The h Bar

Semiclassical

12:16 AM

here's an interesting question: How would one 'derive' the classical HJ from this mindset?

I think the first step is arguably the same: To propose that the velocity is the gradient of some scalar function

I'm not sure where one goes from there tho

bolbteppa

The way they get the classical HJ in that book is fine I think

From the whole $dS = p dq - H dt = \frac{\partial S}{\partial q}dq + \frac{\partial S}{\partial t}d t$ thing

So if $p = mv = \frac{\partial }{\partial q} S = \nabla S$ we have $v = \frac{1}{m} \nabla S$, but they claim in chapter 8 that we should use not $S$ but some other function $\psi$, this seems to just amount to changing the action $S$ which gives $F = ma$ and so seems to just amount to changing $F = ma$

bolbteppa

12:34 AM

@Semiclassical if you find them discussing spin in there along these lines that would be interesting

danielunderwood

2:57 AM

@enumaris there's a setting for how many columns/rows to show if it's that

Secret

3:40 AM

[Random]
$e^{1/\ln(e^e)}=e^{1/\ln(1-(1-e^e))}$

$e^{1-\ln(-1e^e)}=e^{1/\ln(1-(1-e^e))}$

Swapnil Das

5:07 AM

@JohnRennie 'What should I do? My audio system isn't working and the notification says 'No Audio output device is installed'? (Mine is Windows 10, HP laptop)

John Rennie

@SwapnilDas does it work if you plug in headphones?

Swapnil Das

Nope.

I tried to troubleshoot, but the problem couldn't be identified :/

John Rennie

Press Windows-R (the Widows key is to the right of the control key) and type devmgmt.msc and click OK. This should open Device Manager.

Expand Sound, video and game controllers. Does it show your sound card there?

Swapnil Das

Intel(R) Display Audio, yes

John Rennie

In my case it's the Realtek high definition audio

Press Windows-R again and run control then open Sound

You should see something like this

Swapnil Das

5:17 AM

It writes (No audio devices are installed)!

Sir Cumference

Howdy

Swapnil Das

Here it is^^

John Rennie

Hmm, well, that explains the problem. Can you post the screenshot of the Device Manager with the sound section expanded to show the sound devices?

Swapnil Das

Here they are^

John Rennie

Aha, you see the tiny orange triangle on Intel High Definition Audio? That indicates there's a problem.

Swapnil Das

5:26 AM

I was just typing that.

John Rennie

Right click it and choose Properties and see what the properties box says

Swapnil Das

The drivers for this device are not installed. (Code 28)

There are no compatible drivers for this device.

To find a driver for this device, click Update Driver.

Copy pasted.

John Rennie

Weird. Did this just suddenly happen? i.e. it was working fine and then just stopped?

Swapnil Das

Yup

When I try to update the driver it doesn't update.

John Rennie

Has your laptop installed any Windows updates? MS released a batch of updates on Wednesay.

Swapnil Das

5:29 AM

I clicked on Update and Shut down somewhere around wednesday, if I remember.

John Rennie

Did the sound stop working after that, or was itworking after that and stopped working at some later point?

Swapnil Das

Perhaps it did suddenly in b/w some work.

John Rennie

It just seems a coincidence that it stopped around the time Windows was updated. Anyhow, I'd try grabbing the drivers from Intel and see if that helps.

Swapnil Das

Sure, thank you.

John Rennie

Right click on the desktop and choose Display properties

Oh wait, no, sorry I'm getting mixed up.

Swapnil Das

5:35 AM

'Display settings'?

John Rennie

Oops yes, display settings. Then when that opens click Advanced display settings

Then Adapter properties

This is just to identify which drivers you need.

Swapnil Das

There's no such option here ?

John Rennie

That's what it looks like here ...

Swapnil Das

John Rennie

Click on System

Swapnil Das

5:40 AM

There's no advanced display setting :/

John Rennie

It should look like this

Swapnil Das

I got adaptor properties!

John Rennie

Aha HD Graphics 520 is what you need.

This is the link

Swapnil Das

Which one should I choose?

John Rennie

Swapnil Das

5:46 AM

Oh yes, there's only window after that.

John Rennie

Try the top one. I can't test it here because I don't have an Intel audio chip, but you basically just download the installer and run it and see what happens.

Swapnil Das

Sure, lemme try.

user157860

@JohnRennie, Hi, John, is it possible to explain briefly what is Einstein's argument for supposing energy 4/4mc^2 instead of the value of 3/4 found by Wien in 1900 and Hasenhorl in 1904?

John Rennie

I didn't know people had suggested $E = \tfrac{3}{4}mc^2$. That seems a bit odd ...

Swapnil Das

I read some infamous article claiming Einstein a plagiarist which mentioned the above fact.

John Rennie

5:55 AM

Einstein's argument that $E=mc^2$ is well documented and a quick Google will find it, though I don't think it helps very much for modern approaches to SR because we would derive this from the energy-momentum four vector instead.

@SwapnilDas the best way to deal with articles like that is to ignore them. There is an unending stream of nutcases making bizarre claims about relativity, some of them members of this site. You have to just breath deeply and slowly then move on :-)

Swapnil Das

@JohnRennie Ofc. Well the download is now complete.

Contents of the file are now being extracted.

John Rennie

@SwapnilDas cross your fingers, toes, and any other mobile body parts! :-)

user157860

6:19 AM

@JohnRennie, it's not nutcases, see wiki:en.wikipedia.org/wiki/…, M=$/3 E/c^2 , which means E=3/4mc^2, right?

as to the argument, I read his original paper many times, but I do not get the necessity of 4/4

John Rennie

@user157860 I didn't mean that Hasenöhrl was a nutcase. I meant the person writing the article claiming Einstein was a plagiarist is a nutcase.

user157860

@JohnRennie, in effect what he did was just modify the value of 3/4 Hasenhorl had found experimentally to 4/4, that's why I wonder what miraculous theoretical argument could modify an experimental datum

John Rennie

@user157860 I'm not familiar with Hasenöhrl's work, but from the brief description Wikipedia gives it seems a completely different approach to the one Einstein used.

user157860

@JohnRennie, can you say briefly what is the compelling argument that forces one to consider 4/4?

John Rennie

6:38 AM

@user157860 The equation isn't $E=mc^2$, it's $E^2 = p^2c^2 + m^2c^4$. You get the simpler expression when $p=0$.

This comes from the requirement that the norm of the energy momentum four vector be invariant.

user157860

@JohnRennie, that's Dirac's equation, and was devised in 1932, in 1905 couldn't have been Einstein's argument. What was it then?

John Rennie

Einstein's argument was heuristic rather than a proof. Indeed Einstein never really proved it. The details of his argument are easily Googlable. I don't remember it offhand because ... well ... it's only of historical interest these days.

user157860

@JohnRennie, OK thanks, Now concerning my question you put on hold, I did not quote sources because it is simply impossible that passive and active gravitational mass be different.

Anonymous

@user157860 It's improbable that your question will be opened without you quoting sources. These terms don't seem mainstream at all

John Rennie

6:54 AM

@user157860 the problem is your question asks why physicists thought there could be a mass that gravitated but wasn't affected by gravity. But as far as I know physicists have never thought this.

So your question doesn't make any sense.

If you cited a paper from Prof Brainstorm saying this you could then ask what was wrong with Brainstorm's theory. But as it is your question is unanswerable.

user157860

@JohnRennie, if they didn't, well how could they envisage that passive mass could be different from active? I am saying that just supposing that 2 such masses could exist in the same particle is absurd.

I just showed the logical consequence of such a distinction

if i remove that part, will you reopen it?

John Rennie

By all means edit the question, but I honestly can't see what you are asking so I'm unlikely to vote to reopen. Note that I didn't close your question and I don't have the power to reopen it. All I can do is cast a close/reopen vote. The question was closed by a moderator.

user157860

7:10 AM

@JohnRennie, I am asking why the possibility of different active and passive g-mass was ever taken into consideration. Is that a bad question?

John Rennie

@user157860 but you have not provided any evidence that active and passive mass was ever taken into consideration. In any case this seems to be about the history of physics rather than physics itself.

In which case you should ask on the History of Science Stack Exchange rather than here.

Anonymous

@user157860 According to the Wiki article on mass, "Active" and "Passive" are simply two different ways to measure the mass. And indeed, according to experiments they turn out to be equal as expected. I don't quite understand your argument as to why there would be any logical fallacy in case physicists considered two different ways of measuring mass.

Anonymous

Your claim that physicists thought that there could be a particle having only active or only passive mass is nowhere to be found

Anonymous

At max, they could have expected that they'd get different values for them. But that didn't happen.

Anonymous

> Although some theorists have speculated that some of these phenomena could be independent of each other

Anonymous

7:18 AM

Now that might be an interesting topic to ask about ^ (although you'd have to be careful enough to keep it mainstream)

user157860

@Blue, the way mass attracts and is attracted was stated 3 centuries earlier, so, what is the point of two different measures? Distinguishing the two, IMHO, implied that they had considered such possibility, was I wrong? You seem not to exclude that

Anonymous

"Distinguishing the two, IMHO, implied that they had considered such possibility"

Anonymous

No, the second part of the statement doesn't logically follow from the first.

Anonymous

Which is why people are asking you to provide references.

Anonymous

What you could ask instead is: "Wikipedia says 'although some theorists have speculated that some of these phenomena could be independent of each other...'. Why exactly did the theorists feel that passive gravitational mass and active gravitational mass might be produced by different phenomena?"

Anonymous

7:31 AM

As far as I know, asking for clarifications on existing Wikipedia articles is generally on-topic. So probably that question would be allowed.

Anonymous

@user157860 I've suggested an edit to your question on the main site. Check

user157860

7:54 AM

@Blue, thanks also for the upvote

Anonymous

@user157860 No problem. Hopefully, it will get reopened in the current form once it gets through the review queue

Anonymous

@user157860 Yay! QMech reopened it :)

user157860

@JohnRennie, back to mass/energy: if they are exactly equal, mass can never go to infinite, since KE is 1/2 c^2 and if the added mass/momentum has same value, the sum of 1/2+1/4... goes to 2 when the speed is c. Is that right?

btw, thanks for reopening the question

John Rennie

8:16 AM

@user157860 KE is $\tfrac{1}{2} \gamma m c^2$

user157860

@JohnRennie, isn't it $E_m \times (\gamma -1)$

John Rennie

You start with $E^2 = p^2c^2 + m^2c^4$, which gives the total energy. $p$ is the relativistic momentum $p = \gamma m v$.

Then $E - mc^2$ gives you the kinetic energy. However some algebra shows the KE is just $\tfrac{1}{2}\gamma m v^2$.

(I've just realised I put $c$ instead of $v$ in my previous equation - oops)

And since $\gamma \to \infty$ as $v \to c$ that means the KE goes to infinity.

Physics Meta

9:46 AM

-1

Q: Is a question about calculators in SI format according to "what topics can I ask about here"?

I put the following question and immediately received down-vote, thus, deleted it. Could be question is off-topic, could be someone is happy trigger of down-votes. Yes, I known is a strongly practical question but, who doesn't uses a calculator? And I do not see any one that can be configured in...

support

user157860

11:41 AM

@JohnRennie are you sure some algebra works? if v = .0866 c your formula gives (1/2*2*).75, whereas the other gives (2-1=) 1 ($E_m$)

CooperCape

If an observer on earth sees an object flying at at considerable speed ($v$ close to $c$), is the time the observer sees the object take to travel a distance $d$ just $d/v$. I don't see that time dilation/length contraction come into this but I just want to check.

John Rennie

@CooperCape correct

CooperCape

@JohnRennie Okay thanks :).

Jim

11:57 AM

<3 @JohnRennie

What's up in the h bar, physics phriends?

John Rennie

@Jim it tends to be quiet at this time of day

In fact I have to go now so it will be even quieter

Jim

@JohnRennie yeah, this is one of the tragedies of Jim. The h bar is quiet when I've got lots of free time and busy when I have none.

but tragedy+time=comedy. So, really, it's quite funny

Avantgarde

12:51 PM

Hi

Semiclassical

@JohnRennie uhh

KE = E-mc^2 = (gamma-1)mc^2

Emilio Pisanty

@Semiclassical yeah, that looks much more on-target.

John Rennie

@Semiclassical hmm ...

Semiclassical

1:07 PM

The two expressions will agree in the non-relativistic limit but not the ultra-relativistic limit

bolbteppa

@ACuriousMind pages 3 & 4 of spiral.imperial.ac.uk/bitstream/10044/1/31965/2/swieca.pdf link the 11D susy algebra to conserved charges coming from the sugra eom and introduces p-branes as solutions carrying those charges then brings up some issues about topology

2

A: In M Theory, what does the M5 brane look like?

M2 and M5 branes are fundamental objects of M-theory. You can see this from the superalgebra of $D=11$ SUGRA, which is the low energy limit of M-theory. Moreover M2 and M5 couple respectively electrically and magnetically to the RR 3-form $A_3$ of $D=11$ SUGRA. In the compactification limit for...

John Rennie

1:34 PM

(Deleted)

Yashas

What is the intuition for nuclear spin?

Avantgarde

@Yashas Intuition?

Yashas

If the number of neutrons and the number of protons is both odd, the nucleus is said to have an integer spin. Is the spin property different for different particles? i.e. why can't overall half spin of neutrons and overall half spin of protons cancel out?

@Avantgarde Yup. What is it? (sorry for the edit beeps)

Why do spin states split into different energy levels when a magnetic field is applied? Why favor one particular spin over the other? Shouldn't there be some symmtery?

tarit goswami

1:49 PM

It is written in a note of QM that "In classical mechanics . . . observables that can't be represented using position and monmentum are irrelevant". Can anyone explain me why so?

Avantgarde

@Yashas That's ok. My volume was at 0% lol

@Yashas I suppose for the same reason that a charge moving in a circular trajectory (that is, with some angular momentum) experiences force when magnetic field is applied.

Spin is just another type of angular momentum.

@taritgoswami In classical mechanics?

bolbteppa

@taritgoswami because the state of a mechanical system is completely described by the position and velocity of the particle at each moment

tarit goswami

@Avantgarde The full paragraph is - "In classical mechanics, the configuration or state of a system is given by a point $(x, p)$ in the space of coordinates and momenta. This specifies everything else in the system in a fully deterministic way, in that any observable $Y$ that can be expressed as $Y (x, p)$ can be found, and any that cannot is irrelevant."

Avantgarde

Yeah, phase space (dynamics) of your classical system is determined by coordinates and momenta.

Yeah.

Classical mechanics is just that.

In classical mechanics, if you don't know what the coordinates and momenta are, you don't know anything about the system. What else would possess nontrivial information about the system's dynamics?

tarit goswami

@bolbteppa Can you explain what you meant by "completely described"

@Avantgarde What else would possess nontrivial information about the system's dynamics?, yeah I am asking that question. Is there any explanation why we don't care such observable that can't be explained using $(x,p)$?

Avantgarde

2:05 PM

@taritgoswami I can't think of any other 'coordinate' that could supply any possible missing information.

bolbteppa

@taritgoswami in classical mechanics, if you know the coordinates of a particle at one instant of time, that's not enough to know where the particle will be at the next instant of time, to know that one must also know the velocity of the particle, knowing these two things is enough to know where the particle will be at all future times, this is why the equations of motion are in terms of acceleration, because acceleration is completely determined by the position and velocities

Avantgarde

If you know where the particle is, and how it's moving - that's it.

tarit goswami

@bolbteppa Got it :-) Thanks

Anonymous

2:59 PM

@bolbteppa I'm not sure knowing position and velocity are sufficient to know where the particle will be at the next instant of time. In classical mechanics, aren't there edge cases where the force depends on higher order derivatives of position, for example - radiation reaction?

Anonymous

Afaik, we consider position and velocity to be sufficient because in most cases Newton's second (which is a 2nd order ODE) govern's the particle's trajectory.

Anonymous

But there's no axiom in classical mechanics which a priori bans trajectories being governed by higher order differential equations

John Rennie

It's wet and wild in Chester today

Anonymous

I suppose @taritgoswami's question was a bit different: In classical mechanics, position and momenta are well defined at all times and that's sufficient information to characterize the system. There's no additional information you need to characterize a classical system. However, it is not true that knowing position and velocity at one instant is sufficient to determine the position and velocity at the next. You don't know the whole trajectories beforehand.

Anonymous

For that, you need to solve the equations of motion (which are 2nd order differential equations - Newton's law generally but not always)

Anonymous

3:15 PM

@JohnRennie Eh, power cuts suck! But I do enjoy storms (I get to stay home and eat good food) :D I, on the other hand, am sad that a cyclone is my hitting my neighboring state (Odisha) but not West Bengal :P

John Rennie

@Blue I think anyone who has experienced a cyclone would regard the typical UK storm as a bit windy with some rain :-)

I have to walk through the storm to the pub tonight, but that shouldn't be too bad. It's walking back from the pub through the storm that will be trickier :-)

Anonymous

Hehe :P We've had 2-3 cyclones in West Bengal is the past 10 years. There's usually not much damage in the urban areas (and it's sorta like a stronger version of a storm) but the rural areas do suffer :/

Anonymous

@JohnRennie Lol

Anonymous

Ask one of your non-alcohol-drinker friends to drive you back XD

Anonymous

There's always one such friend in a group, no? :D

bolbteppa

3:20 PM

@Blue the reason you can even set up eom in the first place is because solutions of second order ode's are curves with specified initial conditions on where the curve is and how the tangent to the curve starts and looks at all future times, if we ever needed more than positions and velocities then the phase space formulation of classical mechanics would be incomplete, and there are other contradictions you run into if you try as well

and things like the Abraham-Lorentz force I think are related to 'changing forces' i.e. changing electromagnetic fields, involving behavior at more than one point

John Rennie

@Blue nope, we're all drinkers here :-)

It's only about a mile.

Well, it's a mile if you're walking in a straight line. For me after an evening in the pub it's $\sqrt{n}$.

4

Sir Cumference

Mathematica is a spiteful program

@CooperCape If you're taking special relativity, MinutePhysics' videos are fantastic

youtube.com/playlist?list=PLoaVOjvkzQtyjhV55wZcdicAz5KexgKvm

Not a substitute for a textbook, but excellent for intuition

Anonymous

@bolbteppa Sounds interesting but does that mean we cannot represent the path of a classical charged particle (taking into account the AL force) in phase space formulation? I haven't read such a thing before. Do you have any reference?

Anonymous

And what are the other contradictions?

Anonymous

The point I'm trying to get at is that knowledge of $x$ and $p$ for all time are sufficient to uniquely define/characterize a classical system afaik

Anonymous

3:32 PM

In case you're calling it incomplete, what's the extra information that you're hinting at which is needed (apart from $x$ and $p$) to uniquely characterize the system?

bolbteppa

@Blue I'm saying the phase space formulation can account for it, it has to

3

Q: Why doesn't phase space contain acceleration/forces?

I'm watching some Physics lectures on the internet by Leonard Susskind: http://www.youtube.com/watch?v=pyX8kQ-JzHI&feature=BFa&list=PL189C0DCE90CB6D81&lf=plpp_video In this lecture, and also in Wikipedia and other places phase space is described as the space of all the states we need to know to ...

classical-mechanics forces determinism phase-space

74

Q: Why are there only derivatives to the first order in the Lagrangian?

Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded? Is there a good reason for this or is it simply "because it works".

lagrangian-formalism variational-principle action

Anonymous

Thanks, I'll read them

bolbteppa

Insane, you can break Coleman-Mandula in ssb because the resulting symmetry groups may not appear as symmetries of the s-matrix

CooperCape

@SirCumference does it cover Lorentz transformations etc.?

I’ll check it out in a second currently in a lecture

Sir Cumference

@CooperCape Yup, in a pretty enlightening way

CooperCape

3:43 PM

Nice one that’ll be handy thanks

Sir Cumference

@CooperCape Np. Its length contraction/time dilation distinction was a ton clearer than my book.

enumaris

4:17 PM

hmmm

curiosity

Hi everyone, im struggling a bit with scattering theory

we know that we can write the solution of the schroedinger equation as a product of spherical harmonics and a radial function, which satisfies the bessel differential equation

this gives us an asymptotic behavior of $\psi(\vec x) \sim \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta)$

then, various literature claim that there is also another asymptotic form, namely $\psi(\vec{x}) \sim e^{ikz} + f(\theta) \frac{e^{ikr}}{r}$ , which can be obtained from the lippman schwinger equation

then, the two forms are set equal to each other to obtain formulas for the $A_l$ coefficient and the phase shift

but what i dont understand is why we should set these two forms equal to each other

because isn't it true that the plane wave part of the wave function doesnt spread over the whole space, instead, this decomposition should only be valid for $\theta =0$ shouldnt it?

in my opinion we should only set $ f(\theta) \frac{e^{ikr}}{r}= \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta)$ because for any $\theta$ other then 0, the plane wave part has vanished hasn't it?

because in a real scattering experiment, the detector can only detect the transmitted (unaffected) plane wave right behind the target, namely for $\theta =0$

bolbteppa

4:33 PM

@curiosity the second asymptotic form you give is derived in e.g. Griffiths more directly than Lippmann, it comes from a plane wave $e^{ikz}$ scattering off a scattering center and the resulting term being the $f(\theta)$ part, with the wave function a linear combination of these, this is where the $\theta$ comes from and why you can set both forms equal to each other

curiosity

@bolbteppa but isnt it true that in a real physical situation, the plane wave cant extend to infinity in all directions? if we were performing a scattering experiment in the lab, we would only expect the see the contribution of the plane wave for $\theta =0$ and not for the other angles wouldn't we?

we are shooting electrons in form of a plane wave out of a cannon but this stream is confined to a narrow beam right?

not extended over all space and therefore cant reach the other angles

bolbteppa

For all intents and purposes you just treat it as a free particle incoming from infinity and so can treat it as a plane wave since your main concern is the scattering, and in qft it becomes even more important theoretically to work with incoming free particles, you can basically just derive it directly by hand-waving as in here physicspages.com/pdf/Griffiths%20QM/…

curiosity

4:49 PM

but i don't see how the plane wave model is justified for any angle but $\theta =0$ . if we are shooting electrons with an electron gun, and detect the scattered particles, we dont expect any electrons going directly from the gun to the detector

bolbteppa

5:02 PM

Not sure what you mean, maybe the picture on slide 15 will help tcm.phy.cam.ac.uk/~bds10/aqp/lec20-21_compressed.pdf (eye is the detector)

curiosity

yea in that picture, why should the plane wave reach the eye? only the radial wave can reach it

because the eye is like a meter away from the target, and if the angle is greather then zero, the plane wave, which is the narrow electron beam, cant reach it

bolbteppa

In the picture, the plane wave should be infinite

Otherwise it's not a plane wave

curiosity

yes but in reality it isnt

bolbteppa

So you need to use a different function

Abcd

@Blue BTW the height will be independent of the focal length of lens because magnification = v/u

curiosity

5:07 PM

a plane wave would be the result of an infinitely wide electron gun

it cant be true that the results of scattering theory are only valid for such an infinite gun?

bolbteppa

You would need to solve the Schrodinger equation for the particle-gun system to determine the exact wave function coming out

and then call that the incoming wave

It's a model to treat it as a plane wave

Anonymous

@Abcd $m = y_{\text{image}}/y_{\text{object}} = v/u$

Abcd

@Blue Y image doesnt depend on focal length, does it?

curiosity

but what is this model worth if it isnt applicable to any real situation?

why does every textbook teach it?

Anonymous

@Abcd But $v$ does. $v$ and $u$ are related by $f$. Or else you'd have to place the screen at different distances for different lenses

curiosity

5:13 PM

im totally fine with the part of the plane wave model that sais that the wavenumber is sharp

but why should we claim that the plane reaches the detector directly? this would only be true for a very large gun and not for a narrow beam

bolbteppa

How do you know an electron coming out of an electron gun will go straight along the small region, how do you know it wont fire at an angle and go completely outside the small region and never go near the scattering center?

curiosity

and we are always interested in narrow beams afaik

because i have seen electrons shooting out of a gun

and they go straight

its the beams that we use for thing like this: en.wikipedia.org/wiki/Cathode_ray_tube

Anonymous

@Abcd $y_{image}$ does, in fact, depend on the focal length as I explained above. For obtaining an interference we shift around the screen to get the best focused image, usually

Anonymous

So your $v$ needs to change depending on the lens you are using

Anonymous

$u$ and $y_{object}$ are constants on the other hand

Anonymous

5:22 PM

$m$ is a function of $f$ btw

bolbteppa

Well, the idea is to treat it as though it was a particle coming from infinitely far away, this can model the uncertainty in the trajectory due to experimental error, or be a good approximation if the distances are large, more specificity than that and you're missing the point of the difference between scattering problems and just normal full on problems where you solve the Schrodinger equation directly

curiosity

yes infinitely far away is fine, but why would infinitely broad be a decent model for an electron gun?

are you saying we cant apply the plane wave model to an experiment with a narrow electron beam?

with regard to the size of the target, the electrons come from infinetely far away

but not with regard to the detector

and with regard the the size of the target, the plane wave is infinitely spread out

with regard to the position of the detector, the spread of the plane wave is narrow

and when we are making the asymptotic expansion, we write the wavefunction at the place of the detector

bolbteppa

5:39 PM

I'm not sure what that means, but in a scatting problem you have a wave function representing the incoming particle, and a wave function representing the outgoing particle, by linearity you must include both in the total wave function of the system, what you choose as your incoming wave function is up to you,

clearly people choose plane waves for simplicity and since it can roughly model plenty of situations, in a scattering experiment you want your incoming and outgoing wave functions to exist at great distances from the center so that you can basically use incoming plane waves and outgoing waves which take a special form ($1/r$ behavior, e^{ikr}$ plane wave behavior, $f(\theta)$ non-trivial dependence),

this lets you focus on the scattering without extra complications arising from a complicated choice of incoming wave

Ah, another thing, the first form of the wave function you gave is also just a plane wave

[That should be "($1/r$ behavior, $e^{ikr}$ plane wave behavior, $f(\theta)$ non-trivial dependence)" above]

curiosity

we want the incoming plane wave to exist over the whole space where the potential is nonzero, (atomic lenghtscales) but we dont want it to exist at the point of the detector (meters)

isnt that true?

you mean $\psi(\vec x) \sim \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta)$ ? thats not a plane wave

bolbteppa

Wait sorry

Yeah you're right wrong thing

tarit goswami

@Blue thank you, I think to get solution of a 2nd order diff equation, initial value of the variable(here position) and 1st derivative(velocity) is sufficient. Correct me if I am making any mistake.

Anonymous

@taritgoswami Yes, that's correct

tarit goswami

But, my actual question was, why we are treating the observables, that we can't express using x and p, as irrelevant

Anonymous

5:50 PM

@taritgoswami Which observables?

curiosity

@bolbteppa do you see what i mean about the region where the incoming wavefunction should be nonzero? this is really my central point

Anonymous

Example?

tarit goswami

Sorry I don't have any example in hand now :-( but that leads to introduction of wave function in QM as written in the note

Anonymous

Observables in classical mechanics and quantum mechanics are defined differently. Don't confuse them

Anonymous

In classical mechanics, $x$ and $p$ are necessary and sufficient

bolbteppa

5:53 PM

Yeah I see what you mean but I don't see why it matters that it originally was shot from an electron gun at a certain place rather than another if you are going to treat the incoming wave as a free particle

tarit goswami

Yeah that's the cause

@Blue Is there any argument that explains this?

curiosity

it matters because i want to describe an experiment with an electron gun

i dont know wheter i understand you correctly

bolbteppa

If you want to use a different wave function as your incoming particle reflecting the fact it's constrained to a certain region you can try I'm sure, what would the wave function look like if the particle was bounded within a certain region until it comes near the scattering center

curiosity

i think thats the number one application of scattering theory isnt it?

bolbteppa

The number one application is to ignore the source of the particles and just treat the incoming particle as a free particle

curiosity

5:56 PM

what im saying is: every textbook teaches this method with the plane wave and talks about the rutherford experiment and so on

bolbteppa

Yeah

curiosity

so i suspect the method to be applicable for situations like that

Anonymous

@taritgoswami Any other observable in classical mechanics can be expressed in terms of $x$ and $p$. Isn't that reason enough?

tarit goswami

@Blue the cause for defining wave function is that, we can't fully explain the observables with x and p as we have done in classical mechanics

As written in note

curiosity

why would we ignore the source of the particles?

we wanna find out the structure of the atom or some other target

for that purpose, we shoot a narrow beam of electrons on it and measure the scattering amplitude

why should the standard texbook treatment of scattering theory not be applicable to this situation?

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