karl

06-08-2009, 10:05 AM

I put his in a seperate thread so joe and I won't get mixed up!

Karl said: Concerning cincrob on rest mass, I understand him to be saying that he has mass = hf/C^2 only all the time. Not true! This is what I consider rest mass and is valid as a boundary condition only when Φ approaches zero.

cinci said: No, that's not what I said. YES YOU DID! YOU KEEP SAYING WAVE ENERGY IS ALL THE ENERGY AN EM WAVE HAS. YOU USE hf=mC^2 TO DESCRIBE A PHOTON IN TRANSIT.

First, there is no rest mass for a photon. A particle with rest mass cannot achieve the speed c.

Karl: I say: C is natural rest state! I REPEAT THIS

cinci: Whatever that means. Why don't you use the terms that all the real physicists in the world use. If you ever intend to publish your paper, you can't just make up your own terms or worse, use the same terms as everybody else but have different meanings for them.--- I DON'T!!

*******************

cinci: Second the mass of a photon is not hf/c^2 all the time. As the photon moves out of a gravitational field, the frequency goes down and so does the mass.

Karl: I reply , not true for freq. WHY DON'T YOU GIVE ME THE EQUATION THAT DESCRIBES THIS IN TERMS OF Phi.

cinci: Then you have to presume that moving against the force of gravity can be done without energy output, a violation of Newton and the conservation of mass/energy.

***********************

*******************************LET'S QUIT PLAYING AROUND HERE! I AM TALKING ABOUT A BINARY SYSTEM WHERE THE ELECTROMAGNETIC WAVE COMPONENT ACTS AS AN OSCILLATER AND GENERATES A PILOT WAVE (DEBROGLIE WAVE) THRU THE AETHER AND THAT DEBROGLIE WAVE INDUCES ENERGY FROM THE AETHER , BY WAVE INTERACTION, INTO THE PILOT WAVE THUS MAKING THE TOTAL ENERGY GREATER THAN THE EM WAVE ENERGY, THE EXCESS OVER THE EM WAVE ENERGY, hf, RESIDING IN THE DEBROGLIE COMPONENT. INITIALLY, AT EMISSION THE ENERGY IN THE DEBROGLIE WAVE IS 2(hfo -hf), WHERE fo IS THE FREQUENCY A EM WAVE WOULD HAVE IF IT WERE EMITTED IN A GRAVATATIONALLY FREE ENVIROMENT, I.E. Phi EQUAL TO ZERO. THIS IS IMPOSSIBLE , OF COURSE, SO YOU HAVE TO USE THE LIMIT "AS Phi APPROACHES ZERO". ----THE ENERGY IN THE DEBROGLIE WAVE IS DYNAMIC AND THAT IS WHERE YOUR TOTAL ENERGY LOSS COMES FROM AS A PHOTON CLIMBS OUT OF A GRAVITATIONAL WELL.

iM GOING TO STOP HERE BECAUSE IF YOU CAN'T SEE THE POSSIBILITY OF A BINARY SYSTEM, STOCASTIC QM, THEN THER IS NO FURTHER REASON TO TALK ON THIS SUBJECT. THE TOTAL ENERGY EQUATION AND THE FREQUENCY EQUATION ARE IN AGREEMENT WITH EINSTEIN. (I did not say the interpretation of them is! He admitted to not attacking the problem of describing a basic particle. ************************************************** ******

karl replies: This is all that applies, ergo, is correct:

Concerning cincrob on rest mass, I understand him to be saying that he has mass = hf/C^2 only all the time .

cinci: This is a mis-statement of what I said. What I said was that when an atom creates a photon, which happens when an electron falls from a higher state of excitation to a lower one, the photon is created with the same energy regardless of gravitational field strength. As I have consistently said, there is no such thing as rest mass for a photon. So when you state that my position on rest mass is mass = hf/c^2, you are misquoting me.

*************************

Karl: This is what I consider rest mass and is valid as a boundry condition only when Φ approaches zero.

cinci: Again, you need to stop using terms in ways that are clearly different than the way everybody else uses them.

****************

Karl: So,

Total Energy =mC^2 = moC^2[e^ + {(1/c^2) Φ }]

so total mass = m = mo[e^ + {(1/c^2) Φ }] anytime!

cinci: Two questions:

1. What does "e^" mean?

2. Shouldn't the equation have a minus sign?

*******************

Karl: wave energy = hfo[e^ - {(1/c^2) Φ }] and is not dynamic!

cinci: Energy is not dynamic?

**************

Karl: wave frequency = f = fo [e^ - {(1/c^2) Φ }] where hfo=moC^2 only when Φ approaches zero.

dE = Total Energy - Wave Energy for a given Φ and this dE is dynamic and changes with force applied. Any force!

This dE is composed of a pilot wave in the form of a debroglie

wave.

These same conditions apply to the electron. The difference is that the EM wave travels linearly for light and for the electron the EM wave travels in a closed circular orbit!

cinci: Where does this idea come from?

*************

karl says: Read this again. If you did read it you must not comprehend it. Try again! Thats all there is to it in a nutshell. the dE is where you lose or gain energy.

cinci: I did read it every time you posted it. And I say to you that you are proposing that a photon can travel through a gravitational field without loss of energy and that violates at least two laws of physics.

**********************

Karl: I will add for both of you that an EM wave traveling at C is in its's natural rest state. C is just a constant that balances the E/m ratio.

The energy change you keep repeating is thoroughly accounted for here. Oh yes cinci, you are flat wrong on the decrease in frequency that you use to explain the energy loss as the EM wave, or even an electron climbs out of a gravitatonal hole.

cinci: You can't make it wrong just by saying it. You have to account for the photon moving against a gravitational field and you don't.

***************************

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Karl said: Concerning cincrob on rest mass, I understand him to be saying that he has mass = hf/C^2 only all the time. Not true! This is what I consider rest mass and is valid as a boundary condition only when Φ approaches zero.

cinci said: No, that's not what I said. YES YOU DID! YOU KEEP SAYING WAVE ENERGY IS ALL THE ENERGY AN EM WAVE HAS. YOU USE hf=mC^2 TO DESCRIBE A PHOTON IN TRANSIT.

First, there is no rest mass for a photon. A particle with rest mass cannot achieve the speed c.

Karl: I say: C is natural rest state! I REPEAT THIS

cinci: Whatever that means. Why don't you use the terms that all the real physicists in the world use. If you ever intend to publish your paper, you can't just make up your own terms or worse, use the same terms as everybody else but have different meanings for them.--- I DON'T!!

*******************

cinci: Second the mass of a photon is not hf/c^2 all the time. As the photon moves out of a gravitational field, the frequency goes down and so does the mass.

Karl: I reply , not true for freq. WHY DON'T YOU GIVE ME THE EQUATION THAT DESCRIBES THIS IN TERMS OF Phi.

cinci: Then you have to presume that moving against the force of gravity can be done without energy output, a violation of Newton and the conservation of mass/energy.

***********************

*******************************LET'S QUIT PLAYING AROUND HERE! I AM TALKING ABOUT A BINARY SYSTEM WHERE THE ELECTROMAGNETIC WAVE COMPONENT ACTS AS AN OSCILLATER AND GENERATES A PILOT WAVE (DEBROGLIE WAVE) THRU THE AETHER AND THAT DEBROGLIE WAVE INDUCES ENERGY FROM THE AETHER , BY WAVE INTERACTION, INTO THE PILOT WAVE THUS MAKING THE TOTAL ENERGY GREATER THAN THE EM WAVE ENERGY, THE EXCESS OVER THE EM WAVE ENERGY, hf, RESIDING IN THE DEBROGLIE COMPONENT. INITIALLY, AT EMISSION THE ENERGY IN THE DEBROGLIE WAVE IS 2(hfo -hf), WHERE fo IS THE FREQUENCY A EM WAVE WOULD HAVE IF IT WERE EMITTED IN A GRAVATATIONALLY FREE ENVIROMENT, I.E. Phi EQUAL TO ZERO. THIS IS IMPOSSIBLE , OF COURSE, SO YOU HAVE TO USE THE LIMIT "AS Phi APPROACHES ZERO". ----THE ENERGY IN THE DEBROGLIE WAVE IS DYNAMIC AND THAT IS WHERE YOUR TOTAL ENERGY LOSS COMES FROM AS A PHOTON CLIMBS OUT OF A GRAVITATIONAL WELL.

iM GOING TO STOP HERE BECAUSE IF YOU CAN'T SEE THE POSSIBILITY OF A BINARY SYSTEM, STOCASTIC QM, THEN THER IS NO FURTHER REASON TO TALK ON THIS SUBJECT. THE TOTAL ENERGY EQUATION AND THE FREQUENCY EQUATION ARE IN AGREEMENT WITH EINSTEIN. (I did not say the interpretation of them is! He admitted to not attacking the problem of describing a basic particle. ************************************************** ******

karl replies: This is all that applies, ergo, is correct:

Concerning cincrob on rest mass, I understand him to be saying that he has mass = hf/C^2 only all the time .

cinci: This is a mis-statement of what I said. What I said was that when an atom creates a photon, which happens when an electron falls from a higher state of excitation to a lower one, the photon is created with the same energy regardless of gravitational field strength. As I have consistently said, there is no such thing as rest mass for a photon. So when you state that my position on rest mass is mass = hf/c^2, you are misquoting me.

*************************

Karl: This is what I consider rest mass and is valid as a boundry condition only when Φ approaches zero.

cinci: Again, you need to stop using terms in ways that are clearly different than the way everybody else uses them.

****************

Karl: So,

Total Energy =mC^2 = moC^2[e^ + {(1/c^2) Φ }]

so total mass = m = mo[e^ + {(1/c^2) Φ }] anytime!

cinci: Two questions:

1. What does "e^" mean?

2. Shouldn't the equation have a minus sign?

*******************

Karl: wave energy = hfo[e^ - {(1/c^2) Φ }] and is not dynamic!

cinci: Energy is not dynamic?

**************

Karl: wave frequency = f = fo [e^ - {(1/c^2) Φ }] where hfo=moC^2 only when Φ approaches zero.

dE = Total Energy - Wave Energy for a given Φ and this dE is dynamic and changes with force applied. Any force!

This dE is composed of a pilot wave in the form of a debroglie

wave.

These same conditions apply to the electron. The difference is that the EM wave travels linearly for light and for the electron the EM wave travels in a closed circular orbit!

cinci: Where does this idea come from?

*************

karl says: Read this again. If you did read it you must not comprehend it. Try again! Thats all there is to it in a nutshell. the dE is where you lose or gain energy.

cinci: I did read it every time you posted it. And I say to you that you are proposing that a photon can travel through a gravitational field without loss of energy and that violates at least two laws of physics.

**********************

Karl: I will add for both of you that an EM wave traveling at C is in its's natural rest state. C is just a constant that balances the E/m ratio.

The energy change you keep repeating is thoroughly accounted for here. Oh yes cinci, you are flat wrong on the decrease in frequency that you use to explain the energy loss as the EM wave, or even an electron climbs out of a gravitatonal hole.

cinci: You can't make it wrong just by saying it. You have to account for the photon moving against a gravitational field and you don't.

***************************

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