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Author Topic: If Energy is neither created nor used up, where did energy come from?  (Read 13363 times)

Offline PmbPhy

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Regarding energy and mass, I'd like to make a point that people might know subconsciously but not be aware of it consciously. Consider a  body whose relativistic mass m. Then The E in the expression E = mc2 is not the total energy of the body but what I like to refer to as the inertial energy which is the sum of the bodies rest energy and its kinetic energy. The E does not contain the energy of position, i.e., potential energy.

One of the excuses people use against relativistic mass is We don't need relativistic mass because it's the same thing as energy. This is wrong because energy means total energy which includes potential energy. However the E in  E = mc2 is not total energy because it doesn't contain potential energy.
 

Offline Alan McDougall

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Regarding energy and mass, I'd like to make a point that people might know subconsciously but not be aware of it consciously. Consider a  body whose relativistic mass m. Then The E in the expression E = mc2 is not the total energy of the body but what I like to refer to as the inertial energy which is the sum of the bodies rest energy and its kinetic energy. The E does not contain the energy of position, i.e., potential energy.

One of the excuses people use against relativistic mass is We don't need relativistic mass because it's the same thing as energy. This is wrong because energy means total energy which includes potential energy. However the E in  E = mc2 is not total energy because it doesn't contain potential energy.

What then in the case of an matter - antimatter "Clash"? Does that release all the energy (Total)  "held"? in those two opposing forms of matter, converting them into gamma rays? However, in the event just described, "all of the energy"? has just been dissipated into the gamma ray cloud only changed in form.

We are lucky this did not happen in the early universe or we would not be debating this subject, especially in light of the fact that the universe would have then existed as a vast gamma-ray void.

There is no such thing as anti-energy, as some have speculated , energy is energy, the release and consequence usage thereof, can only be reflected as an equation of thermodynamics, in the macro world. At the quantum level it seems to be more mysterious and difficult to define?

There must be a loss somewhere, because it is impossible to convert 100% of energy from one form of it to another. where would that energy have gone?
 

Offline PmbPhy

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Quote from: Alan McDougall
What then in the case of an matter - antimatter "Clash"? Does that release all the energy (Total)  "held"? in those two opposing forms of matter, converting them into gamma rays? However, in the event just described, "all of the energy"? has just been dissipated into the gamma ray cloud only changed in form.
The phrase release the energy is a confusing one since it doesn't have a exact meaning. Typically when people use this phrase they have in mind that photons are energy that's been released or something similar. However that's thinking of photons as "being" energy rather than "having" energy, which is the correct viewpoint. I recommend avoiding that kind of phrasing. It can lead to the wrong idea. For example when a nuke goes off, a lot of the damage that the radiation (in the forum or photons, alpha rays, beta rays, neutrons etc) that is released can do is in the form of particles with very high kinetic energy. So perhaps that's what you may have meant when you say released, i.e. particles with kinetic energy are released?

What is not well known is that when a particle annihilates its antiparticle the result is not always photons. Sometimes its other particles.

Quote from: Alan McDougall
There must be a loss somewhere, because it is impossible to convert 100% of energy from one form of it to another. where would that energy have gone?
It's possible to convert potential energy completely into kinetic energy. It's also possible to convert rest energy completely into radiant energy (in the form of photons).
 

Offline jeffreyH

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Einstein said what he said. "The mass of a body is a measure of its energy-content". Don't blame me if some particle physicist tell you something that flatly contradicts Einstein.

So you are effectively ignoring the gamma factor in there which shows that it is relativistic mass that Einstein is talking about. If you knew your physics you wouldn't make such silly mistakes.

 

Offline timey

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Regarding energy and mass, I'd like to make a point that people might know subconsciously but not be aware of it consciously. Consider a  body whose relativistic mass m. Then The E in the expression E = mc2 is not the total energy of the body but what I like to refer to as the inertial energy which is the sum of the bodies rest energy and its kinetic energy. The E does not contain the energy of position, i.e., potential energy.

One of the excuses people use against relativistic mass is We don't need relativistic mass because it's the same thing as energy. This is wrong because energy means total energy which includes potential energy. However the E in  E = mc2 is not total energy because it doesn't contain potential energy.

Ah - interesting!

Ok - so let me get with this one a bit better.

You say E=mc2 is inclusive of what you call 'inertial energy' and kinetic energy.

KE is calculated as 0.5mv2=KE

So when I see the formula E=mc2, can I assume that the calculation for KE has already been completed?

And that the m in the equation is already complete with the relativistic mass added via the additional KE energy?

And... is this what distinguishes the terms of E=mc2 and e=mc2?
 

Offline JohnDuffield

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Richard Feynman the great physicists, comment that there are no little "Blobs" of energy, nicely answers the question , because his little blob quote informs us that energy it is "Not"  A 'THING" that we find or will ever find in nature.
I'm sorry Alan, but Feynman was at odds with Einstein there. I'm something of a "Feynman fan", but on this, I'm with Einstein.   

Sadly to no avail because people, still seek it here, they seek it there, they seek the irritating illusive blob of non-existing energy everywhere. Sadly for them, in vain, because the"blob of energy" does not exist as a real thing in nature, but a describable mathematical quality of matter, that is useful tool in the physics and thermodynamics.
Energy exists. It is not merely some mathematical quality of matter.


What then in the case of an matter - antimatter "Clash"? Does that release all the energy (Total)  "held"? in those two opposing forms of matter, converting them into gamma rays?
Yes.

We are lucky this did not happen in the early universe or we would not be debating this subject, especially in light of the fact that the universe would have then existed as a vast gamma-ray void.
Gamma-gamma pair production is the reverse of annihilation.

There is no such thing as anti-energy, as some have speculated , energy is energy, the release and consequence usage thereof, can only be reflected as an equation of thermodynamics, in the macro world. At the quantum level it seems to be more mysterious and difficult to define?
It isn't mysterious at all.

There must be a loss somewhere, because it is impossible to convert 100% of energy from one form of it to another. where would that energy have gone?
There is no loss. Energy is the one thing you can neither create nor destroy. 

Energy can only be described as "an equation of thermodynamics" and is not a real separate, tangible material thing that could, "hypothetically, be picked up and held in a persons hands".
Energy is real. It isn't tangible like a suitcase atom bomb is tangible. You can't hold energy in your hands like you can hold a  suitcase atom bomb in your hands. But when the latter detonates and energy is released, you will realise that energy is very real indeed. For about a nanosecond.

The bottom line is this: matter is made of it. You are made of it. 
« Last Edit: 16/06/2016 13:42:34 by JohnDuffield »
 

Offline JohnDuffield

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...However the E in  E = mc2 is not total energy because it doesn't contain potential energy.
I'm afraid it does. Consider a brick in front of you on a shelf. E=mc˛ applies. Now you push the brick off the shelf, whereupon some of its potential energy is converted into kinetic energy as it falls. The brick hits the floor, whereupon the kinetic energy is dissipated. Conservation of energy applies, along with the mass deficit. The brick is again at rest, and E=mc˛ still applies. But now the rest-mass-energy of the brick is less than what it was. When you throw that brick back up onto the shelf you do work on it. You add energy to it. You increase its mass. Note that the Earth is involved in this process, but not much. Momentum is shared equally, but energy is not. Google on collision bullet block.

...that is released can do is in the form of particles with very high kinetic energy
This is a popscience particle-physics view that is at odds with the E=hc/λ wave nature of photons, and at odds with what Einstein said. Radiation does not consist of billiard-ball particles. It consists of photons, and a photon is a wave. When you remove the kinetic energy from the wave, the wave no longer exists. The same is not true for a billiard ball. This is why Einstein said radiation is a form of energy.

« Last Edit: 16/06/2016 14:04:56 by JohnDuffield »
 

Offline PmbPhy

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Quote from: timey
You say E=mc2 is inclusive of what you call 'inertial energy' and kinetic energy.
No. What I said is that Inertial Energy = Rest Energy + Kinetic energy

Quote from: timey
KE is calculated as 0.5mv2=KE
That's the low speed approximation to the kinetic energy, K. Let E0 = m0c2 = the rest energy of the particle where m0 is the proper mass, aka rest mass, of the particle. Let gamma = 1/sqrt{1 - v2/c2}. The expression for the relativistic kinetic energy is then given by K = (gamma - 1)m0c2 = (gamma - 1)E0. If you have a strong math background and would like to follow the derivation for this expression then you can find it on my website at: http://www.newenglandphysics.org/physics_world/sr/work_energy.htm

When following the derivation it's helpful to keep in mind that the kinetic energy is defined by the requirement that the change in kinetic energy equal the work done by the external force acting on the particle. This is called the Work-Energy Theorem.

With all of this we can find the expression for the inertial energy E.

E = K + E0 = (gamma - 1)E0 + E0 = gamma*E0

or

E = gamma*m0c2 = mc2

where m = gamma*m0 = the relativistic mass of the particle.

Quote from: timey
So when I see the formula E=mc2, can I assume that the calculation for KE has already been completed?
Yes, in the sense that I just demonstrated.

Quote from: timey
And that the m in the equation is already complete with the relativistic mass added via the additional KE energy?
The m in E = mc2 is the relativistic mass. I don't know what you mean by "added via the additional KE energy."
« Last Edit: 16/06/2016 14:19:50 by PmbPhy »
 

Offline Alan McDougall

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Richard Feynman the great physicists, comment that there are no little "Blobs" of energy, nicely answers the question , because his little blob quote informs us that energy it is "Not"  A 'THING" that we find or will ever find in nature.
I'm sorry Alan, but Feynman was at odds with Einstein there. I'm something of a "Feynman fan", but on this, I'm with Einstein.   

Sadly to no avail because people, still seek it here, they seek it there, they seek the irritating illusive blob of non-existing energy everywhere. Sadly for them, in vain, because the"blob of energy" does not exist as a real thing in nature, but a describable mathematical quality of matter, that is useful tool in the physics and thermodynamics.
Energy exists. It is not merely some mathematical quality of matter.


What then in the case of an matter - antimatter "Clash"? Does that release all the energy (Total)  "held"? in those two opposing forms of matter, converting them into gamma rays?
Yes.

We are lucky this did not happen in the early universe or we would not be debating this subject, especially in light of the fact that the universe would have then existed as a vast gamma-ray void.
Gamma-gamma pair production is the reverse of annihilation.

There is no such thing as anti-energy, as some have speculated , energy is energy, the release and consequence usage thereof, can only be reflected as an equation of thermodynamics, in the macro world. At the quantum level it seems to be more mysterious and difficult to define?
It isn't mysterious at all.

There must be a loss somewhere, because it is impossible to convert 100% of energy from one form of it to another. where would that energy have gone?
There is no loss. Energy is the one thing you can neither create nor destroy. 

Energy can only be described as "an equation of thermodynamics" and is not a real separate, tangible material thing that could, "hypothetically, be picked up and held in a persons hands".
Energy is real. It isn't tangible like a suitcase atom bomb is tangible. You can't hold energy in your hands like you can hold a  suitcase atom bomb in your hands. But when the latter detonates and energy is released, you will realise that energy is very real indeed. For about a nanosecond.

The bottom line is this: matter is made of it. You are made of it. 

I meant "where has the energy gone" because there is never a 100% conversion of energy from its one form into another it was wrong of me to say "lost" However, you are wrong matter is not a block of energy it is something that contains potential energy as you, yourself pointed out.

Energy is just a mathematical expression of how work is transferred from one body to another and is not a physical material thing at all period!

I wonder if the Higgs Boson  could fit into this debate?

Alan
« Last Edit: 16/06/2016 14:23:17 by Alan McDougall »
 

Offline PmbPhy

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Quote from: Alan McDougall
I meant "where has the energy gone" because there is never a 100% conversion of energy from its one form into another...
On what basis did you arrive at that conclusion? Consider an harmonic oscillator.  The energy of it is given by E = K + V. As the system, e.g. a pendulum, oscillates from one its top position to its bottom position and back starting the cycle over again, the energy shuffles back and forth from potential energy to kinetic energy with a complete conversion of one to the other. I gave another example using pair annihilation of an electron and a positron. Assuming that the particles start from rest and are in contact they will annihilate producing two photons thus converting rest energy into the radiant energy of the two photons. The energy of a photon is 100% kinetic energy so the conversion is between rest energy and kinetic energy. If the particles start off moving towards each other then the system will start off with some kinetic energy. The result will be the same with all of the rest energy being changed to kinetic energy which is part of the final kinetic energy of the two photons. If you place a vane into a bucket of water and let the vane have an initial kinetic energy then the system will eventually settle down with all of the kinetic energy being changed to thermal energy.
 

Offline timey

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Quote from: timey
You say E=mc2 is inclusive of what you call 'inertial energy' and kinetic energy.
No. What I said is that Inertial Energy = Rest Energy + Kinetic energy

Quote from: timey
KE is calculated as 0.5mv2=KE
That's the low speed approximation to the kinetic energy, K. Let E0 = m0c2 = the rest energy of the particle where m0 is the proper mass, aka rest mass, of the particle. Let gamma = 1/sqrt{1 - v2/c2}. The expression for the relativistic kinetic energy is then given by K = (gamma - 1)m0c2 = (gamma - 1)E0. If you have a strong math background and would like to follow the derivation for this expression then you can find it on my website at: http://www.newenglandphysics.org/physics_world/sr/work_energy.htm

When following the derivation it's helpful to keep in mind that the kinetic energy is defined by the requirement that the change in kinetic energy equal the work done by the external force acting on the particle. This is called the Work-Energy Theorem.

With all of this we can find the expression for the inertial energy E.

E = K + E0 = (gamma - 1)E0 + E0 = gamma*E0

or

E = gamma*m0c2 = mc2

where m = gamma*m0 = the relativistic mass of the particle.

Quote from: timey
So when I see the formula E=mc2, can I assume that the calculation for KE has already been completed?
Yes, in the sense that I just demonstrated.

Quote from: timey
And that the m in the equation is already complete with the relativistic mass added via the additional KE energy?
The m in E = mc2 is the relativistic mass. I don't know what you mean by "added via the additional KE energy."

Pete - thanks, although I'm struggling to follow, I am following and would ask that you explain gamma a little.

I left school age 11 and am self taught in everything.  I don't have a formal basis in maths beyond long division but have been studying and catch on real quick.  I understand the principle of maths in relation to formula and geometry, (albeit that this understanding is of a visualist-ic nature), but am having trouble with relativistic mass, and the accumulative tendencies of relativistic mass and gravity.

So if e=mc2 relates to inertial energy and rest mass.  Then I assume that KE is then added as a calculation of 0.5 of that rest mass times velocity squared. (which is what I meant by adding KE to the equation).  For e=mc2 to be inclusive of KE, the result of 0.5mv2 is converted into mass and added to the rest mass to become the 'new' m for the E=mc2 equation...(this is what I suspect your maths are telling me, am I close?)

Then...when potential energy is considered, does this change relativistic mass again?
 

Offline Alan McDougall

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Quote from: Alan McDougall
I meant "where has the energy gone" because there is never a 100% conversion of energy from its one form into another...
On what basis did you arrive at that conclusion? Consider an harmonic oscillator.  The energy of it is given by E = K + V. As the system, e.g. a pendulum, oscillates from one its top position to its bottom position and back starting the cycle over again, the energy shuffles back and forth from potential energy to kinetic energy with a complete conversion of one to the other. I gave another example using pair annihilation of an electron and a positron. Assuming that the particles start from rest and are in contact they will annihilate producing two photons thus converting rest energy into the radiant energy of the two photons. The energy of a photon is 100% kinetic energy so the conversion is between rest energy and kinetic energy. If the particles start off moving towards each other then the system will start off with some kinetic energy. The result will be the same with all of the rest energy being changed to kinetic energy which is part of the final kinetic energy of the two photons. If you place a vane into a bucket of water and let the vane have an initial kinetic energy then the system will eventually settle down with all of the kinetic energy being changed to thermal energy.

My wordings always seem to lack clarity this is what I meant below.

In the practical application of energy conversion at the electricity generating company where I worked for most of my life, there was a constant need for scheduled maintenance of the huge boiler and generators at the Power Station to keep them at maximum efficiency, because wear and tear over time resulted in poorer and poorer energy conversion of this open system and a drop in the efficiency of the machines.

At the most ideal situation the best efficiency of such as a system was in the region of 38%. The rest of the energy being dissipate or wasted out of the cooling towers, condensers and smoke stacks etc, etc.

"Thus the energy had gone somewhere" and this is the core of what I meant when I stated that energy conversion is never 100%. If this were possible we could create a perpetual motion machine .

Through, the conversion of thermal energy into other forms,we reduce the temperature of the system, increasing its entropy, keeping its efficiency at much less than 100%" (even when energy is not allowed to escape from the system).

This is because thermal energy has already been partly spread out among many available states of a collection of microscopic and macro, particles constituting the system.

In such circumstances, a measure of entropy, or dissipation, dictates that future states of an isolated system, must be of at least equal evenness in energy distribution.

In other words, there is no way to concentrate energy without spreading out energy somewhere else, and that is that I what I really meant in my badly worded response?

Regards

Alan.
« Last Edit: 16/06/2016 16:38:55 by Alan McDougall »
 

Offline JohnDuffield

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I meant "where has the energy gone" because there is never a 100% conversion of energy from its one form into another.
There always is. Energy is wasted in an engine or machine, but none of it ever actually disappears. Energy conservation always applies.

However, you are wrong matter is not a block of energy it is something that contains potential energy as you, yourself pointed out.
That potential energy is actually "hidden" kinetic energy, internal to the body. Gravity converts some of this internal kinetic energy into external kinetic energy. When this external kinetic energy is dissipated, you're left with a mass deficit.   

Energy is just a mathematical expression of how work is transferred from one body to another and is not a physical material thing at all period!
Energy isn't just some mathematical expression. Physical material things are made of it. That's the message from Einstein and E=mc˛, and unless somebody has a good explanation as to why Einstein is wrong, I'm sticking with Einstein. I recommend that you do the same.

I wonder if the Higgs Boson  could fit into this debate?
Yes of course. The Higgs boson is quite literally made from the kinetic energy given to the protons in the LHC. It's not totally unlike creating an electron and a positron out of photon kinetic energy in pair production. Note this on Matt Strassler's blog: "The Higgs field, though it provides the mass for all other known particles with masses, does not provide the Higgs particle with its mass".
 

Offline PmbPhy

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Quote from: timey
Pete - thanks, although I'm struggling to follow, I am following and would ask that you explain gamma a little.
gamma is a Greek letter which is set equal to 1/sqrt{1 - v^2/c^2}. It's just a letter which is used to make the formulas look simpler.

Quote from: timey
Then I assume that KE is then added as a calculation of 0.5 of that rest mass times velocity squared.
As I explained in reply #82 that's the non-relativistic expression for kinetic energy. It's only valid for velocities which are much smaller than the speed of light.

Quote from: timey
(which is what I meant by adding KE to the equation).  For e=mc2 to be inclusive of KE, the result of 0.5mv2 is converted into mass and added to the rest mass to become the 'new' m for the E=mc2 equation...(this is what I suspect your maths are telling me, am I close?)
No.

Quote from: timey
Then...when potential energy is considered, does this change relativistic mass again?
As I explained above potential energy, i.e. the energy associated with the position of a particle in a field, is not part of the energy in E = mc^2. However, when a particle is in such a field it is subject to a force. That force changes the speed of the particle which in turn changes the relativistic mass of the particle.
 

Offline jeffreyH

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Timey think of gamma in a similar way to a percentage. Although it isn't it may be useful to think of it in those terms. If we take a stationary object to have 100% mass then if we cause it to move and increase its speed we have to add some percentage to the mass proportional to the increase in speed. The faster the speed the larger the 'percentage' increase. The nearer to light speed the speed of our object is the nearer to an infinite amount of mass it has. Near to infinity however is misleading in this case since defining how near to infinity anything is is not possible. All that can be said is that the greater the increase in the percentage of mass the harder it is to move the object even faster. I have avoided time dilation on purpose here as it wouldn't help. That is a simplified interpretation which hopefully will clarify things for you.
« Last Edit: 16/06/2016 18:56:36 by jeffreyH »
 

Offline agyejy

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As I explained above potential energy, i.e. the energy associated with the position of a particle in a field, is not part of the energy in E = mc^2.

That is not strictly true. The clearest example is the fact that a collection of unbound neutrons and protons has more mass than the same collection bound into a nucleus (assuming the binding energy of the nucleus is negative). Calculating that difference in mass requires that you take into account any electromagnetic forces as well as the attraction of the strong force (and probably the weak force if you want a lot of precision). Going down another layer the mass of the quarks in the protons and neutrons are a very small fraction of the mass of the protons and neutrons. The rest of the mass comes primarily from the fields of the strong force modified again by electromagnetic interactions between the quarks (and probably the weak force to some extent otherwise beta decay is hard to explain). Most of the invariant mass of composite particles like protons and neutrons usually comes from the force fields holding them together (i.e. potential energy). Even a hydrogen atom is slightly (very very slightly) less massive than the combined unbound masses of a proton and an electron.  Even for a single free electron there is still a mass renormalization due to the interaction of the electron with its own field. This is often described as the electron being surrounded by a collection of virtual particles that increase the observed mass.

 

Offline timey

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Timey think of gamma in a similar way to a percentage. Although it isn't it may be useful to think of it in those terms. If we take a stationary object to have 100% mass then if we cause it to move and increase its speed we have to add some percentage to the mass proportional to the increase in speed. The faster the speed the larger the 'percentage' increase. The nearer to light speed the speed of our object is the nearer to an infinite amount of mass it has. Near to infinity however is misleading in this case since defining how near to infinity anything is is not possible. All that can be said is that the greater the increase in the percentage of mass the harder it is to move the object even faster. I have avoided time dilation on purpose here as it wouldn't help. That is a simplified interpretation which hopefully will clarify things for you.

Thanks Jeff - it's s good analogy, one that I've heard before and have grasped the understanding of.

But... to understand 'exactly' how these masses and energies add up mathematically within relativity is my goal.

I am just a tad unclear as to what part of the physical process 'gamma' actual 'is', hence my request.

And my interest, for clarity, lies in the notion that a rock lying on the ground has a rest mass and inertial energy.  It is made up of atoms that have rest mass and inertial energy.  Presumably these atoms have a corresponding De Broglie matter wave.  Pick the rock up and we must add potential energy.  Throw the rock and we must add kinetic energy...

Does the frequency of the atoms that are the matter of the rock increase, and the matter wave decrease with the additional potential energy?

Does the frequency of the atoms that are the matter of the rock increase, and the matter wave decrease with the addition of kinetic energy?
 

Offline timey

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gamma is a Greek letter which is set equal to 1/sqrt{1 - v^2/c^2}. It's just a letter which is used to make the formulas look simpler.

Pete - I do not understand the physicality of 1/square root 1 etc, and what this term relates to within the physical process.  m is mass, v is velocity, e is energy of various types, squaring is a mathematical process, where does the 1 originate from?  What is 1?

As I explained in reply #82 that's the non-relativistic expression for kinetic energy. It's only valid for velocities which are much smaller than the speed of light.

Are you saying that the mathematical process of calculating KE changes above a certain velocity?  If so, what physical process causes the mathematical proportionality of relativistic and non-relativistic velocities to be un-reconcilable within the same mathematical process?
« Last Edit: 16/06/2016 20:00:03 by timey »
 

Offline jeffreyH

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Firstly let's look at v and c that form the fraction v/c that is used in gamma. The value of v can never equal c but must be less than c at all times. We can look at this as v being a percentage of c. In this way we can multiply v by a fraction to represent this. When v = 1/2*c it is half  the speed of light and when v = 99/100*c it is 99% the speed of light etc. So that if v = 1/2*c this is like saying v/c = 1/2.

Since the fraction used in gamma is v^2/c^2 then for the value of 1/2 this becomes 1^2/2^2 which gives 1/4. This is not the end of it though because gamma has the square root of 1 - v^2/c^2 as the denominator. In this case we need to find the square root of 1-1/4. So then we are looking at the square root of 3/4 which approximates to 0.866. The final step is 1/0.866 which translates to a value of 1.1547 approx. So our mass is increased in this case by 115.5% approx at half light speed. If anybody sees an error in my working please point it out.

This is the mathematical description. The physical causes are an entirely different matter. Find that and you will be famous.
 

Offline Alan McDougall

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Firstly let's look at v and c that form the fraction v/c that is used in gamma. The value of v can never equal c but must be less than c at all times. We can look at this as v being a percentage of c. In this way we can multiply v by a fraction to represent this. When v = 1/2*c it is half  the speed of light and when v = 99/100*c it is 99% the speed of light etc. So that if v = 1/2*c this is like saying v/c = 1/2.

Since the fraction used in gamma is v^2/c^2 then for the value of 1/2 this becomes 1^2/2^2 which gives 1/4. This is not the end of it though because gamma has the square root of 1 - v^2/c^2 as the denominator. In this case we need to find the square root of 1-1/4. So then we are looking at the square root of 3/4 which approximates to 0.866. The final step is 1/0.866 which translates to a value of 1.1547 approx. So our mass is increased in this case by 115.5% approx at half light speed. If anybody sees an error in my working please point it out.

This is the mathematical description. The physical causes are an entirely different matter. Find that and you will be famous.

All that is informative but what does it have to do with the question of the thread "If Energy is not created or destroyed where does it come from"?
« Last Edit: 16/06/2016 21:55:50 by Alan McDougall »
 

Offline jeffreyH

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The increases in relativistic mass/energy via the gamma function indicate that some fundamental interaction is at work at the quantum level. Due to these effects being tied very closely to those of gravitation then finding the cause will lead you towards an answer to your question.That is if it can be answered at all.
 

Offline Alan McDougall

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The increases in relativistic mass/energy via the gamma function indicate that some fundamental interaction is at work at the quantum level. Due to these effects being tied very closely to those of gravitation then finding the cause will lead you towards an answer to your question.That is if it can be answered at all.

I agree!; if the answer is ever found it would come from the quantum level, with gravitation playing some vital role in the process
« Last Edit: 16/06/2016 22:43:09 by Alan McDougall »
 

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Offline PmbPhy

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Quote from: agyejy
That is not strictly true. The clearest example is the fact that a collection of unbound neutrons and protons has more mass than the same collection bound into a nucleus (assuming the binding energy of the nucleus is negative).
Hi agyejy. You're incorrect. What I said is precisely true. Please reread it carefully. This time please keep in mind the difference between the potential energy of position of a particle in a field and the internal potential energy of a body due to the interactions between the constituents of the body, i.e. internal potential energy. Wolfgang Rindler explains this in his text Relativity; Special, General and Cosmological - 2nd Ed., (2006). I did you a favor and extracted section 6.3 which is entitled The Equivalence of Mass and Energy. You can download it from my website at: http://www.newenglandphysics.org/other/Rindler_on_potential_energy_and_mass.pdf

On page 113 Rindler writes
Quote
One kind of energy that does not contribute to mass is potential energy of position.
Perhaps if you read it in the context of the entire section it will make more sense to you. Rindler is much better at explaining these things than I am.  :)

And, yes. I'm well-aware of the fact that the internal potential energy of a closed system contributes to the mass of the system. I took that into account when I described nuclear fission on my website. See:
http://www.newenglandphysics.org/physics_world/sr/nuclear_fission.htm

The problem here is that we have different notions of potential energy in mind. In the post above where I explained this I said that it was the potential energy of position that doesn't contribute to mass energy. You appear to have confused that with the internal potential energies between the particles that make up that body.

To be precise, if there is a charged particle, such as an electron, moving in a static electric field then the total energy of the electron is the sum of three forms of energy: kinetic energy, rest mass energy and potential energy. An electron doesn't have internal potential energy . The total energy, which I'll label W, is the sum of those three energies, i.e. W = K + E0+ U where U is the potential energy associated with position.

My intention in raising this point is because I believed that some people might not know this fact.
« Last Edit: 17/06/2016 06:04:07 by PmbPhy »
 

Offline PmbPhy

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Quote from: Alan McDougall
All that is informative but what does it have to do with the question of the thread "If Energy is not created or destroyed where does it come from"?
As I attempted to explain in my first post in this thread, the energy needn't come from anywhere. We can think of it as always having had he value of zero. We can do this because the absolute value of the total energy in the universe doesn't have a physical meaning. What does have a physical meaning is changes in each form of energy. As I'm sure you know the absolute value of the potential energy of a particle, or system of particles, is not physically meaningful. Only gradients in the potential energy function is physically meaningful. Recall that the potential energy of a particle, V, in a field of force F is defined as F = -8319f01cae0e5752ba2e719544659959.gifV. So you can add any constant to V without changing the force on the particle. So you can always set the value of the total energy to zero.

Alan Guth explains something very similar to all of this in his book The Inflationary Universe in Chapter 17 which is entitled Universe Ex Nihilo which starts on page 271. The idea was first speculated by Edward Tryon in a paper which appeared in Nature in 1973. Tyron explained it in terms of the fact that the gravitational energy is negative while mass-energy is positive. Tyron learned from the well-known general relativist Peter Bergmann that in any closed universe the negative gravitational energy cancels the energy of matter exactly. The total energy, or as Guth writes "or equivalently the total mass" is precisely equal to zero.

I believe that answers your question. I'd be more than happy to place chapter 17 of that book into a PDF file and upload it onto my website if anybody would like to read the entire chapter. If so then please let me know.
 

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