Den 06.07.2025 06:27, skrev Bertietaylor:
On Sat, 5 Jul 2025 19:43:23 +0000, Stefan Ram wrote:
The positron turned out to be pretty useful. Just look at the PET
(Positron Emission Tomography) scanner. There's a lot of solid proof
that PET scanners have helped save lives by letting doctors spot
how diseases are moving along and see if treatments are working.
Checked that out. Looks like it works on radioactive injections and
consequent radiation. > Nowhere is it said that positrons are radiated
like say beta rays.
In the case of cancer diagnosis a drug containing Fluorodeoxyglucose
is injected into the bloodstream of the patience.
Fluorodeoxyglucose contains F-18
, which decay via β+ decay.
F-18 with 9 protons and 9 neutrons →
O-18 with 8 protons and 8 neutrons + positron + neutrino
When the ejected positron hits an electron somewhere in the body,
they annihilate and two gamma particles (photons) are created.
e⁺ + e⁻ → γ + γ
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 �
)
(Mass is converted to pure kinetic energy of photons)
These gamma particles are detected by two detectors so a 3D image
is created. Because cancer cells have a higher metabolic rate than
do typical cells, they will appear as bright spots in the image.
Simply calling it pet does not prove the existence of positrons.
Right!
But so does the fact that Positron Emission Tomography scanners work.
-- Paul
https://paulba.no/
On Sat, 5 Jul 2025 19:43:23 +0000, Stefan Ram wrote:
The positron turned out to be pretty useful. Just look at the PET
(Positron Emission Tomography) scanner. There's a lot of solid proof
that PET scanners have helped save lives by letting doctors spot
how diseases are moving along and see if treatments are working.
Checked that out. Looks like it works on radioactive injections and consequent radiation. > Nowhere is it said that positrons are radiated
like say beta rays.
Simply calling it pet does not prove the existence of positrons.
Den 06.07.2025 06:27, skrev Bertietaylor:
On Sat, 5 Jul 2025 19:43:23 +0000, Stefan Ram wrote:
The positron turned out to be pretty useful. Just look at the PET
(Positron Emission Tomography) scanner. There's a lot of solid proof
that PET scanners have helped save lives by letting doctors spot
how diseases are moving along and see if treatments are working.
Checked that out. Looks like it works on radioactive injections and
consequent radiation. > Nowhere is it said that positrons are radiated
like say beta rays.
In the case of cancer diagnosis a drug containing Fluorodeoxyglucose
And how is in the case of a measurement, poor trash:
if a measurement of a property of an observed
object gives a result different than the real
value of the property, the measurement is
a)valid
b)erroneous
c)UUUUUUU!!!UUUUUUU!!!!!!!!UUUUUUUUUUU!!!!! UUUUUUUUUUUUUUUUU!!!!!!!!!!!!!!!!!!!!!!!
PLONK!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Let me guess, the correct relativistic answer
is c, right, poor trash?
Den 07.07.2025 21:58, skrev Maciej Woźniak:
And how is in the case of a measurement, poor trash:
if a measurement of a property of an observed
object gives a result different than the real
value of the property, the measurement is
a)valid
b)erroneous
c)UUUUUUU!!!UUUUUUU!!!!!!!!UUUUUUUUUUU!!!!!
UUUUUUUUUUUUUUUUU!!!!!!!!!!!!!!!!!!!!!!!
PLONK!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Let me guess, the correct relativistic answer
is c, right, poor trash?
Your comments are as well formulated as always.
Well done!
e⁺ + e⁻ → γ + γ
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 Å)
(Mass is converted to pure kinetic energy of photons)
"Paul.B.Andersen" <[email protected]> wrote or quoted:
e⁺ + e⁻ → γ + γ
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 Å)
(Mass is converted to pure kinetic energy of photons)
BTW:
Mass is /not/ changed in this process when the mass of the system
"γ + γ" is being compared with the mass of the system "e⁺ + e⁻".
The total momentum of the system "γ + γ" is 0 when we assume that we
had chosen coordinates in which the total momentum of the system
"e⁺ + e⁻" was 0. So all the energy of "γ + γ" still is mass energy
(zero momentum, non-zero mass).
Only when one (mentally) changes the definition of the system and
looks at one of those "γ" in isolation has one a system with zero
mass and non-zero momentum!
Den 08.07.2025 00:02, skrev Stefan Ram:
"Paul.B.Andersen" <[email protected]> wrote or quoted:Why would you consider γ + γ as a closed system?
e⁺ + e⁻ → γ + γMass is /not/ changed in this process when the mass of the system
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 Å) >>>(Mass is converted to pure kinetic energy of photons)
"γ + γ" is being compared with the mass of the system "e⁺ + e⁻".
"Paul.B.Andersen" <[email protected]> wrote or quoted:
Den 08.07.2025 00:02, skrev Stefan Ram:
"Paul.B.Andersen" <[email protected]> wrote or quoted:Why would you consider γ + γ as a closed system?
e⁺ + e⁻ → γ + γMass is /not/ changed in this process when the mass of the system
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 Å)
(Mass is converted to pure kinetic energy of photons)
"γ + γ" is being compared with the mass of the system "e⁺ + e⁻".
Well, I was talking about a "system", not a "closed system".
Many people go with
E^2 = (mc^2)^2 + (pc)^2
and emphasize that energy and momentum are conserved in a process
"-->", while mass can be converted into "pure kinetic energy".
Now, let's call the values before a process (left of the arrow "-->")
"0" and those after the process (right of the arror "-->") "1".
Then, what the people say would be:
E_1 = E_0 "Energy is conserved"
p_1 = p_0 "momentum is conserved"
m_1 < m_0 "mass is converted into energy"
But these three assumptions are not compatible with
E_0^2 = (m_0c^2)^2 + (p_0c)^2 and
E_1^2 = (m_1c^2)^2 + (p_1c)^2
!
E_1 = E_0 "Energy is conserved"Please explain why these three assumptions are not compatible with:
p_1 = p_0 "momentum is conserved"
m_1 < m_0 "mass is converted into energy"
e⁺ + e⁻ → γ + γ
m₀ = 2m, m₁ = 0
In the case of cancer diagnosis a drug containing Fluorodeoxyglucose
is injected into the bloodstream of the patience.
Fluorodeoxyglucose contains F-18, which decay via β+ decay.
F-18 with 9 protons and 9 neutrons →
O-18 with 8 protons and 8 neutrons + positron + neutrino
When the ejected positron hits an electron somewhere in the body,
they annihilate and two gamma particles (photons) are created.
e⁺ + e⁻ → γ + γ
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 Å)
(Mass is converted to pure kinetic energy of photons)
These gamma particles are detected by two detectors so a 3D image
is created.
Den 06.07.2025 06:27, skrev Bertietaylor:
On Sat, 5 Jul 2025 19:43:23 +0000, Stefan Ram wrote:
The positron turned out to be pretty useful. Just look at the PET
(Positron Emission Tomography) scanner. There's a lot of solid proof
that PET scanners have helped save lives by letting doctors spot
how diseases are moving along and see if treatments are working.
Checked that out. Looks like it works on radioactive injections and consequent radiation. > Nowhere is it said that positrons are radiated
like say beta rays.
In the case of cancer diagnosis a drug containing Fluorodeoxyglucose
is injected into the bloodstream of the patience.
Fluorodeoxyglucose contains F-18, which decay via ?+ decay.
F-18 with 9 protons and 9 neutrons ?
O-18 with 8 protons and 8 neutrons + positron + neutrino
When the ejected positron hits an electron somewhere in the body,
they annihilate and two gamma particles (photons) are created.
e? + e? ? ? + ?
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 �)
(Mass is converted to pure kinetic energy of photons)
These gamma particles are detected by two detectors so a 3D image
is created. Because cancer cells have a higher metabolic rate than
do typical cells, they will appear as bright spots in the image.
Simply calling it pet does not prove the existence of positrons.
Right!
But so does the fact that Positron Emission Tomography scanners work.
-- Paul
https://paulba.no/
If one would try to correct the vagueness of "mass is converted into
energy" by saying, "the mass of 'γ + γ' is smaller than the mass
of 'e⁺ + e⁻'" (thereby saying "of what"), then this would be wrong.
Paul.B.Andersen:> Den 06.07.2025 06:27, skrev Bertietaylor:
In the case of cancer diagnosis a drug containing Fluorodeoxyglucose
is injected into the bloodstream of the patience.
Fluorodeoxyglucose contains F-18, which decay via ?+ decay.
F-18 with 9 protons and 9 neutrons ?
O-18 with 8 protons and 8 neutrons + positron + neutrino
When the ejected positron hits an electron somewhere in the body,
they annihilate and two gamma particles (photons) are created.
e? + e? ? ? + ?
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 �)
(Mass is converted to pure kinetic energy of photons)
These gamma particles are detected by two detectors so a 3D image
is created.
I like to think that the e? and e? don't actually annihilate, but
instead form a tight ball. Since this ball has no charge, it is for all practical purposes undetectable. But then, the evidence that e? and e? completely annihilate into 2 ? photons is compelling, since after all:
m_{e?}c^2 == 9.1094e-31 * 299792458^2 / 1.6022e-19 == 511 keV
A reasonable argument against this is that m_{e?} measurement is controversial, and gamma ray detectors are calibrated assuming that the
e? e? annihilation radiation is 511 keV.
the vacuum state. I.e., "e⁺ + e⁻" is the compound system, the mass of >which differs from the sum of the two components taken in isolation.
Physical evidence is of course primary,
"Paul.B.Andersen" <[email protected]> wrote:
Stefan Ram wrote:>>> E_1 = E_0 "Energy is conserved"
p_1 = p_0 "momentum is conserved"
m_1 < m_0 "mass is converted into energy"
Please explain why these three assumptions are not compatible with:
e⁺ + e⁻ → γ + γ
Well, "energy", "momentum", and "mass" are /vague terms/. One needs
to specify the energy (momentum, mass) /of what/ one refers to.
If one would try to correct the vagueness of "mass is converted into
energy" by saying, "the mass of 'γ + γ' is smaller than the mass
of 'e⁺ + e⁻'" (thereby saying "of what"), then this would be wrong.
m₀ = 2m, m₁ = 0
Well, this is the same point: It's too vague because the system
is not specified. The mass of what system is "m_0"? The mass of
what system is "m_1"? So, to quote Dirac, it's not even wrong!
But, if I assume that m_0 is the mass of "e⁺ + e⁻", then it's
wrong, because when they move in their COG system m_0 > 2m
(assuming m to be the mass of an e⁺ or e⁻ in its COG system)
if they are moving with a non-zero speed in the COG.
And m_1 surely is larger than 2*0 (two times the mass of
a photon). For one illustration: The mass of a gluon is 0.
Yet the kinetic energy of gluons make up for 37 percent of the
/mass/ of a proton. You don't get this value of .37*m_p if you
take the mass of each gluon in isolation and then sum them up.
The mass of the two leptons is 2m and the mass of the two
gamma photons is zero.
How can this be "to vague" and "not even wrong"? :-D
So mass isn't invariant?
So the mass of two photons isn't two times the mass of a photon?
I like to think that the e? and e? don't actually annihilate, but
instead form a tight ball. Since this ball has no charge, it is for all
practical purposes undetectable. But then, the evidence that e? and e?
completely annihilate into 2 ? photons is compelling, since after all:
Huh? Whadahyemean, tight ball?
Actually positronium is a hydrogen-like atom,
with the corresponding structure and spectrum.
(at about half the frequency)
Optical de-excitation radiation can be detected,
and the lifetime depends on the hydrogenlke state.
Jan
To get into this mindset just think about how the mass of an atom
of hydrogen is smaller than the mass of the proton and electron
taken in isolation. This would be another example where the
mass of a combined system differs from the sum of the masses of its >components.
[email protected] (Stefan Ram) wrote or quoted:
To get into this mindset just think about how the mass of an atom
of hydrogen is smaller than the mass of the proton and electron
taken in isolation. This would be another example where the
mass of a combined system differs from the sum of the masses of its
components.
The binding energy of hydrogen is "-13.6 eV".
It is the sum of the potential energy and the kinetic energy
of the electron and is part of the mass energy of hydrogen.
I take this calculation from the World-Wide Web, but might
have added typos when converting it to plain text:
At the mean radius a_B = 0.528 × 10^{−10} m, the potential energy
of the pair is:
1 e^2
U = − ----- --- = −4.36 × 10^{−18} J = −27.2 eV.
4πϵ_o a_B
The kinetic energy is half this value, and agrees with the classical
result for a particle in uniform circular motion:
mv^2 1 e^2 1 1 e^2
---- = ----- ----- => - mv^2 = ----- ---.
a_B 4πϵ_o a_B^2 2 8πϵ_o a_B
So the binding energy is: E = K + U = −13.6 eV.
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