# A Model for the Propagation of Visible Light and Other Rays

Discussion in 'Alternative Theories' started by yaldonTheory, Nov 8, 2016.

1. ### NotEinsteinValued Senior Member

Messages:
1,374
$d = s_y$
$V_B = s_y$
where d is a length (distance), and V is a volume. You have just written: $d = V_B$, in other words: [meter] = [meter]^3. You are wrong; please check your maths again.

Until you define it, which is what you have done. I don't think ether per definition has to be undefined. Heck, the entire idea over a hundred years ago was to discover (and thus "define") it!

Which doesn't preclude it from being an ether model.

So, no absolute reference frame, but an absolute observer? Really, are you going to drag religion into this?
You do know that Newtonian physics can describe the motion of particles from any reference frame, so from infinitely many observers? In fact, perform a Galilean transformation so your observer isn't moving anymore. Your theory breaks down, Newtonian physics continues to operate just fine. Funny that...

I have demonstrated to have better knowledge than you, so yeah...

I think you've confused me and you... I will promise to read more carefully, if you promise to write more carefully.

Well, you haven't given any at the point I stopped reading your text, but you've violated basic Newtonian principles that I know to be a pretty good description of reality. But let's put this to the test: describe (or point me to the derivation of) the hydrogen spectrum using your theory. The perihelion of Mercury. The Michelson–Morley experiment. These are basic observations mainstream science can explain. Can your theory do that too?

I have compare them, and found your model lacking. I in fact reject it based on the fact your cornerstone formula is wrong, and its results are incompatible with basic physical principles (such as the independence of your choice of observer).

I'm sorry you wasted your time, but I wasn't around 10 months ago. But 10 months ago I would also have pointed out your cornerstone formula is wrong. And 10 months from now it'll still be wrong, unless you go and fix it.

In fact, I encourage you to go and fix it!

3. ### yaldonTheoryRegistered Member

Messages:
53
We will tell you what.
According to current mainstream ideas, there is a formula for energy, E=hf. The units for the constant h is a Joule-second ($J\cdot sec$). The units for a Joule is a Newton-meter ($N \cdot m$). The units for a Newton is a KILOGRAM-meter per second squared ($\frac{kg \cdot m}{sec^2}$). There are units for mass in the constant h.
The Yaldon Particle Theory shows why there units for kilograms in the constant h. Formula (2-b) on page 24 of the book provides the correct expression for h, using a momentum (mass-velocity) regarding the energy of a propagated ray.

Math is a logical language, thus we have interpret the formula.
The formula you provide, $Energy=\frac{1}{2}mv^2$, expresses the amount of energy required to move a mass (m) from a velocity of zero to a value of v. The formula we have provided ($Energy=mv^2$) expresses the amount of energy that is available at any moment, as long as a mass (m) moves with a velocity of v.
Do you also think that Albert Einstein should've re-wrote his formula of $E=mc^2$ to $E=\frac{1}{2}mc^2$?

There is nothing wrong with the Physics in The Yaldon Particle Theory. But, there is something wrong when formulas are interpreted incorrectly.

Why are you so skeptical? Do you need more time to become adapted to The Yaldon Particle Theory?

5. ### yaldonTheoryRegistered Member

Messages:
53
Even if you had misread the 'v' as an 'a' you would still be wrong in your assertion to declare the force as a function of time.
F=ma cannot be declared as a function of time. It can be declared as a function of acceleration. F(a)=ma

$\frac{v}{t}=acceleration$
When t=0, then the acceleration will be infinite. Infinite acceleration requires an infinite force. Once again, it is important to interpret formulas correctly.

All of this pertains to the next book, for now we request the discussion to be focused on the phenomena we explain in this current book, The Yaldon Particle Theory: The Explanation of Thermodynamics, Propagation of Rays, Related Phenomena, and the Model of the Atom. Thank you.

We have mentioned that there are stars that are older than the moment that the "big bang" happened on this same thread.
You can take a look at post#17.

According to The Yaldon Particle Theory, starting from page 24, we show how the observable universe will always be spherical with the observer in its center. Equation (2-7) will provide the reason for this.

Updating... or patching the flaws in your models? We notice that lately there has been more patching than updating. The reason for this is due to the lack of having a solid formula acting as a cornerstone (using the term "cornerstone" as a form of imagery) to build an accurate model upon.

Because we know better.
Once again, please take a look at Equation (2-7) on page 27 of The Yaldon Particle Theory. This formula provides a more reasonable and accurate representation of the scope of the observable universe; without having to rely on a "word salad."

7. ### yaldonTheoryRegistered Member

Messages:
53
General Relativity has failed to explain why the speed of light is constant. The Yaldon Particle Theory provides an explanation for the constant speed of light, starting from page 37. Also, The Yaldon Particle Theory doesn't consider the speed of light as the absolute speed (top speed limit) for objects in the universe. Having an absolute speed for objects violates Newton's Second Law of Motion.
Proof is as follows:
$v(t)=F\frac{t}{m}$​
As long as there is a force applied onto a mass, the velocity will have no limit with time.

This is why we are posting on SciForums, to discuss this matter in order to provide a better service to everyone. We have done our best to express ourselves in written and mathematical form.

Please feel free to take any amount of time needed in order to clarify the concerns you may have. Keep in mind that this is a new model, and it will take time for one to adapt.

Since this theory covers all phenomena in the universe, a specification for boundaries must be made.
For example: We want to calculate the speed of a car that is moving on the surface of the Earth. The Earth is revolving around the Sun, and the entire Solar System revolves around the black hole in the center of the Milky Way Galaxy. When calculating the speed of a car moving on the surface of the Earth; we will not factor in the speed of the Earth revolving around the Sun, as well as the speed of our solar system revolving around the black hole. This is the reason as to why we require bounded systems with an observer. In this example, the bounded system is the planet Earth. No philosophy or theology required.

We have already explained to you (and "James R.") how you have misinterpreted the formula $Energy=mv^2$.

As we have stated, this issue will not matter, due to the fact that it will not change any of the formulas in The Yaldon Particle Theory. But, since you so strongly insist, we will add "average" to that particular instance in The Yaldon Particle Theory.

8. ### yaldonTheoryRegistered Member

Messages:
53
The Math in The Yaldon Particle Theory is correct. This theory is not intended for a beginner in Algebra, but we will try to clarify your concern.
Starting from the top of page 5, in The Yaldon Particle Theory, we state:
"The distance (d) will be equal to the average speed ($s_y$) times one second"
Then:
$d=s_y(\frac{meter}{second})\cdot 1(second)$​
One can clearly see that the units for second will cancel out. As a result:
$d=s_y(meter)$​
In this instance, d equals the value of $s_y$. In the same manner, $V_B$ will equal the value of $s_y$.

Then explain to us, from your perspective, what an ether is?
We have explained what a yaldon particle is. We must pull your attention to the fact that each yaldon particle can move independently and freely from one another (with no forces or fields between them). The only instance of a force is upon their collisions with one another.

How one can come to the conclusion of Theology from the statement "...the observer is in motion" is beyond us. Please look again at our example of calculating the speed of a car on the Earth.

Once again, you have arrived at a false conclusion

The Yaldon Particle Theory has not violated any Laws of Newton, and we have shown how General Relativity violates Newton's Second Law of Motion.
The discussion of the emission spectra in low-pressure gases begins on page 32 of The Yaldon Particle Theory.
The Michelson-Morley experiment proves that the yaldon particles are not an ether.

Once again, you have misinterpreted Newton's Second Law. Also, we have already shown you the reason why.

Are you insinuating that you are more knowledgeable than all the people who have viewed the cornerstone (Ft=mv) of The Yaldon Particle Theory for the past 10 months? There is nothing that needs to be fixed with the cornerstone for the yaldon theory.
Ft=mv is a valid statement (formula).

9. ### James RJust this guy, you know?Staff Member

Messages:
31,288
yaldonTheory:

Off-topic, but I'm interested: you keep saying "we" in your posts. Is the yaldon theory the work of a team or an individual? How many people are involved?

Energy is a derived quantity. As you say, its units include mass. Standard physics defined energy in terms of the work done by a force, so the dependence on mass follows from the dependence of force on mass. The yaldon revolution in physics is not needed to explain this.

Yes.

What do you mean by "the amount of energy that is available at any moment"? What kind of energy are you talking about? In what forms do we find this energy, for a point particle of mass m and no internal structure? How can this energy be extracted?

No. Do you think Einstein just guessed at the formula? He didn't; he derived it using the postulates of special relativity. The reason for the lack of the 1/2 in Einstein's formula there (which is about rest energy) is clearly justified, whereas the reason why your kinetic energy is doubled has no apparent justification at this point in our discussion, other than that you made a mistake in your yaldon theory.

The correct parts of yaldon theory that I have seen so far seem to me not to be novel, but to be copies of basic Newtonian physics. And the novel parts of yaldon theory that I have seen tend to be incorrect for various reasons.

Do you need more time to become adapted to accepted physical theories?

Wasn't that precisely the objection that was put to you?

Are you claiming that accelerations and forces can be infinite?

No, it hasn't. The constancy of the speed of light follows directly from the invariance of the laws of physics in different inertial reference frames.

Have you studied any relativity at all? Your problem here seems to be that you are incorrectly applying Newtonian equations to situations that need to be treated relativistically.

Also, where's the observational/experimental evidence for any object travelling faster than the speed of light? Why have we never observed such a thing? Does yaldon theory explain that?

No you haven't. You've made a vague allusion to us leaving something out, but you haven't explained at all. I look forward to your full explanation in an upcoming post.

On that note, I am wondering whether yaldon theory uses calculus at all. Does it? Or can the originators of yaldon theory only cope with algebra?

Not knowing calculus would help to explain why you got the work-kinetic energy theorem wrong.

General relativity replaces Newton's second law of motion, so of course it violates it. The concept of force needs to be generalised in general relativity. If that isn't done then momentum is no longer conserved in an isolated system.

Is yaldon theory equivalent to Newtonian mechanics, or does it make novel predictions that can be tested experimentally?

That cornerstone looks to me like a somewhat incorrect statement about impulse, culled from Newtonian physics. Normally, it would be expressed as

$F\Delta t = m\Delta v$

This is for a finite time interval. In the limit as $\Delta t$ goes to zero, we're just back to $F=ma$, which is of course Newton's second law.

Can you please explain why your formula is revolutionary, when it is really just a thinly disguised version of Newton's second law?

10. ### NotEinsteinValued Senior Member

Messages:
1,374
Sure it can. The fact that you claim it can't shows you fundamentally do not understand the maths (or its notation) used in modern day physics.

Indeed that is important. Alright, tell you what. Since the choice of coordinates is free in Newtonian physics, I choose tomorrow to be t=0. I suggest you start preparing for the end, because according to you, the universe is going to undergo infinite acceleration tomorrow!

In other words, you clearly have no idea what the "t" really stands for. Remember, it's important to interpret the formulas correctly!

We are not allowed to discuss the cornerstone of your yaldon theory just yet? In that case, I'll wait for the next book. If I can't understand the non-mainstream notation you are using, and you are (for now) unwilling to explain it, there is no chance for me to interpret your formulas correctly.

In other words, in order for me to interpret your formulas correctly, you have to first explain them correctly. Your refusal to do so means I would be wasting my time reading the rest of your book.

Point my to any particle in mainstream science that has "average speed" as a property. Go on!

I see no link to any evidence in that post either. Please provide evidence for this claim.

Also, while you're at it, please derive the Cls graph of the CMB from your theory. I mean, if you're going to replace BB theory with something else, you should at least be able to describe that.

I cannot comment on this, because you have given me no chance of understanding it. You're still keeping me stuck at your cornerstone formula.

Well, the way things are looking now, your cornerstone formula is fundamentally wrong (especially at t=0), so you also need to do some updating and patching.

Yeah, it relies on formula-salad. (Also, it's kinda arrogant to think you know better than all scientists.)

Please provide me with a link to text describing why the BB theory's explanation of the BB moment is "word salad". Oh, and I just noticed: "everywhere and anywhere at all times": that's not what BB theory predicts. Only the "everywhere"-part is correct, and only in layman terms. Maybe you need to brush up on what the BB theory really predicts and says before you start throwing rocks around?

11. ### NotEinsteinValued Senior Member

Messages:
1,374
Photons are massless, so you're dividing by zero here too? That's... special.

Please express yourself better, especially when it comes to your cornerstone formula. Leaving its entire discussion and explanation out is bad form.

Please explain what the "t" stand for exactly in your cornerstone formula.

So you are arbitrarily restricting your model to a bounded system, even though you're throwing out other bounded systems. You do know that GR has no problem with unbounded systems, right? Your theory appears to be less applicable than GR!

I haven't even touched that formula, so this argument is non-sense. Please explain to me how I've misinterpreted your cornerstone formula. Specifically, what does "t" stand for?

Thank you for making your definition of a yaldon particle consistent throughout your text; it will improve the readability of it.

12. ### NotEinsteinValued Senior Member

Messages:
1,374
All right, so you are using a non-standard notation. This is very confusing to do without properly describing it in your text. Can you please add to your text that you are "silently dropping/adding factors of "1s where the units demand it", or something along those lines?

How about the medium in which light propgates? The substance that is waving when an electromagnetic wave passes by?

Please look up how to do a Galilean transformation, and apply that to your observer in motion to get to an observer in rest. You fundamentally misunderstand the way observers work in Newtonian physics (let alone GR).

Erm... you are the one constantly dividing by zero? But you're right, I shouldn't boast like that. I withdraw that statement.

Please explain to me how having infinite forces (remember, t=0 !) is not a violation of Newtonian physics.
I do not see any spectrum describe in the section starting on page 32? Could you perhaps do an explicit derivation of the hydrogen spectrum (intensity versus frequency, or something like that), so that we 1-1 can compare with what mainstream science models predict?

You still haven't explained your formula properly. I may be drawing that conclusion prematurely, but for some reason, you seem unwilling (or unable?) to explain why your formula predicts a catastrophic end of the universe at t=0...

No, at most I'm disappointed nobody spotted it earlier. It took me two tries as well! And not getting any (negative) comments for 10 months does not truth make.

And thus the universe endeth, because NotEinstein chose to set t=0 tomorrow.

13. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

Messages:
10,353
Apparently the people that looked at your theory did not pick up on the glaring mistakes in you paper or you did not understand their issues with your paper. This is your lucky day. Now that you have seen the issues you can go about fixing the errors so that your paper can do a better job or reflecting what is observed. Don't forget that you also need to bring your paper into line with the observations about the expanding universe, the invariance of the speed of light, etc. Good luck!

Last edited: Sep 27, 2017
14. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

Messages:
10,353
That's weird since:

$F=m\frac{dv}{dt}$

That certainly looks like force is a function of time, doesn't it?

15. ### yaldonTheoryRegistered Member

Messages:
53
Let us discuss one issue at a time.
Can any of you explain to us how a photon can be massless, when the formula for the energy of a photon has mass in its units?
The formula, E=hf, proves that a photon is false; due to the photon's property of being massless.
Also the formulas for potential and kinetic energy, both, require a mass. Since propagated rays have energy, The Yaldon Particle Theory rejects the notion of a photon; due to a photon being massless, yet possessing a value for energy.
As you say, energy is a derived quantity from mass. This is standard physics, and The Yaldon Particle Theory has no issue with standard physics. Standard physics obeys Newton's Laws. Then we don't need any alternative physics from alternative theories; due to the fact that The Yaldon Particle Theory can explain all phenomena using standard physics. Thank you.

16. ### Dr_ToadIt's green!Valued Senior Member

Messages:
2,193
That would be hands waving, and drool flying about...

YaldonTheory: Another fucking nutbag.

Did you finish grade school, or were you brain-damaged as a result of talking crap like this on a bus?

Yaldon, you should run for political office. You'd do better. Less credibility required, no rigor, and you can lie with impunity.

17. ### exchemistValued Senior Member

Messages:
7,226
This reasoning is false.

For example, consider electrostatic potential energy. That is not a function of mass either: E= (1/4 πε) . qQ/r

Here, the dimension M only appears in the permittivity, ε (Q²M⁻¹L⁻³T²) . Just as it does in Planck's constant (ML²T⁻¹).

So it is perfectly possible to have forms of energy that do not depend on mass.

Thank you.

Last edited: Oct 1, 2017
18. ### NotEinsteinValued Senior Member

Messages:
1,374
Yes indeed, how about one of my points raised days ago?

Interestingly, the equation of the mass of a massless particle also has mass in it. Look! $m=0 kg$. So your question as stated is nonsensical.

Funny how you claim to have a theory explaining just about everything, but you don't even know how the fundamental constants of nature work.

Are you claiming potentials and kinetic energy don't exist for massless particles?

And here I was, thinking that energy was a conserved quantity derived through Noether theorem, with no mass being required at all. I guess you and me have very different ideas of what "standard physics" is!

19. ### yaldonTheoryRegistered Member

Messages:
53
We find that this thread has been derailed; due to us being quoted out of context and supplemented with straw man logical fallacies.

For example our last post was:
And "NotEinstein" response is:
Clearly, one can see our original statement being dissected and supplemented with several straw man logical fallacies.
This behavior has devalued the integrity of the entire thread.

20. ### NotEinsteinValued Senior Member

Messages:
1,374
Discussing your theory is derailing this thread? Then why did you post it on a public discussion forum in the first place?

Then point them out. If I quoted you out of context, I will apologize and correct it. If I used straw man fallacies, I will apologize and correct them.

Then point them out. Again, I will apologize and correct them.

And if other members committed them, I will try to get them to apologize and correct them.

I would argue that the thread has indeed been devalued, but by your dodging of questions. Multiple people have now pointed out your cornerstone formula is flawed, and you haven't adequately responded to that at all.

Right now, from where we stand, your theory lies in shambles, and you are here only complaining about thread derailment.

21. ### exchemistValued Senior Member

Messages:
7,226
But you can still reply to my post 114, can you not?

This was a short and objective response to the question you yourselves posed in post 112, pointing out that electrostatic potential energy also does not involve mass and thus one cannot conclude from the absence of mass in E=hν that it must be wrong.

I await your response with interest.

22. ### arfa branecall me arfValued Senior Member

Messages:
5,638
I think there might be a fundamental misunderstanding of photon mass, and the notion that mass is "derived from" energy or vice-versa.

Photons have zero rest mass, which in no way says photons have zero mass; if photons were truly massless they would have zero kinetic energy and there are plenty of experiments that show they have nonzero kinetic energy and nonzero momentum. Further, photons interact with gravity (the bending of light), which implies they do have mass. (But not Newtonian mass!)

Then there's the fact that electrons (with nonzero mass) interact by exchanging photons, which alters the electron's momentum. I think Alphanumeric once explained here why the notion that photons are massless isn't a proper one, and that explanation included plenty of equations. It kind of revolves around photons having zero rest mass but not ever being at rest (because of those nasty laws of physics, in particular special relativity).
Paraphrasing a certain well-known rock group:
"If ye don't have mass, how can ye have any momentum? Ye can't have any momentum if ye don't have any mass!"

Yaldon theory looks suspiciously like a nothing-burger.

Last edited: Oct 4, 2017
23. ### exchemistValued Senior Member

Messages:
7,226
I think we've had discussions on this forum before about the pros and cons of working in terms of "relativistic mass", vs. rest mass. There is a case for both, though modern practice seems to be to reserve the term "mass" for rest mass.

Personally I dislike treating the energy of a photon as "kinetic", since the Newtonian formula 1/2 mv² can't be applied, and nor can even the more complicated relativistic one, as you get a denominator of zero when v=c. For a photon, E=pc (from the energy-momentum relation E² = m²c⁴ + p²c²) . Applying de Broglie's relation, p = h/λ, and bearing in mind that c=νλ, one gets E = h/λ . νλ = hν.

In the context of this thread, I think it is more important to disabuse this Yaldon person or persons of the notion that all forms of energy are dependent on mass. The example I gave of electrostatic potential energy was intended to illustrate this.

origin likes this.