StringJunky Posted September 4, 2012 Share Posted September 4, 2012 (edited) No, but not all particles can be observers, either. Photons for example have no frame of reference and do not observe anything. Would it be right to say: you cannot observe from that which you are observing with. Is this why it is illogical to talk about a photon's FOR? Edited September 4, 2012 by StringJunky Link to comment Share on other sites More sharing options...
michel123456 Posted September 5, 2012 Share Posted September 5, 2012 I don't understand why the speed of light is involved in all this. If light travels at a certain fixed speed, why should that affect the time when something happens. For example, suppose at this present moment, a star 10,000 light-years away is exploding in a supernova. We on Earth in 2012, won't see it for another 10,000 years, as it will take light that long to reach the Earth. But surely that doesn't mean the event hasn't happened. Or are we supposed to think - no event happens until it's been seen by us on Earth. Isn't that a bit Geocentric? Instead of "geocentric" you could use the word "relative". Every observer has its own present. This relative present is placed where the observer is. Anywhere else, it is not his present. Actually what an observer calls "the present" looking around him is in fact the sum of events he can observe at a simultanate time he calls "now'. Physically all these simultanate events belong to his past. The only simultanate event that is not in his past is the event of his own existence exactly where the observer is. So, to "us" (meaning by "us" an humanoid living on planet Earth in some 10.000 future years), the supernova will appear to explode then. For another observer anywhere else in the Universe, the supernova will appear to explode at another time depending on the distance. Only to an observer upon the supernova it appears to happen "now". So that's relative, that's not geocentric. You could call that "observocentric" . Link to comment Share on other sites More sharing options...
swansont Posted September 5, 2012 Share Posted September 5, 2012 I don't understand why the speed of light is involved in all this. If light travels at a certain fixed speed, why should that affect the time when something happens. For example, suppose at this present moment, a star 10,000 light-years away is exploding in a supernova. We on Earth in 2012, won't see it for another 10,000 years, as it will take light that long to reach the Earth. But surely that doesn't mean the event hasn't happened. Or are we supposed to think - no event happens until it's been seen by us on Earth. Isn't that a bit Geocentric? You have to go an extra step into this. Ignoring, for the moment, general relativity effects, we can synchronize clocks in any reference frame using the method described by Einstein in his original SR paper. This accounts for the delay from the speed of light. All clocks read noon at noon, even though I might not get a signal from your clock for one second. I merely adjust for that delay in the calculation, because everyone in our frame will have a well-established delay of t = d/c. Similarly, all calendars will read 2012 when the supernova happens, even though we won't know about it until 12012. But until we observe it, we can't discuss the event. A problem arises when we try and incorporate a second reference frame, and special relativity has to be applied: time dilation and length contraction appear, and agreement of simultaneity for events that aren't co-located is lost. Some events can appear in either order, depending on your reference frame. In physics, you mention the reference frame and your location, and what your clock reads. Other observers will have a different time tag to what they see. Past, present and future are qualitative expressions that can only be guaranteed to have meaning in your own frame, so they have limited usefulness. Link to comment Share on other sites More sharing options...
IM Egdall Posted September 5, 2012 Share Posted September 5, 2012 I don't understand why the speed of light is involved in all this. If light travels at a certain fixed speed, why should that affect the time when something happens. See link http://www.youtube.com/watch?v=b_PtnzqxEFQ THis is one of many web demonstrations of Einstein's "light clock" thought experiment. The speed of light is unaffected by the speed of the observer or the speed of the source of the light. In other words, you always measure the same speed for a beam of light (in a vacuum), no matter what your speed. The video shows how this principle leads to the slowing of the passage of time with relative motion. Hope it helps answer your question. Link to comment Share on other sites More sharing options...
md65536 Posted September 6, 2012 Share Posted September 6, 2012 (edited) Actually what an observer calls "the present" looking around him is in fact the sum of events he can observe at a simultanate time he calls "now'. Physically all these simultanate events belong to his past. Why would one call "the present", events that "belong to one's past"? I don't. You're talking about one's "past light cone", which is of course considered to be one's relative past. Yes, it's what one experiences in a given moment. There's no confusion of the two, given a finite speed of light. A distant event observed in the present is a past event. A distant present event can't yet be experienced. To confuse things though, if we speak about events without specifying a reference frame, then two events that are outside each other's light cones (both past and future cones), are separated by a space-like interval, in which case "Generally, the events are considered not to occur in each other's future or past." [http://en.wikipedia....e-like_interval]. Observation events or experiential events imply an observer which implies their reference frame. --- Back to the original question... I think it might be best understood with mathematical concepts? All of the following is based on wikipedia entries... it is only my understanding of it, which is lacking!, but it's not hard to look up the entries. Perhaps someone can tell me if the following makes any sense! A "moment" --- as in a set of simultaneous events --- is frame-dependent, so let's consider a given observer's frame, with the observer as an object. The observer has a world line which represents all of the observer's locations through spacetime, in order. It is like a sequence of all its possible "now"s. At any point (event) on that world line, a moment "now" including that point can be represented by a Cauchy surface... that is tangential to the world line? Now is, when: At the proper time of the event, and where: anywhere on the surface. Note: "when" is according to a specific clock, and I'm using the observer's clock, but it could be expressed instead using any clock whose world line intersects the Cauchy surface, with "when" being the time on that clock at the point of intersection...blegh! It all depends on choice of frames and clocks, and it doesn't matter which is used as long as it is specified. The set of all Cauchy surfaces tangential to the observer's world line essentially "slices up" 4-d spacetime into a set of moments, each considered to be a "now" that is simultaneous with an event on the observer's world line, where it intersects the surface. (Does that make the set of surfaces a foliation of spacetime? Or is it not a foliation due to relative simultaneity, which can result in multiple such Cauchy surfaces intersecting the same distant event? ie. the surfaces are not generally parallel?) At each moment on the world line there is also a light cone. Inside the past light cone is all past events that can influence "now" (ie. events that can be known about, now), and inside the future light cone is all events that can be influenced by "now". At http://en.wikipedia....wiki/Light_cone, the image illustrates the distinction between the "present surface" and "presently visible past lightcone". Similarly, all calendars will read 2012 when the supernova happens, even though we won't know about it until 12012. But until we observe it, we can't discuss the event. I think this might be less confusing to note that we're talking about unpredictable events here. We can discuss predictions of events, based on information we've received up until now, but we can't discuss any new information from an event until we observe it. I don't think there's any physical difference between predicting past, present, or future events outside your light cone. Edited September 6, 2012 by md65536 Link to comment Share on other sites More sharing options...
LaurieAG Posted September 7, 2012 Share Posted September 7, 2012 (edited) I suppose if you lived in a 2D universe a 2D light cone might work but we need to work out what we are actually observing in 3D in the present before we start adding time dimensions, past or future and extrapolating back to 2D light cones. Consider a source that is rotating around a galactic centre and an observer at a location exactly one galactic year (one complete galactic rotation of the source) away. The photons continuously stream from the source as it rotates and travel directly to the observer at c. At the present time of the observation, 1 galactic year has elapsed and the source has rotated back to its original starting position: (1) is there a continuous stream of photons connecting the observer to the rotating source if the source continues to rotate and emit consistently and the path is never blocked or interfered with during the entire period of the rotation? (2) If you believe this physical photon stream exists in the present: (a) is it a perfectly straight line? (b) is it a curved line? (regardless of whether the source is moving directly away from or towards the observer, just to keep it simple)? Edited September 7, 2012 by LaurieAG Link to comment Share on other sites More sharing options...
md65536 Posted September 7, 2012 Share Posted September 7, 2012 (edited) A light cone is a 3d surface with a 4d interior. A simple intersection of an observer's (past) light cone and a (past) hyperplane "moment" is a sphere (2d surface), corresponding to all the events simultaneous with that "moment" that the observer now sees. I think it's intuitive enough. In flat space, everything from a moment 1 year ago that I can see right now happened along a sphere around me, events that are a distance of 1 lightyear away. The union of all possible spherical "observable moments" makes a light cone surface, which from an observer's perspective in a single moment appears exactly as what I see: a volume of space where different distances correspond to different times in the past. And a hyperplane "now" corresponds to what I can't see: A volume of space all at the same time, according to my frame of reference. (1) is there a continuous stream of photons connecting the observer to the rotating source if the source continues to rotate and emit consistently and the path is never blocked or interfered with during the entire period of the rotation? (2) If you believe this physical photon stream exists in the present: (a) is it a perfectly straight line? (b) is it a curved line? The light is measured discretely, not continuously... each photon is independent and not connected with the others. The source is moving, and each photon will travel on its own path. So there's no single physical "stream" along which all of the photons travel. The paths are geodesics, so basically "straight lines through curved space". I would say yes to both (a) and (b), depending on who observes the line. I think that to the receiver of the photons, the lines are straight. Edited September 7, 2012 by md65536 Link to comment Share on other sites More sharing options...
LaurieAG Posted September 8, 2012 Share Posted September 8, 2012 (edited) Hi md65536, The light is measured discretely, not continuously... each photon is independent and not connected with the others. The source is moving, and each photon will travel on its own path. So there's no single physical "stream" along which all of the photons travel. Questions one and two are mutually exclusive and your answers indicate that you are not discussing the real world in the present. Say you had an observation point that was stationary relative to the centre of rotation of the source. (1.1) would you observe this photon stream if you made continuous discrete observations from this location? The paths are geodesics, so basically "straight lines through curved space". I would say yes to both (a) and (b), depending on who observes the line. I think that to the receiver of the photons, the lines are straight. Euclid had a similar theory but his was reversed i.e. the light shot out of the eye directly to the stars on opening your eyes at night. Please note that the attached image shows the path of light flowing from the source during one complete galactic rotation from the perspective of the source i.e. path = 1,4 4,3 3,2 2,1 and 1,0. Consider 1,4 as the photons from location 1 after 4 quarters of rotation and 4,3 as the photons from location 4 after 3 quarters of rotation etc down to 1,0 the original start point of the rotating source at the present time after 1 complete rotation. The correct order of the light coming into the observer is 1,4 2,3 3,2 4,1 and 1,0 and this change just requires flipping the paths shown along the vertical 1,0 to 1v axis. Edited September 8, 2012 by LaurieAG Link to comment Share on other sites More sharing options...
LaurieAG Posted September 8, 2012 Share Posted September 8, 2012 The point I was trying to make is that, regardless of what appears in relative space, the continuous streams of photons emitted from 2 rotating galactic sources in euclidian space will be curved and the only mass involved in the curvature of this light path is the mass of the two sources that allows them to rotate around their common galactic centre. On the other hand, in relative space at the present, the 'observer' would 'see' the curved photon path as a straight line quanta only under certain conditions. Link to comment Share on other sites More sharing options...
michel123456 Posted September 8, 2012 Share Posted September 8, 2012 Hi md65536, Questions one and two are mutually exclusive and your answers indicate that you are not discussing the real world in the present. Say you had an observation point that was stationary relative to the centre of rotation of the source. (1.1) would you observe this photon stream if you made continuous discrete observations from this location? Euclid had a similar theory but his was reversed i.e. the light shot out of the eye directly to the stars on opening your eyes at night. Please note that the attached image shows the path of light flowing from the source during one complete galactic rotation from the perspective of the source i.e. path = 1,4 4,3 3,2 2,1 and 1,0. Consider 1,4 as the photons from location 1 after 4 quarters of rotation and 4,3 as the photons from location 4 after 3 quarters of rotation etc down to 1,0 the original start point of the rotating source at the present time after 1 complete rotation. The correct order of the light coming into the observer is 1,4 2,3 3,2 4,1 and 1,0 and this change just requires flipping the paths shown along the vertical 1,0 to 1v axis. You should use a dynamic graph to show what happens. Make a .gif The source emits photons all the time. The observer catches photons that where emitted some time ago. It is like the arrow of a hunter trying to catch a flying bird: the hunter has to shoot in a direction where the bird will be. In our case, the hunter (the source) ejects arrows (photons) in all directions all the time, so that the bird (the observer) is hit continuously. The photons travel in "straigt lines" (read geodesic). Link to comment Share on other sites More sharing options...
md65536 Posted September 8, 2012 Share Posted September 8, 2012 (edited) the continuous streams of photons emitted from 2 rotating galactic sources in euclidian space will be curved There's no continuous stream. Imagine shooting bullets out the side of an accelerating car (a Yardie Lobo, say). If you "connect the dots" of the bullets, it will make a curve, even though each of the bullets is moving in a straight line (ignoring gravity and wind drag). Similarly, even in flat space, your "stream of photons" can be curved because the photon source is moving. However each photon is moving in a straight line. There is no physically meaningful "stream"; it is only what you get when you consider all the photons together. However, there is nothing connecting one photon to another. They all behave independently, in their own individual null-geodesic "stream". I take back what I said about geodesics being "straight lines in curved space", because I guess that's taking the idea of geodesics far too literally, and is a misconception of mine, and I also think that my belief that a null geodesic is straight for an observer is not shared by mainstream science. I guess you would say: the path of a photon is only curved where space is curved. Edited September 8, 2012 by md65536 Link to comment Share on other sites More sharing options...
LaurieAG Posted September 10, 2012 Share Posted September 10, 2012 I guess you would say: the path of a photon is only curved where space is curved. I only showed the paths of all the photons that could only travel from the source directly in a straight line to the observation position. The source continually emitted photons during one complete galactic rotation and this path was never obscured by anything during that time. The paths of these photons are only curved because the sources that emitted them are rotating around a galactic centre. Link to comment Share on other sites More sharing options...
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