Balloon vs SR

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Mar 2019
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math vs physics

Anyone aware why traditional Lorentz transformation and the "mass - space" eqaution lead to two different results?
 
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authentic

Oh? I can't find the "mass - space" equation in any authentic textbook. Should science limit only to authentic textbook? ** haha...
 
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X4 Explanation of Mass Oscillation of Perfect Neutral Basic Particle (PNBP)

Actually, in situation of traditional 3D space SR, concerning no interaction, the constant speed movement can be treated as of neutral situation in 4D space. If an object is moving with a speed v in 3D space, then in 4D space on the right side, V1 = +1v while in the 4D space on the left side V2 = -1v. The 4D space speed V = V1 + V2 = 0. Or in an intuitive way, the speed in 3D frame is positive, then the speed in the 4D frame on the right side is positive while the speed in 4D frame on the left side is negative. Positive plus negative = 0.
It means the object does not move in 4D space (in the attached picture, z axis ignored for convenience of watching).
In fact, this 4D space analysis has no contradiction with the mass-speed equation eliminates the movement element. Their same purpose is trying to discover something more basic.
1.In traditional 3D space SR, according to the mass-speed equation M = γM0, it seems that speed is active while mass is passive.
2.The mass-space equation L ∝ 1 / M can be transformed as M ∝ 1/ L. The latter reflects 3D space is active while mass is passive. This semi-tramp nickname the 3D space as “mass space”. It reflects the mutual effect between 3D space and mass.
With regard to the 4D space equation X = X4x, once upon a time, we set out from the 3D space element x to research the aspect of uncertainty/wave character of cosmos and consider the variation of X4 state lead to the variation of 4D space X. Actually, in just the aspect of certainty/particle character, if three elements X, X4, x are all variants, there is possibility that the equation establishes too.
In X4 theory, for a perfect neutral basic particle (PNBP), the states of “not anti”, “anti”, “superposition of not anti and anti”/ neutral are all itself (simply as X4 = +*, -*, ±*). One idea is that it can oscillate between these three states.
If the variation of X4 state and the 4D space X can lead to the variation of 3D space x (exactly say the size of the PNBP L or denoted as r, the concept of size of particle mentioned by htam9876 in the thread “model of unit electrical charge”), that means the mass of the PNBP will oscillate.
Of course the mass will not increase or decrease itself. It is transferred between the PNBP and another particle it approaches. It is just the approach to another specific X4 = +*, or X4 = -*, or neutral particle (or not approach other particle) that the X4 state of the PNBP vary. When the PNBP is moving in the natural environment, the chance of encountering those three situations should be fair.
The oscillation between those three X4 state is digital. If it can leads to the variation of both the 4D space X and 3D space x, they will be a course, and then the mass oscillation will be a course too. Only a short abstract of the conjecture is talked here.
(Note: One example of transference of mass between two particles can be found in the “two body issue” in relevant lecture in QM. This respect of analysis put aside temporarily here. In X4 Theory, if proton is defined as “not anti” basic particle, then the electron will be “anti” basic particle. The concept of X4 basic particle was mentioned by rabbit htam9876 in the thread “X4 model of unit electrical charge”.)
If the PNBP is referred to the neutrino, this semi-tramp would not say “objection”.


Li Qiang Chen
August 27th, 2019
 

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Mar 2019
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1/3

"The early findings were associated with the solar neutrino problem. That is, the measured neutrino flux at the Earth's surface, from the Sun, was 1/3 of..."
This words are abstracted from an astrophysicist's post in the lounge.
This semi-tramp's humble explanation is very simple: every three solar neutrinos has three different experiences during the course of moving in the natural enviroment before they reaches the detector on the Earth. The chance is even.
 

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relevant

One previous post of rabbit htam9876 is relevant. So, this semi-tramp post it here too. One model, woo...Bounce.
.................
X4 Explanation of the μ Hydrogen Phenomenon
When the electron of a Hydrogen atom is replaced with a μ particle, the atom turns to be μ Hydrogen. For detailed information, see relevant publications.
There is a phenomenon occurs: the radius of the proton in μ Hydrogen atom contracted. Why?
Below we try to use the conclusion in X4 Theory of Photon Ⅲ to explain it.
The radius of the proton in Hydrogen atom is r1 ≈ 8.768×10ˆ-16m, while the radius of the proton in μ Hydrogen atom is r2 ≈ 8.418×10ˆ-16m.
The mass of an electron is Me ≈ 9.1×10ˆ-31kg, while the mass of a μ particle is Mμ ≈ 200 ×9.1×10ˆ-31kg.
The mass of a proton is Mp ≈ 1.6726×10ˆ-27kg.
The reduced mass of the electron in Hydrogen atom is:
μ1 = Me Mp/ ( Me+ Mp) ≈ (9.1×10ˆ-31) ×(1.6726×10ˆ-27)/ [(9.1×10ˆ-31)+(1.6726×10ˆ-27)] ≈ 9.0951×10ˆ-31kg
The increased mass of the proton in Hydrogen atom is:
M1 = Mp+ Me -μ1 ≈ (1.6726×10ˆ-27)+ (9.1×10ˆ-31)- (9.0951×10ˆ-31) ≈1.6726×10ˆ-27kg
The reduced mass of the μ particle in μ Hydrogen atom is:
μ2 = Mμ Mp/ ( Mμ+ Mp) ≈ (200×9.1×10ˆ-31) ×(1.6726×10ˆ-27)/ [(200×9.1×10ˆ-31)+(1.6726×10ˆ-27)] ≈ 0.1641×10ˆ-27kg
The increased mass of the proton in μ Hydrogen atom rise to be :
M2 = Mp+ Mμ –μ2 ≈ (1.6726×10ˆ-27)+ (0.182×10ˆ-27)- (0.1641×10ˆ-27) ≈1.6905×10ˆ-27kg
r1/r2 ≈(8.768×10ˆ-16)/ (8.418×10ˆ-16) ≈ 1.0416
M2/ M1 ≈(1.6905×10ˆ-27)/ (1.6726×10ˆ-27) ≈ 1.0107
Basically in conformity with the relationship r∝1/M in X4 Theory of mass space.

Conclusion: It’s the rise of the increased mass of the proton leads to the contract of the radius of the proton in μ Hydrogen atom.
 

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Mar 2019
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not X4 question

That dragon stands on the ground and watch the Sun. He got a round ball. Gravity effect of the Sun in all directions is even.
If that dragon flys toward the Sun in constant speed, will he get an oval ball instead? If it is, then gravity effect of the Sun in all directions is not even. That means physical rules are different in two inertial frames?
 

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Firstly they will not be (perfect) inertial frames (varying gravitational acceleration),
But that is perhaps being picky about a minor detail.

I think that the principle argument to your scenario is that;
compression effects only occur due to the observer being in a different (inertial) frame to the actor.

The Actor will see the ball as round, because it is in the same (inertial) frame as he is,
the Observer will see the ball as oval, because he is in a different (inertial) frame.

Note: Actor as in the person who performs an action, not as someone who is on stage performing Shakespeare.

(There should really be the word "acter", as different from actor, but the dictionaries say that there isn't)
 
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Thank you. Professor woodpecker. Although i don't know exactly what your "actor" or "observer" mean., I feel you are talking something more physical than that obscure dragon.
 

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Lorentz vs dragon

This discussion is somewhat relevant, so this semi-tramp posts it in this thread.
That dragon posted a very good answer to my question in another thread: “The answer to your question is that the velocity component in the x direction is equal to c. This is from Huygen's principle.” (details see that dragon’s thread “common misconception in physics Ⅱ”)
This semi-tramp calls it the “Huygens-dragon principle”.
………………………………………..
Next research it further step with the traditional Lorentz transformation of speed in 3D space.
Assume the rest inertial frame is S, the moving inertial frame is S’, the moving speed of S’ is u, here u ≠ c for simplicity. (see the attached picture)
According to the “Huygens-dragon principle”, in the rest inertial frame S: cx = cy = cz = c
In the moving inertial frame S’:
In along direction: cx’ = (cx – u) / (1 – u cx / c²) = (c - u) / (1 – u/c)
In cross direction: cy’ = cy / (1 – ucx/ c²) γ = c / (1 – u/c) γ
In vertical direction: cz’ = cz / (1 – ucx/ c²) γ = c / (1 – u/c) γ
(Note: γ is the 3D space Gama factor)
We can not surely derive cx’ = cy’ = cz’ = c
That’s to say physics rule is not equivalent in different inertial frames.
Lorentz vs that dragon, who’s correct?
That dragon of course would like that Lorentz is correct.
What funny is that this semi-tramp can prove in his own way ** theory that it’s that dragon who is correct. It is very simple.
……………………………………….
According to the mass-space equation, the contraction of physical space is the same situation in all directions. Then in the moving inertial frame S’: cx’ = cy’ = cz’ = (c - u) / (1 – u/c)
Next, is analysis in 4D space.
X = X4*χ=∞*χ= ∞
So, the four dimension spacial coordinate in the world of the released photon is X≡∞. (The four dimensional space of X4 state = ∞ represents the world of the released photon.)
That is to say any point of three dimensional space is the same in the eyes(world) of the released photon.
In another word, the constant velocity movement in three dimensional space does not effect (u ≡ 0 ) in the eyes(world)of the released photon.
Then: cx’ = cy’ = cz’ = c
................................
Why a step forward in geometry/math/physics has to be wrong? haha...
 

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