Would like to know what energy is beyond its definition

[jado]

what is energy? i would like to know what energy is beyond it's definition.
simply the ability to do work is not enough. what is the nature of this entity?
what is the relation of energy beyond its interaction with matter?

 

[hilo1222]

i would say energy is any thing that shows a reaction in physical nature

 

[mysearch]

what is energy? i would like to know what energy is beyond it's definition.

I realize that the following quote by Richard Feynman doesn’t really answer your question, but it might help put some others into perspective:

Energy is not something perceptible, but a quantity that is calculated using a set of rules. The concept of energy remains open to future modification and, in effect, I can't say what energy is, I simply know how to calculate it effects`. Richard Feynman

If Feynman was unsure of the concept, you might not get a simple definite answer, so have taken the opportunity to raise some additional issues and questions related to energy and the cosmology model that I would also be interested in getting some clarification.

First, we might consider Einstein’s equation E=mc^2, which highlights a relationship between energy and rest mass. The equivalence of inertial mass and gravitational mass are linked by Newton’s 2nd law and his universal laws of gravitation, i.e. F=m_I*a=GMm_G/r^2. With reference to the energy orientation of this discussion, we might differentiate the gravity equation to define the potential gravitational energy, i.e. U=-GMm/r. Note, the negative sign.

The definition of relativistic mass is more controversial these days, but can be initially introduced as m’=\gamma m. Of course, we should not forget Planck’s definition E=hf and given that [h] and [c] are both constant, does this suggest some equivalence between mass and frequency?

In addition to rest mass energy, it may also be useful to confirm whether all other forms of energy can be described in terms of positive kinetic energy and negative potential energy? The sign convention defines potential energy ranging from a maximum of zero to a negative minimum. As such, a radially free-falling mass [m], starting from rest, at an infinite radius [r] from a central mass [M] has no kinetic energy and no potential energy. As mass [m] falls towards [M] it acquires positive kinetic energy and an equal amount of negative potential energy. Therefore, the conservation of energy is maintained in the sense that mass [m] starts out with the maximum potential energy, i.e. zero, which is then convert to positive kinetic energy balanced by an equal amount of negative potential energy.

The reason for highlighting these points is to raise some additional issues in terms of a very simple cosmological model that only contains 2 objects, i.e. our mass [M] and [m].

If the space between [M] and [m] expands, might we assume it still acquires potential gravitational energy? Where did this expansive energy come from? Must it have a positive sign to conserve the apparent net gain of negative potential energy? Can the expansive energy be described in terms of either potential of kinetic energy?

If our simple model collapses back towards a big crunch, might we assume that the net energy was zero? However, having inserted the unknown expansive energy into our model, at the point of crunch, we appear to have the contents of our universe, i.e. [m], accelerating towards the ‘central’ singularity with a net gain of positive kinetic energy and countered by a proportional increase in negative potential energy. Would positive and negative energy forms simply cancel at the big crunch? This question has some relevance with respect the idea of the universe being a quantum fluctuation!

If our model universe doesn’t collapse, does it imply that the universe requires and maintains a net gain of energy?

 

[yuiop]

If our model universe doesn’t collapse, does it imply that the universe requires and maintains a net gain of energy?

My hunch is that the universe always maintains net zero energy.

This would require some aspect of gravity to considered as a form of negative energy.
An interesting discussion on the subject can be found here. http://www.mathpages.com/home/kmath613/kmath613.htm


On the subject of energy, see if you can figure out why when a particle and an antiparticle annihilate each other so that each has its positive rest mass completely converted into kinetic energy, the total rest mass of the two photons produced is the same as the initial positive rest mass energy of the particles.

 

[jado]

stop looking at the effect, see that which is the effect. what particle what anti particle? look beyond this.

 

[mysearch]

Response to #5

stop looking at the effect, see that which is the effect. what particle what anti particle? look beyond this.

Hi Jado: Clearly, the discussion of energy so far is not what you were looking for. Do you already have an answer in mind that is based on scientific principles or is it more philosophical in nature? From my perspective, I was conscious that this is a cosmology forum, which is why I tried to move the discussion in this direction.

 

[mysearch]

Response to #4

Hi Kev: Thanks for the indirect link to http://www.mathpages.com/
What a thought-provoking source of material, by Kevin Brown, who appears to be a somewhat anonymous character as far as trying to track down his academic background. As an aside, I picked up the following quote that might go some way to describing my own motivation:J

Why can't somebody give us a list of things that everybody thinks and nobody says, and another list of things that everybody says and nobody thinks? Oliver Wendell Holmes

I read the article referenced with interest and I suspect that I will need to read it a few times to understand all the points being made. However, the thought processes behind Maxwell’s ideas about negative energy are both insightful and educational. Hadn’t really thought about the anomaly of like-charges leading to repulsion, while the analogous concept of similar ‘mass-charge’ leads to attraction, unless you consider the idea of negative energy. Was also interested in the idea behind the following quote:

so the electric displacement is somewhat like stretching a little spring at each point in space.

The reason for highlighting this point relates to a question raised about the work of Gabriel LaFreniere, who simulates space resonance waves by modelling each point in space in a similar way. Have you ever looked at this idea, although any discussion of this idea is possibly beyond the remit of this forum?
https://www.physicsforums.com/showpost.php?p=1836008&postcount=4

#4: On the subject of energy, see if you can figure out why when a particle and an antiparticle annihilate each other so that each has its positive rest mass completely converted into kinetic energy, the total rest mass of the two photons produced is the same as the initial positive rest mass energy of the particles.

On the basis that the particle and antiparticle have opposite charge, the annihilation has to maintain the conservation of charge, i.e. net zero charge. As I understand it, along similar arguments, the particle and the antiparticle annihilate into two photons, because of the conservation of spin, i.e. the two photons have net zero spin. This just leaves the issue of mass-energy, which I assume is resolved via the following equation: E = mc^2 = hf. However, given that there are two photons, I am not sure whether the energy distribution is always symmetric, i.e. f=mc^2/2h?

#4: My hunch is that the universe always maintains net zero energy.

Based on my assumptions, which might be wrong, let's now follow your hunch. In #3, I speculated whether we could consider kinetic energy being positive and potential gravitational energy being negative in order to conserve energy under radial free-fall. By following this line of thought, it might be assumed that the expansion of the universe required an initial input of positive energy that was balanced by the implicit negative gravitational potential linked to this expansion. Within this simplistic model, I imaged this balance might be restored at a big crunch, but raised the question of a violation of idea behind the universe starting as a quantum fluctuation, if the universe never collapsed. However, the idea of the big crunch also raised a picture of all the particle/photonic mass in the universe, ever increasing in positive kinetic energy and negative potential, as it converged back towards a singularity. Can we assume this energy just cancels and would the quantum process, assumed to have generated the fluctuation in the first place, exist outside our definition of the universe?

On a more practical note, baryogensis describes the process of matter-antimatter annihilation in the earlier universe and suggests 1 billion+1 particles for every 1 billion antiparticles, does this account for the estimate of there being +2 billion photons for every particles? For reference, this and a few other questions were raised in the following thread:
https://www.physicsforums.com/showthread.php?t=235046

 

[yuiop]

Hi Kev: Thanks for the indirect link to http://www.mathpages.com/
What a thought-provoking source of material, by Kevin Brown, who appears to be a somewhat anonymous character as far as trying to track down his academic background.

Glad you like it. It is one of my personal favourites ;) The site is a mine of useful information and insightful comment.

...
On the basis that the particle and antiparticle have opposite charge, the annihilation has to maintain the conservation of charge, i.e. net zero charge. As I understand it, along similar arguments, the particle and the antiparticle annihilate into two photons, because of the conservation of spin, i.e. the two photons have net zero spin. This just leaves the issue of mass-energy, which I assume is resolved via the following equation: E = mc^2 = hf

This is what I was getting at. The rest mass energy of a photon can be expressed in terms of the energy momentum relationship is:

m_o^2 = (hf/c)^2 - (p)^2

where p is the mometum of the photon. Because by definition p=hf/c it is obvious that the rest mass of a single photon is zero. What is not so obvious that a system comprising of a pair of photons of equal energy going in opposite directions have a net momentum of zero and so the net rest mass of the 2 photons is not zero:

m_o^2 = (2*hf/c)^2 - 0

So in this case, the system of two photons which individually have no mass have that quality called mass that we normally assign to "solid" particles. This is real mass in that a box of photons weighs more than an empty box and a cloud of photons can be a source of gravity. This gives an indication of how ephemeral the concept of mass is. The same is true of energy. A bullet shot from a rifle appears to have lots of kinetic energy but from the point of view of someone co-moving with the bullet the bullet has zero kinetic energy. You might point out that the bullet has rest mass from any observers perspective, but then again so does a pair of photons and photons are nothing but pure momentum energy and no mass. So basically, energy is just a point of view!

...

However, given that there are two photons, I am not sure whether the energy distribution is always symmetric, i.e. f=mc^2/2h ?

AFAIK the distribution should be symmetric because a particle can only annihilate with its own specific tailor made antiparticle, although I have not seen that explicitly stated. For example I do not think a positron could annihilate a proton.

...
Based on my assumptions, which might be wrong, let's now follow your hunch. In #3, I speculated whether we could consider kinetic energy being positive and potential gravitational energy being negative in order to conserve energy under radial free-fall. By following this line of thought, it might be assumed that the expansion of the universe required an initial input of positive energy that was balanced by the implicit negative gravitational potential linked to this expansion. Within this simplistic model, I imaged this balance might be restored at a big crunch, but raised the question of a violation of idea behind the universe starting as a quantum fluctuation, if the universe never collapsed. However, the idea of the big crunch also raised a picture of all the particle/photonic mass in the universe, ever increasing in positive kinetic energy and negative potential, as it converged back towards a singularity. Can we assume this energy just cancels and would the quantum process, assumed to have generated the fluctuation in the first place, exist outside our definition of the universe?

I like to think to think the big crunch bounces back and starts the next big bang. Its a bit like a pendulum how a pendulum works. Potential energy is converted into kinetic energy as it swings down and then the reverse happens and kinetic energy is converted to potential energy as it swings back up. This converting of energy from one form to another can continue indefinitely as a cyclic process in an ideal system and I like to think a universe is an ideal system because no energy can escape from the universe. I think the main objection to a cyclic universe is the 2nd law of thermodynamics but I do not think that is a clear cut issue here. For example thermodynamics predicts that a cloud of hot atoms in space would expand and cool but gravity has opposite ideas and collapses the gas and heats it up instead to form a star. Sometimes gravity goes even further and collapses the gas to a black hole and Hawking showed that a black hole has a very high entropy and so there is no contradiction with thermodynamics insisting entropy always increases. The problem is Hawking also predicted that blacks eventually evaporate. If the formation of a black hole is an example of increasing entropy then there is a paradox because the evaporation of a black hole would be a case of spontaneous reduction in entropy. I personally believe that the situation is resolved by thinking of gravitational entropy as being the opposite of thermal entropy and the net entropy (thermal +gravitational) is always zero. Its a bit like the pendulum again but substitute the words "gravitational entropy" for potential energy and "thermal entropy" for kinetic energy. The system just oscillates back and forth converting gravitational entropy to thermal entropy and back again in an everlasting cyclic process. At the moment the available evidence suggests the universe is accelerating and can never collapse again. Roger Penrose is currently working on a cyclic model where towards the end of our current cycle, everything ends up in black holes which is followed by a period where all the black holes evaporate. His idea requires that neutrons eventually decay like everything else and that the final stage ends up with a radiation dominated universe with no massive particles. At that stage the next universe kicks off without collapsing again. In the model he is working on the entropy is always increasing. It is sort of counter intuitive to me because it seems to require a bottomless pit of potential energy reserves. (maybe that is exactly what the vacuum energy is?). An increase in thermal entropy is usually a loss of potential energy. A hot object next to a cold object has potential energy and entropy works towards bringing the system to thermal equilibrium so that there is no potential energy left. This is the basis of the "heat death" idea of the universe where the universe ends up as cold low-density lifeless expanse. To me, one alternative based on the Penrose idea is that the late epoch of one cycle is a universe with no matter and just the remnants of the radiation red shifted to a very low energy state then perhaps re-collapse will be possible because the gravitation equation for a radiation dominated universe is different from a matter dominated universe. However someone would have to do the math ;)

...
On a more practical note, baryogensis describes the process of matter-antimatter annihilation in the earlier universe and suggests 1 billion+1 particles for every 1 billion antiparticles, does this account for the estimate of there being +2 billion photons for every particles? For reference, this and a few other questions were raised in the following thread:
https://www.physicsforums.com/showthread.php?t=235046

That seems reasonable. I guess any cyclic model would have to be able to have some sort of anti-baryogenis stage just during the big crunch, where remnant photons from the last stage are blue shifted to high energies and produce particle-antiparticle pairs again.

 

[yuiop]

Hi Jado,

This post from the Relativity forum might be more the sort of thing you are looking for.

"How space time hold matter in orbit? by contact?"
This by no means anybody can guess even now

- Einstein said "There is no sense in regarding matter and field as two qualities quite different from each other ... Could we not reject the concept of matter and build a pure field physics? We could regard matter as the regions in space where the field is extremely strong. A thrown stone is, from this point of view, a changing field in which the states of the greatest field intensity travel through space with the velocity of the stone" -
mystery lies in the fecund soil of quantum electro/chromo area but Einstein from 1930 till his death tried to establish his belief through only math without any result.

 

[mysearch]

Latex Display Problem?

Hi Kev: This is just a test because my browser is having trouble displaying all of the Latex code embedded in your reply #8. So wanted to do a test before actually replying to the technical content of your post by embedded the same latex code in this test post.

Your first Latex block is embedded in the second quote, i.e. E = mc^2 = hf.
The actual display corresponds to the following file:
https://www.physicsforums.com/latex_images/18/1837325-2.png
However, if I try to access this link directly I get the following error message:
“Sorry the page your looking for can't be found.”
I have cut and pasted the following Latex code from my post #7: f=mc^2/2h, which seems OK and corresponds to the following file:
https://www.physicsforums.com/latex_images/18/1836992-1.png
This also displays OK directly.

Have you got a similar problem or is it just my system? Apologises for disrupting the thread with a technical support issue but was unsure where to post sucg problems, i.e. is there a technical support channel in PF?

P.S. Does anybody else see this problem?

 

[mysearch]

Response to #8:

#8: What is not so obvious that a system comprising of a pair of photons of equal energy going in opposite directions have a net momentum of zero and so the net rest mass of the 2 photons is not zero:

Had a bit of trouble with this statement. I don’t want to belabour the point, but would like to highlight the essence of my confusion. Understand that we can assign a kinetic mass to a photon by virtue of E=mc^2=hf such that m_k=hf/c^2. We can also adapt the general energy equation for a photon:

E^2 = m_o^2c^4 + p^2c^2 such that m_o = E^2/c^4-p^2/c^2

If we assume momentum of a photon \rho=m_k*v=(hf/c^2)*c=hf/c and substitute into the equation above, we agree that the rest mass of a photon is zero. Presumably, if momentum is conserved during a collision between a particle and anti-particle, I would have thought that the net momentum of the 2 photons has to match the original momentum of the particles? If this were a head-on collision, i.e. net zero momentum, then I would also assume that the 2 photons are ejected in opposite directions, i.e. net zero momentum. From a conservation of energy perspective, we might approximate the total energy of the mass particle in terms of rest mass and kinetic, ignoring any potential gravitational energy for now:

E_T=m_oc^2+1/2mv^2

On the basic of a symmetric split across the 2 photons emitted, each photon would have an energy of E_T=hf, i.e. corresponding to the particle energy, which could again be interpreted in terms of a photon kinetic mass, but not as a rest mass. I understand that the velocity associated with kinetic energy is relative to the observer, as presumably is the frequency of the photon.

#8: I like to think to think the big crunch bounces back and starts the next big bang.

Are you making this statement on the basis that you believe the universe has net zero energy? Again, as I understand the arguments so far, I believe the concept of inflation not only solves the issues of the horizon and flatness problems, but drives \Omega_T to unity and would lead to the conclusion of an open-flat universe that doesn’t collapse. The interpretation of measurements seems to conform that \Omega_T must be very close to 1. It was this position that led me to ask whether such a universe would require a net positive amount of energy?

https://www.physicsforums.com/showthread.php?t=235046

As a final point, have you read the article referenced in the link above. It appears to forward the idea of a finite mass universe within an infinite spatial universe.

 

[jado]

re#9
yes, it makes sense. I think we are going to have too see this from a unique perspective, ie .if a coin is flipped what are the chances that it will land head and tale at the same time? most likely never. But what if it could. i see the field as the coin landing as head and tale at the same time. not matter and field, but energy and space. if there is a an excess amount of energy in a given space then matter is created. this may relate to quantum mechanics as well. energy and space encompassing within each other. one static and the other dynamic.

 

[George Jones]

Have you got a similar problem or is it just my system? Apologises for disrupting the thread with a technical support issue but was unsure where to post sucg problems, i.e. is there a technical support channel in PF?

P.S. Does anybody else see this problem?

I could PM you this, but maybe a public post is better. Yes there was a problem with Latex, which I think now is fixed. The bottom forum on the main Physics Forums page is Forum Feedback & Announcements; this would be a good place to report such problems.

 

[mysearch]

I could PM you this, but maybe a public post is better. Yes there was a problem with Latex, which I think now is fixed. The bottom forum on the main Physics Forums page is Forum Feedback & Announcements; this would be a good place to report such problems.

George: thanks for the feedback. Did try out your suggestion just to see how it works:
https://www.physicsforums.com/showpost.php?p=1838760&postcount=1

 

[yuiop]

Had a bit of trouble with this statement. I don’t want to belabour the point, but would like to highlight the essence of my confusion. Understand that we can assign a kinetic mass to a photon by virtue of E=mc^2=hf such that m_k=hf/c^2. We can also adapt the general energy equation for a photon:

E^2 = m_o^2c^4 + p^2c^2 such that m_o = E^2/c^4-p^2/c^2

If we assume momentum of a photon \rho=m_k*v=(hf/c^2)*c=hf/c and substitute into the equation above, we agree that the rest mass of a photon is zero. Presumably, if momentum is conserved during a collision between a particle and anti-particle, I would have thought that the net momentum of the 2 photons has to match the original momentum of the particles?

Yes, it does. In the centre of momentum frame of the original two particles the net momentum is zero and after annihilation the net momentum of the two photons is still zero and it is in fact observed that the two photons are emitted in opposite directions.

If this were a head-on collision, i.e. net zero momentum, then I would also assume that the 2 photons are ejected in opposite directions, i.e. net zero momentum. From a conservation of energy perspective, we might approximate the total energy of the mass particle in terms of rest mass and kinetic, ignoring any potential gravitational energy for now:

E_T=m_oc^2+1/2mv^2

On the basic of a symmetric split across the 2 photons emitted, each photon would have an energy of E_T=hf, i.e. corresponding to the particle energy, which could again be interpreted in terms of a photon kinetic mass, but not as a rest mass. I understand that the velocity associated with kinetic energy is relative to the observer, as presumably is the frequency of the photon.

You are talking about individual photons here while I am talking about the pair of photons as a system. Each photon has zero rest mass but as a pair they have positive rest mass.

The momentum of a single photon is as you mentioned \rho=m_k*v=(hf/c^2)*c=hf/c The momentum is a vector with direction and magnitude and the direction is given by the c in hf/c. For the two photons the total momentum is hf/c + hf/(-c) = 0 and it follows that the rest mass of the two photons as a system is \Sum(m_o) = \sum(E^2/c^4)- \sum(p^2/c^2) =2*E^2/c^4[/tex]<br /> <br /> <blockquote data-attributes="" data-quote="mysearch" data-source="post: 1837940" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> mysearch said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Are you making this statement on the basis that you believe the universe has net zero energy? Again, as I understand the arguments so far, I believe the concept of inflation not only solves the issues of the horizon and flatness problems, but drives \Omega_T to unity and would lead to the conclusion of an open-flat universe that doesn’t collapse. The interpretation of measurements seems to conform that \Omega_T must be very close to 1. It was this position that led me to ask whether such a universe would require a net positive amount of energy?<br /> <br /> <a href="https://www.physicsforums.com/showthread.php?t=235046" class="link link--internal">https://www.physicsforums.com/showthread.php?t=235046</a><br /> <br /> As a final point, have you read the article referenced in the link above. It appears to forward the idea of a finite mass universe within an infinite spatial universe. </div> </div> </blockquote><br /> Well the available evidence suggests total Omega is slightly greater than unity suggesting a finite universe. I am not sure I am too impressed with that article. It starts with an assumption of dark energy to explain how their dust universe model works. I would prefer an explanation for the dark energy.<br /> <br /> P.s. Sorry about the latex problem. I noticed it and tried to fix it at the time, but no joy and thought it was just a problem with my pc.

 

[mysearch]

Response to #15

For the two photons the total momentum is hf/c + hf/(-c) = 0 and it follows that the rest mass of the two photons as a system is \Sum(m_o) = \sum(E^2/c^4)- \sum(p^2/c^2) =2*E^2/c^4

Interesting logic. However, I am not sure that you can mix the idea of a scalar energy and vector momentum within the same equation, which appears to only be addressing scalar energy. Coming at the problem from a different perspective, we have already agreed that [m_o=0], at least, in the isolated case and that:

E^2 = m_o^2c^4 + \rho^2c^2; if I now substitute \rho= hf/c, but only consider the magnitude, we get:

E^2 = m_o^2c^4 +|(hf)|^2, i.e. I am only considering the scalar energy component in this context.

As such, m_o^2 = (E^2 - |(hf)|^2)/c^4, but E=hf, which still suggests, at least to me, that the rest mass of the combined system is zero.

 

[yuiop]

Interesting logic. However, I am not sure that you can mix the idea of a scalar energy and vector momentum within the same equation,

You can and this case you must. Momentum is definitely a vector quantity. When working out the result of particle collisions in Newtonian or Relativistic terms the energy is always scalar and momentum always has a direction that must be taken into account. In relativity rest mass is an invariant quantity and it cannot change for a total isolated system, (although it can change for a single decaying particle.) During an elastic collision the rest mass of individual particles do not change. If you do the calculations for a relativistic elastic collision and treat the moment as a scalar quantity the rest mass of the system and particles is not conserved and nor is the total energy of the system if there are any negative velocities involved. The equations simply do work if you do not treat momentum as a vector.

The required equations are here: http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

If you look at the energy terms in the calculations the velocities are scared and so the energy is independent of direction. The velocities of the momentum terms are not squared and the direction becomes important. The energy-momentum relationship applies to everything and photons are no exception.

 

[mysearch]

Response to #17

Hi Kev: if you have got time I would like to get to the bottom of this misunderstanding. First, can we simplify the discussion to a single example, i.e. particle and antiparticle annihilating? As such, I will characterise the collision under discussion as follows:

2m_oc^2 + E_{k(+)} + E_{k(-)} \rightarrow 2hf

This equation describes the rest mass energy and the kinetic energy of our original particles coming together in a head-on non-relativistic collision to produce 2 photons of equal frequency, i.e. energy. I believe we have agreed that in isolation, a photon has no rest mass, only a kinetic mass, i.e. m_k=hf/c^2[/itex]. However you have made the following statement, which is the issue of confusion:<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <i>#8: What is not so obvious that a system comprising of a pair of photons of equal energy going in opposite directions have a net momentum of zero <b>and so the net rest mass of the 2 photons is not zero</b>:</i> </div> </div> </blockquote><br /> As understood, this statement is based on your equation:<br /> \Sum(m_o)^2= \sum(E^2/c^4)- \sum(p^2/c^2) =2*E^2/c^4<br /> <br /> Now the momentum of a photon is given by \rho= hf/c, but because we have 2 photons and a collision where the conservation of momentum is required to be net zero, it reduces the equation above to the form shown with the implication that the 2-photon system has a rest mass. However, if I take the equation above and substitute for the photon energy (E=hf) it appears to open up another interpretation:<br /> <br /> \Sum(m_o)^2 = 2*E/c^2 = 2*(hf)/c^2<br /> <br /> This form seems to be equivalent to E=mc^2=hf, which reduces to m=hf/c^2 as cited above as kinetic mass [m_k] of a photon, not its rest mass. As such, I see no disagree over your following statement other than the earlier inference of a photon rest mass rather than what I prefer to call kinetic mass.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <i>So in this case, the system of two photons which individually have no mass have that quality called mass that we normally assign to &quot;solid&quot; particles. This is real mass in that a box of photons weighs more than an empty box and a cloud of photons can be a source of gravity. This gives an indication of how ephemeral the concept of mass is. The same is true of energy. A bullet shot from a rifle appears to have lots of kinetic energy but from the point of view of someone co-moving with the bullet the bullet has zero kinetic energy. You might point out that the bullet has rest mass from any observers perspective, but then again so does a pair of photons and photons are nothing but pure momentum energy and no mass. So basically, energy is just a point of view!</i> </div> </div> </blockquote><br /> On the basis of E=mc^2=hf, where [c] and [h] are constant, the inference is that [m] and [f] have some sort of equivalence. If so, any relativistic effect on mass, which changes energy with respect to a given frame of reference, must also change [f]. Am I still missing any other subtlety in the argument?

 

[yuiop]

Hi Kev: if you have got time I would like to get to the bottom of this misunderstanding. First, can we simplify the discussion to a single example, i.e. particle and antiparticle annihilating? As such, I will characterise the collision under discussion as follows:

2m_oc^2 + E_{k(+)} + E_{k(-)} \rightarrow 2hf

This equation describes the rest mass energy and the kinetic energy of our original particles coming together in a head-on non-relativistic collision to produce 2 photons of equal frequency, i.e. energy. I believe we have agreed that in isolation, a photon has no rest mass, only a kinetic mass, i.e. m_k=hf/c^2[/itex]. However you have made the following statement, which is the issue of confusion:<br /> <br /> <br /> As understood, this statement is based on your equation:<br /> \Sum(m_o)^2= \sum(E^2/c^4)- \sum(p^2/c^2) =2*E^2/c^4<br /> <br /> Now the momentum of a photon is given by \rho= hf/c, but because we have 2 photons and a collision where the conservation of momentum is required to be net zero, it reduces the equation above to the form shown with the implication that the 2-photon system has a rest mass. However, if I take the equation above and substitute for the photon energy (E=hf) it appears to open up another interpretation:<br /> <br /> \Sum(m_o)^2 = 2*E/c^2 = 2*(hf)/c^2<br /> <br /> This form seems to be equivalent to E=mc^2=hf, which reduces to m=hf/c^2 as cited above as kinetic mass [m_k] of a photon, not its rest mass.

<br /> <br /> In a way you have answered the question of this thread. Energy is just a conversion factor. The energy equivalent of the rest mass of the two photon system happens to be same as the energy equivalent of the kinetic energy of the two photons. It does not mean the rest mass IS the the kinetic mass. <br /> <br /> This simple analogy might make it clearer. The kinetic energy of a 20 tonne train moving at 60 kph is equivalent to the (non nuclear) energy content of a bag of potatoes.<br /> <br /> That does not mean a 20 tonne train IS a bag of potatoes.<br /> <br /> <blockquote data-attributes="" data-quote="mysearch" data-source="post: 1840263" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> mysearch said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> On the basis of E=mc^2=hf, where [c] and [h] are constant, the inference is that [m] and [f] have some sort of equivalence. If so, any relativistic effect on mass, which changes energy with respect to a given frame of reference, must also change [f]. Am I still missing any other subtlety in the argument? </div> </div> </blockquote><br /> I think you are onto something profound and interesting here which is probably relates mass/energy to time dilation. I will have to think about it some more.

 

[mysearch]

Response to #19

The energy equivalent of the rest mass of the two photon system happens to be same as the energy equivalent of the kinetic energy of the two photons.

Thanks for the response. On the basis on the quote above, I believe you have resolved my confusion. However, I still need to decide whether a photon is a train or potato.

I think you are onto something profound and interesting here which is probably relates mass/energy to time dilation. I will have to think about it some more.

While possibly outside the scope of this forum, you might want to reflect on the implications of why a particle has both a Compton and deBroglie wavelength, but a photon only has a Compton wavelength, i.e.

Compton \lambda = h/mc
deBroglie \lambda = h/mv

The Compton wavelength comes directly from E=mc^2=hf=hc/\lambda and is often described in terms of fundamental limit on measuring the position of a particle, taking quantum mechanics and special relativity into account.

Note: What positional resolution can you give to a wave?

While the de Broglie wavelength is inversely proportional to the momentum of a particle and the frequency is directly proportional to the particle's kinetic energy. A pseudo derivation can be considered in terms of our equation:

E^2 = m_o^2c^4 + \rho^2c^2 = m_o^2c^4 + m^2v^2c^2

For a photon, [m_o=0] and [v=c], so the equation reduces to the form:

E^2 = m^2c^2c^2 where m_?=hf/c^2=h/c\lambda

This takes us back to Compton’s equation. For a particle at rest, i.e. [v=0], the equation reduces to the form of rest mass energy:

E^2 = m_o^2c^4 which we might wish to equate to E=hf

This again leads back to Compton’s equation. However, in our last case, we will assume that the particle velocity [v] is approaching [c] allowing the form of the equation to reduce to:

E^2 \rightarrow m^2v^2c^2 then equating E=mvc=hf=hc/\lambda; Note: assume m_o&lt;&lt;m_k

We arrive at deBroglie’s equation, but what physical interpretation can be inferred? It seems to suggest that a particle has 2 associated wavelengths when [v>0], i.e. the Compton wavelength by virtue of its rest mass and the deBroglie wavelength by virtue of its velocity [v], so how might this wave-duality exists along side the wave-particle duality of quantum physics? One possible interpretation is to picture the Compton-carrier wave being modulated with the deBrogie wave. However, I will terminate at this point because I recognise that I am beginning to cross the line into speculation, but would be interested in your thoughts.

 

[mal4mac]

If Feynman was unsure of the concept...

Feynman was not unsure *about* the concept. Although, certainly, he does not know what energy 'is' in terms of fundamental building blocks! See lecture I 4-1. Energy is simply the name for a numerical quantity that does not change in a natural process. It cannot be described as anything concrete.

You might say a concrete atom is composed of concrete protons, neutrons, and electrons. But you cannot decompose energy into concrete building blocks. To quote the man "we have no knowledge what energy is. We do not have a picture that energy comes in little blobs... there are formulas for calculating some numerical quantity and when we add it all together it gives "28" -- always the same number. It is an abstract thing in that it does not tells us the mechanisms or reasons for the various formulas."

So we have an object on a shelf with potential energy E = mgh, it drops of the shelf and h (height) decreases. But amazingly if we work out mgh + 1/2mv^2 then we get the same number E! So we can use this number (the total energy) to make useful calculations.

So the energy is defined in terms of more fundamental quantities, but because energy can be transformed in many ways you can't define it as being *just these* quantities. For instance you can't say 'energy is mgh' because sometimes it's 1/2mv^2, sometimes the sum of these, and sometimes something else entirely.

So energy is a concept built on shifting sands. And like shifting sand dunes it is not a constantly existing, concrete object. It's a number, an abstraction. Stretching the metaphor to breaking point, it like the number of shifting sand dunes in a desert if some magic djin had made that number constant. A magic number indeed, but no real "blob" that you can pick up and show to mum.

 

[mysearch]

Response to #27

My comment about Feynman was intended to be somewhat retrospective and a general attempt to response to the original question raised in this thread, i.e.

what is energy? i would like to know what energy is beyond it's definition.

However, there seem to be a somewhat contradictory element in your arguments:

Feynman was not unsure *about* the concept…..
…So energy is a concept built on shifting sands. And like shifting sand dunes it is not a constantly existing, concrete object.

Personally, `a concept built on shifting sand` seems to suggest a degree of uncertainty in the concept. While I couldn’t find the exact source of my quote from post #3, here is another source that put it into better context:

One of the best elementary explanations of energy was given by Nobel laureate Richard Feynman (Feynman et al 1963, pp 4-1 to 4-80). Energy is not something perceptible but a quantity that has to be calculated using a comprehensive set of rules. Feynman used a metaphorical story about a mother counting a kid’s toy blocks. An underlying principle, conservation of blocks, leads Mother to look in more and more places (do more experiments) and to find new ways to account for blocks (invent more theory) to make sure that the conservation principle works. That kind of explanation is nothing like a traditional schoolbook definition. Feynman’s introductory statement is about a thousand words long even before he gets to specific examples. His “definition” differs from formal texts in another very important way: the concept remains open to future modification, even falsification. In effect, he says “I don’t know what energy is but if you have plenty of time I can teach you how to calculate it”.

I guess it not in the nature of these discussions to replicate the details that can be found in most physics textbooks, but it might be worth tabling a few issues that others may wish to correct or clarify in the context of cosmology or general relativity. Based on the assumption that we can calculate the amount of energy associated with a system, maybe we can define a few rules of energy ‘accountancy`, e.g.

- Energy is an attribute of a system, which may consist of one or more objects.

- Broadly, the conservation of energy states that energy cannot be created or destroyed, it can only be changed from one form to another or transferred from one body to another, but the total amount of energy remains constant.

- So when the energy of a local system increases (or decreases) there is a corresponding decrease (or increase) outside the local system.

Of course, in the context of cosmology is it not necessary understood whether the system, i.e. the universe, is open or closed. However, we might still want to rationalise the multitude of energy forms as follows:

- There are only two basic kinds of energy: kinetic energy (KE) and potential energy (PE).

- Kinetic energy is associated with motion.

- Potential energy is associated with interactions between objects, e.g. gravitational

- Potential energy can be defined only for a system consisting of two or more parts, i.e. the system requires structure and, as such, an isolated particle cannot have potential energy……..etc.

So in this context I would like to table the following question:

Is dark energy a pressure per unit area or energy per unit volume?

I ask this question because, in a gas, the pressure is normally associated with kinetic collisions of particle.s In contrast, in a fluid, pressure is often defined as energy per unit volume. So is this expansive energy of the universe still kinetic in nature, i.e. positive? Is it offset by gravitational potential energy, i.e. negative? If not, what happens to the idea of the conservation of energy?

 

[jado]

I, guess what i am looking for is, what is it that the universe is made of, the basic the most fundamental part of it. the only thing i can think of is space and energy. space and energy because this are two entities which can be detected but can not be broken down into smaler part. I, suppose i am sugesting space is much like mobius strip, with space and energy being the two components representing front and back or rathe simotanusly being frant and back.

 

[throng]

I, guess what i am looking for is, what is it that the universe is made of, the basic the most fundamental part of it. the only thing i can think of is space and energy. space and energy because this are two entities which can be detected but can not be broken down into smaler part. I, suppose i am sugesting space is much like mobius strip, with space and energy being the two components representing front and back or rathe simotanusly being frant and back.

I do get what you're driving at!

The universe is made of energy. We see light, matter, movement etc as energy forms yet cannot describe energy as energy its primary state.

If the primary state of the universe was one of inertia, it is natural to assume some nature of force acted in order to perpetuate motion.

So we seek the nature of an energy that exists in inertia. (even before space/time)

I think we could theoretically deduce some basic qualities of this primary Energy.

Its energy is equal to total energy in the universe, by laws of energy conservation. In so saying it was the cause and the universe (being forms of energy) is the effect.

I guess that's as far as I go, though I find the nature of primal energy interesting.

I have explored this question in philosophy and spiritualism and asking "what is the nature of energy" is like asking "what is the nature of god". Maths and words just loose meaning.

It should be more accurately defined, Not many inquire into this.