A Model for the Propagation of Visible Light and Other Rays

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

  1. exchemist Valued Senior Member

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    6,623
    How do you account for electrostatic potential energy? This is not a function of mass but of charge, of course.
     
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  3. yaldonTheory Registered Member

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    Did you completely ignore our mathematical proof from post #138.
    You might not be able to see this... but the math can.
     
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  5. yaldonTheory Registered Member

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    As we have stated before, these fields are from the dynamic arrangement of yaldon particles, and yaldon particles have a mass \(m_y\)
     
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  7. exchemist Valued Senior Member

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    OK so how do you derive the formula for the potential energy: E= (1/4 πε) . qQ/r, which is not a function of mass?
     
  8. arfa brane call me arf Valued Senior Member

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    5,365
    Nope. I got stuck at the part that claims a mass with constant velocity has potential energy. Potential energy relative to what other mass?

    Besides, your claim contradicts every physics textbook I've ever read. Please explain why I should set aside centuries of physics and accept your claim about potential?
     
  9. yaldonTheory Registered Member

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    53
    This is later works, for now let us keep the discussion focused to the 77 pages in The Yaldon Particle Theory: The Explanation of Thermodynamics, Propagation of Rays, Related Phenomena, and the Model of the Atom.

    The potential energy will turn into kinetic energy while the mass begins to have a change in its velocity.
    Anyone who understands The Yaldon Particle Theory will not want an explanation from any other theories.
     
  10. exchemist Valued Senior Member

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    6,623
    So yet another example of a basic phenomenon you cannot yet account for. Funny, it seems every time I ask you to account for something important in physics, you can't do it until later.

    The reason I raise electrostatic potential is because you made a claim that E=hν must be false as mass did not appear in it.

    I then pointed out that neither does E= (1/4 πε) . qQ/r. Do you believe this formula is correct, or do you think it must be false too as mass does not appear in it either? Surely if your logic for E=hν is sound, you must also dismiss this formula too. Yes?
     
  11. arfa brane call me arf Valued Senior Member

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    5,365
    Nope. A mass with constant velocity already has a defined kinetic energy, according to those centuries of physics I refer to.
    A change in velocity is an acceleration, moreover. An acceleration means there is a force acting on the mass, where does the force come from?
    Yawn.
     
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  12. yaldonTheory Registered Member

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    53
    We state that a photon is false, due to its property of having no mass yet possessing a value for energy.
    We have no issue with the constant h, or the formula E=hf. The Yaldon Particle Theory accurately declares h with the formula (2-b) on page 24. Here is the formula:
    \(E_p=(M_g\cdot v_p \cdot t_p)\cdot f\)​
    The value contained within the parenthesis is h.

    It appears to us that are committing slander, show us a direct quote where we state your above allegation of:
    Once again, we state that a photon is false. Here is our direct quote:
    Please be mindful of attributing false claims to our statements.

    So we take it then that you are satisfied with The Yaldon Particle Theory's explanation for all of these phenomena:
    ...and you are only waiting on the explanation for electrostatic potential energy, yes?
    We wonder, should we explain electrostatic potential energy; will you have another phenomenon that you'll wish for us to define? For example: the gravitational field, magnetic field, as well as the electric field are beyond the current scope of this work.
    Don't you realize that you are deterring the topic of this thread away from the 77 pages within this .pdf?
    https://drive.google.com/file/d/0B-vl7kjDEZ8HV2lZN1dId3JpcUU/preview

    Newton's Laws have been misinterpreted. Kinetic energy only applies to a mass with a variable velocity, which requires a force in order to change this constant velocity. For a mass that has no change in its velocity, it will have potential energy.
    Once again, these comments do not pertain to our current work. Thank you.
     
  13. arfa brane call me arf Valued Senior Member

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    5,365
    By you, or by Newton?

    Please think carefully about your answer.
    So kinetic energy is equivalent to force, or is that a misintepretation of Yaldon's Laws?
    Not in the universe I live in.

    Your theories are based on nothing more than bland statements, most of which you haven't got any supporting evidence for. That is, your theories predict nothing useful, and don't suggest any kind of experiment. So your theories are not science, but daydreaming, in which you become the hero who saves science from its ignorance I assume.

    Are you wearing your undies on the outside though? That's very important.
     
  14. origin Trump is the best argument against a democracy. Valued Senior Member

    Messages:
    9,993
    Yaldon Theory:

    This does not look right.
    \(\int\limits_0^{v_{max}} F(v)~ dv = \frac{m}{t}\int\limits_0^{v_{max}}v~dv \)

    You are treating t as a constant and it is a variable, you cannot move it to the left of the integral sign if it is a variable. Since your are integrating with respect to velocity over the time of 0 to Vmax it certainly seems like a variable.

    What do you think?
     
  15. yaldonTheory Registered Member

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    53
    The formula for that integral has come to a correct representation for kinetic energy. If that formula for the integral \((\int\limits_0^{v_{max}} F(v)~ dv = \frac{m}{t}\int\limits_0^{v_{max}}v~dv)\) was wrong, then the result would be wrong. The resulted expression \(E_k=\frac{1}{2}mv^2_{max}\) is correct.

    And t is a constant, it is a fixed amount time that is required to move a mass from zero to \(v_{max}\).

    Here is the proof once more with extra lines added for clarity:
    For Kinetic Energy:
    \(F(v)=\frac{m}{t}v\)
    \([F(v)=\frac{m}{t}v]\cdot dv\)
    \(\int\limits_0^{v_{max}}F(v)\cdot dv=\frac{m}{t}\int\limits_0^{v_{max}}v\cdot dv\)
    \(Fv_{max}=\frac{1}{2}\frac{m}{t}v^2_{max}\)
    \(F[v_{max}\cdot t]=\frac{1}{2}mv^2_{max}\)
    Where \(v_{max}\cdot t = s_{max}\)
    \(F\cdot s_{max}=\frac{1}{2}mv^2_{max}\)
    Where \(F\cdot s_{max}=E_k\)
    \(E_k=\frac{1}{2}mv^2_{max}\)
    For Potential Energy:
    \(F(s)=m\frac{v}{t}\)
    Where \(m\frac{v}{t}\) is a constant value; then the velocity will also be a constant \((v_c)\).
    After multiplying both sides of the above equation by ds and taking the limit from zero to \(s_{max}\):
    \(F(s)\int\limits_0^{s_{max}}ds=m\frac{v_c}{t}\int\limits_0^{s_{max}}ds\)
    \(F\cdot s_{max}=m\frac{v_c}{t}\cdot s_{max}\)
    Where \(F\cdot s_{max}=E_p\)
    \(E_p=mv_c\frac{s_{max}}{t}\)
    Where \(\frac{s_{max}}{t}=v_c\), since both are properties of the same mass (m)
    \(E_p=mv^2_c\)
    When \(E_k=E_p\) for the same mass (m)
    \(\frac{1}{2}mv^2_{max}=mv^2_c\)
    Then:
    \(\frac{1}{2}v^2_{max}=v^2_c\)​
    Or:
    \(v^2_{max}=2v^2_c\)​
     
  16. arfa brane call me arf Valued Senior Member

    Messages:
    5,365
    Rubbish.

    Newton's laws of motion have allowed real scientists to launch satellites into orbit, and send probes deep into the solar system.
    How does that represent a misinterpretation? You don't have a theory, you have a fantasy.
    There is no zero velocity, there is however relative velocity. You haven't understood this at all; your theory is bollocks.

    Mods please move this thread to the Cesspool where it belongs.
     
  17. NotEinstein Registered Senior Member

    Messages:
    548
    Clearly you are using the variable \(t\) incorrectly. Do you mean \(\Delta t\)?

    Clearly you are using the variable \(t\) incorrectly. Do you mean \(\Delta t\)?

    Clearly you are using the variable \(t\) incorrectly. Do you mean \(\Delta t\)?
     
  18. NotEinstein Registered Senior Member

    Messages:
    548
    So if the answer is correct, the derivation must also be? That is wrong on so many levels, I don't even know where to begin!
     
  19. exchemist Valued Senior Member

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    6,623
    I responded to your challenge to show "how a photon can be massless, when the formula for the energy of a photon has mass in its units".

    The formula obviously doesn't have mass in its units (the units are those of energy), but, giving you the benefit of the doubt, I assumed you must have meant that M appears in the dimensions of energy.

    And so I gave an example of another formula, for a different type of energy, which (obviously) also has M in its dimensions even though it is also massless. Thus illustrating the flaw in your reasoning.

    And now you are trying to wriggle out of dealing with this point.
     
  20. origin Trump is the best argument against a democracy. Valued Senior Member

    Messages:
    9,993
    Doing math that violates basic principles to get the 'right answer' is not math.

    Oh my goodness! You do not seem to understand what a constant is and what a variable is. That is going to kill your math operations.

    If I use your definition of a constant then integration is now so easy a 10 year old can do it. Here is why:
    In the integral you wrote time goes from 0 to time tmax, as velocity goes from v to vmax. Lets just say tmax is 10 sec., so the t goes from 0 - 10. You are saying this is a constant because it is a fixed change in t.
    Likewise for velocity it goes from v to vmax. Lets say vmax is 10 m/sec., so v goes from 0 - 10 m/sec. So using your definition of a constant, velocity is a constant because the change in velocity is fixed.
    In other words everything is a constant using your definition.
    So the integral of force, (mass x acceleration) is: \(1/2 (ma)^2\).
    Your definition of a constant means that no matter what is integrated you simply integrate a constant, put 1/2 in front of the constants and square them, done!

    Another way to look at what you are doing is t0 try and properly integrate using your definition of constant time.
    F=ma =( m/t)v.
    \(\int{\frac{m}{t}} v~dv\)
    mass and time are constants according to your definition (of course mass really is a constant)
    \(\frac{m}{t}\int v~dv\)
    But time is a constant!! so we have to move that out of the integral.
    \(\frac{m}{t^2}\int s~ds\) where s is a distance
    and it just gets worse from there...

    here is why time is NOT a constant:
    \( a =\frac{dv}{dt} \) Acceleration is a change in velocity over a change in time.

    Another major problem is that you are integrating force trying to get energy which is wrong.

    Energy is not the integral of force!
    Energy is force applied over a distance.
    E=f *s So you should be integrating like this:

    \(\int ma~ds\) energy is the integral of force over a change in distance.

    You have made several basic math errors it appears to me. Sorry.....
     
    Last edited: Oct 11, 2017
  21. yaldonTheory Registered Member

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    53
    It is apparent that you have made contradictory remarks to one another, since you do not have a solid foundation to stand upon.

    If you are so satisfied with Newton's Laws, how come you didn't complain when the user on SciForums known as "Not Einstein" stated that Newton's Laws are an "approximation." Here is a direct quote of "NotEinstein" stating this:
    The Yaldon Particle Theory claims that Newton's laws of motion are sound with no approximation within them, this is why real scientists who accept the Laws of Newton are able to launch these satellites and probes into space. This only proves that our cornerstone formula is valid. Thank you for your help.

    Here is another contradictory remark between "arfa brane" and "James R."
    And "James R." stated:
    The staff member on SciForums known as "James R." seems to agree that a mass will move from a velocity of zero.

    And how many times we must repeat ourselves when we state that the velocity of the mass is within a bounded system, we had stated this already in post#104.
    And here we are, once again, finding ourselves teaching the fundamentals of Algebra and Calculus in order to prove that 1+1=2.
    The equation \(F(v)=\frac{m}{t}v\) declares that F is a function of v only. When this equation is plotted on a Cartesian coordinate system, it will appear as shown in the following image: https://drive.google.com/file/d/0B-vl7kjDEZ8HRzdJRzJoRlROdHM/view

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    The slope of the line in the image above will be \((\frac{m}{t})\), and the shaded area underneath this line will be equal to \((F\cdot v_{max})\).
    Then:
    \(F\cdot v_{max}=\frac{v_{max}\cdot\frac{m}{t}v_{max}}{2}\)
    Where \((\frac{m}{t}v_{max})\) is the value of the force, F, when \(v=v_{max}\)
    \(F\cdot v_{max}=\frac{1}{2}\frac{m}{t}v^2_{max}\)
    \(F\cdot (v_{max}\cdot t)=\frac{1}{2}mv^2_{max}\)
    Where \((v_{max}\cdot t)=s_{max}\)
    \(F\cdot s_{max}=\frac{1}{2}mv^2_{max}\)
    Where \(F\cdot s_{max}=E_k\)
    Then:
    \(E_k=\frac{1}{2}mv^2_{max}\)​
    Finding the formula for kinetic energy really is that easy. Real Math and Science does not care for the complications brought about by the useless jargons and terminologies.

    In the same manner, the equation \(F(s)=\frac{m}{t}v\) declares that F is a function of s only.
    \(F(s)=\frac{m}{t}v\) where s is the distance.
    In the above equation F is a function of s only, then the term \(\frac{m}{t}v\) represents a constant value for the force, F.
    For example: When F(s)=5 Netwons, it can be plotted on a Cartesian graph as seen in the following image: https://drive.google.com/file/d/0B-vl7kjDEZ8HNzNKZ3I3V2FFRW8/view

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    According to the image above, the \(Work=5\cdot s_{max}\)

    But, in this instance \(F(s)=\frac{m}{t}v\); and this can also be represented on a Cartesian graph as seen in the image below: https://drive.google.com/file/d/0B-vl7kjDEZ8HSVFCT0UtY0s5ODg/view

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    Then:
    \(E_p=s_{max}\cdot \frac{m}{t}v\)
    Where \((\frac{m}{t}v)\) is a constant value of the force, F.
    \(E_p=mv\frac{s_{max}}{t}\)
    Where \(\frac{s_{max}}{t}=v\)
    Since the term \((\frac{m}{t}v)\) is a constant value for the force, then the constant value of v will be \(v_c\):
    \(E_p=mv^2_c\)​

    The proof is in the Math. Once again, this proof for kinetic/potential energy is not part of the topic of this thread. Especially when "arfa brane" already admits that, "Newton's laws of motion have allowed real scientists to launch satellites into orbit, and send probes deep into the solar system."
    There is nothing wrong with the Laws of Newton, as well as the cornerstone for The Yaldon Particle Theory; and both can be applicable for large and sub-atomic phenomena. Thank you.
     
  22. arfa brane call me arf Valued Senior Member

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    5,365
    No. Scientists who understand what Newton's laws are, understand them as an approximation. It's like how we understand Avogadro's number is an approximation--not known to an exact value but close enough for practical purposes.
    You don't have a cornerstone formula or a valid formula, you have an incorrect formula.
    Please give an example of an object with mass and a velocity of zero. And please stop misrepresenting other people at this site.

    Your equations are just wrong, you don't know how to do real calculus. I wouldn't hire you to write a program, if that's the quality of your work--it's rubbish and you're a big fraud.
     
  23. origin Trump is the best argument against a democracy. Valued Senior Member

    Messages:
    9,993
    It is really best to know a little math prior to trying to teach it.

    Since you did not respond to my post can I assume you now realize that your following integration is not correct?

    \(\frac{m}{t}\int v~dv\) (time is a variable so it cannot be moved out of the integral)

    If by chance you are not convinced, go to the nearest community college and ask any math or physics teacher they will clear up your confusion in no time. Or you could ask anyone who has taken a calculus course...

    As far as algebra...
    Unfortunately that declaration would be wrong since velocity is a function of time. For some reason have broken down acceleration to velocity /time. So in your strange equation F is a function of v and t.

    \(F(v,t)=\frac{m}{t}v\)

    Since velocity is a function of time how in the world do you think force is not a function of time?

    I understand that you have put a lot of work into this but at some point you have to (or at least should) face the reality that you have a mish mash of incorrect or misleading equations and ideas. I am sure everywhere you go with this you get the same feed back. Time to start listening to that feed back IMO.

    Hey, at least that Yaldon emblem thingy looks professional!

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    Last edited: Oct 13, 2017

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