Science > Physics > Compton wavelength and Schwinger's vacuum breakdown
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Science > Physics |
| User: |
"John C. Polasek" |
| Date: |
02 Sep 2006 08:29:58 AM |
| Object: |
Compton wavelength and Schwinger's vacuum breakdown |
A paper by Melissinos arXiv:hep-ph/9805507 discusses Schwinger's idea
of vaccum breakdown.
Schwinger conjectures that breakdown will occur and generate showers
of electrons and positrons, if an electric field E_c is so strong
that an electron traversing a Compton wavelength L_c would generate
m_ec^2.
E_c*L_c*e = mc^2 Eq. 10
But Melissinos (and others) define
L_c = hbar/mc
instead of
L_c = h/mc.
Is there a reason for this? This leads to field intensity E_c such
that
and
E_c = m^2c^3/e*hbar = 1.3x10^18 V/m Eq. 11
Using h instead of hbar would make the field lower by 2pi. This
version of the CWL does not seem justified.
John Polasek
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| User: "FrediFizzx" |
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| Title: Re: Compton wavelength and Schwinger's vacuum breakdown |
02 Sep 2006 12:56:14 PM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:4p1jf2tu2r0at5gligr499eeqbloagmppg@4ax.com...
A paper by Melissinos arXiv:hep-ph/9805507 discusses Schwinger's idea
of vaccum breakdown.
Schwinger conjectures that breakdown will occur and generate showers
of electrons and positrons, if an electric field E_c is so strong
that an electron traversing a Compton wavelength L_c would generate
m_ec^2.
E_c*L_c*e = mc^2 Eq. 10
But Melissinos (and others) define
L_c = hbar/mc
instead of
L_c = h/mc.
Is there a reason for this? This leads to field intensity E_c such
that
and
E_c = m^2c^3/e*hbar = 1.3x10^18 V/m Eq. 11
Using h instead of hbar would make the field lower by 2pi. This
version of the CWL does not seem justified.
John, hbar/mc is the rationalized Compton wavelength. It is reduced by
a factor of 1/2pi. The bar thru the lambda_C symbol in the paper means
lambda_C/2pi. The use of hbar comes from Dirac. Probably because of
his quantization condition and the HUP using h/4pi. Most all advanced
physics texts and papers are using hbar and not h.
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
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| User: "srp" |
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| Title: Re: Compton wavelength and Schwinger's vacuum breakdown |
03 Sep 2006 09:17:12 AM |
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John C. Polasek a écrit :
A paper by Melissinos arXiv:hep-ph/9805507 discusses Schwinger's idea
of vaccum breakdown.
Schwinger conjectures that breakdown will occur and generate showers
of electrons and positrons, if an electric field E_c is so strong
that an electron traversing a Compton wavelength L_c would generate
m_ec^2.
E_c*L_c*e = mc^2 Eq. 10
But Melissinos (and others) define
L_c = hbar/mc
instead of
L_c = h/mc.
Is there a reason for this? This leads to field intensity E_c such
that
and
E_c = m^2c^3/e*hbar = 1.3x10^18 V/m Eq. 11
Using h instead of hbar would make the field lower by 2pi. This
version of the CWL does not seem justified.
John Polasek
Hi John
To me, this simply means that he is using the amplitude of the
compton wavelength. Given that the wavelength is L_c, then
of course, the transverse amplitude of such a wavelength
will be L_c / 2 pi.
The trouble is that some sources name the Compton wavelength
amplitude "rationalized compton wavelength" and sympolize it
with L_c just the same or even plain "compton wavelength", still
symbolizing with L_c, even though it mathematically is
compton wavelength divided by 2 pi, or in plain talk,
the compton wavelength amplitude.
Simple confusion on terms. The use of either h or hbar is
the telltale.
André Michaud
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