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Groups > sci.physics.relativity > #380199
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Newsgroups | sci.physics.relativity |
| Subject | Re: The Dingleberry Anti-Gravity, Perpetual Motion Machine. |
| Date | 2016-03-26 14:30 +0100 |
| Organization | PointedEars Software (PES) |
| Message-ID | <56F68EE7.9070008@PointedEars.de> (permalink) |
| References | (10 earlier) <nd37vi$77j$1@gioia.aioe.org> <1703637.umW7bC727F@PointedEars.de> <nd4vt9$15ia$1@gioia.aioe.org> <56F60964.40408@PointedEars.de> <313cfbp86ninf6nrr5aiaihemtosk0hvkg@4ax.com> |
wrote:
> On Sat, 26 Mar 2016 05:00:36 +0100, Thomas 'PointedEars' Lahn
> <PointedEars@web.de> wrote:
>> wrote:
>>> On 26/03/2016 7:45 AM, Thomas 'PointedEars' Lahn wrote:
>>>> The explanation is that you have the misconception that mass
>>>> constitutes a solid object. [
] What the layman [
] calls a
>>>> massive object is [
] actually made up of constituents held
>>>> together by fundamental interactions (also, forces): [
]
>>>
>>> Yes we know all that. Matter is made of smaller bits of matter, all
>>> of which have mass. ..and part of the mass is associated with
>>> binding energy. That doesn't tell us anything.
>>
>> It does not tell *you* anything. It tells *me* that there is more to
>> matter than mass, and that assuming otherwise is the cause of your
>> misconception about how electronpositron annihilation works.
>
> It tells you nothing other than the fact that it does happen.
Incorrect.
It tells me that because both mass and energy are properties of an object,
and theory predicts, as experiment confirms, that they are equivalent, the
interacting particles have both mass and energy, that the total energy of
the particle comes from its mass m *and* its momentum p, and that this
energy can be converted into other forms of energy upon interaction in such
a way that the quantity of total *energy* of this system is *conserved*.
So that if the produced photons are moving relative to the center-of-
momentum frame of the interaction – and they *are* *both* moving relative to
it *at the speed of light* –, each of them must have momentum, therefore can
have zero mass but not zero total energy, which is why the positron *cannot*
have negative mass.
The annihilation reaction is:
e⁻ + e⁺ → γ + γ
The total energies must be
for the electron: E_e⁻ = √((m_e⁻ c²)² + (p_e⁻ c)²)
for the positron: E_e⁺ = √((m_e⁺ c²)² + (p_e⁺ c)²)
for one photon: E_γ = E(m_γ = 0) = p_γ c = ℏ k × c = ℎ f∕c × c = ℎ f
Therefore:
E_e⁻ + E_e⁺ = 2 E_γ
√((m_e⁻ c²)² + (p_e⁻ c)²) + √((m_e⁺ c²)² + (p_e⁺ c)²) = 2 ℎ f
Why at least *two* photons? Because total momentum must *also* be a
conserved quantity, and as the electron and the positron interact (when they
collide, if you will) they have opposite velocities v⃗, therefore opposite
momentum p⃗, in the center-of-momentum frame (i.e., for an observer for
which we define that they are at rest in the collision point):
p⃗_e⁻ = m_e⁻ v⃗_e⁻∕√(1 − (|v⃗_e⁻|∕c)²) > 0 as v⃗_e⁺ > 0
p⃗_e⁺ = m_e⁺ v⃗_e⁺∕√(1 − (|v⃗_e⁺|∕c)²) < 0 as v⃗_e⁺ < 0
So if total momentum is to be conserved, at least two photons have to be
produced that have opposite momentum as well. And indeed, one observes that
at least two photons *are* produced; when exactly two, they are going off in
opposite directions from the collision point. (This is the *physical* basis
of the PET scan in medicine.)
p⃗_e⁻ + p⃗_e⁺ = 0⃗ = p⃗_γ₁ + p⃗_γ₂ ⇒ p⃗_γ₁ = −p⃗_γ₂ ⇒ v⃗_γ₁ = −v⃗_γ₂
<https://en.wikipedia.org/wiki/Annihilation#Electron.E2.80.93positron_annihilation>
None of this works out if m_e⁺ < 0 as you claim. But it works out perfectly
if not only m_e⁺ > 0, but m_e⁺ = m_e⁻.
Which is – as *you are refusing to answer the question put to you* in the
other subthread –
What makes an antiparticle different from the particle it is
the antiparticle of?
– consistent with the *actual* definition for “antiparticle”:
<https://en.wikipedia.org/wiki/Antiparticle>
“Corresponding to most kinds of particles, there is an associated
antimatter antiparticle with the *same* mass and opposite *charge*
(including electric charge [‘q’]).”
(emphasis and [comment] mine).
So, as you agreed that “the positron is the antiparticle of the electron”,
the positron *cannot* have negative mass or any other mass than the mass of
the electron *for that reason, too*.
Because the electron does _not_ have negative, but *positive* mass, and
neither negative mass nor any other mass than the mass of the electron would
be “the same mass” as the mass of the electron.
So, *in fact*, contrary to your claim:
m_e⁺ = m_e⁻
q_e⁻ = −1 e
q_e⁺ = +1 e
q_e⁺ = −q_e⁻
See also: <http://www.wolframalpha.com/input/?i=electron+vs.+positron>
>>>> [
] when you consider that we now know that 98.7 % to 99 % of the
>>>> mass of protons and neutrons [
] does _not_ come from their
>>>> constituent quarks [1], but from the space between the quarks, from
>>>> the *energy* of *empty space* [2], and that atoms are mostly empty
>>>> as well¹, it should not be at all surprising that matter can
>>>> change into pure radiation. And it should not be surprising that
>>>> where there is so much empty space, radiation can pass through
>>>> most massive objects.
>>>
>>> Yes we know all that.
>>
>> The truth is: Now that I have told *you*, *you* know it. Because
>> people who know this would not claim that the positron could have
>> negative mass.
>
> Well since no definition exists for either positive or negative mass you
> seem mumbling nonsense.
Oh for crying out loud. *You* *were* *the* *only* *one* *here* claiming
that the positron could have “negative mass”, and now you are saying there
would be “no definition” for “negative mass” in the first place? What were
you blathering about, then?
Since a definition exists for (positive) mass, namely in classical mechanics
that it is “the quantity of matter” that is the cause of a force, gravity,
that attracts objects with (positive) masses m₁ and m₂ to one another [1] –
F⃗₁₂ = −G m₁ m₂ ^r⃗₁₂∕|r⃗₁₂|² –, what follows is that negative mass must
cause a force with opposite direction, antigravity (if you will), that
repels an object with negative mass m₁ from an object with positive mass m₂.
Since in relativity, mass is defined in terms of the energy E₀ that an
object has at relative rest – m = E₀∕c² –, it means that positive mass
follows from an object that has positive rest energy, while negative mass
follows from an object that has negative rest energy.
So, out of ignorance, *you* have been “mumbling nonsense” instead.
_____________
[1] Isaac NEWTON (1713/1726). “The Mathematical Principles of Natural
Philosophy”. Translation by Andrew MOTTE (1792). Book Ⅰ, Def. Ⅰ
and Ⅴ.
<https://books.google.com/books?id=Tm0FAAAAQAAJ&printsec=frontcover&dq=Newton+mathematical+principles+Motte&lr=&as_drrb_is=b&as_minm_is=12&as_miny_is=1720&as_maxm_is=12&as_maxy_is=1800&num=20&as_brr=3>
>>> It does not tell us anything about the transformation from mas to
>>> energy.
>>
>> Mass is not transformed to energy. *Only* *you* have the
>> misconception that it would be, that those two would be independent
>> properties and quantities.
>
> You have no asnwer. I dout if you have ever even considered the that the
> question exists.
The answer is, as is known to *us* – not you – since more than a century (1905),
E = √((m c²)² + (p c)²)
which reduces to
E = m c²
for objects at relative rest (p = 0).
<https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence#Applicability_of_the_strict_mass.E2.80.93energy_equivalence_formula.2C_E_.3D_mc2>
>>> What we really need to know is more information about the physical
>>> nature of fields.
>> What *you* really need to know is how to consult introductory
>> resources on physics.
>
> You don't understand the question so you can hardly mark comments about
> its answer.
You wish. Your problem is that you are asking a nonsensical question (“How
can mass turn/change/be transformed into energy?”), and expect an answer to
*this* that makes sense. So it is *your* misconception, *your* mistake; it
is not one of the *few* people who are *still* replying to you.
EOD
PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."
(from: WolframAlpha)
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