Ask any question you want about Physics

a reply to: BASSPLYR

These things are where i go cross eyed, SU(1), SU(2),and SU(3) are all Lie Groups, I'm not going to pretend to understand it anymore, as It was a long time since I studied it. They are useful in particle physics as they form the basis of many calculations/theory. This includes forming conjugations, symmetries, and spontaneous symmetry breaking

Best thing I personally can do is throw this link your way... for the next lecture, you can just change the numbers

hepwww.rl.ac.uk...

It would take me a while to look at them and understand it, I did a similar Group theory course previously and not sure how i passed to be honest lol... Whenever i see Gamma matrices it still makes me want to cry

a reply to: ErosA433

thank you very much Eros! I'll check out the link and see if I can make any sense of it. the SU(2)'s in particular is what I'm trying to focus on. I'm thinking I've got my work cut out for me this weekend with this subject. lots of rereading the same paragraph over and over going "wait! what? maybe if I read it a fourth time I'll get it.."

I got a question

Say you have 2 or more stars in a binary, trinary or whatever configuration system. One of those stars goes nova, can the other, regular stars in relatively close proximity survive that? or what happens to them?

I'm going to assume something like a neutron star or magnetar would be relatively untroubled, but I'm talking about regular, main sequence stars. Their stability is tenous at best, so when faced with a massive external pressure wave, something must happen to them?

sorry to overload you guys, especially when you're trying to enjoy the weekend. but I gotta another group of questions. going back to QED.

can someone describe in layman's terms Cavity QED?

What exactly do they mean when they (smart phd people ) in regards to Cavity QED use terms like "reflective cavity." And, "cavity resonant with the atomic transition."

I understand in general CQED is about bombarding atoms inside a reflective chamber or cavity with photons (usually microwaves) while simultaneously dialing the resonance in so that it's in accord with the atomic transition, whatever that is, and being able to manipulate properties of the atomic material in said cavity. to make quantum transistors or something? (ok, probably really off on that last part for all I know)

so.....manipulate atomic properties? like what do they mean, altered or manipulate in what way?

I understand that this process is being looked into for quantum computing.

finally, on a related subject that I'll probably need to understand to get this whole CQED thing is how the Jaynes-Cumming Model works and what exactly are Vacuum Rabi Oscillation Cycles?

any help on this topic?

Can anyone explain more about these 2 numbers, and why they threaten the end of Physics?

The two most dangerous numbers in the universe are threatening the end of physics

a reply to: MasterAtArms

Depends upon many factors such as mass and proximity,

The first thing that would happen is that they would get a huge burst of neutrinos. About 99% of the energy of a Supernovae is in neutrinos. Without looking at reaction types here, that could have some detrimental effect on the stars nearby, in terms of fusion at the core, but not enough to really cause say a failure of fusion. Next would come the photon and material energy as material is pushed away.

The photons would cause rapid heating of the outer envelope of the stars, this would cause them to puff out i think, so you would expect the star to become eggshaped or puffy facing the nova... all this wont matter too much (since most material wont have time to puff out anywhere) as the ejecta would smash into it.

Now, this is where size matters.

A large star might have its outer regions ripped off from the shockwave, but otherwise be unchanged, it might loose many solar masses if its a large star, possibly even change spectral type.

A smaller star however might have enough material removed that it simply dies itself... might be left with a puffy cloud that cools into a dwarf star.

Interesting question, and I hope some others weigh in too, these are my own ideas based just upon my old Astronomy notes

a reply to: oletimer

These two numbers are not dangerous or threatening the end of anything.

1) Higgs field strength.

Simple, from the person they interviewed, he believes the field should be either off... or on... and if it is on, you go and look at your theories tuned to represent all the data we have relating to higgs coupling and your number is huge.

You look at a particle believed to be the first order excitation of the higgs field and you look at what field strength pops out of that... and its very small.

Meaning : The theory is missing something, or the data is missing something.

2) Dark Energy

It is a rather large unknown, and remains the subject of a lot of speculation and interest in the physics world today. Thought to be a measurement of the elasticity of the universe, it is essentially the energy contained within pure vacuum much like problem number 1, Theorists worked out the energy that should be contained within vacuum based on our current theories and... the number is huge... BUT when you look at the data from astronomy, it is actually very small...

Meaning : The theory is missing something, or the data is missing something.

Cliff knows his stuff, though i think this is kind of a case media misrepresenting the context of what he said. What he really said is that, there are a few places we worry about in particle physics today because theory and data do not match, and these are topics of hot study... BUT these could be areas of physics which defy study in our own lifetimes and such seem like impossible tasks.

The media latched on and made it sound like "End of the physics world, we already know everything we can, lets pack up and go home"

Which is quite simply... not the case.

originally posted by: Peeple
a reply to: Arbitrageur

I know it wasn't physics, but still a big:
Thank You!

I didn't know how helpful my answer was but you're welcome, thanks for the feedback.

originally posted by: greenreflections
In case where single photon is released from photon gun in space, would my wide big, enormous in size detector catch that same single photon or it will arrive as a wave with the most energetic impact being on bulls eye?

That depends partly on the design of your photon gun, but let's call it a laser and look at the measurements of a student who studied where photons from the laser arrived:

laser.physics.sunysb.edu...


The photon has a certain probability of being detected at various distances from the axis or focal point of the laser which is a Gaussian or you may have heard it called a "bell curve". It's most likely to be detected at the central peak of the "bell curve" but and you can see it can be detected in many other places, but the further from the center, the lower the probability of detecting it.

If I can interpret one of your questions as asking if this distribution gets wider at a greater distance from the laser, the angles remain the same and as a result then yes the further from the laser the wider this distribution.

I would think that photon released will start to spread wider as it travels and detector reacts on its arrival as the wave, meaning multiple cells of detector register it with most impact at the place of the detector where I aimed it.

Is there experimental evidence to support this? In experiments I've seen, one photon gets one detection at one "cell" of the detector, so I've not seen evidence to show multiple cells of a detector are detecting the same photon.

While the wave equation predicts the probability of finding the photon in a certain area spreads out with distance from the source, I don't know if this means the photon itself spreads out as I don't claim to understand what the wave function truly represents in reality before the photon is detected.

you will say that you did not understand any of what I asked. Am I right?))

I understand a lot of your question but one thing is not clear. With photons you have electromagnetic waves where the electric and magnetic fields vary, and you also have the wave function which is used to predict the probability of where the photon is most likely to be detected. When you say "on its arrival as the wave" I don't know if you're referring to electromagnetic waves or the wave function.

When the photon is finally detected, all of its electromagnetic energy is measured in the one cell that detects it as far as I know but if you have experiments showing otherwise I'll take a look. I can't say how spread out the EM radiation was before that because we can't measure that without affecting what we're trying to measure. I do think the wave function is spread out as it's an accurate predictor of where the photon is likely to be detected, but the physical meaning of the wave function has been debated.

Does the quantum wave function represent reality?

Using the wave function, physicists can calculate a system's future behavior, but only with a certain probability. This inherently probabilistic nature of quantum theory differs from the certainty with which scientists can describe the classical world, leading to a nearly century-long debate on how to interpret the wave function: does it representative objective reality or merely the subjective knowledge of an observer?

If it's been debated for nearly a century you can bet the answer to the debate is not simple.

a reply to: greenreflections

In case where single photon is released from photon gun in space, would my wide big, enormous in size detector catch that same single photon or it will arrive as a wave with the most energetic impact being on bulls eye?

I would think that photon released will start to spread wider as it travels and detector reacts on its arrival as the wave, meaning multiple cells of detector register it with most impact at the place of the detector where I aimed it.

first of all... there is no such thing as photon in real life.
a photon, is an mathematical construct, that is used to calculate the "force interactions" in a debility theory of someone...

Arbitrageur said...

The photon has a certain probability of being detected at various distances from the axis or focal point of the laser which is a Gaussian or you may have heard it called a "bell curve". It's most likely to be detected at the central peak of the "bell curve" but and you can see it can be detected in many other places, but the further from the center, the lower the probability of detecting it.

YEAH... (face palm)
excuse him... he is just parroting something he read on the net

first you need to understand is the fact, that any detector is made of atoms.
atoms with electrons...

how it works is, or better said, the MS theoretical explanation is the "photoelectric effect"
explained in short...
radiation is quantized, energy is transferred by some mystical construct called photons

forget it !

the moment an charged particle moves, it radiates EM field.
E instantly... M with the speed of C
soo...
if the electrons in the detector are reacting to a certain radiation vector change in M field ( E is insignificant at those distances ), they get "pushed" into a different direction they would usually move ( vibrate ).
now, if the wave comes in, the electrons in the detectors "have to be lucky" to change the path in such way that they fly off of the atom, otherwise they don't.
this elections in the "right condition " is the moment the detector will give you a read. ( potential difference )

so.. if you have a plane with billions of the detectors, like a screen used by the double slit experiment, and a EM wave front comes, not every one of the atoms is in "the position" to react properly ( instantly ) to give you a "dot on the screen" just some...
they call it, the WAVE collapse ( face palm )

there is no point like particle moving with the speed of C form the emitter to the receiver !!!
change in EM field does that, but NOTHING physically travels from A to B
there is also no energy transfer, there is just a path correction

a reply to: Arbitrageur
2 Jan 2016

No reason to throw insults at me when you're the one who doesn't understand the subject matter.

insults ??

I'm talking facts, you are parroting BS from the net !
Do you read it and think about it or just parrot ???

Yes the CMB photons all move at C, but the way to determine if you're at rest with respect to the CMB isn't by comparing your speed to theirs, it's by looking in various directions for red or blue shift.

here wiki

To determine the redshift, one searches for features in the spectrum such as absorption lines, emission lines, or other variations in light intensity. If found, these features can be compared with known features in the spectrum of various chemical compounds found in experiments where that compound is located on Earth

are you telling us we know the features in the spectrum of various chemical compounds found in the Big Bang ???
is there any reference to it ???

originally posted by: Arbitrageur

??

you are kidding right ?

this is just some dots moving around !

have you ever seen gas particle moving around ?
...

YEH.. that's what I think !

a reply to: greenreflections

Man, from what I was talking about above, QM does not have to be what you are arguing about. My thinking was such that it does not matter what exactly holds rubber ball in the original geometrical shape in one piece. Name them any way you want but these forces that keep the shape of rubber sphere in original round shape are existing albeit not important in how it is achieved in my case.

yeah.. sorry...
inertia...
change in position is a straggle...

a reply to: Arbitrageur

KrzYma was also asking questions about some early experiments which used emulsion stacks to find particle tracks so I don't see any logic in trying to explain away LHC detector technology without also explaining tracks in emulsion stacks. However I'm still not sure KrzYma knows what an emulsion stack is either.

I don't know what emulsion stacks are ??
are you insulting me ??

are you suggesting LHC is using emulsion stacks ???
NO... and you know they don't !

let me tell you something, and if you listen you may get it right... ( yeah... the prophet )

all detectors are made of matter... atoms
they react to EM fields.

billions of protons have a "certain" potential charge

if you push billions against billions or even against whatever the target is... there is only one outcome
EM disturbance... ripples

look at this vid and how electric field is acting... even without "electron transfer"
and think about LHC detectors...


as you hopefully can see in this vid about 0:50 to 1:10 the electric field is acting at a distance
don't tell me the LHC detectors works in a different way then measuring potential difference, there is no other way

... there is no mystical forces, NO woodo, no ghosts !

ripples is what they detect... nothing significant at all

waste of "potential"

originally posted by: KrzYma
first of all... there is no such thing as photon in real life.
a photon, is an mathematical construct, that is used to calculate the "force interactions" in a debility theory of someone...

Maybe you're thinking of "virtual photons", they might be mathematical constructs for all I know. But photons have well-studied observable properties so I don't know why you'd call those mathematical constructs, unless as I suspect you're not one of the people who has "well studied" said properties.

originally posted by: KrzYma
are you telling us we know the features in the spectrum of various chemical compounds found in the Big Bang ???
is there any reference to it ???

If you want to know absolute redshift of a distant galaxy, it's nice to know the spectral shift, but the CMB is a different story. The CMB is uniform so when we see it "redder" in one direction and "bluer" in another that's due to our motion. It's all very red shifted, but slightly less redshifted in the direction we're moving toward and slightly more redshifted in the direction we're moving from.

Also note that one paragraph from a wiki article is usually not a comprehensive description of the phenomenon, and it's certainly not in this case.

originally posted by: KrzYma
don't tell me the LHC detectors works in a different way then measuring potential difference, there is no other way

... there is no mystical forces, NO woodo, no ghosts !

ripples is what they detect... nothing significant at all

waste of "potential"

Asked you before... please describe how you think LHC detectors operate and how the signals appear...

lets compare notes... i did work on a device very similar to the tracker in the ATLAS detector, maybe we can compare notes

a reply to: ErosA433

There's a lot of different detector so where to start.😞

a reply to: dragonridr

Well roughly broken down there are only about 5 fundamental types of detector used in ATLAS, the breakdown might have lots of different names, but in the end they are all similar types in different configurations.

The point really is wanting to know this... how does a typical detector at the LHC 'detect' a high energy gamma

From nearest the interaction point we have
The Inner Detector composed of
- Pixel Detector
- Semiconductor Tracker
- Transition Radiation Tracker
Calorimeter system composed of
- Liquid argon electromagnetic calorimeters
- Tile calorimeters
- Liquid argon hadronic end-cap
Muon Spectrometer

I can break these down as follows, The inner detector are various forms of silicon strip like detectors in different form, but very similar operation principle. The transition radiation tracker is an interesting one

The calorimeters all operate on the same principle of slowing a particle to have it deposit energy within the material which can be read out in different ways...

So that might be vague, but is designed to be vague, soooo not aimed at you there dragonridr, but at KrzYma...

A high energy Gamma is produced, how is it detected?
A high energy Electron is produced, how is it detected?
A high energy Hadron is produced, how is it detected?

a reply to: Arbitrageur

laser.physics.sunysb.edu...



The photon has a certain probability of being detected at various distances from the axis or focal point of the laser which is a Gaussian or you may have heard it called a "bell curve". It's most likely to be detected at the central peak of the "bell curve" but and you can see it can be detected in many other places, but the further from the center, the lower the probability of detecting it.

thanks for your answer and diagram.
with the diagram I dont understand completely what ''detector position'' means. It is relative to what? Just the distance away from source of photon emission? And what (mils) means? Over all it looks to me that it represents light intensity dependence of the distance of detector from the sourse. Or detector moving left or right of the center?

and you said ''most likely to be detected'' when axis ''Energy intensity'' on the diagram clearly says just what it does. It is distance correlation with light intensity and not chance of photon to be detected. )) I know, I am repeating myself..))

Advise how I read that diagram correctly, please.

originally posted by: KrzYma
a reply to: greenreflections

Man, from what I was talking about above, QM does not have to be what you are arguing about. My thinking was such that it does not matter what exactly holds rubber ball in the original geometrical shape in one piece. Name them any way you want but these forces that keep the shape of rubber sphere in original round shape are existing albeit not important in how it is achieved in my case.

yeah.. sorry...
inertia...
change in position is a straggle...

and while changing position it gets potential energy too. So, inertia should some how be equal to potential energy. No?

originally posted by: greenreflections
with the diagram I dont understand completely what ''detector position'' means. It is relative to what? Just the distance away from source of photon emission? And what (mils) means? Over all it looks to me that it represents light intensity dependence of the distance of detector from the sourse. Or detector moving left or right of the center?

I'm not sure why you couldn't just click the link above the diagram which answered all your questions, I really have to copy/paste that for you because you don't want to click the link?

laser.physics.sunysb.edu...

The intensity distribution of a typical laser beam was measured by moving a photodetector on a translational stage in increments of mils (1/1000th of an inch). The detector was mounted behind a hair-sized (100 micrometers diameter) pinhole, which was located 145 cm away from the output end of the laser. The laser used was a 10 mW Melles-Griot He-Ne laser (wavelength 632.8 nm). The light intensity is proportional to the electrical current produced by the photodetector, which was measured with a milli-ammeter.

So the distance from the source is constant and the detector is moved from side to side. It shows the light is already spread out at 145cm from the source, and another diagram at that source shows how that "bell curve" continues to spread out at greater distances from the laser (it doesn't spread out for a short distance but after that the angle of spreading is about the same):

and you said ''most likely to be detected'' when axis ''Energy intensity'' on the diagram clearly says just what it does. It is distance correlation with light intensity and not chance of photon to be detected.

In this case I have no reason to suspect there is any significant difference between the two. How do you get more intensity without more photons? You don't. The energy of each photon is known with a monochromatic light source as E=hf, and it's based on the color (or frequency, f) of the light. So you can calculate the number of photons from the total energy for each cell on the bell curve graph by knowing the energy of each photon, though you might need to allow for the detector being less than 100% efficient due to some small losses.

Then if you want to know the probability of finding the photon at any "cell" in the histogram, divide the area of that "cell" by the total area under the histogram and that's your probability from the experiment. You could make it a little more accurate by calculating the actual Gaussian distribution and using that to calculate the areas/probabilities instead, however that graph is a fairly smooth looking histogram so you wouldn't gain a lot of accuracy for that extra work. You would gain some accuracy by considering a radially symmetrical 3-D model instead of this 2-D slice but for conceptual purposes it's probably easier to work with the 2D slice (which is not exactly 2D because of the pinhole used in the experiment).

In order to detect individual photons, you need a very sensitive detector, and a very attenuated light source. Such techniques have been employed in double slit experiments to eliminate the possibility that the particles are interfering with other particles, by sending the particles through the slits one at a time. You can see an example of such an experimental setup here:

Single Photon Interference

That setup actually counts individual photons.