Anyone know where einstien put the resistance?

[GORDON HERMA]

I have a question for those of you that are versed in physics. I am kind of a hobbiest and really enjoy reading theories without the equations. As i read about Einstiens equation (e=mc2) I wondered where the mass that is lost due to resistance during the conversion into energy is represented. (according to realitivity if i kick a ball it looses mass when it is in motion) If this is located in this equation could u please reply to this with a bit of explanation for a lost person. I have an equation i tried to work out but i dont know if its correct. ------- --- 2 2

-----------------------------------------------------------------------------E = M-(M/R*C)C

 

energy equals mass minus (mass divided by resistance multiplied by the constant squared), multiplied by the constant squared.

 

Does this make any sence to anyone ? Any info is greatly appreciated 8)

Thanks Gordon....

[gagsrcool]

Hi,

 

Mass is ofcourse lost but not at all cases. Mind it, if mass is lost by doing so, then we would be violating Dalton's law that mass cannot be created or destroyed but can only be changed from one form to another. There are very exceptional cases such as matter-antimatter collisions and radioactivity where mass is lost. Lost to energy. That is it gets converted to energy. So, energy must also be a form of mass, if we are not to violate Dalton's laws.

 

I could help you better if you give your equations along with the method opted to derive it.

 

gagsrcool

[insane_alien]

the ball gains mass in the form of kinetic energy when you kick it. energy is then lost due to friction with the air(i hope this is the resistance you were talking about) and hence the ball loses this mass. when the ball stops it will be exactly the same mass as before you kicked it. overall no mass is lost and no mass is gained.

[swansont]

No rest mass is gained, period. If you want to use archaic definitions for mass, you should define them at the outset. Otherwise I can say that it changes color when you kick the ball ^{_{but that I've redefined things so that color = speed.}}

[GORDON HERMA]

First i would like to say thanks for the replies and nice to meet you all.

 

Like insane alien said the ball has the same mass when it slows back down. I meant lost more along the lines of one perspective from a state of rest.

To explain what i meant in my original post i wrote 2 explanations.

 

#1 If we were to measure the ball from a state of rest, the ball in motion would be a smidgen less in mass. That smidgen has been converted to kiknetic energy. However the force keeping the form of the ball is greater then the force causing the conversion. So as the ball slows back down the kinetic energy is converted back bit by bit untill the ball reaches its previous state of rest. Now if the force were increased to the point where it approached the speed of light, eventually the mass losses most of its form that is measureable from the state of rest, and is only represented to us as a smidgen of matter that is not converted as it wizzes by. This smidgen is what "R" is supposed to represent. This is how i got to my conclusion Gagsrcool. The way i see it is; if the speed of light can not be surpassed, then there will always be a portion of mass that resists the conversion. So the equation is either missing "R" or i dont fully understand the conversion. Which brings me to the reason i posted. I need to know..hehe

 

Here is the other way to look at it...

 

#2 If the ball were pushed within a field of everyday air, the force keeping the form of the air will cause resistance. If the ball is pushed fast enough that resistance will cause a conversioun of mass into heat energy. (Just like orbital re-entry). If pushed fast enough though a burn up occurs. This burning breakdown of the mass is converted into energy. After all the mass that can be converted at this point from force, is convered, there still exists a portion of mass that did resist the conversion and was left behind in the form of ashes. These ashes also represent "R" along with the smidgen of matter that is left measureable from a state of rest.

 

Hope this makes more sense of what i was asking in my first post.

Is that portion of the mass that represents "R" missing from or not part of the equation. If so i would love to understand it. Thanks in advace for any help.

Gordon...

[swansont]

If we were to measure the ball from a state of rest, the ball in motion would be a smidgen less in mass.

 

 

No. The rest mass is unchanged, and even in the (ugh) relativistic mass analysis, the energy of motion comes from outside the ball — energy is added to the system. (That changes, however, if you look at a self-propelled object, but it also gets more complicated to analyze.)

[GORDON HERMA]

Again i must say i appreciate the discussion very much, and thanks for taking the time to reply to someone that is learning.

No. The rest mass is unchanged

You are right about the rest mass. I guess the way i was thinking about it was, the faster in relativistic motion the ball is, the more the phase (kind of like expanded matter) of the mass has changed along with it the permeability from my state of rest. I guess I meant less permeability of the same matter = less mass in the matter. Just in a different sense.

Can the strong nuclear force binding the form of the ball account for the ball converting back as it returns to a state of rest?

 

If so, there has to be a threshold past a certain speed where the speed force is greater then the strong nuclear force. This would cause a breakdown in the mass, and leave behind "ashes" of some sort(Proof of "R"). Below the speed of breakdown the ball could be converted back to its state of rest and mass because the strong nuclear force is greater. Just like a tug of war...but once one team is in the mud its over. Team Strong Nuclear Force and Team Motion..hehe

Just like the everyday air resistance in #2 that is caused by the opposing strong nuclear force that keeps the air in form (aside from motion i.e. wind), the ball also has this resistance in the form of strong nuclear force keeping its form. If a breakdown of the ball in a field of air occurs, can a breakdown of the ball occur within a field of itself being expanded around the smidgen in #1 that resisted conversion when it reaches that threshold?. The breakdown starting from the smidgen that resisted the conversion (expansion/permeability of mass) in the relativistic point of view in #1.

 

and even in the (ugh) relativistic mass analysis, the energy of motion comes from outside the ball

All conversions absorb some kind of applied force not of itself, into itself to get them rolling. After this happens are the ball and force still considered seperate? Even in a state of rest Strong Nuclear Force is being used to keep the form. Could this be possible?

 

All of this is what i based my equation on. Just trying to figure out if i got it right? E=M-(M/R*C^2)C^2

Gordon...

[swansont]

You are right about the rest mass. I guess the way i was thinking about it was' date=' the faster in relativistic motion the ball is, the more the phase (kind of like expanded matter) of the mass has changed along with it the permeability from my state of rest. I guess I meant less permeability of the same matter = less mass in the matter. Just in a different sense.

Can the strong nuclear force binding the form of the ball account for the ball converting back as it returns to a state of rest?

 

If so, there has to be a threshold past a certain speed where the speed force is greater then the strong nuclear force. This would cause a breakdown in the mass, and leave behind "ashes" of some sort(Proof of "R"). Below the speed of breakdown the ball could be converted back to its state of rest and mass because the strong nuclear force is greater. Just like a tug of war...but once one team is in the mud its over. Team Strong Nuclear Force and Team Motion..hehe

Just like the everyday air resistance in #2 that is caused by the opposing strong nuclear force that keeps the air in form (aside from motion i.e. wind), the ball also has this resistance in the form of strong nuclear force keeping its form. If a breakdown of the ball in a field of air occurs, can a breakdown of the ball occur within a field of itself being expanded around the smidgen in #1 that resisted conversion when it reaches that threshold?. The breakdown starting from the smidgen that resisted the conversion (expansion/permeability of mass) in the relativistic point of view in #1.

 

 

All conversions absorb some kind of applied force not of itself, into itself to get them rolling. After this happens are the ball and force still considered seperate? Even in a state of rest Strong Nuclear Force is being used to keep the form. Could this be possible?

 

All of this is what i based my equation on. Just trying to figure out if i got it right? E=M-(M/R*C^2)C^2

Gordon...[/quote']

 

 

Nope. This is mostly gibberish, from a physics standpoint.

[GORDON HERMA]

Could you please eleborate it for me swansont. The gibberish is because i dont know the correct terminology. I am sorry i cant present this as clearly as i would like. I guess my final question is where did einstien put the missing mass that does not get converted in his equation. If this is represented in the equation E=MC^2 could u please eleborate.

 

If we can not surpass the speed of light then, there is always a portion that does not convert to energy.

 

There are always ashes left behind when i convert firewood to energy.

 

Radioactive materials are always left over when they convert atoms in Nuclear Power Plants (I know the majority of the materials were seperate but they did become radioactive).

 

All this excess comes from the original mass in all these situations but i see no representation for it in the equation E=MC^2. Which is supposed to be mass converted into energy. If they use the mass numbers that they do to calculate their equations then the final results are off by an ashes "R" amount because the equation figures the conversion of the mass in its entirety. If there are ashes left over then there has to be some mass that resists the conversion and should not be calculated as converted.

 

From a purely mathmatical standpoint "Could this account for the left over or "missing" mass "R" that resisted the conversion and was left as ashes or radioactive materials? I know that this stuff was originally part of the matter before the conversion. Does this make any sense?

 

E=M-(M/R*C^2)C^2

 

As a metaphore it seems to me like the equation is a word that has a silent e that is not sounded out.

 

Final Question....

Gordon Herman

[insane_alien]

The strong nuclear force does not hold together a football its electromagnetic forces. strong nuclear force dissapears outside the nucleus of an atom.

 

If we can not surpass the speed of light then, there is always a portion that does not convert to energy

 

nope we see reactions that to a complete mass convertion all the time its called annihilation and involves an anti matter particle and a matter particle colliding and turning into 2 gamma photons(energy) no particles are left.

 

There are always ashes left behind when i convert firewood to energy.

 

This isn't a nuclear reaction. however, if you took the reactants(including the oxygen, match, anything else you need) and got an exact measurement of the mass and took the products after they cooled down and weighed them to the same accuracy there would be a small discrepancy. the lost mass is the energy that was stored in the chemical bonds of the wood and oxygen.

 

Radioactive materials are always left over when they convert atoms in Nuclear Power Plants (I know the majority of the materials were seperate but they did become radioactive).

 

this is because it is not a conversion. just splitting the atom. the mass lost is energy stored in the binding of the atomic nuclei by the strong force.

 

From a purely mathmatical standpoint "Could this account for the left over or "missing" mass "R" that resisted the conversion and was left as ashes or radioactive materials?

 

no. there is no resistance.

 

Does this make any sense?

 

not a bit.

 

As a metaphore it seems to me like the equation is a word that has a silent e that is not sounded out.

 

err... my mind is turning to mush just thinking about what this could possibly mean.

[swansont]

You have to account for all of the energy and all of the mass inany reaction you consider. In fission, for example, the products have less mass than the reactants, and the mass difference shows up and kinetic energy of the fission producs and photons.

 

The actual equation is E² = m²c⁴ + p²c²

 

Motion shows up in the momentum term as part of the total energy. For an object at rest it reduces to E = mc²

[GORDON HERMA]

The strong force does give the football its characteristics of mass in the form of a conglomerate of atoms. The collective is held together by the electomagnetic but the resistance comes from the collective mass. So a combination of both i would say.

The strong nuclear force does not hold together a football its electromagnetic forces. strong nuclear force dissapears outside the nucleus of an atom.

 

 

I would consider a photon a particle,just a particle out of phase with certain

matter. I read a book where they quoted einstien

saying something to the effect of "What would light be like if i could match its speed? Could i hold it?" I think this one is still open for debate. It is possible to catch a single photon on a photographic film. Just my take on it, but i dont know the accepted view.

nope we see reactions that to a complete mass convertion all the time its called annihilation and involves an anti matter particle and a matter particle colliding and turning into 2 gamma photons(energy) no particles are left.

 

 

I agree it is not a nuclear reaction, but the conversion at this primal level still shows one of the reasons i think that there is some matter that resists conversion. I'm sure measurement is possible and that small discrepancy is exactly what I am trying to get a lock on.

This isn't a nuclear reaction. however, if you took the reactants(including the oxygen, match, anything else you need) and got an exact measurement of the mass and took the products after they cooled down and weighed them to the same accuracy there would be a small discrepancy. the lost mass is the energy that was stored in the chemical bonds of the wood and oxygen.

 

 

Conversion is a loose term here. It was an atom containing energy and then it was left over particles void of that energy. That is a conversion. This process did require resistance in order for matter to change its form. It was applied force vs the atoms mass. With every split atom there are left over particles. Just like the firewood. Then we get to build Nevada Mountains.

this is because it is not a conversion. just splitting the atom. the mass lost is energy stored in the binding of the atomic nuclei by the strong force.

 

 

If a particle exists there is some kind of resistance..

Photon vs photovoltaic cell ...the photon looses and breaks down leaving behind an imprint on it

firewood vs heated oxygen.....the firewood looses and breaks down leaving ashes

atom vs applied force.....the atom looses and breaks down leaving radioactive materials

All of these are being resisted and are converting a paticle into energy. When all is said and done in each instance there is something left behind.

no. there is no resistance.

 

 

If a particle exists it retains potential resistance whithin itself when interacting with other particles. If mass is converted in whatever way there always seems to be some left over particles. Where does E=MC^2 represent that loss. Granted the loss can vary given different states that the matter interacting in.

 

More or less temperature = more or less resistance when dealing with firewood.

 

More or less wavelenght = more or less resistance when dealing with photons.

 

More or less mass = more or less resistance when dealing with atoms.

 

not a bit.

 

 

About the metaphore, if no one ever did the equation with the mass taken out for resistance, and always got the same amounts how would anyone know that the number was off by the amount held within the resistance.

 

Gordon

[GORDON HERMA]

For an object at rest it reduces to E = mc2

If nothing can acctually be in a state of rest how do we determine the values of a state of rest when dealing with all of the accountable forces, without taking into account the resistances that are tied to any kind of movement from any kind of force? If we have mass we have resistance and force inherent. Even in a state of rest we have resistance from force interaction, but we still exclued it from the equation, why?

 

Mind you im dealing with this haveing no educational background except my own reading habbits.

Gordon

[insane_alien]

If nothing can acctually be in a state of rest how do we determine the values of rest when dealing with all of the accountable forces, without taking into account the resistance that are tied to any kind of movement from any kind of force? If we have mass we have resistance and force inherent. Even in a state of rest we have resistance from force interaction, but we still exclued it from the equation, why?

 

if we take it from the frame of reference of the system then it stays E=mc^2

 

the resistance comes from the collective mass.

 

i have seen no equation relating resistance and mass. there is no relation. its not as if 1 kg = 3 ohms

 

I would consider a photon a particle,just a particle out of phase with certain

matter.

 

This is close to reality. a photon is a particle, this is true. it is just a particle that can only travel at c(speed of light) because it is massless. its not out of phase with anything. phase is more associated with waves at this scale.

 

I agree it is not a nuclear reaction, but the conversion at this primal level still shows one of the reasons i think that there is some matter that resists conversion.

 

No it shows that there is no interaction taking place that results in conversion. No matter was even consumed in the reaction. If you took all the matter (number of atoms) before and after it would be the same number.

Just because energy exhibits mass does not mean it becomes matter.

 

If a particle exists there is some kind of resistance.

 

Not always true.

 

Photon vs photovoltaic cell ...the photon looses and breaks down leaving behind an imprint on it

 

yes it imparts its energy to move a single electron to one side or the other.

 

firewood vs heated oxygen.....the firewood looses and breaks down leaving ashes

 

An unstable situation resulting in the atoms rearranging themselves into more stable arrangements.

 

atom vs applied force.....the atom looses and breaks down leaving radioactive materials

 

nah the atom just accelerates. it doesn't always leave radioacive materials. stable products are also formed when atoms split/decay

 

All of these are being resisted and are converting a paticle into energy. When all is said and done in each instance there is something left behind.

 

I don't see where the resistance is. also these are only a few examples. what about matter/antimatter annihilations. the ultimate proof of E=mc^2. also no products with rest mass.

 

If a particle exists it retains potential resistance whithin itself when interacting with other particles.

 

so ehh where did you get this from. You say you are uneducated in these matters so you must have read it somewhere or came to a weird conclusion from something. Can you post a source.

 

If mass is converted in whatever way there always seems to be some left over particles.

 

Matter/ anti matter annihilations leave no massive particles. only 2 photons.

 

Where does E=MC^2 represent that loss

 

It doesn't because there is no loss. if you take the change in mass and apply E=mc^2 you will get the energy released from the interaction.

 

More or less temperature = more or less resistance when dealing with firewood.

 

More or less wavelenght = more or less resistance when dealing with photons.

 

More or less mass = more or less resistance when dealing with atoms.

 

once again there is no resistance.

 

About the metaphore, if no one ever did the equation with the mass taken out for resistance, and always got the same amounts how would anyone know that the number was off by the amount held within the resistance.

 

but there is nothing held by the resistance because there is no resistance. even if this were true the equation still holds since when applying E=mc^2 to fission because you only take the change in mass and ignore the remaining mass and you get the energy out. not everything is a total mass to energy conversion because the energy released is not from matter. it is potential energy. the energy itself has mass but there are no energy particle per se.

[GORDON HERMA]

No matter was even consumed in the reaction.

Where did the firewood go?

 

 

 

If you took all the matter (number of atoms) before and after it would be the same number.

Nothing can disappear completely, i did not mean it that way, some one stated Dalton's Laws earlier in the thread. It just changes form.

 

 

 

no interaction taking place that results in conversion

The firewood is now heat and light beacause of the interaction.

 

 

 

if we take it from the frame of reference of the system then it stays E=mc^2

 

How can we define the reference of the system at a state of rest, if mass is using force to keep shape?

 

 

 

its not out of phase with anything.

\

Just a loose meaning from my end meaning it only reacts with certain forms of mass. I. E. It can go through a translucent film and my hand can not.

 

 

 

its not as if 1 kg = 3 ohms

1kg of what thought, if you put a live wire into 1kg of anything there is resistance.

 

 

 

i have seen no equation relating resistance and mass.

This is the basis for my questions.

 

 

 

I hope you dont think im harping on you alien, it just seems so clear to me and i am not totally sure how to explain my thoughts. You guys have helped me a ton with this thread, the more i read the better i understand the aspects involved. I know that my idea seems a bit off to people that know their stuff and i expected difficulty with my ability to understand all of this. Thanks for haveing patients with me though.

 

Gordon..

[GORDON HERMA]

Quote:

If a particle exists it retains potential resistance whithin itself when interacting with other particles.

 

 

so ehh where did you get this from. You say you are uneducated in these matters so you must have read it somewhere or came to a weird conclusion from something. Can you post a source.

 

Its a deduction on my part didnt find it anywhere. The way i think about it is if there were particles that had no resistance and no interaction with other particles of any kind, we would not know about them from observations.

 

 

but there is nothing held by the resistance because there is no resistance.

The resistance i mean is between the force used to sustain the mass(strong) and the force that is used to convert mass into energy. The second force is introduced and i understand why its not in the equation at a state of rest, but the first is tied at the hip with mass itself. You cannot have one without the other.

 

Gordon

[insane_alien]

Where did the firewood go?

 

Didn't go anywhere its chemical arrangement merely changed. I'm good a chemistry so i'll explain this in chemical terms since it is a chemical reaction. Also, firewood is perhaps a bit complex for this(loads of different chemicals in it) so I will use the combustion of methane.

 

If we take the reaction formula for the combustion of methane:

 

CH4 + 2O2 --> CO2 + 2H2O + energy

 

We see that no matter has been consumed as it has the exact same number of atoms on either side. CH4 and O2 have unstable bonds that have quite a bit of energy stored in them. this energy manifests itself as MASS but NOT MATTER. it seems you have these definitions slightly confused. when the atoms rearrange themselves to CO2 and 2H2O the energy is emitted in the form of kinetic energy and photons. Since the energy has left the system there is a small reduction in MASS but not MATTER. all the atoms are still there just like in firewood. the carbon dioxide will escape if it is not a closed system(which is likely) and may lead to the appearence of mass and matter loss but the gas is merely in the atmosphere.

 

Nothing can disappear completely, i did not mean it that way, some one stated Dalton's Laws earlier in the thread. It just changes form.

 

This is true but absolutely nothing disapeared at all.

 

The firewood is now heat and light beacause of the interaction.

 

Yes but NO MATTER was converted. It was just one form of energy being changed into another form of energy(potential into heat and light).

 

Just a loose meaning from my end meaning it only reacts with certain forms of mass. I. E. It can go through a translucent film and my hand can not.

 

This is another example of confusing terms here. MASS is not always MATTER. Here are some links explaining them.

http://en.wikipedia.org/wiki/Mass

http://en.wikipedia.org/wiki/Matter

 

1kg of what thought, if you put a live wire into 1kg of anything there is resistance.

 

Lets say mercury at 4K. At this temperature there is no resistance. This is electrical resistance and has little to do with mass --> energy conversions.

 

I hope you dont think im harping on you alien, it just seems so clear to me and i am not totally sure how to explain my thoughts. You guys have helped me a ton with this thread, the more i read the better i understand the aspects involved. I know that my idea seems a bit off to people that know their stuff and i expected difficulty with my ability to understand all of this. Thanks for haveing patients with me though.

 

You show that you are learning(Which is a good thing, means you are not a crackpot) and are asking some good questions. To discuss these things wit someone who shows even a tiny bit of understanding and learning is always good.

 

Its a deduction on my part didnt find it anywhere. The way i think about it is if there were particles that had no resistance and no interaction with other particles of any kind, we would not know about them from observations.

 

You do not need resistance for interaction. The earth is interacting with the sun but there is no resistance between them. An electron will interact with the nucleus of an atom but there is no resistance.

 

The resistance i mean is between the force used to sustain the mass(strong) and the force that is used to convert mass into energy. The second force is introduced and i understand why its not in the equation at a state of rest, but the first is tied at the hip with mass itself. You cannot have one without the other.

 

Strong force does not sustain most of the mass. particles can have mass and not be affected by the strong force. There is no second force that i am aware of although swansont may be able to answer it better than me.

Mass does not mean that the strong force is necessarily there.

 

Keep up the good questions.

[GORDON HERMA]

Thank you again alien this has been most benificial for me. In the process now of learning most of the terms associated with physics. I got a book to help me with some of the meanings. I'll see you around the boards i think i like it here, some very good reading.

[insane_alien]

Glad i could help.

 

I got a book to help me with some of the meanings

 

Wikipedia is a REALLY good place to start. its got all sorts of stuff about all the sciences and is free. I wouldn't start buying books until your trying to learn the more advanced aspects of whatever your studying.

 

Wikipedia

[GORDON HERMA]

Wikipedia

 

This site is great..one of my new favorites. Very detailed information and its free!

 

Gordon..

[GORDON HERMA]

First I want to say thanks to all that replied to me when I posted the other day on this topic, I did not fully understand the terms and methods behind what I was trying to convey. In regards of that I did some homework and posting on learning the way to present myself. I came up with a better description and method to help me better convey what I meant. Below is a new description of what I originally meant, along with the method I used to arrive at this theory.

 

 

 

Mass, and the force that bind it together are bound at the hip. You can not have one without the other. Einstein does not account for the force binding the mass together in a state of rest in his equation E=MC^2. If mass can not pass the speed of light, then there is always a portion that does not get “converted“. I think the reason for this is because there is a binding force already causing some energy within it at a state of rest. The more mass in the equation, the more energy that is “converted”. The more mass you have, the smaller the amount of force being left behind because more is being used to sustian/bind the mass.

 

In conclusion there is a binding force that is causing energy within the mass in a state of rest. This is being mistaken as converted mass in the original equation E=MC^2. Since that Energy cannot be converted because its already Energy , there is an amount of mass that should not be calculated (is resisted) in the totality. It would seem to me that you cant effect one without effecting the other or leave one out without leaving the other. The energy is there in the totality, not because it was converted, but because it was allready energy.

 

This is like the word with a silent “E” that I was trying to describe in a metaphor about Einstein’s equation

Here is the method…

 

A= Actual Converted Energy

E= Total Energy

C= The constant of the speed of light

O= “Phase/Oscillation” (A Very Loose Meaning)

R= Resisted Binding Force/Mass

 

First to find the baseline Phase/Oscillation “O” of the mass we need to do the equation

MC=O

Second we need to evenly distribute “O” among the original mass at rest, so we need to do the equation.

M/O=R

Third we must subtract "R" from the original mass because it was allready energy

Then last we must multiply it by the “C^2” for the conversion of the rest.

 

This should give us the Actual Converted Energy or “A”

The equation looks like this

 

A=M-(M/O)C^2

or

A=M-R(C^2)

 

If you compare 2 different amounts of mass with this equation using a percentage of “A” with E=MC^2, “E” as the base 100% of the same mass, you will see the scaling pattern for this hidden counterpart.

 

Thank you and I look forward to reading the replies, if this is off please post. I love these boards!

Gordon Herman…

[insane_alien]

First off, WOW! this is a much better description than you started off with, well done. Now to buisness.

 

Mass, and the force that bind it together are bound at the hip.

 

There are a few sub atomic particles that have mass that exist without the strong force. neutrionos. remarkable unreactive.

 

Einstein does not account for the force binding the mass together in a state of rest in his equation E=MC^2

 

He didn't need to as the potential energies these forces cause would be measured in the mass and if the matter is completely converted to energy (annihilation) then this mass(energy) will be released as well. E=mc^2 is merely an equation which relates mass and energy.

 

If mass can not pass the speed of light, then there is always a portion that does not get “converted“

 

Not sure how this works in your head. Can you explain it a bit more please.

 

The more mass in the equation, the more energy that is “converted”.

 

Its more of the more matter the more energy that is stored as mass.

 

In conclusion there is a binding force that is causing energy within the mass in a state of rest

 

Yes but this energy manifests itself as mass. if you took this energy away but kept the matter there somehow. the matter would appear less massive.

 

Since that Energy cannot be converted because its already Energy

 

Maybe its time to introduce you to one of the more mindboggling concepts of this equation. Mass and energy are the same thing. basically the equation is saying Energy=Mass multiplied by some constant(which works out to be c^2)

 

The energy is there in the totality, not because it was converted, but because it was allready energy.

 

Again Energy is Mass.

 

This is like the word with a silent “E” that I was trying to describe in a metaphor about Einstein’s equation

 

Yeah ehhh... still don't get that one.

 

First to find the baseline Phase/Oscillation “O” of the mass we need to do the equation

MC=O

 

You'll need to explain this a bit more.

 

A=M-(M/O)C^2

or

A=M-R(C^2)

 

OK using this equation i will now work out the difference caused (note: i'm doing this on the fly so i on't know if there will be meaningful results.

 

now in the fission of U-235(m=3.902995712694255e-25 kg)is hit by a neutron(m=1.66053886 e-27 kg) total mass =3.919601101294255e-25 kg

the atom fissions into Kr-92(m=1.3602052167944938e-25 kg), Ba-141(m=2.3413597926e-25 kg) and 3neutrons(m = 4.98161658e-27 kg) total mass = 3.7513811751944938e-25 kg

 

difference in mass = 1.682199260997612e-26kg

 

according to einstein the energy released per atom should be (1.682199260997612e-26 * 299,792,458^2) 1.5118852974888513201329405194757e-9 J which is what we see. and i don't have enough time to finish right now so i'll be back later.

 

EDIT: Found some more time.

 

right now for your equation:

MC=O ; 1.682199260997612e-26 kg * 299,792,458 m/s =5.04310651e-18 kg m s^-1

 

M/O=R ; 1.682199260997612e-26/5.04310651e-18= 3.335640954e-9(too little time to work out units will do it later)

 

A=M-R(C^2); 1.682199260997612e-26 - 3.335640954e-9(299,792,458^2) = -299792458.2

 

oh dear we have a negative energy. something must have crapped up. anyone want to check my maths cos i hope it wasn't me since i have a maths exam tomorrow.

 

You probably need to work on this a bit more if its going to stand up in any way shape or form. got to go now.

 

Sources for values and fission products:

http://en.wikipedia.org/wiki/Nuclear_fission

http://en.wikipedia.org/wiki/Speed_of_light

http://www.ivv.fraunhofer.de/ms/ms-exact_masses.html

http://en.wikipedia.org/wiki/Amu

[GORDON HERMA]

hello alien.. thanks for the reply.

 

There are a few sub atomic particles that have mass that exist without the strong force. neutrionos. remarkable unreactive.

Most do travel right through things, but on occasion there is a reaction. I cant remember off hand where i read that, but it does happen.

 

He didn't need to as the potential energies these forces cause would be measured in the mass and if the matter is completely converted to energy (annihilation) then this mass(energy) will be released as well. E=mc^2 is merely an equation which relates mass and energy.

Not sure how this works in your head. Can you explain it a bit more please.

In my head i cant see how there can be annihilation if nothing can surpass the speed of light. Now because we are using the squared speed of light for the conversion, i dont see how we can use them for annihilation. This just makes it seem like there is always a representation of mass that should not convert.

 

Its more of the more matter the more energy that is stored as mass.

I mean the bigger "M" is, the bigger "E" is in E=MC^2. Basicly the same thing.

 

Yes but this energy manifests itself as mass. if you took this energy away but kept the matter there somehow. the matter would appear less massive.

Yeah ehhh... still don't get that one.

 

 

Quote:

First to find the baseline Phase/Oscillation “O” of the mass we need to do the equation

MC=O

I think this is where i went wrong in trying to convey my theory, in the math -- However it seems like there is mass on some level being misrepresented in the conversion becuase of permeability/phase. I guess im viewing it like light from the sun in an analogy, If you take a portion of light you have xrays,visible light,ect all in that portion. Just at different wavelenghts. I guess im viewing mass(light) the same ways because of permeability/phase(wavelenght).

The equation MC=O is just to start a "baseline" type of point to begin a scale based on the permeaility/phase factors within light.

 

 

You'll need to explain this a bit more.

 

 

Quote:

A=M-(M/O)C^2

or

A=M-R(C^2)

A= Actual Conversion Energy

E= Energy

C= The constant of the speed of light

O= “Phase/Oscillation” (I guess im viewing it like light from the sun in an analogy, If you take a portion of light you have xrays,visible light,ect all in that portion. Just at different wavelenghts. I guess im viewing mass(light) the same ways because of permeability/phase(wavelenght)

R= Resisted Binding Force/Mass (The amount that could not be converted because of phase/oscillation vs. strong force/energy being used in a state of rest.)

 

I misplaced the math and did division between the mass and oscillation/phase

try this one. seems alot simpler if i dont add my own characters, the first one is without them..hehe

mc^2=e

mc=o

e-o=r

rc^2=a

a-r = True?

 

(mc^2 -mc)c^2 - (mc^2-mc) = True? <----Is this equation allready used ?

 

or

 

(e-o)c^2 -(e-o) = True?

 

or

 

a-(e-o) = True?

 

or

 

a-r = True?

 

 

True*100(/E)= the %

If you compare 2 different amounts of mass with this equation using a percentage of “A” with, “True” as the base 100% of the same mass, you will see the scaling pattern for this hidden counterpart.

Assuming all numbers are correct.

 

The total mass of the fission products from a single reaction is less than the mass of the original fuel nucleus, and the excess is released as energy via Einstein's relation E=mc2.

Does more mass mean there is more stong force inherent...or strong force being used. The numbers above are based on strong force being used. I.E.- no matter how much mass there is, only a certain amount of strong force is alowed because of the matter.

 

oh dear we have a negative energy. something must have crapped up. anyone want to check my maths cos i hope it wasn't me since i have a maths exam tomorrow.

Your numbers are good alien. It was my equation that was off. I hope it makes more sense now. I look forward to your replies.

 

Gordon Herman...

[insane_alien]

Most do travel right through things, but on occasion there is a reaction. I cant remember off hand where i read that, but it does happen.

 

This is true. But they do not interact through the strong force. They only interact through the weak force and maybe gravity.

 

Now because we are using the squared speed of light for the conversion

 

Although the constant is = to the speed of light, its just a number. it doesn't matter if the particle can't go faster than the number or if it can't even go faster than 1 m/s its merely the relation of energy to mass.

 

I mean the bigger "M" is, the bigger "E" is in E=MC^2. Basicly the same thing.

 

Yup basic algebra. But i still think your definition between matter and mass are a little bit fuzzy. it can be a hard concept to grasp sometimes. you can have mass without matter but you can't have matter without mass.

 

oes more mass mean there is more stong force inherent...or strong force being used. The numbers above are based on strong force being used. I.E.- no matter how much mass there is, only a certain amount of strong force is alowed because of the matter.

 

Ehh no there is the same amount of strong force its just that it cannot interact between the fission products as they are too far away. when you pull(or push depending on how you look at it) the fission products out of the reach of the strong force of a particle the energy used (which gets stored as potential energy) gets released and in the case of muons actually goes to form 2 quarks(well a quark and its antiquark). in a uranium atom the fission products move out of the strong force's potential well of hundereds of particles so lots of energy gets released in light and kinetic energy.

 

Your numbers are good alien. It was my equation that was off. I hope it makes more sense now. I look forward to your replies.

 

I'll get on to those tomorrow unless someone else does them for me. Bed time for me i think my mind is numb from studying.

[GORDON HERMA]

mean there is more stong force inherent...or strong force being used

Ehh no there is the same amount of strong force

Sorry about that i was correcting myself in mid-type..hehe. What i meant by that is the strong force is the same, but a larger amount is being used with larger amounts of mass. If the theory and the numbers are correct each ("particle"?) has only a certian amount of strong force it can draw from at any given time. Which leads me to think that if a particle only uses half the possible energy, then that half should be left out of the conversion equation because its allready energy. Hence the discrepency i was trying to nail down of that "Used" amount "R" that should not be calculated. I understand that the total amount is all interchangeable in the equation e=mc^2 between the mass and energy, but if the numbers are right the truley converted is a smidgen less.

 

 

Although the constant is = to the speed of light, its just a number. it doesn't matter if the particle can't go faster than the number or if it can't even go faster than 1 m/s its merely the relation of energy to mass.

Yes, but i think the very reason that number is used is because of its particle motion "properties"(I'm not sure what word is supposed to be used there but properties was the closest i could come in my vocabulary).

 

I can't do the percentage comparison of the actual mass of another, but i can use the one mass you listed in your previous post. I did the equation adding .1 to the mass amount and the numbers worked out for a comparison, but i did not list them only the mass for the on you posted.

These are the ones you posted.

 

 

M = 1.682199260997612e-26

C = 299,792,458 m/s 89875517873681764

E = 1.5118852974888513201329405194757e-9

O = 5.04310651300257633610296e-18

R = 1.511885292445744807130364183372e-9

A = 135881473.62416411826056096476634

True Energy = 135881473.6241641167486756723206

 

mc^2=e

mc=o

e-o=r

rc^2=a

a-r = True Energy

 

True * 100(/a)=

 

Percentage - 99.999999999999998887349943946415%

 

100% - 99.999999999999998887349943946415%

=

0.00000000000000111265005605359% is the percetage of force that should not be calculated within e=mc^2 at this amount of mass.

 

The larger the mass the smaller the amount of strong force left, but i think this number can never reach annihilation because of the nature of the speed of light.

 

(mc^2 -mc)c^2 - (mc^2-mc)=True

The reason i subtracted mc from both is to equalize the oscillations because they are both at the same (level)state of rest.

 

Gordon ..

P.S. I edited my last post a bit becuase of errors, but this one is as clear as i can present it. I'll work on another reformed write up tonight. Thanks for taking time away from your studies to help me work through my presentation on this, I would never have got to this point without your help. Hopefully i can finish my method explanation before tomorrow.