The h Bar

7:00 PM

To really do the calculation, though, you have to have the complete diagram. Or a good (mathematical) justification for leaving out part of it.

Arc676

So, just to summarize, can you have a diagram that starts or ends with a photon and tell a physicist that the mass of the particles used to create the photon are in its kinetic energy?

David Z

There are things called factorization theorems which justify how you can draw parts of Feynman diagrams and do certain kinds of meaningful calculations with them.

ManishEarth

@DavidZ oh, right

David Z

@Arc676 I'm not quite sure what you mean there

ManishEarth

@Arc676 that would be a partial diagram then, as DZ says

@DavidZ think pair production. Is it an OK diagram?

Arc676

7:02 PM

For example, if I draw a diagram with an electron-pair that annihilates into a photon, can I say that it's complete by saying that the mass of the leptons is in the photon's energy?

David Z

No

Arc676

Or a photon that pair-produces an electron-pair saying that the mass comes from the energy of the photon?

I'm guessing neither that is complete

David Z

@Arc676 Think about it from the center-of-mass frame of the electron and positron. In that frame, the net momentum before the reaction is zero, but after the reaction is nonzero.

@Arc676 Again, in that case, the net momentum after the reaction is zero but before the reaction is nonzero.

Arc676

I don't know what center of mass is...

David Z

Well, it's really center of energy

ManishEarth

7:03 PM

@DavidZ wait, it would be the reverse, no?

In the center of mass frame, the e and p have zero net momentum

But the photon has momentum in all frames

David Z

Well, whichever state (initial or final) has the electron and positron, there is zero net momentum.

ManishEarth

ah yes that

now it makes sense

(same thing that Chris said, but I thought over it more after that)

I think I've already gone through this before and gotten myself straightened out, but seem to be rusty.

David Z

@Arc676 Anyway, if you don't know what center of energy is, it's probably way premature for you to be learning about Feynman diagrams

Arc676

I'll google it

So you CANNOT have a diagram that starts or ends with a photon

David Z

But the gist is that I'm talking about a reference frame in which the electron and positron have zero net momentum.

user54412

7:05 PM

i've seen authors distinguish between "center of mass" (origin of your coordinates is such that the masses times vector displacements sum to 0) vs. "center of momentum" (origin is not important, but probably coincides with center of mass for ease; velocity is chosen such that (3-)momenta sum to 0)

ManishEarth

@DavidZ CoE is CoP , or ...?

Arc676

What's the difference between center of mass, energy, and momentum

David Z

@Arc676 You cannot have a diagram that starts or ends with a single photon. Unless it's a really boring diagram that just has one photon doing nothing.

Arc676

A diagram with no reaction, that is

ManishEarth

@DavidZ can't the photon itself split?

David Z

7:06 PM

I was about to correct myself ;-)

user54412

at first I was going to counter with a photon -> e+e- virtual pair -> photon...

David Z

You can have a diagram where a photon splits into a virtual electron-positron pair and then turns back into a photon. The point is, if you start with a photon, you have to end with a photon, and vice-versa.

ManishEarth

Yep

user54412

@DavidZ but if that photon is on-shell, is it possible?

Arc676

@ManishEarth You said e and p can represent electron and positron. Doesn't p represent proton?

user54412

7:07 PM

as in, will the electron/positrons be too off shell?

David Z

@Arc676 I'm not sure of the exact technical definitions of these things. I'm just talking about a reference frame in which the initial particles have zero net momentum.

ManishEarth

@Arc676 no, that was me trying to type less

e and e+

David Z

@ChrisWhite You can certainly draw the diagram and calculate the amplitude, and I think it will come out to be nonzero, but I'm not sure.

Arc676

@ManishEarth Ah...

ManishEarth

@ChrisWhite what's too off shell?

isnt off shell ... off shell?

user54412

7:08 PM

@ManishEarth i have no idea :P

tpg2114

@ManishEarth Sorry, my wireless likes to crap itself randomly lately and I just now got back on... I was around for that conversation, yet

ManishEarth

np

user54412

but it just seems that finding any finite position in E-p space for a typically massive particle to behave like it has no mass...

Arc676

Does anyone know a site with an explanation for center of energy and momentum or should I ask here?

user54412

maybe i just need to write down an actual equation

ManishEarth

7:09 PM

@Arc676 here is fine

David Z

The whole E^2 - p^2 = m^2 relation goes out the window for virtual particles

ManishEarth

CoMomentum is the frame where net P is zero

Arc676

@DavidZ Are you sure about that?

ManishEarth

Dunno what CoE is, suspect it to be the same thing

user54412

@ManishEarth CoMomentum? that's a new term for me

ManishEarth

7:09 PM

In the context of Feynman diagrams I've seen CoMo being used

Arc676

Center Of Momentum @ChrisWhite

ManishEarth

@ChrisWhite Center of Momentum :P You just used it

user54412

ohhhh

ManishEarth

I would call it CoP, but that gets confusing

David Z

@Arc676 Yep. E^2 - p^2 can have any value for a virtual particle. But in general, the greater the difference between E^2 - p^2 and the m^2 of the actual particle, the less likely the reaction is.

ManishEarth

7:10 PM

and CoM is Center of Mass.

user54412

reminds me of one particular intro to SR book that insists on saying "momenergy" for 4-momentum

ManishEarth

@ChrisWhite hahahaha sounds like something I would do

Arc676

@DavidZ If the formula is E^2=(mc^2)^2+(pc)^2 shouldn't you mention c somewhere in that equation?

David Z

@ChrisWhite confirmed that photon -> e+ e- -> photon is a necessary correction to the photon propagator

ManishEarth

@Arc676 natural units, c=1

Arc676

7:11 PM

Natural units?

David Z

Natural units

In physics, natural units are physical units of measurement based only on universal physical constants. For example the elementary charge e is a natural unit of electric charge, and the speed of light c is a natural unit of speed. A purely natural system of units is defined in such a way that some set of selected universal physical constants are each normalized to unity; that is, their numerical values in terms of these units are exactly 1. While this has the advantage of simplicity, there is a potential disadvantage in terms of loss of clarity and understanding, as these constants are then...

Arc676

So when exactly can you use c=1

David Z

Feel free to put in the c's to make the units work out, if you like.

@Arc676 Whenever you're using units such that the numerical value of the speed of light is 1.

ManishEarth

@Arc676 anytime, as long as you remember that this constrains your unit system

Arc676

If c=1 what is OUR speed when we walk...?

ManishEarth

7:13 PM

See, we have 3-5 fundamental units (mass, length, time, current, temperature)

Arc676

I'm guessing this numerical value is only used when one measures really fast objects right?

@ManishEarth and they are kg, m, s, A, ºC right?

ManishEarth

If we set G,c,hbar to 1, that fixes a system of units for mass, length, time. Setting epsilon_0 and k_B to zero fixes the other two if necessary

@Arc676 those are the SI units

David Z

@Arc676 $1.4 \times 10^{-9}$

ManishEarth

But I can pick a unit system where $G=c=\hbar=k_B=\varepsilon_0=1$

Arc676

Is a formatted formula supposed to appear? I just get plain text

and a lot of slashes and dollar signs...

ManishEarth

7:14 PM

Nope, I use an extension

Arc676

Ah

David Z

68

A: Should chat have TeX support?

I will leave the original post for historical reference, but as mentioned in the Update below, all four bookmarks are located on this installation page. There are four bookmarks: start ChatJax installs MathJax and starts a loop that renders $\LaTeX$ as needed. This is intended for use in chat, ...

ManishEarth

github.com/dvdfreitag/SE-Chat-MathJax

4

Q: MathJax in chat (ChatJax offshoot)

This is an offshoot of ChatJax, which enables MathJax along with mhchem on chat. Copy the text below: javascript:(function(){if(window.MathJax===undefined){var%20script=document.createElement("script");script.type="text/javascript";script.src="https://d3eoax9i5htok0.cloudfront.net/mathjax/lat...

Arc676

@DavidZ On the topic of before, if you CAN'T have a diagram in which the final state weighs more than the initial state, you can only have a Higgs as an intermediate particle right?

Or can you begin the diagram with the Higgs?

ManishEarth

@Arc676 why can't we tree up a bunch of lighter ones to create a Higgs?

David Z

7:18 PM

@Arc676 When the initial state particles are moving very fast, i.e. they have a lot of kinetic energy, there can be enough energy to create a real/on-shell/m=125GeV Higgs boson.

ManishEarth

Higgs will decay though

David Z

The Higgs boson itself is unstable, so it will decay quickly whether it's real or virtual.

But if your Higgs boson is on shell, you can validly draw a Feynman diagram which starts with a Higgs boson.

ManishEarth

but can we draw one that ends with an H?

because decay

user54412

@DavidZ in the same sense as you can validly draw a diagram beginning/ending with a free neutron, right?

Arc676

I thought you said you cannot have a "real" complete diagram in which it produces a particle with more mass by using the energy of the others

David Z

7:20 PM

@ChrisWhite Yeah, that sounds about right

Arc676

@ManishEarth That would be a diagram that doesn't show the last reaction, therefore incomplete

David Z

@Arc676 No, I didn't say that. Well, maybe I did by accident way back, but I would have been wrong. If you remember which message it was exactly, I could clarify.

ManishEarth

thought so

user54412

@ManishEarth i feel there's a time reversal symmetry you're trying to break there

ManishEarth

@ChrisWhite The force is strong with this one

David Z

7:21 PM

@ManishEarth You can, if you take the final state Higgs boson to be on shell. You would then find that the amplitude is zero unless the initial state particles have 125 GeV in the center of momentum frame.

Arc676

@DavidZ Not sure which message, but I think it came up when we were talking about the initial or final photon

@ManishEarth Who is "this one"

ManishEarth

@DavidZ So we can have a feynman ending with an undecayed product if it is on shell?

@Arc676 CW

David Z

@Arc676 Yeah, I don't quite remember either, but I think I may have made an incorrect statement back there somewhere. The point is, it's energy that matters, not mass.

ManishEarth

As in, feynman diagrams can end with muons and all?

Arc676

@ManishEarth You referred to the person you were directing the message to in third person...?

David Z

7:22 PM

@ManishEarth Yes. Of course that doesn't mean the product will sit there forever. It will decay, you just don't have to show that in the Feynman diagram.

ManishEarth

@Arc676 it's a quote from Star Wars

Arc676

@DavidZ That makes sense since they're interchangeable

ManishEarth

@DavidZ ah, thanks. That's what I originally meant when talking about "not having to show the full reaction"

Arc676

@ManishEarth I know that part, but you added the @, so it's like calling the person you're talking to "him"

user35386

What do you think Ludovic is really asking in this series of questions: physics.stackexchange.com/questions/81366/…, physics.stackexchange.com/questions/78916/…, physics.stackexchange.com/questions/82095/…, physics.stackexchange.com/questions/83088/torque-on-object

ManishEarth

7:23 PM

@Arc676 Oh, that's because I used the message-reply. Wanted to link messaged

user35386

Oh, and physics.stackexchange.com/questions/83370/…

Arc676

@ManishEarth Ah

David Z

@Arc676 you might find this interesting reading

ManishEarth

@Articuno Ludovic? I don't think anybody understands his questions

user35386

okay, that is relieving

Arc676

7:24 PM

It seems this Ludovic has built up a reputation for hard-to-understand questions

ManishEarth

though...they probably should be closed then

David Z

Yeah, I was going to say, it seems like several people understood them well enough to upvote them

ManishEarth

@Arc676 hard-to-understand similar fluid dynamics questions with big pictures that somehow my brain doesn't let me read in full

Arc676

How do you have a planet inside water (first link)

ManishEarth

O_0

Arc676

7:32 PM

Just to summarize the earlier topic, I understood that you can have basically ANY particle in the initial state, as long as the final state has equal or inferior mass, correct?

David Z

@Arc676 Energy. Not necessarily mass.

Arc676

Right, energy. How do you measure the energy of each particle?

David Z

Actually that depends on how you define mass. If you define it as the total of the masses of the individual particles in the initial state, then you can definitely have higher-mass particles in the final state. But if you define the mass of the initial state as the energy of the initial state in the reference frame where those particles have zero net momentum, then you cannot have more energy in the final state.

Arc676

So can I have two gluons produce a quark-pair that interact and create a Higgs boson (which later decays into a quark pair)?

David Z

Yep

Arc676

7:40 PM

And how would I mathematically prove this?

David Z

Draw the Feynman diagram, calculate the corresponding amplitude, and show that it's nonzero

The process is known as gluon fusion, or vector boson fusion

Arc676

I've never calculated "amplitude" I followed a non-mathematical and very basic tutorial (yes, I know, not the best way to learn physics, but I'm also busy with other things)

Another idea: we know how to draw vector fusion, how do I calculate exactly how many Z bosons the up-quarks can emit? They obviously cannot emit an infinite amount of Z bosons and an infinite amount of Higgs bosons, because that would be an incredibly obvious violation of mass conservation (and consequently, due to their interchangeability, energy conservation)

David Z

No, remember, there is no such thing as mass conservation in particle physics

Arc676

So energy conservation, how do I prove that I cannot have an infinite amount of Z bosons and therefore higgs bosons?

How do I know how many times the same two quarks can accomplish vector fusion

In fact, how do I know vector fusion can be achieved at all? How can two up-quarks react and have a final state of 3 up-quarks and one anti-up-quark

David Z

Um... wait, I'm a little confused. There's no need for up quarks to be involved.

Vector boson fusion looks like this:

(not the bottom right diagram)

Arc676

7:49 PM

Wait, I thought vector fusion was only the diagram on the top right

the one with the quarks Z bosons, and the Higgs in the final state

Even if the three (i.e. not the bottom right) are all called vector fusion, I meant the one on the top right

How do I prove "this quark can only emit x Z bosons, and therefore only x Higgs bosons can be created"

David Z

You don't, really, because in principle the quark can emit any number of Z bosons. But the probability decreases sharply as the number of emissions increases.

Arc676

Wait, so I can essentially spend the rest of my life drawing a diagram that never ends in which I produce infinite matter in a very unlikely situation? Wouldn't that violate energy conservation since each Higgs can decay into a quark-pair, each with its own energy?

David Z

Not necessarily. Remember, the masses of the particles only matter for the initial state and the final state. So as long as in the end, you wind up with some combination of particles that has a total energy equal to the total energy you started with, there's some probability that the reaction will happen.

(well, subject to all the other conservation laws of course)

Arc676

So it's actually possible to have two quarks, produce 100 Higgs bosons, have them decay into 100 quarks pairs, and end with 202 up quarks and 100 anti-up-quarks each with a minute amount of energy?

David Z

Yep

Gigi10012

7:59 PM

I have one question: For example i have triange (a = 5 and b = 4) and i want to calculate C ^ 2 using gmn(metric tensor) * dy(r) * dy(s). C ^ 2 will equal to (kronecker_delta(mn) * sum((dx(m)/dy(r)) * (dx(x) / dy(s)) * dy(r) * dy(s)), and then it will equal to (kronecker_delta(mn) * sum(dx(m) * dx(n)), and When I will put numbers, it will equal to 5 ^ 2 + 4 ^ 2 yes? Am I correct?

David Z

Well, possible in the sense that you could draw that Feynman diagram, and it would make a nonzero contribution to the total amplitude

Arc676

@DavidZ Wow...

David Z

@Gigi10012 5^2 + 3^2? Did you mean 5^2 + 4^2? In any case someone else might have to respond more carefully to that.

Gigi10012

@DavidZ thanks

Arc676

I'm gonna go now, I'll c you all tomorrow!