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Would like to know what energy is beyond its definition

  1. Aug 15, 2008 #1
    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?
     
  2. jcsd
  3. Aug 15, 2008 #2
    Re: energy

    i would say energy is any thing that shows a reaction in physical nature
     
  4. Aug 15, 2008 #3

    mysearch

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    Re: energy

    I realise that the following quote by Richard Feynman doesn’t really answer your question, but it might help put some others into perspective:
    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 [tex]E=mc^2[/tex], 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. [tex]F=m_I*a=GMm_G/r^2[/tex]. With reference to the energy orientation of this discussion, we might differentiate the gravity equation to define the potential gravitational energy, i.e. [tex]U=-GMm/r[/tex]. Note, the negative sign.

    The definition of relativistic mass is more controversial these days, but can be initially introduced as [tex]m’=\gamma m[/tex]. 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?
     
  5. Aug 15, 2008 #4
    Re: 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.
     
  6. Aug 16, 2008 #5
    Re: energy

    stop looking at the effect, see that which is the effect. what particle what anti particle? look beyond this.
     
  7. Aug 16, 2008 #6

    mysearch

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    Response to #5

    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.
     
  8. Aug 16, 2008 #7

    mysearch

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    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
    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:
    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
    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: [tex]E = mc^2 = hf[/tex]. However, given that there are two photons, I am not sure whether the energy distribution is always symmetric, i.e. [tex]f=mc^2/2h[/tex]?
    Based on my assumptions, which might be wrong, lets 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
     
  9. Aug 16, 2008 #8
    Re: Response to #4

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


    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:

    [tex]m_o^2 = (hf/c)^2 - (p)^2 [/tex]

    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:

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

    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!


    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.


    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 ;)

    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.
     
    Last edited: Aug 16, 2008
  10. Aug 17, 2008 #9
    Re: energy

    Hi Jado,

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

     
  11. Aug 17, 2008 #10

    mysearch

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    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. [tex]E = mc^2 = hf[/tex].
    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: [tex]f=mc^2/2h[/tex], 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?
     
  12. Aug 17, 2008 #11

    mysearch

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    Response to #8:

    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 [tex]E=mc^2=hf[/tex] such that [tex]m_k=hf/c^2[/tex]. We can also adapt the general energy equation for a photon:

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

    If we assume momentum of a photon [tex]\rho=m_k*v=(hf/c^2)*c=hf/c[/tex] 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:

    [tex]E_T=m_oc^2+1/2mv^2[/tex]

    On the basic of a symmetric split across the 2 photons emitted, each photon would have an energy of [tex]E_T=hf[/tex], 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.

    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 [tex]\Omega_T[/tex] 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 [tex]\Omega_T[/tex] 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.
     
    Last edited: Aug 17, 2008
  13. Aug 17, 2008 #12
    Re: energy

    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.
     
  14. Aug 17, 2008 #13

    George Jones

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    Re: Latex Display 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.
     
  15. Aug 18, 2008 #14

    mysearch

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    Re: Latex Display Problem?

    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
     
  16. Aug 18, 2008 #15
    Re: Response to #8:

    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.

    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 [tex]\rho=m_k*v=(hf/c^2)*c=hf/c[/tex] 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 [itex] \Sum(m_o) = \sum(E^2/c^4)- \sum(p^2/c^2) =2*E^2/c^4[/tex]

    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.

    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.
     
  17. Aug 18, 2008 #16

    mysearch

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    Response to #15

    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 [tex][m_o=0][/tex], at least, in the isolated case and that:

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

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

    As such, [tex]m_o^2 = (E^2 - |(hf)|^2)/c^4[/tex], but E=hf, which still suggests, at least to me, that the rest mass of the combined system is zero.
     
  18. Aug 18, 2008 #17
    Re: Response to #15

    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.
     
  19. Aug 19, 2008 #18

    mysearch

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    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:

    [tex] 2m_oc^2 + E_{k(+)} + E_{k(-)} \rightarrow 2hf[/tex]

    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. [tex]m_k=hf/c^2[/itex]. However you have made the following statement, which is the issue of confusion:
    As understood, this statement is based on your equation:
    [tex] \Sum(m_o)^2= \sum(E^2/c^4)- \sum(p^2/c^2) =2*E^2/c^4[/tex]

    Now the momentum of a photon is given by [tex]\rho= hf/c[/tex], 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:

    [itex] \Sum(m_o)^2 = 2*E/c^2 = 2*(hf)/c^2[/itex]

    This form seems to be equivalent to [tex]E=mc^2=hf[/tex], which reduces to [tex]m=hf/c^2[/tex] as cited above as kinetic mass [tex][m_k][/tex] 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.
    On the basis of [tex]E=mc^2=hf[/tex], 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?
     
  20. Aug 19, 2008 #19
    Re: Response to #17

    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.

    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.

    That does not mean a 20 tonne train IS a bag of potatoes.

    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.
     
  21. Aug 20, 2008 #20

    mysearch

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    Response to #19

    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. :smile:
    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 [tex]\lambda = h/mc[/tex]
    deBroglie [tex]\lambda = h/mv[/tex]

    The Compton wavelength comes directly from [tex]E=mc^2=hf=hc/\lambda[/tex] 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:

    [tex]E^2 = m_o^2c^4 + \rho^2c^2 = m_o^2c^4 + m^2v^2c^2 [/tex]

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

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

    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:

    [tex]E^2 = m_o^2c^4[/tex] 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:

    [tex]E^2 \rightarrow m^2v^2c^2 [/tex] then equating [tex]E=mvc=hf=hc/\lambda[/tex]; Note: assume [tex]m_o<<m_k[/tex]

    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:eek:, but would be interested in your thoughts.
     
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