I've never seen polarization, it's very confusing.
Reading Ref. 1 I admit I am a little lost in some places. I was hoping that someone in this area could explain the basic premise of integrating the Lie bracket and further is it connected to nlab's Quantomorphism group? If this is so, can we 'derive' a quantum system by Lie integration of a Poi...
Let be $(M^{n+1},g)$ a spacetime (Lorentz manifold, connexe and time-oriented), $n\ge 2$, and $S\subset M$ a null hypersurface (codim $S=1$ and the restriction of $g$ to each tangent space $T_p S$ is degenerate).
If $K$ is a null vector field of $S$, show that the integral curves of $K$ are nul...
For example, I want to write an answer to the topic:
Be nice with the comments
that has the address:
"meta.physics.stackexchange.com/questions/7243/be-nice-with-the-comments"
If I press "log in" the effect is zero. Why?
I am redirected to the same link:
without any warning like "you are no...
