Re: Renormalization in QED

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 07/19/04 

Date: Mon, 19 Jul 2004 17:07:57 +0200

kenseto wrote:
> "Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in message
> news:cd6b2v$fgt$1@news.urz.uni-heidelberg.de...
>

[snip]

>>>>Please give an example for when and where a "new kind of energy" was
>>>>invented *by people working in QM*. That was your original claim
>>>>above...
>>
>>I notice that you *still* have not given an example for that.
>
>
> How about virtual particles pop in and out of the vaccuum? Does that require
> any energy?

No. These particles always "pop out of the vacuum" at least in pairs.
One of the particles of the pair (or of the triple, or more) has
positive energy, the other negative energy. The total energy is zero.

And even if it not were zero, this still would not be "a new kind of
energy". This would be quite ordinary rest energy, kinetic energy,
electromagnetic energy etc.

[snip]

>>>>>BTW why isn't the W and Z bosons detected frequently
>>>>>sice they are involved in the weak interactions such as the decay of a
>>>>>neutron?
>>>>
>>>>Please tell me *how* one would go on in detecting the W-bosons
>>>>involved in the decay of a neutron. I see no way to do that.
>>>
>>>
>>>NO you tell me why not.
>>
>>I said I see no way how one could do that. What more can I say???
>
>
> Why can you detect the W bosons in the accelerators and not the W bosons
> in neutron decay?

See below. In the accelerators, one *also* does not detect the W bosons
directly!

>>>Why can you detect the W bosons in the
>>>accelerators and not the W bosons involved in the neutron decay?.
>>
>>Err, in the accelerators, the W bosons are *also* not detected directly.
>>One only detects their decay products. So this is very similar to
>>the neutron decay example above.
>
>
> So the W bosons were never detected.

They are not *directly* detected!

> The products can be interpreted as
> that the peeled-off down quark interacts with the free S-Particles to
> give rise to the decay products. So no W bosons are needed.

How does this hypothesis explain that the distribution of the decay
products shows a peak at the W boson mass?

[snip]

>>>So what is supersymmetric partners invented for?
>>
>>Thanks for admitting essentially that you do not know this - in other
>>words, you did not know what you were talking about.
>>
>>To answer your question: mostly due to aesthetic reasons. Symmetry
>>is a powerful principle to construct theories - and it turns out
>>that using supersymmetry, renormalization gets easier. (what irony!
>>look at the title of the thread!)
>
>
> Yet another worthless epicycle. :-)

Supersymmetry was *not* invented to cure a problem with the theory
- so how can you call it an epicycle? Aren't epicycles things which
are invented soley to rescue a theory when problems occur?

Bye,
Bjoern

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