# do the particles ever collide in QED

Discussion in 'Physics & Math' started by Arlich Vomalites, Jan 15, 2015.

1. ### QuarkHeadRemedial Math StudentValued Senior Member

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Of course he can't. H doesn't even know the Bohr conjecture for himself, and posts references and a silly picture that are at best misleading

Moreover he has refused my challenge to derive the Schrodinger equation, probably because he cannot.

This in spite of having been provided with all the necessary information! Hell, this is an assignment given to 1st year Chemistry undergrads

3. ### Farsight

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It isn't some billiard ball thing. It isn't a point particle. It isn't some thing that has got a field. Its field is what it is.

Meh, you're splitting hairs. Why have you bolded the word particle?

Yeah yeah, see the picture above, where n = 1, 2, 3 etc.

It's like what Wiki says: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves. The lowest possible energy an electron can take is therefore analogous to the fundamental frequency of a wave on a string. Higher energy states are then similar to harmonics of the fundamental frequency." Changing an orbital is something like changing gear.

I can't I'm afraid. We're dealing with energy-momentum waves wherein energy E=hc/λ or E=hf, momentum p=hf/c, and you divide by c again for mass m=hf/c² which you can rewrite as m=E/c². This is just a measure of resistance to change-in-motion for a wave in a closed path. Of course I can point you at say hyperphysics if that helps:

Maybe you need to start appreciating that a "particle" like the electron is a wave in a closed path?

5. ### QuarkHeadRemedial Math StudentValued Senior Member

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Some expert!!
I am sure you sincerely think I need that "help". Believe me, I do not

Like an orbit, say? You once again contradict yourself, are you too blinkered to see it?

7. ### Farsight

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I gave a good enough answer.

No I can't. You derive it. Show me how it's done. You will of course be aware that I know of papers such as http://arxiv.org/abs/physics/0610121 and http://vixra.org/pdf/1206.0055v2.pdf.

If it's so easy to derive the Schrodinger equation, you won't have a problem showing me, now will you?

Edit:

You're no expert. Your physics knowledge is poor. Every time I expose this you cut and run.

No, not like an orbit. The wave is in a closed path because it's moving through itself displacing itself. It isn't going round something else.

8. ### exchemistValued Senior Member

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I say, people, if you want a laugh, I really do recommend the second reference in Farsight's post above! This Chinese author confidently asserts the Schroedinger equation must be wrong, because it predicts orbitals that are not spherically symmetrical for the hydrogen atom, (i.e. some are dumbell shaped etc). It seems he or she has (a) never come across spherical harmonics and (b) has never spoken to a chemist! I'd like to see a model of chemistry in which there are no p orbitals, for instance. Would be a bit of an, ahem, challenge to the theory of chemical bonding.

Quite amazing. I don't know where Farsight dug this one up, but full marks for finding it. Astonishing what papers manage to get written, if nothing else. I wonder who the author's supervisor is, or was?

….or maybe the publication date was 1st April…..

9. ### PhysBangValued Senior Member

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No, you gave no answer. You admitted that you have no equation and you have, currently, only one picture that you admit vaguely represents your idea. You "answer" is that you are sure that you can eventually find a picture that matches the details of an electron because you have a mystical experience of the whole of physics that no scientist has ever experienced, despite your inability to do the relevant physics.

Again, citations to vixra just cement a reputation as a crank. Despite your recent efforts to remove your listing as an "internet crackpot", you are doing all you can to justify the appellation.
Since you have no equation, we have no way of verifying this claim. It is demonstrably not science.

10. ### QuarkHeadRemedial Math StudentValued Senior Member

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No, none whatever, but you won't read it, so why should I bother?

Oh shittikens......

First the Schrodinger equation is an energy eigenfunction on some state vector/function $\psi$.We seek a Hamiltonian operator H = K + U, where K is kinetic energy and U is potential energy. One writes (in skeletal form) $H\psi = \epsilon |\psi\rangle$ Note that H is an operator, and that whenever an operator can be written as the sum of terms, these terms must themselves be operators

Classically momentum for an object of mass $m$ is defined as $p=mv$ where $v$ is velocity of our particle

Classically kinetic energy is defined as $K=\displaystyle \frac{1}{2}mv^2$

Without even thinking, I can combine these 2 equalities as $\displaystyle K= \frac{(mv)^2}{2m}= \frac {p^2}{2m}$

Now turn to the Bohr conjecture for a bound "orbiting" particle with mass $m$ and an orbit radius of $r$.

This states that the allowable angular momenta are given by $\displaystyle mvr=n\frac{h}{2 \pi}$. Loosely speaking, since we don't believe in "orbits" the radius is irrelevant, and therefore (from a previous post of mine) so is the integer$n$

Thus $\displaystyle mv = p = i \frac{h}{2 \pi}$ where I have included the imaginary factor as we are doing quantum mechanics (actually there is more to it than that!).

So putting this all together one gets for the kinetic energy operator $\displaystyle K=-\frac{h^2}{8 \pi^2 m}$ And since as a convenient shorthand we have that $\displaystyle \frac{h}{2 \pi} \equiv \hbar$ we arrive at $\displaystyle K = -\frac{\hbar^2}{2m}$

Since this is second order, I seek a "direction". I look no further than $\displaystyle\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ which we recognize as the Laplacian $\nabla^2$

I now seek the potential energy operator U. Since this depends only on position {x, y, z} I end up with

$\displaystyle [-\frac{\hbar^2}{2m} \nabla^2 + U(x,y,z)]\psi = \epsilon |\psi \rangle$

Last edited: Feb 5, 2015
11. ### Farsight

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Huh? You haven't derived Schrodinger's wave equation. You "turn to the Bohr conjecture" but promptly discard r and n, so all you done is said momentum p = angular momentum ћ, then you've chucked in an i and a minus and the Laplacian and the potential energy. This is desperately unconvincing, Quarkhead. You're just juggling with terms, you have no concept of any of the underlying reality. And then you reject all concept of the underlying reality!

12. ### QuarkHeadRemedial Math StudentValued Senior Member

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Farsight, you really are unhinged! I gave my version of how to derive the Schrodinger equation in response to your confession you did not know how to do it

Now you tell me it is wrong! It is like saying "I am colour blind, but that is not red"

13. ### Farsight

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I'm not unhinged. Your "derivation" was rubbish. You really don't know what you're dealing with. After everything I've told you! Sigh. You can lead a PhD to knowledge, but you can't make him think. But let's see if we can:

Try to understand that we are dealing with a wave that exhibits energy-momentum. E is a distance-based measure of this, p is a time-based measure, and you switch from the former to the latter by dividing by c, which is distance over time. Then there's another measure called m, where we divide by c again. And all the while, what we're dealing with is a wave. When that wave is moving linearly we talk about E=hf and p=hf/c or p=h/λ. When that wave is moving in an "orbital" we can talk about E=mc² and m=E/c² or m=hf/c² or m=h/λc. This "orbital" or closed path can be thought of as a circle with a diameter 2π times the radius. We can envisage that it's comprised of n wavelengths, and we can start with n=1. But we know that the electron is a spin ½ particle associated with two 360 degree rotations. So we can think of a double-loop orbital where the circumference is half the wavelength, which is 2π the diameter. And then we recall that one way of expressing the dimensionality of action h is momentum x distance, so we know that momentum p = ћ isn’t ideal. A distance has gone missing, some distance associated with ћ or h. What is it that’s common to all photons regardless of wavelength?

And why is that an electron has a Compton wavelength of 2.46 x 10⁻¹²m? And only that?

Last edited: Feb 5, 2015

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More comedy.

15. ### krash661[MK6] transitioning scifi to realityValued Senior Member

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i am lrrr, ruler of the planet omicron persei 8

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16. ### QuarkHeadRemedial Math StudentValued Senior Member

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You may think it a laughing matter. I certainly do not

Look, I do not mind in the least being denigrated, but what we all have to bear in mind is that, for better or worse, a lot of people use the internet as an educational tool.

And when some arrogant oaf presents a science subject in a totally mangled form, but with an air of seeming authority it is certain to mis-lead those whose only desire is to learn real science

People such as Farsight do a massive disservice to education in general bur especially to the young. I think we should all take this very seriuosly

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Totally agree.
Just as a reminder this same Farsight also claimed late last year, that he was about to rewrite 20th/21st century cosmology, and had formulated a TOE.
Yet he still sees the need to infest this forum along with the other handful of nutty alternative pushers, that are unable to get a look in anywhere else to sprout their nonsensical take on cosmology.

18. ### OnlyMeValued Senior Member

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One of the things you have told "us", is that you don't know the math. (BTW I don't do the math anymore myself, but I do respect those who still can or just take the time to share it!)

The last bit you put in italics above, would seem to apply in spades to some who don't have PhDs! Several others in this discussion have tried to present you with knowledge.., so far, though I am sure you could call it thinking, I haven't seen you make much sense.

The last bit of your post I redacted for emphasis below,

What credible referce did you get all that from? The language I mean, none of that sounds very scientific or convincing for that matter. If not a credible reference, perhaps you could just correct the math, you are objecting to!

19. ### exchemistValued Senior Member

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Yes you are right, we can all have a laugh, but there will be people put off coming here to learn and discuss, as a result of this stuff. (By the way, I have not forgotten the discussion about spin operators, but I'm having to do quite a bit of revision before I can make offer any useful comments or questions.)

20. ### Dr_ToadIt's green!Valued Senior Member

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I beg your pardon, any and all who may have taken my post in an unintended way.

I meant more of his comedy of errors, omissions and outright lies in support of claims that change by the minute, at some un-knowable whim of God, I suppose. God agrees with Farsight, right? Sure!

21. ### krash661[MK6] transitioning scifi to realityValued Senior Member

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2,973
it is broken. can you make it go ?

22. ### zgmcRegistered Senior Member

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789
I bet Geordi can fix it.

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23. ### QuarkHeadRemedial Math StudentValued Senior Member

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No, my post was not directed specifically at you. Not at all.

I merely used your quote as a springboard for my rant. And do I feel better for having had it? Not really, but what business is it of mine how this board regulates its affairs?