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Questions concerning Bhabha scattering

  1. May 10, 2012 #1


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    I am trying to get a better understanding of what aspects of an electron-positron collision can be physically observed as opposed to mathematically inferred. In part, my confusion is based on a number of different sources, e.g. see links below, which seem to adopt different approaches to describing the same thing, i.e.
    As a basic starting point, see attached diagram below, I am assuming that the electron-positron (p1,p2) pair on the left might be described as an initial state that can be observed. Likewise, the electron-positron (p3,p4) pair may be liken to some final state that can also be observed. So my first questions are:
    • Are (p1+p2)=(p3+p4) representative of initial and final observable states, such that the conservation of momentum and charge have to be maintained?
    • Does the probability of any final observable outcome depend on the energy of the initial state, e.g. does the probability of collision also have to account for the collision cross-section, if defined as a function of energy?
    While the initial diagram attached makes no speculation as to what permutation of interactions might exist within the collision zone, the Wikipedia page on Bhabha scattering alludes to 2 basic permutations, labelled ‘annihilation’ and ‘scattering’. However, from this description and the cross reference to Feynman diagrams:
    • Is it is true say that annihilation and scattering reflect only the main possibilities of many other potential permutations; although other permutations may have diminishing probability amplitudes associated with each additional vertex coupling, which are not observable?
    In the first Wikipedia diagram:
    Graphically, we appear to have an electron [e-] and positron [e+] colliding and annihilating at [t1,x], which results in a photon being created, which then ‘transmutes’ back into an electron and positron at [t2,x]. As such, the photon seems to exists for [t2-t1] but travels no distance.
    • In the specific case of electron-positron annihilation, can the photon continue to exist as it doesn’t seem to violate any obvious conservation laws and would appear to have one less coupling factor? – see question regarding whether this is a virtual photon below.
    • What quantum process describes a photon of energy changing back to an electron-positron pair, i.e. it doesn’t appear to result in a lower energy state?
    In the second Wikipedia diagram:
    We appears to have a positron [e+] scattering at a point [x1,t], while the electron scatters at a point [x2,t]. Based on a left-to-right timescale, it seems to suggest zero time, but a separation [x2-x1].

    • Is this sort of physical interpretation simply inappropriate, if so, what are these diagrams actually trying to convey by alluding to time-distance axes?
    • Does the photon in either diagram have any observable existence, i.e. does it have to be described as a virtual photon and does it have to comply with all conservation laws?
    • Is there any physical evidence that the virtual photon is ‘really’ created or is this simply an inference of the maths?
    • In terms of the scattering process, does the physical separation [x2-x1] essentially correspond to the dynamic energy cross section of the collision?
    • Does scattering require more initial state energy than annihilation?
    I realise that this is a lot of questions, but was hoping that different members might be able to clarify one or two of the issues raised.

    Attached Files:

  2. jcsd
  3. May 10, 2012 #2


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    Yes, momentum and charge must be conserved at each step in the diagram.

    It's usually presented the other way around: the collision cross-section is computed in terms of the probability of collision (the matrix element squared that the page describes how to compute). The energy dependence is explicitly described in term of the Mandelstam variables used at http://en.wikipedia.org/wiki/Bhabha_scattering

    Yes, there are additional diagrams involving the same initial and final states that could be considered due to including different types of interactions. In this case, the probability for those events to occur is smaller than the "leading-order" annihilation and scattering diagrams, so they give a smaller contribution to the total cross-section. If one needed to compare to a precise measurement, one would need to include those corrections to find agreement.

    No, the photon in the diagram has an energy [itex]E[/itex] and momentum [itex]\vec{p}[/itex] that satisfies

    [tex] E^2 - |\vec{p}|^2 = s,[/tex]

    which is the center-of-mass energy of the initial electron-positron pair. A real photon must satisfy

    [tex] E^2 - |\vec{p}|^2 = 0,[/tex]

    because it is massless. This is essentially what distinguishes a real particle from a virtual one. The virtual particle does not satisfy the "mass shell condition" that it's 4-momentum satisfy [itex]|\mathbf{p}|^2 = m^2[/itex].

    The electromagnetic interaction couples a photon to an electron. This is described by each individual vertex in that diagram. Depending on how you arrange the in and out states of the vertex, you can get an electron and positron in the initial state and a photon in the final state, or an electron in the initial state and an electron and photon in the final state.

    Energy is conserved at every vertex, but a diagram cannot represent a complete process unless all of the initial and final state particles can satisfy the mass shell condition I described above.

    The computation of the matrix elements are being written in momentum space, so you don't really see it, but the quantum amplitude really integrates over all vertex points. So the matrix element is really

    [tex] \mathcal{M} = \int d\mathbf{x}_1 d\mathbf{x}_2 M(\mathbf{x}_1 \mathbf{x}_2).[/tex]

    The momentum space expressions given there are what you'd get by using plane wave solutions for the wavefunctions and using standard formulas for Fourier transforms.

    Note the need to integrate over all space is an inherently quantum mechanical idea that's tied to the interpretation of the wavefunction.

    The photon is virtual for the reasons given before. We cannot measure its existence by virtue of that fact.

    The cross section depends on the directions and magnitude of the initial electron and positron. This is a bit mysterious when written in terms of the Mandalstam variables [itex]s,t,u[/itex] but it's a standard exercise to convert this to an expression for more intuitive variables.

    In a sense. If you imagine an electron and positron at rest a small distance apart, the mutual interaction will pull them together and they will annihilate rather than scatter. They must have significant kinetic energy to overcome the potential energy of attraction.

    This is not really what's being described by the Bhabha process though. Bhabha scattering already assumes that the particles scatter (we have an electron and positron in the final state) and computes the cross section for the scattering process, which we could call [itex]\sigma(e^+e^-\rightarrow e^+e^-)[/itex]. This is an example of what's called an exclusive cross section. Since we exclude other possible final states. The inclusive cross section would be [itex]\sigma(e^+e^-\rightarrow \mathrm{anything})[/itex], which would include real annihilation processes where we have just photons in the final state.
  4. May 11, 2012 #3


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    Really appreciated the knowledgeable answers provided to all my questions. While I need to think about and read around some of the issues raised, I had a couple of immediate points I would like to clarify, if possible:
    I had read that Feynman rules require the conservation laws to be applied at each vertex, but I also read somewhere that the conservation laws do not always apply to virtual particles. For example, can a free electron emit a (virtual?) photon, such that from the rest frame of the electron, energy is not conserved? However, at an even more basic level, I am not that sure that I understand how a free electron emits a photon, as I thought this was restricted to electrons in atomic orbitals – is this another virtual attribute?
    Thanks for the Wikipedia reference to Mandelstam variables, which I hadn’t heard of, so this is an issue for me to review in more detail. However, I raised the question based on an article – as per first link in post#1. The suggestion appeared to be that the cross-section was proportional to the inverse square of energy. If so, and based on your later comments, high energy would appear to increase the probability of scattering, as opposed to annihilation. Therefore, wouldn’t the probability of scattering also be a function of the energy of the initial state? Is this point simply accounted for by the momentum of the inital state?
    Again, I need to review your response in more detail, but wanted to check the relativistic basis of your equations, i.e. [itex]E^2=m^2c^4+p^2c^2[/itex]. So based on this equation, you are saying that a ‘real’ photon has zero rest mass such that [itex]E^2-|\vec{p}|^2=0[/itex], but a ‘virtual’ photon has some component - is this some aspect of rest mass? However, I think I need to look at the issue of ‘on shell and off shell’ in more detail, e.g. http://en.wikipedia.org/wiki/On_shell_and_off_shell
    By way of a somewhat philosophical point, do you think virtual photons have any physical existence, which are simply beyond our ability to detected, or are just mathematical requirement of QFT?

    Again, really appreciated the helping hand. Thanks
  5. May 11, 2012 #4


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    The Feynman rules do require the conservation laws at the vertices. But there are two features that put virtual particles at odds with the conservation of energy and momentum. First, as mentioned is the fact that virtual particles are "off shell." They have the wrong energy and momentum when compared to their mass. The mass shell condition is usually interpreted as a conservation law of its own: if a particle has a given momentum, you can use the mass shell condition to determine the energy it must have. For virtual particles, this rule is violated.

    Another (related) feature comes up when we study higher order corrections, which inevitably involves so-called loop diagrams. A loop diagram is a diagram, as the name suggests, where there is a closed loop of particle lines, like this one:


    This diagram describes the process where an electron with 4-momentum [itex]p[/itex] emits a photon of momentum [itex]l[/itex] and then reabsorbs it a bit later. The Feynman rules tell us to conserve momentum, so in the intermediate state the electron has momentum [itex]p -l[/itex]. In the final state, it has momentum [itex]p[/itex] again.

    We had to introduce the momentum [itex]l[/itex] because of energy conservation, but energy conservation does not fix [itex]l[/itex] to any specific value. In fact the Feynman rules go on to say that we must integrate over all possible values of [itex]l[/itex]. We then see, even more dramatically, how far from being on mass shell the intermediate particles are. The photon mass shell condition is [itex]|l|^2=0[/itex], but in this diagram
    [itex]|l|^2[/itex] actually runs over all values from [itex]0[/itex] to [itex]\infty[/itex].

    Maybe the best way to put it is that, when dealing with virtual particles, energy and momentum are conserved in a strict mathematical sense. But because the mass shell condition is violated, normal physical intuition regarding the conservation laws does not hold.

    When you distinguish scattering from annihilation, do you mean the individual diagrams listed on http://en.wikipedia.org/wiki/Bhabha_scattering#Solution? What I was trying to say before was that, for any process that we're actually computing, the probability is given by the square of the matrix element, so we don't have to try to use the cross section to answer the question. From http://en.wikipedia.org/wiki/Bhabha_scattering, the square of the matrix element for the scattering diagram is

    [tex] P_\mathrm{scatt} \sim \frac{s^2+u^2}{t^2},[/tex]

    while that for annihilation is

    [tex] P_\mathrm{annih} \sim \frac{t^2+u^2}{s^2}.[/tex]

    I've dropped numerical factors that are common to both processes.

    These results were computed in the limit where the energy of the electrons is large compared to their mass, so it turns out that we can approximate [itex] s+t+u\approx 0[/itex]. Then

    [tex] \frac{P_\mathrm{annih}}{P_\mathrm{scatt}} \approx \frac{t^4+t^2(s+t)^2}{s^4 +s^2(s+t)^2}.[/tex]

    The convenient limit to consider is the one where the center-of-mass energy of the particles, [itex]s[/itex] is large, but the momentum transfer between them, [itex]t[/itex] is held fixed, so the ratio [itex]t/s[/itex] is small. Then

    [tex] \frac{P_\mathrm{annih}}{P_\mathrm{scatt}} \approx \left(\frac{t}{s}\right)^2 \rightarrow 0, ~~~~\frac{t}{s}\rightarrow 0.[/tex]

    So yes, at large energies, the scattering diagram is a larger contribution than the annihilation diagram.

    Note that the probabilities (and even the cross section itself) are not solely just functions of the initial momenta. This is because the amount of momentum transferred is a variable. The final state momenta can take a range of values while still satisfying total energy and momentum conservation.

    No, we don't say that virtual particles have a different mass than a real particle of the same type. We just say that the virtual particle is off-shell and doesn't satisfy that formula. Part of the Feynman rules involves including factors (propagators) that depend on the mass of the particle. For virtual particles, we use the physical mass in these expressions, not some fictitious mass.

    If we say that something is only physical if we can directly measure it, then virtual particles are not real. However, virtual particles are just real particles that are off of their mass shell. This is allowed by the Heisenberg uncertainty principle for energy and time. In between measurements, otherwise real particles can become virtual and we can use Feynman diagrams to describe the processes by which that can happen, like the photon loop correction to the electron in the picture above.

    So I would say that we cannot ever detect a virtual particle, but we understand the nature of what a virtual particle is too well to just say that it is some mathematical bookkeeping convenience.
  6. May 12, 2012 #5


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    Again, thanks for the additional information. I am now trying to work through some of the maths and physics implied in some of the concepts raised. While I guess this might take some time, I just wanted to just table some issues, as I go, in the hope that if I go ‘off-shell’ in some of my assumptions, somebody might be able to put me back ‘on-shell’:redface:.

    On and Off Shell?
    As far as I can determine the terminology of ‘on-shell’ and ‘off-shell’ appears rooted in the relativistic equation [itex]E^2=m^2c^4+p^2c^2[/itex]. Any particle that conforms to this equation is said to be on-shell, e.g.

    [itex]E^2 - p^2c^2=m^2c^4[/itex] :real particle
    [itex]E^2 - p^2c^2=0[/itex] :real photon

    So, by way of a summary of current understanding: a virtual particle is said to be off-shell because it does not comply with these equations, i.e. a massless virtual particle, such as a photon, can have (+/-) mass and described as off mass shell by having a ‘mass’, which might be better described as ‘borrowed energy’ within the confines of the energy/time variant of the uncertainty principle. So while the conservation of energy and momentum must apply at the vertex of a Feynman diagram, the virtual photon associated with either annihilation or scattering can assume an off-shell mass/energy.
    Does this also touch on the issue of bare parameters and renormalisation?

    Mandelstam variables?
    At the moment, I am just trying to understand how these variables relate to any physical aspect of the collision/scattering to which they are associated, i.e. the s-channel appears to be associated with annihilation, while the t-channel seems to be associated with scattering. In the case on annihilation between relativistic particles, i.e. let E=pc, such that I would have thought the total energy/momentum, predicated on the vertex laws of conservation, could be expressed as [itex]E_1+E_2=E_3+E_4[/itex] and [itex]p_1+p_2=p_3+p_4[/itex]. So, from the perspective of energy conservation at the annihilation vertex, would the resultant energy of the photon [itex]E=hf=E_1+E_2[/itex] ignoring any deviation within the uncertainty principle? If so, could the momentum of the photon also be initially quantified as [itex]k=p_1+p_2[/itex]. However, in the case of scattering, would the photon energy at the vertex would be [itex]E=hf=E_1-E_3[/itex], while the momentum would be [itex]k=p_1-p_3[/itex]? In contrast, the Wikipedia definition of the Mandelstam variables (s) and (t) is:

    [itex]s=\left(p_1 + p_2 \right)^2 = \left(p_3 + p_4 \right)^2 [/itex] :annihilation
    [itex]t=\left(p_1 – p_3 \right)^2 = \left(p_2 + p_4 \right)^2 [/itex] :scattering

    Not sure why the definition of (s) and (t) involves the square of the total energy or momentum other than reflecting its roots in the relativistic equation [itex]E^2=m^2c^4+p^2c^2[/itex]? As yet, I am also not sure I understand the Wikipedia rationalisation of (s) and (t) at the high-energy limit, e.g.

    [itex]s=\left(p_1 + p_2 \right)^2 = (p_1)^2+(p_2)^2+2p_1.p_2 \approx 2p_1.p_2[/itex]

    As yet, it is not totally clear to me as to how the Mandelstam variables provide any additional insights into the description of Feynman diagrams over and above the definition of momentum and energy derived from the relativistic equation. However, the answer may well be in the various description of annihilation and scattering in terms of (s,t,u) as described in post#4, which I am still working on.

    Feynman Rules and other stuff?
    While I have just got a copy of Mandl and Shaw, plus a few free references, I have found the maths heavy going for a beginner like me. Feynman’s QED book and lectures have helped me gain some initial insights to the basic ideas involved, while an article by Alex Nelson: Notes on Feynman Diagrams has provided a starting point regarding the actual calculations behind the Feynman diagrams in the form of 2 toy models, i.e. tree and loop.
    This seems a logical argument, although classical free electron cannot even emit or absorb photons. At this stage, QFT seems to predicated on so many mathematical abstractions and approaches that it is sometimes difficult to know what to make of it all. Thanks
    Last edited: May 12, 2012
  7. May 12, 2012 #6


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    On the contrary, every particle that has ever been detected was virtual. It is the "real" particle that is a mathematical fiction! A virtual particle is a solution of the inhomogeneous wave equation that is emitted and absorbed at points A and B a finite distance apart. A real particle is a solution of the homogeneous wave equation that exists forever without ever being emitted or absorbed, and as such is an idealized limiting case.
  8. May 12, 2012 #7


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    I agree with this summary, but I prefer the borrowed energy concept.

    The mass that we actually measure in experiments is a renormalized mass called the pole mass. This pole mass is what appears in the mass shell condition for on-shell particles, even though it is the bare mass that appears in the Feynman rules. It's kind of hard to explain precisely why all this works without explaining how renormalization in QFT works in technical detail. In brief, virtual particles are allowed to go off-shell whether we're using the bare or renormalized masses.

    I think that your confusion is the Mandelstam variables are defined in terms of the 4-momenta. If you look back at my other posts, I tried to distinguish the ordinary 3-momenta with an arrow. In the formulas for the Mandelstam variables above

    [tex]s=\left(p_1 + p_2 \right)^2 = (E_1+E_2)^2 - | \vec{p_1} - \vec{p_2}|^2, [/tex]

    with similar expressions for the other combinations.

    It is the 4-momenta that appear in that formula above, so we can use

    [tex](p_1)^2=m_1^2,~~~(p_2)^2 =m_2^2,[/tex]

    because the external particles are on shell. Then

    [tex] s = 2p_1\cdot p_2 + m_1^2 + m_2^2 = 2p_1\cdot p_2 \left( 1+\frac{m_1^2 + m_2^2}{2p_1\cdot p_2 }\right) .[/tex]

    In the high energy limit all energies and 3-momenta are much larger than the masses, [itex]E,|\vec{p}|\gg m[/itex], so we ignore the ratio on the RHS.

    Learning QFT requires a strong background in quantum mechanics, especially perturbation theory, and electrodynamics, especially special relativity. Its going to be the case that too that authors will expect a certain level of familiarity with general physical ideas from other areas, as well as some mathematical sophistication. I don't know your background, but you are asking good questions, so you probably have a good idea of what prerequisites you would want to review before diving into QFT proper.
  9. May 12, 2012 #8


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    I can sympathize with this to some extent, but there is a great deal of background that goes into properly appreciating any full attempt at answering the question. It's actually worse than this, because no one really understands why we can ignore Haag's theorem.

    My original reply to mysearch's question started with a discussion of the measurement problem in QM and it just degenerated from there. I chose not to go down that path because my own understanding is not sufficient to avoid the esoteric in trying to remain rigourous. I don't really get the impression that the people that study QM foundations agree on everything either.

    While strictly accurate, your reply really doesn't shed any light on why measured particles are on shell. If you have any insight on why it's a better answer to give someone whose just trying to learn QFT, I'd be happy to listen. There seems to be a some happy medium between giving practical (incomplete, but not strictly incorrect) explanations and giving rigorous (correct, but nevertheless incomplete or opaque) explanations that not everyone agrees on. I tend to think it's better to err on the side of explaining what we do understand than emphasizing the esoteric details that we do not really understand.
  10. May 12, 2012 #9


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    fzero, I don't think it's really as deep as all that. It's an illusion, a consequence of the fact that our particle detectors have a finite resolution. Particles we measure appear to be on the mass shell because the distance between their emission and absorption points is large (but finite), and all the probability amplitudes have canceled out except for ones very near the mass shell, exceeding the ability of our detectors to resolve. They're still virtual, but seem real, because we can no longer tell the very slight difference.
  11. May 12, 2012 #10


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    That is a good way of looking at it, which should have been more obvious to me. Thanks.
  12. May 13, 2012 #11


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    Hi, you raised a issue regarding educational background that I would like to respond having noted your role as a science advisor and homework helper. First, your help in this thread has been much appreciated and very useful, as QFT has/is proving to be a very difficult subject to self-learn from textbooks and articles in isolation. While I am a science graduate, my degree was in applied physics and so long ago it was probably written on papyrus:frown:. Basically, I am now retired and use some of my time trying to filling in the gaps in my education and the PF mission statement has been very supportive of my goals:

    "Our mission is to provide a place for people (whether students, professional scientists, or others interested in science) to learn and discuss science as it is currently generally understood and practiced by the professional scientific community."

    As such, I rely heavily on the PF forum to help out when I get stuck on a subject, although by its nature, the PF can be a ‘hit or miss’ method that often depends on member interest and qualifications to answer specific issues. While many appear to like the brevity of a sound-bite response, the value of such responses may only be appreciated by those who already know the subject. I mention this, not as a criticism, as each member is entitled to their own approach, but by way of explanation as to why I have appreciate the balanced level of detail provided in response to my questions and why I use this thread, and others, to detail my own understanding for future reference. Many thanks
    Last edited: May 13, 2012
  13. May 13, 2012 #12


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    As do I. To be honest, I struggle to understand the maintenance of the particle paradigm in much of QFT semantics. I appreciate that at some scale, the particle model is more convenient and may actually make more sense, but as a fundamental description of nature, particles that have no defined substance makes no sense to me. At some level, it seems that we must be left with only waves, fields and energy, or even bits of string!, such that mass is only a form of energy density. However, I accept that I am not really qualified to make such statements and possibly Bill_K has another more informed perspective based on his earlier posts.
    I hadn’t heard of the ‘pole mass’ and thought that the concept of ‘mass’ must already have more than enough definitions – but clearly I was wrong on this point as well. However, if I have understood the description above, based on the on-shell condition aligning to the relativistic equation discussed in early posts, does the pole mass essentially align to the measured rest mass in the frame of reference of the measurement? While I haven’t really read too much into the mathematical details of renormalisation, I thought the bare mass resulted from the self-energy surrounding charged particles, which renormalisation limits to some finite and measurable value, i.e. does the pole mass equate to the rest mass or do relativistic factors still have to be taken into consideration.
    Your right. I will try to brush up on the concepts of 4-vectors in general before posting any further questions, but thanks for the clarification, it was a useful pointer.
    Last edited: May 13, 2012
  14. May 14, 2012 #13


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    It is accepted that the issue of 4-momenta is possibly not the focus of this sub-forum, but the issue was originally raised in connection to understanding the Mandelstam variables in post #5, to which Fzero responded in post #7 as follows:
    Before considering 4-vectors, in general, I wanted to try to establish an initial descriptive model, which is essentially classical, so that the conservation of energy may be considered as a scalar quantity, while the conservation of momenta is considered as a vector quantity. However, this model is analogous to an annihilation collision between 2 particles, as the conservation of energy and momentum would appear ‘relatively’ unambiguous between the particles and virtual photon in this case. It is also assumed that the centre-of-mass of this model can be used as the frame reference for all variables. As such, the inbound positron and electron particles are assigned energy values (E1) and (E2) and momenta (p1) and (p2) respectively. While it is assumed that we are dealing with relativistic particles, energy is a scalar quantity, such that combined energy of the inbound positron and electron might 'initially' be assumed to be converted into the virtual photon in the form: E=hf=E1+E2. In contrast, momentum is a vector quantity, which we might initially describe in terms of a 2D model, such that the combined momentum of the positron and electron would be the sum of [x,y] components, e.g. the y-components might cancel, leaving only the x-components of momentum being conserved in the photon? In terms of vector geometry:

    [itex]k^2= \left(p_1 \right)^2 + \left(p_2 \right)^2 – 2 \left(p_1 p_2 \right) cos \theta [/itex]
    [itex]k^2= \left(p_1 \right)^2 + \left(p_2 \right)^2[/itex] when θ=90

    While the idea of an analogous billiard ball collision might be inappropriate, it is only intended as a reference model for some of the ideas that still seem to apply, if the conservation of energy and momentum are applicable at the collision vertex within the Feynman annihilation diagram. For example, if we view the Feynman annihilation diagram referenced as a space-time diagram, we appear to end up with the suggestion that the virtual photon only exists in time, albeit briefly, as there is no implied propagation in space. Just for reference, the scattering diagram appears to suggest the opposite, i.e. the virtual photon propagates in space in zero time. As such, it is not really clear as to how momentum is transferred to the virtual particle, especially if the mechanism actually involves some form of quantum wave superposition.
    While I accept the statement above, I am also assuming that mathematical sophistication still has to be rooted in understanding, not rote learning, and therefore I really wanted to try and establish whether my understanding of 4-vectors is correct and therefore the following statements are not assertions, but rather points for clarification:

    [itex]4-vector=[scalar, \left(vector \right)][/itex]
    [itex]4-position~~[\vec{x}] =[ct, \left(x, y, z \right)][/itex]

    Given the reference above to special relativity, we might describe 4-position in terms of a Minkowski spacetime diagram:

    [itex]s^2= \left(x_0 \right)^2 + \left(x_1 \right)^2 = \left(x_0’ \right)^2 + \left(x_1’ \right)^2[/itex]

    Here (x0,x0’) are the generalisation of time (ct), while (x1,x1’) represents the 1D separation in space in 2 different frames of reference, such that the spacetime interval or proper time in invariant in all frames of reference. We might simply highlight that the invariance of a spacetime interval is linked to the Lorentz factor [itex]( \gamma )[/itex]. Extending the 4-vector concept to velocity and momentum:

    [itex]4-velocity ~~[\vec{v}] = \gamma[c, \left(v_x, v_y,v_z \right)][/itex]
    [itex]4-momentum ~~[\vec{p}] = \gamma m_0[c, \left(v_x, v_y, v_z \right)][/itex]

    In the case of 4-momentum, we might rearrange the definition of (p0), such that it is linked to energy (E):

    [itex] p_0 = \gamma m_0 c[/itex]
    [itex] c p_0 = \gamma m_0 c^2=E[/itex]
    [itex] p_0 =E/c[/itex]

    At this point we can defined 4-moment as:

    [itex] \vec{p} = [E/c, \gamma m_0 \left(v_x, v_y, v_z \right)][/itex]
    [itex] \vec{p} = [E/c, \gamma \left(p_x, p_y, p_z \right)][/itex]

    We might now interpret this equation in two ways. First, by analogy to a Minkowski spacetime diagram, which was shown to lead to an invariant quantity called the spacetime interval. As such, if we plotted energy against momentum, the implication is that we might end up with another invariant quantity, although this might be best presented in terms of 4-momentum constrained to 1D:

    [itex] \vec{p}.\vec{p} = \left(p_0 \right)^2 + \left(p_1 \right)^2 = \left( \gamma m_0 c \right)^2 + \left(\gamma m_0 v \right)^2 = \left(m_0 c \right)^2[/itex]
    [itex] \left(p_0 \right)^2 + \left(p_1 \right)^2 = \left( E/c \right)^2 + \left( p \right)^2 = \left(m_0 c \right)^2[/itex]
    [itex] E^2 = \left( m_0 c^2 \right)^2 + \left(pc \right)^2 [/itex]

    In essence, we see the encapsulation of the relativistic energy equation within the definition of 4-momentum, where rest mass energy is the invariant quantity in all frames of reference. If this understanding of 4-momentun is correct, it should be possible to reconcile it with the following statement from post#2:
    It is assumed that [itex] E^2 - |\vec{p}|^2 = s[/itex] is using natural units, such that [c=1]? If so, this equation suggests that (s) is analogous to the invariant rest mass energy term in the relativistic energy equation above. As such, it encapsulates a descriptive difference in a virtual photon compared to a real photon, which has no rest mass? The equation [itex]|\mathbf{p}|^2 = m^2[/itex] is explained in terms the earlier derivation:

    [itex] |\mathbf{p}|^2 = \vec{p}.\vec{p} = \left( E/c \right)^2 + \left( p \right)^2 = \left(m_0 c \right)^2[/itex]

    Again, if [c=1], we have the relationship suggested between 4-momentum and the rest mass? However, making reference back to original descriptive model above, it still appears that the annihilation collision between an electron and positron requires the conservation of energy at the vertex, i.e. E=E1+E2. While a ‘real’ photon would defined this energy in terms of E=hf, a virtual photon has some notional concept of rest mass energy, as defined by ? As such, there appears to be some ‘ambiguity’ as to the definition of momentum in terms of a virtual photon as opposed to a real photon, i.e. [itex](p= \gamma m_0 v)[/itex] versus [itex](p=E/c)[/itex]?

    While I realise that people might not have the time or inclination to wade through all this, any clarifications or corrections would be welcomed. Thanks
  15. May 14, 2012 #14


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    Terms like "defined substance" probably apply best to extended objects. "No defined substance" seems to imply properties of point particles rather than any quantum mystery, at least to a first approximation.

    The whole wave-particle duality is a good means of explaining the need for quantum mechanics. However, once you learn QM, the distinction should really be blurred. Elementary particles behave like point particles, the wave nature is a consequence of the probability amplitudes that quantum mechanics uses to describe them. There's no distinction between particle and wave, the distinction is really between the quantum description and the classical limit.

    The pole mass is essentially the mass measured in the rest frame of the particle in question. When you see some histogram referring to the discovery of a new particle, you're typically looking at a bell curve of events centered at an energy that corresponds to the mass of the particle. This is the pole mass.

    The bare mass corresponds to the parameter that we put into the Lagrangian. Because of the way the renormalization procedure deals with divergences, it turns out that the bare parameters themselves must have an infinite part. Even the finite part of the bare parameters are not themselves physical, since there is still some freedom to choose what is called a "renormalization scheme." While the details are exceedingly technical, the lesson to take away is that when the renormalization scheme is chosen, finite observables like scattering amplitudes emerge. By comparing enough of these to experiments, we use physical data to fix the unknown parameters of the theory (like the finite part of the bare parameters).

    There actually is implied propagation in space. These diagrams are really being drawn in momentum space. As I said in an earlier post, we could have just as well done the computation in position space, where would we actually have to integrate over the positions of the interaction vertices. Because of the way the Fourier transform works, such an analysis would reduce to the momentum space Feynman rules that are usually presented.

    The "suggestion that the virtual photon only exists in time" is an artifact of both the diagram being drawn in momentum space and it's being drawn symmetrically. The diagram drawn with the photon at an angle is perfectly equivalent. In fact, the only difference between the "annihilation" and "scattering" diagrams is encoded in the way that the external particles are assigned to be "in" or "out" states. Therefore we get different expressions depending on how the 4-momentum of the photon is written in terms of the external momenta. This is more apparent in the nomenclature also used on the wikipedia page of [itex]s,t,u[/itex] "channels," which makes this distinction precise in terms of the Mandelstam variables.

    Yes, I was using natural units with [itex]c=1[/itex]. Also, I used [itex]\mathbf{p}[/itex] to refer to the 4-momentum, so your last equation needs to be corrected accordingly.

    There is no particularly good reason to correlate the deviation from the mass shell condition with a mass for the virtual particle. This is why I emphasized the "borrowed energy" concept as being more appropriate. The mass shell condition is simply not obeyed by a virtual particle, we can't use it to assign a fictitious mass. There are good physical reasons like gauge invariance that prevent us from concluding that a virtual photon has a nonzero mass.

    While it is convenient to define a formula for relativistic momentum in terms of the velocity for a massive particle, the fact that the formula breaks down for massless particles, leads one to conclude that momentum is far more fundamental of a concept than velocity. This formula is useful for introducing special relativity since it appeals to classical intuition, but it is almost never used outside of classrooms. It is far more useful to specify (and measure) momenta instead of velocities.
  16. May 15, 2012 #15


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    I have appreciated your insights, and patience. You have raised a number of issues throughout this thread that I now recognise I need to understand better, possibly a lot better. I have found a couple of video lectures, e.g. David Tong and Prof.V.Balakrishnan, which may help. However, the biggest problem I am having is finding a good maths-physics textbook that will help me with the step function between QM and QFT. Any suggestions welcomed. Thanks
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