This post has Hawking's idea of QFT in a spherical spacetime (time has no beginning and no end)
in this QFT, you cannot talk about an evolution from an initial wavefunction to a final wavefunction
nevertheless, this universe does have a wavefunctional (surprisingly, Hawking uses wavefunctionals as the fundamental description of the universe.)
the wavefunctional is given by a path integral formula. This path integral isnt being integrated from an initial to final state. but it is a natural generalisation of that to a periodic time
@MoreAnonymous if u approximate a small GR region by SR (by choosing appropriate co ordinates), ur error will be something like $O(x^2)$. I mean the Taylor expansions of the two metrics will agree upto at least the first derivative.
Can anyone help me understand how in this here: https://quantummechanics.ucsd.edu/ph130a/130_notes/node486.html the D.E looks , the way it looks initially. I have derived the D.E in a similar form, but mine contains the complex i and the sign before the mass m is minus
@imbAF That resource uses a non-standard definition of $\gamma^\mu$ where your missing $\mathrm{i}$ is absorbed into the definition of the $\gamma$-matrices, see the definition on quantummechanics.ucsd.edu/ph130a/130_notes/node480.html
@MoreAnonymous it's just Riemann normal coordinates
you should already know this because it has been mentioned to you several times in this chat
In fact, you discussed essentially the same question - how to arrive at the Minkowski metric - here a few months back and you were already told normal coordinates are the answer
I don't understand why you now come back and demonstrate no awareness of this technique in your question(s)
@MoreAnonymous I don't know what you mean by "method 2" but Ryder was just talking about normal coordinates. Which you asked him to post as an answer. And now you claim that's not what you're confused about???
So here's 2 ways I know of obtaining the special relativistic limit from general theory of relativity (at a point in spacetime).
The equivalence principle
Taking the derivative of soley the stress energy tensor and the metric setting equal to the metric to Lorentz metric.
$$ \nabla_\mu T^{\mu...
@MoreAnonymous sure, but Ryder's answer, which mentions "choosing appropriate coordinates" is clearly not about any of that but about choosing normal coordinates.
(also your question as written makes once again no sense to me because it lacks any details of your thought process and uses notation in confusing ways but I'm tired of litigating that every time)
@ACuriousMind hey again. wanted to make sure of something.
In the following: $S[x(t)] = \int_{t1}^{t2} x(t) dt$,
we say $S$ is functional because it expects us to provide it with a function of t and put it instead of $x(t)$ and it will give us number. For example, we could provide it with $2t+5$.
Then we have action functional in the lagrangian such as:
$S = \int_{t_1}^{t_2} L(x(t), \dot x(t), t) dt$,
Now, $L = \frac{1}{2}m\dot x^2 - mgx$ and putting it inside the $S$ would yield:
@Chemistry You can, of course, view the expression $S = \int L(x(t),\dot{x}(t),t)\mathrm{d}t$ as a functional in both $L$ and $x$, i.e. $S[L(q,v),x(t)]$, which becomes a number only when fed both a concrete $L$ and a concrete $x$. This is the same as your "function of a function of a function", the equivalence between "function of two functions" and "function of a function and a function" is called currying.
We do not usually think about $S$ in that way because in our actual physical applications we want to vary $x$ but we do not want to vary $L$ - the $L$ is fixed for a particular physical system, so it is not relevant to consider it as an input explicitly.
After yesterday's talk, I thought that notation $L(x(t), \dot x(t))$ is a function of only $t$ and it shouldn't contain $x(t)$. I thought that this notation is the same as $f(g(x))$ which is only a function of $x$ and it doesn't have $g(x)$ in it explicitly, but for action, it seems like $S$ contains $x(t)$ explicitly in the integral
What I mean is we should have the following notation: $S = \int_{t0}^{t1} L[x(t), \dot x(t)] dt$
Yeah, but tell me why $S = \int_{t0}^{t1} L(x(t), \dot x(t)) dt $ is even correct. This notation $f(g(x))$ means it has $x$ explicitly and not $g$, but notation $\int_{t0}^{t1} L(x(t), \dot x(t)) dt $ then must have $t$ explicitly and not $x(t)$
that's why the best thing is to use $S = \int_{t0}^{t1} L[x(t), \dot x(t)] dt$ with the $[]$ and not $()$
what is wrong with saying that L(x(t), \dot x(t)) only has t explicitly. are u worried that we wont later be able to take partial derivatives of this wrt x and v?
well, yesterday, you said: $()$ means a composite functions. So If I have $f(g(x))$, then that means the final $f$ function is a function of $x$ and not $g(x)$. It only contains $x$.
Then, i look at action and in the integral, I see the composite function notation $L(x(t), \dot x(t)$ which says that it contains $t$ explicitly, it's a function of $t$ and not $x(t)$. and then if so, this becomes wrong as I can't vary $x(t)$ which I need to. So If we had written $L[x(t), \dot x(t)]$ in the integral, this would say that $L$ itself became a functional such as it contains $x(t)$ expicitly and no…
@Chemistry Given $f: B\to C$ and $g: A\to B$ you can view $f(g(x))$ as $A\to C, x\mapsto f(g(x))$ but you can also view it as $[A,B]\to [A,C], g\mapsto (x\mapsto f(g(x))$ or even as $[B,C]\times [A,B]\to [A,C], (f,g)\mapsto (x\mapsto f(g(x))$ or...
the problem is the notation is ambiguous without stating what function you actually mean, which things are inputs and which are fixed
@ACuriousMind then you are saying that notation $()$ in general is confusing. and it's not a notation of only composite functions
and can be interpretted as one wishes
but yeah, i think I get the final picture. in the integral of action, we end up having $x(t)$ explicitly so it's a functional of action of $x(t)$ the same way as $S[x(t)] = \int_{t0}^{t1} x(t)dt$ and we start to vary
I recently asked this question which got closed because its answer was based on opinions. But that was exactly the point of my question, and hence the usage of tag of "convention". So I am not sure why it got closed.
@Chemistry the thing being integrated is x(t), which is only a function of t. the result of that integral is a functional that takes a function x(t) as input
we're on the same page . action would look like this - $S = \int_{t0}^{t1} \frac{1}{2} m\dot x(t)^2 - mgx(t) dt$ and this is functional. it expects you to give it $x(t)$
but you don't know what $x(t)$ you give, so you start assuming to give some assumed $x(t)$ which is true path
I think I got that part.
my question is this. Ok, in the beginning, L was not a function of $t$, but just function of $q, v$ (independent variables) and then in the action, they become functions of $t$.
How did people decide that $L$ in the beginning shouldn't be a function of $t$ ? did they derive euler lagrange and when they saw euler equations are partial derivatives with respect to $q$, they said ah, if L had been a function of $t$, we wouldn't be able to take partial derivatives, so let's just have $L$ as only a function of independent variables ?
yeah, but in the beginning, why did they decide that in the L, $q$ and $\dot q$ must not have been a function of t
what was the reason
I like this answer, but I don't get something:
On one hand, let us first consider the role of the Lagrangian. Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ is a function of both the instantaneous position $q(t_0)$ and the instantaneous velocity $v(t_0)$ at the instant $t_0$. Here $q(t_0)$ and $v(t_0)$ are _independent_ variables. Note that the (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ does not depend on the past $t<t_0$ nor the future $t>t_0$. (One may object that the velocity profile $\dot{q}\equ…
the thing I'm asking is when they came up with Lagrangian, L = K - U, did they know the euler lagrange at that time ? if they did, then thats why they said that in the $L$, $q$ and $\dot q$ shouldn't be. a function of $t$. I got no other reason
let's say you just hypothesize a lagrangian L(t). and u try to find stationary paths of $\int L(t) dt$. this wont yield u Newton's second law @Chemistry
L(t) is just a generic function. there's nothing to work with there
in that case, if $q$ and $\dot q$ depend on $t$, you still get euler lagrange, but you can't apply it to $L$ because if $q$ is a function of t, you cant deerivative L with $q$. Thats what I was meaning
lemme summarise : if u hypothesize a least action principle $\int g(t) dt$, u can only get to the EL eqn when u assume g(t) is of the form L(x(t), dot x(t)). and in the EL eqn, this L will show up and u will b taking its partial derivatives
and if you assumed that $x$ and $\dot x$ are functions of $t$ in the $L$, you won't be abe to take partial derivatives
so that is one reason why they might have decided that L in the beginning is a function of 2 independent variables which don't depend on $t$. makes sense ?
what I was meaning is imagine they said that L is a function which contains two variables $q$ and $\dot q$.
if we hypothesize that $q$ and $\dot q$ is a function of $t$, you say that $L$ in the end would be a function of only t and you then can't take partial derivatives on it.
my take was that since we got L = 10q + 20\dot q in which $q$ and $\dot q$ are functions of $t$, but we don't know what functions, so we got $q$ and $\dot q$ in $L$, we still wouldn't be able to take derivatives. Whats the derivative of $10q + 20\dot q$ with respect to $q…
@Chemistry Yes, I'm saying if you really want to properly express what's going on, you need to start writing down the (co)domains of the functions in correct mathematical style. Trying to argue about this just within the notation of writing functions like $f(x)$ will never solve the problem of this particular topic being confusing.
On one hand, let us first consider the role of the Lagrangian. Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ is a function of both the instantaneous position $q(t_0)$ and the instantaneous velocity $v(t_0)$ at the instant $t_0$. Here $q(t_0)$ and $v(t_0)$ are _independent_ variables. Note that the (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ does not depend on the past $t<t_0$ nor the future $t>t_0$. (One may object that the velocity profile $\dot{q}\equiv\fra…
Qmechanic's point does not seem completely correct to me. e.g. the Dirac equation is first order but its lagrangian still treats $\partial _{\mu} \psi$ as indepedent variables
so the Lagrangian treating derivatives as independent does not have to do with the freedom in choosing initial conditions independently
i may be missing something
and ofc when L=, let's say, dot x +V(x), then again the evolution eqn is first order but we still have to treat dot x as an independent variable while taking partial derivatives
Again, the classical equation of motion is generically 2nd order. It does not matter if you can construct specific unphysical situations where it degenerates to a first-order equation.
i do know that. but, so far, it is just a co-incidence for this particular case that the number of independent variables in the Lagrangian= number of independent initial conditions @ACuriousMind
hmmm.. i guess i know what u r saying. the second order eqn is used as a motivation to hypothesize the form L(x, dot x)
but we should make it clear that the partial derivative treating dot x as independent to x does not have to do with the independence of x and v in the initial conditions (just in case u misunderstood this) @Chemistry
Qmechanic is not saying this. there are counter examples like the Dirac eqn @Chemistry
@RyderRude . I think i get it. it's just like you got 2 reasons why L treatts them independently. one reason is strong, second one not that strong, but still helps enforce it
The argument really is: In classical point particle mechanics the equations of motion are 2nd order, so the full specification of a physical state (that can generically serve as the initial condition) consists of two independent data points $q$ and $v$
@RyderRude I don't know how often to say this: That the framework has points where it degenerates does not invalidate the generic argument. Some systems are not well-modeled by Lagrangian mechanics in this way and that's fine.
@ACuriousMind it says V'(x(t))=0 as the EL equation. So this perfectly makes sense as a condition to classify stationary trajectories. and it does not have to model a physical system because the principle of least action is, first and foremost, a mathematical principle
its a theorem deriving EL eqn by stationarising the action. it is purely mathematical
this should not be upto debate. it is a co-incidence that the number of initial conditions in Newtonian mechanics = number of independent variables of the Newtonian lagrangian. if u r saying that this is more than a co Incidence, u have to give any mathematical result that gives this connection between these two
or it must be the case that EL eqn does not provide the stationary trajectories of L= dot x +V(x)
if either of those things is true, then i am wrong
@RyderRude You're just drawing the wrong conclusions here. The conclusion isn't "Lagrangian mechanics have situations where there's no solutions for some initial conditions $(q,v)$", the conclusion is that Lagrangians where this happens are just inadmissible as Lagrangians in Lagrangian mechanics even if you can formally put them into the action and look at their E-L equations.
Physics is not mathematics, just because you can put something into a formula physics doesn't have to accept it
3
and careful formalizations of Lagrangian mechanics will usually state constraints on the Lagrangian, up to outright demanding it be of the form $\dot{q}^2 + V(q)$ to avoid such pathologies
@Chemistry Think you're getting confused about something easy. $L(x,y,z) = \sqrt{1+z^2}$ is some function, some 3D surface embedded in 4D $(x,y,z,w)$ space, with $w = L(x,y,z)$. If we set $y = y(x)$ and $z = y'(x)$ then we're restricting ourselves to a curve on this surface, where once we know the $y$ component, the component in the other direction, $z = y'(x)$, is fixed. So $S[y] = \int_0^a L(x,y(x),y'(x))dx = \int_0^a \sqrt{1+[y'(x)]^2}dx$ is a functional,
the 'arc length' functional, giving the arc length of some $y(x)$. A function like $L(x,y,z) = f(x,y)\sqrt{1+z^2}$ leads to the functional $S = \int_0^a f(x,y(x)) \sqrt{1+[y'(x)]^2}dx$ which is integrating $f$ over the arc length. Different curves $y$ produce different results. Only difference in mechanics is that we use $t$ as our parameter, i.e. $S = \int_0^a L(t,x(t),x'(t))dt = \int_0^a \sqrt{1+[x'(t)]^2}dt$. Just ask how things apply to arc length and you should be able to figure it out
@ACuriousMind Could you explain something to me, which seems counter intuitive. In the FW transformations at one point (during the consideration of this type of transformation), it is said that $\vec \alpha e \vec A$ is of order $O(\vec p^2/m)$. How do we come up with this?
@imbAF I don't know what you're talking about; you haven't even confirmed if by "FW" you really mean Foldy–Wouthuysen, and you cannot assume that anyone else knows what specific steps are in whatever text you're looking at
I am computing the loop corrections to the electromagnetic energy of an electron in an electromagnetic field, i.e. the first integral. After a few calculations it is reduced to the other integral (red arrow). Now there is this sentence underline in blue. What does that mean?
The only currents I know are either the $e\overline{\Psi}\gamma^\mu\Psi$ appearing in the lagrangian or the one in $(19)$ (which at lowest order would again be $e\overline{\Psi}\gamma^\mu\Psi$)
I mean, he makes it sound like "the first is the electric interaction as you can recognize yourself and now we show that the second is the magnetic interaction"
good point, I forgot to write it. There would also be an electric dipole terms otherwise
The second part should be called the "dipole" part, yeah
Let's cast it in this way. How does the text infer that the "convection current" is not relevant for our discussion and we can only focus on the second part?
That's the Klein-Gordon current applied to plane waves which satisfy the Klein-Gordon equation, but they are electrons, so he's probably just using a different name to indicate the distinction
exactly, that's the KG current written with spinors
Spinors satisfy the KG equation, so... it makes sense
It may sound like a stretch here but from this perspective it looks like they are separating the "scalar" part (which would be related to $e\vec{E}$ somehow) from the dipole part
Which is just the meaning of the Gordon decomposition after all, isn't it?
@Mr.Feynman spinors satisfy KG only if there is no EM field. The coupling to EM field is different between KG and Dirac, and that causes the SR correction to Bohr energies to be in different directions.
@imbAF FW is basically useless in the grand scheme of things so I would suggest not putting too much effort into it.
hello -- i have been doing a problem which involves solving the free particle schrodinger equation with various boundary conditions imposed. first, i used the Dirichlet condition, then i used a periodic boundary condition $\psi(0) = \psi(L)$ and $\psi'(0) = \psi'(L)$. then, [...]
[...] part of the problem is to add up the first five energy levels and to compare the sum between the two. the solution explains that if we add up more and more energies, the fractional difference should become negligible. i am wondering why this should be true? i showed that it seems to be true at least for the first five, but i dont see why i should expect the energy sums to be the same? [...]
[...] this might be related to my second question, which is what it means physically to impose these conditions on a free potential situation?
@ACuriousMind okay i see -- you mean because, for instance, 0 and 2pi are the same point, so the wave function must have the same value at them for it to make physical sense?
@Relativisticcucumber difficult to say. I very much like Primosale, which a type of Pecorino. Then, I love Grana Padano, Parmigiano Reggiano (very similar, but don't tell that to people from Emilia-Romagna). Last but not least, my beloved Gorgonzola
@Relativisticcucumber That one you'd usually do to model infinite walls at $0$ and $L$ - "wavefunction can't penetrate the walls so has to be zero there"
@Mr.Feynman Parmigiano (as Parmesan) is like the cheese to put on your pasta here (though it of course typically will not be authentic Parmigiano and just some hard cheese like it)
@ACuriousMind Sure, we do that in Italy as well. Just, there is a difference between Parmigiano and parmigiano(=parmesan) as far as I notice: Parmigiano Reggiano is not just a type of cheese but a brand, that's why I used the upper case. You can also eat that in pieces, instead of putting it on past. Then, some colloquially call parmigiano any hard cheese they put on pasta (which could be Parmigiano)
Grana Padano is a different brand that is (almost) equal to Parmigiano
I'm not really sure about the word "parmesan" being generic, but it would be like when one calls "Scotch" a generic tape because the Scotch brand is so iconic
well i think i found the problem. he says in his notes " In the thermodynamic limit the choice of boundary condition is irrelevant.", but now i think he just meant this for this problem
my dad is one of those olde guy who would just keep boiling water to kill germs. he would already have to full jugs and still continue to boil, so that the kettle is full with his cooling water and I would not have anything to boil water for tea.
In the end he realised that I would rather not touch any drop of the plain water. Only if he leaves me a kettle to make tea, would I drink anything at home in a week.
@ACuriousMind i ran out of the scottish ones. this is suntory. too smooth.
my dad can have espresso and immediately go back to sneeppuu
and while, yes, caffeine is more concentrated in tea leaves than in coffee beans, you dont actually consume the leaves (unless in Japanese ground-up leaf powder), and so you get much less caffeine intake by tea than by coffee, typically
I'm going over this article by Davies, in which he derives the form of the energy-momentum tensor (emt) in 2D spacetime assuming a non vanishing trace anomaly.
He considers a metric of the form
$$
ds^2=C(u,v) du dv
$$
then argues that:
the trace $T^\mu_\mu$ should be functions of quadratic deriv...
So here's 2 ways I know of obtaining the special relativistic limit from general theory of relativity (at a point in spacetime).
The equivalence principle
Taking the derivative of soley the stress energy tensor and the metric setting equal to the metric to Lorentz metric.
$$ \nabla_\mu T^{\mu...
I recently asked this question which got closed because its answer was based on opinions. But that was exactly the point of my question, and hence the usage of tag of "convention". So I am not sure why it got closed.
