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Asymmetric photon interaction with gravity?

  1. Oct 19, 2013 #1

    bahamagreen

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    Assumptions:

    1] I'm an observer at inertial rest
    2] Light is going c
    3] Gravitational interaction can't exceed c

    I'm going to use "photon" and "graviton" as shorthand to pose the questions initially. Maybe correcting the form or assumptions of the questions will provide the answer...

    The photon is moving past me at c, and if another photon was to be following the first one, I would not observe it catching up to the first one.

    If the photon could send a graviton backwards to a mass behind it, that mass might receive it, but my observation must be that the mass couldn't send a graviton that could catch up to the photon.

    Likewise, the photon might intercept a graviton from a mass in front of it, but my observation must be that the photon could never send a graviton forward to a mass in front of it.

    This makes the photon appear to my observation to have the inability to complete an exchange with masses either behind or in front of it, even though the photon is both sending and receiving gravitons, there is not a "proper" exchange. The photon is receiving from the mass in front and sending to the mass behind, and neither of the masses is enjoying a complete exchange.

    Questions:

    1] Does an exchange interaction require sending and receipt completion for both parties to the exchange?

    2] How do photons exchange gravitons with masses directly behind and in front if both photons and gravitons go c?

    3] Do light cones apply here? Light cones have their causal regions in their interiors, but what I'm imagining is something like a light cone that describes the photon's interaction with gravitons... the interiors would be defining the regions where the gravitons from masses could (front cone) and could not (back cone) ever intercept the photon...?

    Thanks for any assistance on this.
     
    bahamagreen, Oct 19, 2013
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  3. Oct 19, 2013 #2

    Dale

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    We don't have a working theory of quantum gravity, so your question, as posed, cannot be answered yet. However, if you modify it to be "gravitational wave" instead of "graviton" and "pulse of light" instead of "photon" then you have an answerable question.

    In general relativity the spacetime produced from massless radiation is called a pp-wave spacetime:
    http://en.wikipedia.org/wiki/Pp-wave_spacetime
    http://arxiv.org/abs/gr-qc/0410006
     
    Dale, Oct 19, 2013
  4. Oct 19, 2013 #3

    bahamagreen

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    Thanks,

    I looked through both references and some others but could make little progress through the terminology. A large part of the problem is that I'm not even sure what I'm questioning yet. I have found some terms with respect to Lorentzian geodesics that seem very suggestive. Can you tell me if any of the following might be what I'm asking about?

    -connectedness
    -nonreturning property
    -pairs of congugate points
    -compactly imprisoned ends
    -causal disprisoning
    -nonimprisonment of future and/or past endless causal curves
     
    bahamagreen, Oct 19, 2013
  5. Oct 19, 2013 #4

    The_Duck

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    Yes, but let's be clear what this means. It means that *changes in the gravitational field* can't propagate at speeds faster than c.

    You'd be better off not thinking about gravity working by "exchanging gravitons." It does in a sense, but to usefully reason in that way you have to know about the mathematical details that give rise to this description. In particular the "virtual gravitons" that are exchanged can in some sense travel at infinite speeds.

    You're better of thinking in terms of fields. Objects with energy set up gravitational fields around them, and nearby objects move in response to those gravitational fields. If an object moves, the resulting change in its gravitational field propagates outward at c.

    For the above-mentioned reasons I'm going to use the field language instead of talking about gravitons. I'm guessing the underlying question here is "suppose a photon is traveling away from a massive object. Does the photon feel the gravity of the object, and does the object feel the gravity of the photon?"

    The answers are yes and yes. As it travels away from the object, the photon is traveling through the object's gravitational field, which already exists, so nothing needs to "catch up" to the photon in order for the photon to feel the object's gravity. The object also feels the gravity of the photon: the photon's gravitational field is changing in time because the photon is moving, but it does have gravity. You're right that you should think about light cones: all objects in the future light cone of the photon will feel the effects of its gravity.

    I guess another question you are asking is "suppose a photon is travelling towards an object. Does the object feel the photon's gravity?" The answer is no: objects can only feel the gravity of objects within their past light cone.
     
    The_Duck, Oct 19, 2013
  6. Oct 19, 2013 #5

    bahamagreen

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    Thanks,

    I was using "asymmetry" is two ways, one in the sense that the photon's interaction was geometrically different from all the other particles, and second, that this interaction was asymmetric in the forward and backward direction of travel.

    The first sense would be disposed - all particles only feel gravity from sources in their back cones - and the second sense confirmed... but, then if I take this:

    "As it travels away from the object, the photon is traveling through the object's gravitational field, which already exists, so nothing needs to "catch up" to the photon in order for the photon to feel the object's gravity."

    wouldn't this apply to the forward cone as well? If I changed it to be...

    "As it travels toward the object, the photon is traveling through the object's gravitational field, which already exists..."

    It seems that the photon could feel the gravity of the mass in front of it, but the mass could not feel the gravity of the photon approaching it.

    This was why I asked about what constitutes a "proper" exchange - whether both parties needed to have sent and received for the influence to have occurred (or equivalent language for the field interaction).
     
    bahamagreen, Oct 19, 2013
  7. Oct 21, 2013 #6

    bahamagreen

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    Modifying "graviton" to "gravitational wave" restricts the question from asking about gravitational interaction to only asking about changes in gravitational interaction, does it not?

    Surely the intent of the original question (aside from its clumsy form) is answerable.

    The_Duck, your answer has confused me more... I hope it is clear that all observations are being made from inertial rest.
     
    bahamagreen, Oct 21, 2013
  8. Oct 21, 2013 #7

    PeterDonis

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    No. If you're going to think about interactions as exchanges of particles, then each individual particle exchange is one-way: the "source" emits a particle (and recoils--it changes momentum so that overall momentum is conserved at emission), and that particle is later absorbed by the "receiver" (which changes momentum as a result of the absorption, so that overall momentum is conserved at reception). The momentum changes in the source and receiver are what constitute the observable "interaction".

    Of course any interaction that's observable macroscopically will be made up of zillions of individual particle exchanges; and if there are two objects interacting, each one will be "source" in about half the exchanges and "receiver" in the other half, so the observable interaction is two-way. But each individual particle exchange is one-way.

    Virtual gravitons, which are what the photons would exchange to affect each other gravitationally, don't have to go at c; they can go faster than c.

    More precisely, virtual particles in general do not have to be "on-shell"; that means their mass, momentum, and energy don't have to be related the way the mass, momentum, and energy of a real particle are, by the "mass shell" equation ##E^2 - p^2 c^2 = m^2 c^4##. (For gravitons--and photons, for that matter--##m## is zero, but the equation still holds; it just says ##E = pc##.) The observed speed of the particle is ##v = pc / E##, so a real "on shell" particle will have ##v < c## if ##m > 0##, or ##v = c## if ##m = 0##. But a virtual particle doesn't have to obey those restrictions; it can have ##v > c##.

    Yes, but in order to see how they apply, you have to stop thinking of interactions as exchanges of particles. (Which is a good thing, since that viewpoint has a number of significant limitations, one of which is that it easily misleads people into asking the sorts of questions you're asking.) Instead, you have to think of them as fields on spacetime, as The_Duck said, or, in the case of gravity, you can think of it as curvature of spacetime. Any mass or energy that is present acts as a source of the field (or as the source of spacetime curvature).

    If you adopt this view, then you can show something very useful: if I want to know what the field is at a given event in spacetime, I only have to know what sources are present in the past light cone of that event. Nothing outside the past light cone of an event can affect the field at that event.

    So, for example, if I want to know the gravitational field affecting a particular photon's motion at a particular event, I only have to look in the past light cone of that event. The Einstein Field Equation encodes this fact; so when people talk about solutions of the EFE, they are talking about particular spacetimes in which we know the configuration of the sources and have used this knowledge, through solving the EFE, to arrive at a complete description of the field at every event, such that that field is determined completely by the sources in the past light cone of that event. In the case you describe, with multiple photons moving in the same direction, that solution is the pp-wave spacetime that DaleSpam linked to.
     
    PeterDonis, Oct 21, 2013
  9. Oct 22, 2013 #8

    bahamagreen

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    Thanks,

    Your answer to #1 (if keeping with particles and exchange) suggests that the photon would be able to receive gravitational exchange particles from ahead but not from behind... but...

    Your answer to #2 recommends that the exchange mechanics cannot support this line of thinking because the exchange of virtual particles includes >c exchanges. So I'm looking at the field approach...

    Your answer to #3 has me continuing to wonder; I'm still missing things:

    1] If the virtual particle exchange idea is not the preferred mode, did not their >c mechanics arise to account for some observations or requirement? That is, would not the preferred field mechanics have to include an account of the same observations or requirements (not the observation of >c but the need to have something that accounts for the same reason the >c was attributed to virtuals in the first place)?

    Or another way to ask it; the invoking of light cones does seem to eliminate the influence of >c virtual particles... does the field replace any >c mechanics for which the virtuals were posited in the first place?

    2] If the field is present in space-time, why are its sources only in the past light cone? Does not a mass in front of the light curve space-time in the front light cone, and is not that curvature's influence present at the event (apex of the light cones)? I'm not understanding how light can be gravitationally "blind" to its forward light cone if the sources in the forward light cone curve space-time in the forward light cone. I don't know the math words for it, but is not the curvature caused by a source radially symmetric without any asymmetry across the apex of the light cones?
    Or let me ask it this way... is it that light is only "blind" to curved space in the forward light cone (how, if the event is in regionally curved space-time?), or is the curvature not present in the forward light cone? The later would be the asymmetry that would have curvature behind the event and none in front...?

    3] I'm not understanding some fundamental things about light cones and space-time curvature? I'm going to take a look at it some more.
     
    bahamagreen, Oct 22, 2013
  10. Oct 22, 2013 #9

    PeterDonis

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    If you mean, do we have to allow virtual particles to move faster than c in the math because we've actually observed some interactions to occur faster than c, the answer is no.

    It is true that we allow virtual particles to move faster than c in the math because that's the only way that modeling interactions as particle exchanges can be made to match the data. But if that shows an issue with anything, it's an issue with modeling interactions as particle exchanges: it's basically telling you that that model is fundamentally wrong, even if it works as an approximation in some cases.

    Of course the field model has to account for the same observations. But it does so in a way that makes causal structure--which events can causally influence which other events--a lot clearer, by making it obvious that causal influences can't travel faster than light. That is also true in the virtual particle model, but it's much harder to see that it's true in that model.

    Read what I said again, carefully. I said that the field at a given event is determined completely by the sources present in the past light cone of that event. Obviously, to account for the field at every event in spacetime, you have to include the sources at every event in spacetime, because (at least on certain highly likely assumptions about spacetime structure) every event is in the past light cone of some other event. But if you're looking at the field at some particular event, that field doesn't "know" about any sources outside the past light cone of that particular event.

    (Remember that an "event" is a point in spacetime: a given location in space, at a given instant of time. If you stay at the same location in space over time, you may measure changes in the field as new sources come into your past light cone--i.e., as you move to later events along your worldline, whose past light cones include more of spacetime than the past light cones of earlier events along your worldline did.)
     
    PeterDonis, Oct 22, 2013
  11. Oct 23, 2013 #10

    bahamagreen

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    Thanks,
    I'm still failing to see this...

    I see how the light cones limit what can reach and influence the event for influences that mediate through space-time... their paths need to be within the past light cone.

    What I don't see yet is why this would apply to gravitation as space-time curvature... the curvature is throughout the space-time region containing the event - its static component is already "in place". The event seems like it it already being influenced by sources because the event is influenced by the local curvature, even for sources in the front light cone and in the "elsewhere" regions outside of both light cones.

    How is the local curvature at the event not causally continuous with the curvature all around it? I can see how a change in the curvature might need to come through the past cone, but the static component of the curvature seems like it is already in place. What I'm interpreting from your description is that the space-time is only curved in the past light cone and either flat or nonexistent elsewhere...?

    In other words, if the event is measuring X amount of curvature, are you saying that all of that X is from the past light cone and none from the other space-time around the event? I've been thinking that matter sources curve space-time independent of events - that the events are embedded within the local space-time and the degree of curvature is a local value at the event, that value coming from the regional space-time, some of which is in the forward light cone and the elsewhere region. I'm not seeing how the identification of an event could make the regional curvature not contribute to the local curvature at the event...?
     
    bahamagreen, Oct 23, 2013
  12. Oct 23, 2013 #11

    Dale

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    No, it isn't. We don't have any theory which has "gravitons". How could you answer a question about flubnubitz without a theory of flubnubitz?

    The scientific community is trying to obtain a theory which has gravitons, but it hasn't achieved that goal yet.

    We do have a theory of gravity which can answer questions about gravitational interactions in terms of various fields (curvature and stress energy). It can describe both interactions and changes in interactions, but neither of those involve gravitons. As stated, the original question is not answerable.
     
    Dale, Oct 23, 2013
  13. Oct 23, 2013 #12

    yuiop

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    It is.
    Consider the set of all objects that have always been stationary with respect to a given observer. The set of events that constitute where they all are "now" is outside the past and future light cone of our observer.When their wordlines are extended into the past, they must intercept the past light cone no matter how far away they are. The set of events where these wordlines intercept the past light cone represents the static curvature that our observer responds to.

    Now consider the set of all objects that are moving with respect to our observer. With the exception of objects moving towards the observer at the speed of light, the wordlines of all these objects must also intercept the past light cone, so the observer is aware of them, but sees where they were at some point in the past, rather than where they are "now". This set of events constitutes the dynamic part of the curvature at the observer's location. Of course, if an object that was stationary with respect to the observer suddenly starts moving and if this change event is outside the past light cone, then the observer will not be aware (gravitationally or otherwise) of the change, until some time in the future.

    Elsewhere, the curvature at each event is determined by its own past light cone.I think this might be the key point you are missing.
    Yes.
    Generally speaking, the future light cone has no influence on the gravitational curvature at a given event except perhaps in the most cutting edge quantum gravitation theories which are still a work in progress.
     
    yuiop, Oct 23, 2013
  14. Oct 23, 2013 #13

    bahamagreen

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    Thanks, that gives me a little flicker of illumination.
    I don't understand it yet, but I'll keep at it.
     
    bahamagreen, Oct 23, 2013
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