SciForums.com > Science > Physics & Math > Einstein's Relativity PDA View Full Version : Einstein's Relativity Post ReplyCreate New Thread RJBeery03-31-10, 10:58 AMIn chapter 7 of Einstein's book "Relativity" he says ...the Dutch astronomer De Sitter was also able to show that the velocity of propagation of light cannot depend on the velocity of motion of the body emitting the light. This is obviously a pretty basic concept in SR. Then in chapter 9, "THE RELATIVITY OF SIMULTANEITY", he is explaining why lightning flashes A and B, at the front and back of a high velocity train, may not appear to strike at the same time for all frames (specifically for a man X standing on the embankment vs a man Y in the train). However, the REASON he gives for this discrepancy bothers me... Now in reality he [man Y] is hastening towards the beam of light emitted from B whilst he is riding on ahead of the beam of light coming from A. This is confusing to me. Replace the lightning flashes with light bulbs, and put the light bulbs on the train. It should now be obvious that the passenger does not "hasten towards" nor "ride ahead of" the incoming light. Even if the answer lies somewhere in a train length contraction it seems that the mid-point of the train (M') between either the lightning flashes or the train bulbs, where man Y resides, should not be in contention and therefore the time to reach that mid-point should be equal. What am I missing? Pete03-31-10, 11:16 AMReplace the lightning flashes with light bulbs, and put the light bulbs on the train. It should now be obvious that the passenger does not "hasten towards" nor "ride ahead of" the incoming light. I don't follow. To me, it seems clear that in the embankment frame, the passenger is moving toward light from the front of the train, and moving away from light from the rear. Janus5803-31-10, 01:10 PMAs Pete pointed out, this is as determined from the Embankment frame. The light is rushing toward the embankment observer at c from both both A and B, and the train observer is rushing towards B and away from A. Thus, according to the Embankment observer, the light from B hits the train observer first. This one thing both observers must agree upon.( If they didn't, you could set up a situation where an actual physical contradiction existed) Thus the train observer must see flash B before flash A. Since, as you point out, he is located an equal distance from the ends of the train, the light coming from them takes an equal amount of time to reach him, The only way he could see the flashes at different times is for them to have originated at different times. The lightning strikes take place at the same time in the Embankment frame, but at different times in the Train frame. RJBeery03-31-10, 02:22 PMYes I can follow your logic, and maybe my gripe is just with Einstein's particular explanation. He says "in reality, Y is hastening towards the light", but this presumes an absolute reference frame when we all know that Y's reality is just as valid as X's. Also, he says the following: Just when the flashes* of lightning occur, the point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity of v of the train. The footnote (denoted by the asterisk) reads "As judged from the embankment." This means that the photons have already traveled to M. If M and M' are co-located and the flashes occur "when" X sees them, Y shall also see them; therefore they would agree on simultaneity, correct? Maybe the translation is incorrect, and by saying "when the flashes occur" what is meant is that the flashes have just occurred yet the photons have not traveled? However, we are now back to the issue of presuming a privileged or absolute reference frame to assign "when" these spatially separated events occur... phyti03-31-10, 03:26 PMYes I can follow your logic, and maybe my gripe is just with Einstein's particular explanation. He says "in reality, Y is hastening towards the light", but this presumes an absolute reference frame when we all know that Y's reality is just as valid as X's. Also, he says the following: The footnote (denoted by the asterisk) reads "As judged from the embankment." This means that the photons have already traveled to M. If M and M' are co-located and the flashes occur "when" X sees them, Y shall also see them; therefore they would agree on simultaneity, correct? Maybe the translation is incorrect, and by saying "when the flashes occur" what is meant is that the flashes have just occurred yet the photons have not traveled? However, we are now back to the issue of presuming a privileged or absolute reference frame to assign "when" these spatially separated events occur... He is using the speed of light relative to the observer, c-v, c+v, as if the ground is the fixed frame. He uses the fixed frame in specifying the simultaneity convention. Strange when SR denies an absolute frame, but once you've done that, you can eliminate the fixed frame and express the relations with each observers relative values. Consider the addition of velocities formula, it just transforms the relative v to a 3rd fixed frame. Someone once asked "how do you know what's relative, unless you know what's absolute?" I have no answer for that. "when the flashes occur" Here he's equating the perception of the event to the occurrence of the event. His material is at times difficult to follow. RJBeery03-31-10, 03:40 PMHis material is at times difficult to follow. I guess that's my problem. I thought I was internalizing SR fairly well before I read his book but now I'm confused by his explanations. Here he's equating the perception of the event to the occurrence of the event. Agreed. So, then, what is your response to my statement that X and Y are co-located at the perception of the event and therefore concur on simultaneity? Jack_03-31-10, 06:06 PMIn chapter 7 of Einstein's book "Relativity" he says This is obviously a pretty basic concept in SR. Then in chapter 9, "THE RELATIVITY OF SIMULTANEITY", he is explaining why lightning flashes A and B, at the front and back of a high velocity train, may not appear to strike at the same time for all frames (specifically for a man X standing on the embankment vs a man Y in the train). However, the REASON he gives for this discrepancy bothers me... This is confusing to me. Replace the lightning flashes with light bulbs, and put the light bulbs on the train. It should now be obvious that the passenger does not "hasten towards" nor "ride ahead of" the incoming light. Even if the answer lies somewhere in a train length contraction it seems that the mid-point of the train (M') between either the lightning flashes or the train bulbs, where man Y resides, should not be in contention and therefore the time to reach that mid-point should be equal. What am I missing? You are missing nothing. Jack_03-31-10, 06:08 PMAs Pete pointed out, this is as determined from the Embankment frame. The light is rushing toward the embankment observer at c from both both A and B, and the train observer is rushing towards B and away from A. Thus, according to the Embankment observer, the light from B hits the train observer first. This one thing both observers must agree upon.( If they didn't, you could set up a situation where an actual physical contradiction existed) Thus the train observer must see flash B before flash A. Since, as you point out, he is located an equal distance from the ends of the train, the light coming from them takes an equal amount of time to reach him, The only way he could see the flashes at different times is for them to have originated at different times. The lightning strikes take place at the same time in the Embankment frame, but at different times in the Train frame. You are ignoring the fact in the frame of the train, the train is not moving. It cannot be moving to and away from the light This is a common logic error where one takes a particular frame as preferred. AlphaNumeric03-31-10, 08:07 PMIf you have three things colliding simultaneously in a given frame then in a different frame this must also be true. This is in contrast to the fact if they collide with one another in pairs, rather than all at once, then different frames will disagree on precisely when these 3 pair collisions occur. If the person on the train sees lightning hitting the front and the back at the same time then light from each strike has reached him simultaneously. Someone on the embankment will see this happen as well, the light reaches the person on the train at the same time. Since its clear from the embankment's point of view that the light from the back of the train has travelled further you can see that from the embankments point of view the lightning hits the back of the train first. The problem with explaining and understanding all of this is that, as someone commented earlier, its important not to cling to some notion of universal 'now'. We're so used to seeing things happen almost the moment they happen that we forget this kind of thing isn't always true. You can see a jet fly by and then 10 seconds latter its engine noise follows. If you've ever played golf you'll have seen someone on another hole hit their ball and then a few seconds later there's the crack of the club hitting the ball. As such you always have to refer to the frame you're considering very clearly. The different version of the train/embankment setup is the one where the guy on the train flashes torch beams towards the front and the back. From his point of view they hit the front and back simultaneously (which is equivalent to him seeing them at the same time on their return) but the embankment says the one which moves towards the back initially is the first to hit the mirrored ends. Despite that they still agree that both reach the guy in the middle at the same time. This follows by the symmetry of the set up. If the train is L = 2cT in length and the man is in the middle he'll see the light beams B_{1} and B_{2} cover an outward journey of distance cT and then return journeys of length cT again. This takes 2T units of time to do. From the embankment's point of view the back of the train is closing on the backwards moving light B_{1} while the front is moving away from B_{2}. Hence it seems like the backwards moving light reflects first. But now the man is receeding from the reflected B_{1} in the same way the front was for B_{2}. Likewise the man closes on the reflected B_{2} in the same way the back did for the shone B_{1}. Hence you find that B_{1} does two journeys of time T_{1} and T_{1}' such that T_{1}+T_{1}' = 2cT and B_{2} does two journeys of time T_{2} = T_{1}' and T_{2}' = T_{1}, so T_{2}+T_{2}' = 2cT as well. The specific expression for the T_{j}^{(\prime)} depends on the relative velocity which I can't be bothered to work out on the back of an envelope. Pete03-31-10, 08:48 PMI guess that's my problem. I thought I was internalizing SR fairly well before I read his book but now I'm confused by his explanations. It's not the clearest work. Some vocabulary is a little odd (that "in reality" is a good example), and he didn't have tools like minkowski diagrams and four-vectors. Much better descriptions exist now. Janus5803-31-10, 09:07 PMYou are ignoring the fact in the frame of the train, the train is not moving. It cannot be moving to and away from the light No, I'm not. I specifically said " Since, as you point out, he is located an equal distance from the ends of the train, the light coming from them takes an equal amount of time to reach him,.." It is the very fact that the light does travel at the same speed from the ends of the train in the Train frame that leads to the conclusion that the lightning strikes took place at different times in that frame. Here's two animations showing what happens according to frames. The first is the embankment frame and the second is the train frame. http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul1.gif http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul2.gif Pete03-31-10, 09:17 PMHi janus, You might want to peruse this thread to see where Jack is coming from. I've linked to the post that best describes the key sticking point. Jack's initial description of his argument is earlier in the thread. RJBeery03-31-10, 09:37 PMThe different version of the train/embankment setup is the one where the guy on the train flashes torch beams towards the front and the back. From his point of view they hit the front and back simultaneously (which is equivalent to him seeing them at the same time on their return) but the embankment says the one which moves towards the back initially is the first to hit the mirrored ends. Despite that they still agree that both reach the guy in the middle at the same time. This follows by the symmetry of the set up. If the train is L = 2cT in length and the man is in the middle he'll see the light beams and cover an outward journey of distance cT and then return journeys of length cT again. This takes 2T units of time to do. From the embankment's point of view the back of the train is closing on the backwards moving light while the front is moving away from . Hence it seems like the backwards moving light reflects first. But now the man is receeding from the reflected in the same way the front was for . Likewise the man closes on the reflected in the same way the back did for the shone . Hence you find that does two journeys of time and such that and does two journeys of time and , so as well. The specific expression for the depends on the relative velocity which I can't be bothered to work out on the back of an envelope. So the answer to the following paradox (which is really what I'm after in this thread): Given any light source, there are infinite frames within which the light appears to spherically radiate and this causes a contradiction. Is the following: Given any frame A observing a light, that light appears to spherically radiate from the source; the light is calculated (and observed?) NOT to spherically radiate for any other frame B unless the "round trip" back to the B-frame inertial light source is calculated which indeed shows that the light has traversed in all directions an equal distance for an equal amount of time. In other words, if you strictly define "spherical radiation of light" as being photons moving the same distance for the same amount of time (and you require a return to the frame's emission point) then light IS INDEED calculated and observed to be spherically radiating from any arbitrary frame for all other frames. Does that sound right? James R03-31-10, 10:28 PMRJBeery: However, the REASON he gives for this discrepancy bothers me... Now in reality he [man Y] is hastening towards the beam of light emitted from B whilst he is riding on ahead of the beam of light coming from A. "In reality" is a bad choice of words. He should have written "In the frame of the observer watching the train go past". It's important to remember that Einstein's was the first explanation of the relativity of simultaneity. It can hardly be expected to be the clearest or the best. We've had 100 years to work out how best to teach it. This is confusing to me. Replace the lightning flashes with light bulbs, and put the light bulbs on the train. It should now be obvious that the passenger does not "hasten towards" nor "ride ahead of" the incoming light. Again, the problem is that the language used was not clear. In the train frame, the observer is not "hastening" towards or away from either source of light - he's not going anywhere. In the frame of the observer watching the train, however, the passenger is certainly "hastening" towards the point on the track at which the bulb was located when the light was emitted. phyti: He is using the speed of light relative to the observer, c-v, c+v, as if the ground is the fixed frame. No, he isn't. He is taking the speed of light to be c in both reference frames; this one of the postulates of special relativity. Someone once asked "how do you know what's relative, unless you know what's absolute?" I have no answer for that. The answer is: everything is relative. There is no absolute. Here he's equating the perception of the event to the occurrence of the event. No he isn't. You're misunderstanding. His material is at times difficult to follow. I agree there are more complete and clear explanations available. Einstein makes no errors but his is not always the easiest exposition to follow. I don't know why people insist on reading Einstein when they have many excellent textbooks available that teach the material in a 21st century manner with lots of nice diagrams etc. Some university introductory texts are in their 12th edition. The way they teach this stuff has been revised and revised, particularly to address common misconceptions of beginning students of relativity. Jack_04-01-10, 04:52 PMIf you have three things colliding simultaneously in a given frame then in a different frame this must also be true. This is in contrast to the fact if they collide with one another in pairs, rather than all at once, then different frames will disagree on precisely when these 3 pair collisions occur. If the person on the train sees lightning hitting the front and the back at the same time then light from each strike has reached him simultaneously. Someone on the embankment will see this happen as well, the light reaches the person on the train at the same time. Since its clear from the embankment's point of view that the light from the back of the train has travelled further you can see that from the embankments point of view the lightning hits the back of the train first. The problem with explaining and understanding all of this is that, as someone commented earlier, its important not to cling to some notion of universal 'now'. We're so used to seeing things happen almost the moment they happen that we forget this kind of thing isn't always true. You can see a jet fly by and then 10 seconds latter its engine noise follows. If you've ever played golf you'll have seen someone on another hole hit their ball and then a few seconds later there's the crack of the club hitting the ball. As such you always have to refer to the frame you're considering very clearly. The different version of the train/embankment setup is the one where the guy on the train flashes torch beams towards the front and the back. From his point of view they hit the front and back simultaneously (which is equivalent to him seeing them at the same time on their return) but the embankment says the one which moves towards the back initially is the first to hit the mirrored ends. Despite that they still agree that both reach the guy in the middle at the same time. This follows by the symmetry of the set up. If the train is L = 2cT in length and the man is in the middle he'll see the light beams B_{1} and B_{2} cover an outward journey of distance cT and then return journeys of length cT again. This takes 2T units of time to do. From the embankment's point of view the back of the train is closing on the backwards moving light B_{1} while the front is moving away from B_{2}. Hence it seems like the backwards moving light reflects first. But now the man is receeding from the reflected B_{1} in the same way the front was for B_{2}. Likewise the man closes on the reflected B_{2} in the same way the back did for the shone B_{1}. Hence you find that B_{1} does two journeys of time T_{1} and T_{1}' such that T_{1}+T_{1}' = 2cT and B_{2} does two journeys of time T_{2} = T_{1}' and T_{2}' = T_{1}, so T_{2}+T_{2}' = 2cT as well. The specific expression for the T_{j}^{(\prime)} depends on the relative velocity which I can't be bothered to work out on the back of an envelope. This is not a timing issue. As the OP has noted, light proceeds spherically from the light emission point in the frame. I got you to finally admit this fact. Now, when light hits at B, that is an emission point in M and that is one in M'. The emission point rides with the frame. This is SR. Same goes for A. Hence, the stationary train view has two light spheres proceeding toward M'. These two static light sphere to M', were both equidistant light emissions when M and M' were co-located. That is the OP's point. AlphaNumeric04-01-10, 05:01 PMI got you to finally admit this fact.While I can't prevent you peddling your ignorance in other forums at least try not to tell flat out lies. Yes, you disagree with me (and everyone else) as to how special relativity works. That doesn't mean you can lie with abandon. Provide a link to a post of mine where I state that a light cone doesn't map to a light cone under a Lorentz transformation. In fact I've been telling you that repeatedly because you claim you've found a transformation which doesn't map the light cone to the light cone. In contradiction to the algebraic identities which define the Lorentz transformations. Provide a link or retract your claim. This is SR.You have made it clear you're not a person to trust when you make such statements. Jack_04-01-10, 05:53 PMWhile I can't prevent you peddling your ignorance in other forums at least try not to tell flat out lies. Yes, you disagree with me (and everyone else) as to how special relativity works. That doesn't mean you can lie with abandon. Provide a link to a post of mine where I state that a light cone doesn't map to a light cone under a Lorentz transformation. In fact I've been telling you that repeatedly because you claim you've found a transformation which doesn't map the light cone to the light cone. In contradiction to the algebraic identities which define the Lorentz transformations. Provide a link or retract your claim. You have made it clear you're not a person to trust when you make such statements. Originally Posted by Jack_ This is not a timing issue. As the OP has noted, light proceeds spherically from the light emission point in the frame. I got you to finally admit this fact. AN: I have already said that I agree that the two frames associated to the rigid spheres will disagree on the question of "Where is the centre of the photon sphere?". http://www.sciforums.com/showpost.php?p=2508258&postcount=95 Now, I have to assume the light sphere is spherical in all frames. If you disagree I will toss that logic from Einstein in your face. If the light emission point in the frame is the center of the light sphere, as you confessed, and all frame emission points diverge, then there exists an infinite number of unique light spheres. :xctd: Now, if you want to claim LT preserves the spherical nature of the light sphere in the stationary frame after any time t, when mapped to the moving frame, I will provide the necessary education otherwise. In any event, I just showed you cannot even remember what you say. AlphaNumeric04-01-10, 06:53 PMNow, I have to assume the light sphere is spherical in all frames. No, you don't have to assume it, it follows by the definition of Lorentz transformations. The link you provide does not have me saying they aren't spheres. It has me saying they are spheres but their origins in different frames do not agree. In fact you've linked to the post where I explain that the region internal to the light cone is warped by the Lorentz transformation and that while a LT maps the light cone to a light cone it warps the distribution of points on the light cone. In this thread (http://www.sciforums.com/showthread.php?t=100356&page=2) I draw pictures showing the light cone to be spherical in both frames. I also provided the algebra to describe it here (http://www.sciforums.com/showpost.php?p=2507562&postcount=41). I also explain why the light cone maps to itself by using space-time intervals here (http://www.sciforums.com/showpost.php?p=2509323&postcount=104) and provided a diagram here (http://www.sciforums.com/showpost.php?p=2508375&postcount=97). You later provided a diagram which said what I'd already told you. The fact you didn't realise this shows you either don't understand or don't even bother to read what people say to you. And I've [url=http://www.sciforums.com/showpost.php?p=2509577&postcount=110]previously (http://www.sciforums.com/showpost.php?p=2509062&postcount=103) explained why your interpretation of your results are wrong. You fail to realise that you haven't arrived at some new algebraic result, you have arrived at something well know and then you've jumped to an incorrect conclusion. If the light emission point in the frame is the center of the light sphere, as you confessed, and all frame emission points diverge, then there exists an infinite number of unique light spheres.And there's your incorrect conclusion. There's only ever one light cone in any given frame. The fact different frames disagree on the centre of the sphere is immaterial. All they need worry about is what is in, on or outside of the light cone. And they all agree on that. You fail to realise time and again that you're not telling me something I haven't worked out or providing some new insight. You're coming up against a counter intuitive result and simply declaring it wrong. Now, if you want to claim LT preserves the spherical nature of the light sphere in the stationary frame after any time t, when mapped to the moving frame, I will provide the necessary education otherwise.I already did, here (http://www.sciforums.com/showpost.php?p=2509323&postcount=104). You've seen me post that already and yet you continue to pretend I haven't provided you with any retorts. All points on the light cone map to being on the light cone. That is ALL that matters. You're assuming that the centre of the light cone in one frame must map to the centre of the light cone in another. This is wrong. You're making up your own definitions, you're constructing strawmen. A point on the light cone maps to a point on the light cone. You don't need to make any reference to the centre of the light cone because it transforms differently to the light cone itself. You're claiming it should transform in the same manner but that's not the case. The reason the light cone maps to itself is because all frames see light moving at the same speed, c. If your argument were correct, that the centres should not change either than you're claiming that all frames see all speeds as unchanged by Lorentz transformation. If the light emission point in the frame is the center of the light sphere, as you confessed, and all frame emission points diverge, then there exists an infinite number of unique light spheres.And there's your incorrect conclusion. There's only ever one light cone in any given frame. The fact different frames disagree on the centre of the sphere is immaterial. All they need worry about is what is in, on or outside of the light cone. And they all agree on that. Go on, walk me through the algebra of substituting x' = \gamma(x-vt) and t' = \gamma(t - vx/c^{2}) into ds^{2} = -c^{2}(dt')^{2} + (dx')^{2}. What does it evaluate to if -c^{2}dt^{2} + dx^{2} = 0? What is the physical implication, given that all points on the light cone, in a particular frame, satisfy -c^{2}dt^{2} + dx^{2} = 0. In case you can't get it you find -c^{2}dt^{2} + dx^{2} = 0 = -c^{2}(dt')^{2} + (dx')^{2} For ds^{2}=0 you have \frac{dx}{dt} = c by rearrangement. And likewise \frac{dx'}{dt'} = c. This arises for only speed v=c, as only light speed is frame independent. If we use your logic then since all concentric circles insider the light cone map to themselves you find that if \frac{dx}{dt} = v then a Lorentz transformation takes this to \frac{dx'}{dt'} = v. Your logic requires that ALL speeds are frame independent, not just light speed. Clearly this is not the case and arises by the warping of the region inside the light cone by the transformation. Just draw a few diagrams and its easy to see. That's yet another reason why your logic is flawed. Remember that the Lorentz transformations are DEFINED such that they leave ds^{2} unchanged. If you claim you have a transformation which doesn't do this then its not a Lorentz transformation. Your entire argument rests on the assumption your definition of a light cone is the correct one. It doesn't matter where the light cone was emitted from at some point in the past, you work out where it is in different frames at any particular moment by transforming the photons, as they are the physical things. The centre of the light sphere has no physical properties and has no role to play at all in any considerations. Even if it were accepted that SR is inconsistent with its treatment of photon sphere centres it would have absolutely no implications for any physical predictions because the centre has no physical role to play. You can't get relativity to say "Frame 1 says A causes B while Frame 2 says B causes A". That would be a contradiction. All you've got is you don't like the nonintuitive structure of a theory you don't understand. n any event, I just showed you cannot even remember what you say. No, you showed you don't understand what people say and you don't understand what SR says. Instead you make up your own revisionist memories. I've just linked to plenty of posts of mine where I've already addressed the points you raise. Several times. So you claiming I can't remember what I said is a little hypocritical. Jack_04-01-10, 07:57 PMNo, you don't have to assume it, it follows by the definition of Lorentz transformations. The link you provide does not have me saying they aren't spheres. It has me saying they are spheres but their origins in different frames do not agree. In fact you've linked to the post where I explain that the region internal to the light cone is warped by the Lorentz transformation and that while a LT maps the light cone to a light cone it warps the distribution of points on the light cone. In this thread (http://www.sciforums.com/showthread.php?t=100356&page=2) I draw pictures showing the light cone to be spherical in both frames. I also provided the algebra to describe it here (http://www.sciforums.com/showpost.php?p=2507562&postcount=41). I also explain why the light cone maps to itself by using space-time intervals here (http://www.sciforums.com/showpost.php?p=2509323&postcount=104) and provided a diagram here (http://www.sciforums.com/showpost.php?p=2508375&postcount=97). You later provided a diagram which said what I'd already told you. The fact you didn't realise this shows you either don't understand or don't even bother to read what people say to you. And I've [url=http://www.sciforums.com/showpost.php?p=2509577&postcount=110]previously (http://www.sciforums.com/showpost.php?p=2509062&postcount=103) explained why your interpretation of your results are wrong. You fail to realise that you haven't arrived at some new algebraic result, you have arrived at something well know and then you've jumped to an incorrect conclusion. And there's your incorrect conclusion. There's only ever one light cone in any given frame. The fact different frames disagree on the centre of the sphere is immaterial. All they need worry about is what is in, on or outside of the light cone. And they all agree on that. You fail to realise time and again that you're not telling me something I haven't worked out or providing some new insight. You're coming up against a counter intuitive result and simply declaring it wrong. I already did, here (http://www.sciforums.com/showpost.php?p=2509323&postcount=104). You've seen me post that already and yet you continue to pretend I haven't provided you with any retorts. All points on the light cone map to being on the light cone. That is ALL that matters. You're assuming that the centre of the light cone in one frame must map to the centre of the light cone in another. This is wrong. You're making up your own definitions, you're constructing strawmen. A point on the light cone maps to a point on the light cone. You don't need to make any reference to the centre of the light cone because it transforms differently to the light cone itself. You're claiming it should transform in the same manner but that's not the case. The reason the light cone maps to itself is because all frames see light moving at the same speed, c. If your argument were correct, that the centres should not change either than you're claiming that all frames see all speeds as unchanged by Lorentz transformation. And there's your incorrect conclusion. There's only ever one light cone in any given frame. The fact different frames disagree on the centre of the sphere is immaterial. All they need worry about is what is in, on or outside of the light cone. And they all agree on that. Go on, walk me through the algebra of substituting x' = \gamma(x-vt) and t' = \gamma(t - vx/c^{2}) into ds^{2} = -c^{2}(dt')^{2} + (dx')^{2}. What does it evaluate to if -c^{2}dt^{2} + dx^{2} = 0? What is the physical implication, given that all points on the light cone, in a particular frame, satisfy -c^{2}dt^{2} + dx^{2} = 0. In case you can't get it you find -c^{2}dt^{2} + dx^{2} = 0 = -c^{2}(dt')^{2} + (dx')^{2} For ds^{2}=0 you have \frac{dx}{dt} = c by rearrangement. And likewise \frac{dx'}{dt'} = c. This arises for only speed v=c, as only light speed is frame independent. If we use your logic then since all concentric circles insider the light cone map to themselves you find that if \frac{dx}{dt} = v then a Lorentz transformation takes this to \frac{dx'}{dt'} = v. Your logic requires that ALL speeds are frame independent, not just light speed. Clearly this is not the case and arises by the warping of the region inside the light cone by the transformation. Just draw a few diagrams and its easy to see. That's yet another reason why your logic is flawed. Remember that the Lorentz transformations are DEFINED such that they leave ds^{2} unchanged. If you claim you have a transformation which doesn't do this then its not a Lorentz transformation. Your entire argument rests on the assumption your definition of a light cone is the correct one. It doesn't matter where the light cone was emitted from at some point in the past, you work out where it is in different frames at any particular moment by transforming the photons, as they are the physical things. The centre of the light sphere has no physical properties and has no role to play at all in any considerations. Even if it were accepted that SR is inconsistent with its treatment of photon sphere centres it would have absolutely no implications for any physical predictions because the centre has no physical role to play. You can't get relativity to say "Frame 1 says A causes B while Frame 2 says B causes A". That would be a contradiction. All you've got is you don't like the nonintuitive structure of a theory you don't understand. No, you showed you don't understand what people say and you don't understand what SR says. Instead you make up your own revisionist memories. I've just linked to plenty of posts of mine where I've already addressed the points you raise. Several times. So you claiming I can't remember what I said is a little hypocritical. And there's your incorrect conclusion. There's only ever one light cone in any given frame. The fact different frames disagree on the centre of the sphere is immaterial. All they need worry about is what is in, on or outside of the light cone. And they all agree on that. You post all junk. Exactly how are you going to make one light sphere with an infinite number of origins? I want to see this math. Stay on task and let's operate on this. If you step here mathematically or anyone, I will force all of you into a contradiction. Now put up or shut up. phyti04-01-10, 10:13 PMI guess that's my problem. I thought I was internalizing SR fairly well before I read his book but now I'm confused by his explanations. Agreed. So, then, what is your response to my statement that X and Y are co-located at the perception of the event and therefore concur on simultaneity? Read the section where Einstein defines simultaneity for an arbitrary frame of reference. It's a relative simultaneity for each frame, thus the moving train description of events will be different than the ground description. RJBeery04-01-10, 11:59 PMRead the section where Einstein defines simultaneity for an arbitrary frame of reference. It's a relative simultaneity for each frame, thus the moving train description of events will be different than the ground description. Technically speaking, according to his description, both observers would concur on the simultaneity of the lightning strikes since they are taking the measurement when observer X sees them (and, by his own description, observer Y is co-located with X at that time). I am not questioning SR, I am complaining about his explanation of it, which others have pointed out may be lacking due to the fact that he was the first to describe it. AlphaNumeric04-02-10, 07:36 AMYou post all junk.Excellent response. You demand I give 'math proofs' and when I do so you don't retort any of them. Exactly how are you going to make one light sphere with an infinite number of origins?Explain to me what precisely you didn't get about my previous posts comments about space-time intervals. I asked you a direct question and you ignored it. I want to see this math.I just provided it. Obviously you need to have your hand held through this one. Frame 1 has coordinates (t,\underline{x}}). Light is associated with null 4-vectors p^{\mu} such that p^{\mu}p_{\mu} = 0. For the space-time interval you get ds^{2} = dx^{\mu}dx_{\mu} = dx^{\mu}dx^{\nu}\eta_{\mu\nu} = -dt^{2} + d\underline{x}\cdot d\underline{x}. Frame 2 has coordinates (t',\underline{x}'}). Light is associated with null 4-vectors q^{\mu} such that q^{\mu}q_{\mu} = 0. For the space-time interval you get (ds')^{2} = (dx')^{\mu}(dx')_{\mu} = (dx')^{\mu}(dx')^{\nu}\eta_{\mu\nu} = -(dt')^{2} + d\underline{x}'\cdot d\underline{x}'. Let Frame 2 be moving with velocity \underline{v} with respect to Frame 1. Thus to go from Frame 1 coordinates to Frame 2 coordinates we apply a Lorentz boost in the direction \frac{\hat{v}}{|v|}. We denote this by the matrix \Lambda^{\mu}_{\nu} and sync the origins of the two frames such that (t,x) = (0,0) = (t',x') coincide initially. We will make use of the definition of \Lambda, that \Lambda^{\top} \cdot \eta \cdot \Lambda = \eta or in a given set of coordinates \Lambda^{\rho}_{\mu} \eta_{\rho\sigma} \Lambda^{\sigma}_{\nu} = \eta_{\mu\nu}. A null 4-vector in Frame 1, p^{\mu}, transforms to P^{\mu} \equiv \Lambda^{\mu}_{\nu}p^{\nu}. We compute its norm, P^{\mu}P_{\mu}. Inserting its definition we have P^{\mu}P_{\mu} = P^{\mu} \eta_{\mu\nu} P^{\nu} = \Lambda^{\mu}_{\tau}p^{\tau} \eta_{\mu\nu} \Lambda^{\nu}_{\sigma}p^{\sigma} = p^{\tau} \Big( \Lambda^{\mu}_{\tau}\eta_{\mu\nu} \Lambda^{\nu}_{\sigma} \Big) p^{\sigma} . By definition of the Lorentz transformations the terms in the brackets simplify : \Lambda^{\mu}_{\tau}\eta_{\mu\nu} \Lambda^{\nu}_{\sigma} = \eta_{\tau\sigma}. Therefore P^{\mu}P_{\mu} = p^{\tau}\eta_{\tau\sigma}p^{\sigma} = p^{\mu}p_{\nu}. Therefore p^{\mu} and P^{\mu} have the same norm. Therefore if p^{\mu}p_{\mu} = 0 then P^{\mu}P_{\mu} = 0. If p^{\mu} is null then its Lorentz transformation image, P^{\mu} is also null, aka on the light cone. I haven't had to restrict what particular Lorentz transformation I'm considering since they all obey the same defining property, they leave the metric unchanged. Thus I've proven that regardless of what Lorentz transformation you apply a null vector maps to a null vector aka a point on the light cone goes to a point on the light cone. Now no doubt you're struggling to follow this algebra so I'll do a particular case. We'll consider a 2d setup ie Frame 1 has coordinates (t,x) and Frame 2 has coordinates (t',x') and they are related by a Lorentz boost of speed v along the x axis. Therefore we have the standard formula for the coordinates : x' = \gamma(x-vt) and t' = \gamma(t - vx) (having set c=1 for less cluttered algebra. A null space-time interval is ds^{2} = -dt^{2} + dx^{2} in Frame 1 and in Frame 2 (ds')^{2} = -(dt')^{2} + (dx')^{2}. We have (t',x') in terms of (t,x) so we'll consider how the second interval changes. To do that we need to work out the dx' and dt' using the Lorentz transformations. dx' = \frac{\partial x'}{\partial x}dx + \frac{\partial x'}{\partial t}dt} = \gamma dx - \gamma v dt dt' = \frac{\partial t'}{\partial x}dx + \frac{\partial t'}{\partial t}dt} = -\gamma v dx + \gamma dt Putting them into (ds')^{2} = -(dt')^{2} + (dx')^{2} we have : (ds')^{2} = -( -\gamma v dx + \gamma dt)^{2} + ( \gamma dx - v\gamma dt)^{2} = -\Big( \gamma^{2}v^{2} dx^{2} + \gamma^{2}dt^{2} - 2v\gamma^{2} dx dt \Big) + \Big( \gamma^{2} dx^{2} + v^{2}\gamma^{2}dt^{2} - 2\gamma^{2} dx dt \Big) = -\gamma^{2}\Big( 1 - v^{2} \Big) dt^{2} + \gamma^{2} \Big( 1 - v^{2} \Big) dx^{2} = -dt^{2}+dx^{2} The final equality makes use of \gamma^{-2} = 1-v^{2}. Thus the specific Lorentz transforms used, ie a boost along the x axis by speed v, leave the space-time interval invariant, irrespective of what v is. Thus a null vector maps to a null vector and a point on the light cone maps to a point on the light cone. QED. By the way, this exercise is exactly one which 1st year physics students in my university covered in their first term. It's little more than just algebraic substitutions. It's that simple. And yet you haven't managed to grasp it. If you step here mathematically or anyone, I will force all of you into a contradiction.Again, who precisely are you trying to convince. This stuff is so basic and so simple to work through that anyone doing physics covers it and even people who don't do physics can work through the algebra (provided you give them the dx' and dt' expressions as they require the person to know differentiation otherwise). You claim there's a contradiction between although the light cone of Frame 1 is mapped to a light cone in Frame 2 their centres do not map to one another. That is not a contradiction. In fact its a requirement else all velocities would be frame independent, which really would be a contradiction. Now put up or shut up.I've 'put up' before. I've provided plenty of explanations, diagrams and algebra. Now I've done it again. You really seem to be struggling to grasp that this stuff is easy. Its second nature to anyone whose done a physics degree. You make the mistake of assuming that because many of these results are new or confusing to you then everyone else is in the same boat. You haven't done some amazing new result which no one has ever thought of, you've just done a bit of algebra everyone whose done physics has done and you've come up against a result which confuses you. Happens to everyone. I have had it happen plenty of times to me. But unlike you I went away and read and read and then asked people for explanations which I actually listened to and thought about. Each and every time I've realised what my mistake was. Its a shame you can't just swallow your pride and accept you don't understand something. It'd have been a lot easier to do before you started declaring to all and sundry you've toppled relativity. It now seems you've got too much pride invested in your easily refutable claims to be willing to man up and say "I'm wrong". Jack_04-02-10, 05:41 PMExcellent response. You demand I give 'math proofs' and when I do so you don't retort any of them. Explain to me what precisely you didn't get about my previous posts comments about space-time intervals. I asked you a direct question and you ignored it. I just provided it. Obviously you need to have your hand held through this one. Frame 1 has coordinates (t,\underline{x}}). Light is associated with null 4-vectors p^{\mu} such that p^{\mu}p_{\mu} = 0. For the space-time interval you get ds^{2} = dx^{\mu}dx_{\mu} = dx^{\mu}dx^{\nu}\eta_{\mu\nu} = -dt^{2} + d\underline{x}\cdot d\underline{x}. Frame 2 has coordinates (t',\underline{x}'}). Light is associated with null 4-vectors q^{\mu} such that q^{\mu}q_{\mu} = 0. For the space-time interval you get (ds')^{2} = (dx')^{\mu}(dx')_{\mu} = (dx')^{\mu}(dx')^{\nu}\eta_{\mu\nu} = -(dt')^{2} + d\underline{x}'\cdot d\underline{x}'. Let Frame 2 be moving with velocity \underline{v} with respect to Frame 1. Thus to go from Frame 1 coordinates to Frame 2 coordinates we apply a Lorentz boost in the direction \frac{\hat{v}}{|v|}. We denote this by the matrix \Lambda^{\mu}_{\nu} and sync the origins of the two frames such that (t,x) = (0,0) = (t',x') coincide initially. We will make use of the definition of \Lambda, that \Lambda^{\top} \cdot \eta \cdot \Lambda = \eta or in a given set of coordinates \Lambda^{\rho}_{\mu} \eta_{\rho\sigma} \Lambda^{\sigma}_{\nu} = \eta_{\mu\nu}. A null 4-vector in Frame 1, p^{\mu}, transforms to P^{\mu} \equiv \Lambda^{\mu}_{\nu}p^{\nu}. We compute its norm, P^{\mu}P_{\mu}. Inserting its definition we have P^{\mu}P_{\mu} = P^{\mu} \eta_{\mu\nu} P^{\nu} = \Lambda^{\mu}_{\tau}p^{\tau} \eta_{\mu\nu} \Lambda^{\nu}_{\sigma}p^{\sigma} = p^{\tau} \Big( \Lambda^{\mu}_{\tau}\eta_{\mu\nu} \Lambda^{\nu}_{\sigma} \Big) p^{\sigma} . By definition of the Lorentz transformations the terms in the brackets simplify : \Lambda^{\mu}_{\tau}\eta_{\mu\nu} \Lambda^{\nu}_{\sigma} = \eta_{\tau\sigma}. Therefore P^{\mu}P_{\mu} = p^{\tau}\eta_{\tau\sigma}p^{\sigma} = p^{\mu}p_{\nu}. Therefore p^{\mu} and P^{\mu} have the same norm. Therefore if p^{\mu}p_{\mu} = 0 then P^{\mu}P_{\mu} = 0. If p^{\mu} is null then its Lorentz transformation image, P^{\mu} is also null, aka on the light cone. I haven't had to restrict what particular Lorentz transformation I'm considering since they all obey the same defining property, they leave the metric unchanged. Thus I've proven that regardless of what Lorentz transformation you apply a null vector maps to a null vector aka a point on the light cone goes to a point on the light cone. Now no doubt you're struggling to follow this algebra so I'll do a particular case. We'll consider a 2d setup ie Frame 1 has coordinates (t,x) and Frame 2 has coordinates (t',x') and they are related by a Lorentz boost of speed v along the x axis. Therefore we have the standard formula for the coordinates : x' = \gamma(x-vt) and t' = \gamma(t - vx) (having set c=1 for less cluttered algebra. A null space-time interval is ds^{2} = -dt^{2} + dx^{2} in Frame 1 and in Frame 2 (ds')^{2} = -(dt')^{2} + (dx')^{2}. We have (t',x') in terms of (t,x) so we'll consider how the second interval changes. To do that we need to work out the dx' and dt' using the Lorentz transformations. dx' = \frac{\partial x'}{\partial x}dx + \frac{\partial x'}{\partial t}dt} = \gamma dx - \gamma v dt dt' = \frac{\partial t'}{\partial x}dx + \frac{\partial t'}{\partial t}dt} = -\gamma v dx + \gamma dt Putting them into (ds')^{2} = -(dt')^{2} + (dx')^{2} we have : (ds')^{2} = -( -\gamma v dx + \gamma dt)^{2} + ( \gamma dx - v\gamma dt)^{2} = -\Big( \gamma^{2}v^{2} dx^{2} + \gamma^{2}dt^{2} - 2v\gamma^{2} dx dt \Big) + \Big( \gamma^{2} dx^{2} + v^{2}\gamma^{2}dt^{2} - 2\gamma^{2} dx dt \Big) = -\gamma^{2}\Big( 1 - v^{2} \Big) dt^{2} + \gamma^{2} \Big( 1 - v^{2} \Big) dx^{2} = -dt^{2}+dx^{2} The final equality makes use of \gamma^{-2} = 1-v^{2}. Thus the specific Lorentz transforms used, ie a boost along the x axis by speed v, leave the space-time interval invariant, irrespective of what v is. Thus a null vector maps to a null vector and a point on the light cone maps to a point on the light cone. QED. By the way, this exercise is exactly one which 1st year physics students in my university covered in their first term. It's little more than just algebraic substitutions. It's that simple. And yet you haven't managed to grasp it. Again, who precisely are you trying to convince. This stuff is so basic and so simple to work through that anyone doing physics covers it and even people who don't do physics can work through the algebra (provided you give them the dx' and dt' expressions as they require the person to know differentiation otherwise). You claim there's a contradiction between although the light cone of Frame 1 is mapped to a light cone in Frame 2 their centres do not map to one another. That is not a contradiction. In fact its a requirement else all velocities would be frame independent, which really would be a contradiction. I've 'put up' before. I've provided plenty of explanations, diagrams and algebra. Now I've done it again. You really seem to be struggling to grasp that this stuff is easy. Its second nature to anyone whose done a physics degree. You make the mistake of assuming that because many of these results are new or confusing to you then everyone else is in the same boat. You haven't done some amazing new result which no one has ever thought of, you've just done a bit of algebra everyone whose done physics has done and you've come up against a result which confuses you. Happens to everyone. I have had it happen plenty of times to me. But unlike you I went away and read and read and then asked people for explanations which I actually listened to and thought about. Each and every time I've realised what my mistake was. Its a shame you can't just swallow your pride and accept you don't understand something. It'd have been a lot easier to do before you started declaring to all and sundry you've toppled relativity. It now seems you've got too much pride invested in your easily refutable claims to be willing to man up and say "I'm wrong". This is all nice. This is about the train experiment. Let's see that proof of yours from the stationary train frame. Now, let me help you. The light emission points in the train frame will not be A and B. They will be A' and B' stationary to M' or the SR clock sync method fails. Once you correctly determine the correct space time interval, you will find M' must be struck by the light simultaneously. Here is the problem, when you take a frame stationary, here is what you must do. Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the stationary system.'' http://www.fourmilab.ch/etexts/einstein/specrel/www/ Do you understand this? AlphaNumeric04-03-10, 04:26 AMWell done on not responding to anything I said. What precisely do you not understand about my demonstration that null vectors map to null vectors under a LT? It demonstrates that the light cone goes to the light cone. This is counter to a claim you have made. I'd like you to respond to it. All you're doing is trying to change the subject. I've quoted bits of your replies and I've responded to them. Can't you do the same? Is all you're capable of just avoiding anything anyone says? You keep saying "Oh my points still stand, SR is dead" but if all you do is avoid replying to any retorts people make then you're in denial and its plain for anyone following your posts to see. You keep demanding people answer your questions and when I do you ignore my questions and ignore anything I've said. I respond directly to points you raise. Can't you do the same? If you're so confident shouldn't you have nothing to hide? I've offered to help you submit your work to a journal. I've offered to raise the level of discussion beyond Calculus 101. I've provided pictures, equations, explanations, you refuse to discuss any of them. Instead you want people to reply so you can monologue a bit more. You aren't interested in what people have to say about your claims, you just want to claim some more. If you can't reply to direct questions don't expect other people to reply to yours. You claim you've killed SR but if you're unwilling to reply to anything anyone says and you're always changing the subject whenever anyone treds too close to things you don't understand. AGAIN, if you think I'm 'too primitive' then why do you want my comments? Either you think I'm intelligent enough to be worth listening to or you're simply wanting more attention. Please answer this question. If you don't then I'll simply keep repeating it until you do answer it. As I said, don't expect people to answer your questions if you can't answer theirs. Rational discourse is the foundation of science but you wouldn't be aware of that.... QuarkHead04-03-10, 08:02 AMObviously you need to have your hand held through this one. {.......}Very nice explanation Alpha. However the words "pearls" and "swine" spring to mind CptBork04-03-10, 08:32 AMWhen a guy comes along and states that he knows Relativity is wrong a priori because "it's in his way", you know he has no interest in any sort of rational dialogue; it becomes more like preaching some sort of silly cult religion, I guess we can call it the Church of Jack. RJBeery04-03-10, 10:55 AMJack, I've read your proposed solution to the SR problems we've been discussing (i.e. Absolute Motion). Wouldn't absolute motion imply that we could detect redshift/blueshift from an inertial light source? AlphaNumeric04-03-10, 11:15 AMVery nice explanation Alpha. However the words "pearls" and "swine" spring to mindThanks. You'd think that Jack would respond to the maths he keeps demanding people provide. He claims to have taught vector calculus to undergraduates but I'm dubious about that so I thought I'd provide both the general tensor derivation and a specific case in terms of coordinates that the light cone is Lorentz invariant. No response, no comment, no retort. And then he has the laughable hypocrisy to demand people answer his questions, retort his comments. :shrug: He still hasn't answered my repeated question of why he wants to have me reply to his posts when he considers me 'too primitive' and obviously not a 4.0 student like him (so he says) and that I don't understand relativity (despite having more qualifications and more published work than him and wanting to raise the discussion to the level of fibre bundles). If I'm so ignorant and crap at physics why does he want me to comment on his 'work'? He's ignored it previously and yet he craves it. Sounds a bit of an attention whore. QuarkHead04-03-10, 01:20 PMYou see what I fail to understand about this forum is this. Say I read a pop-sci book about, say, relativity or string theory, to take two recent examples. I believe I am sufficiently intelligent to know when things are being horribly over-simplified, even though I do not not know enough about either subject to detect what it is exactly. So I might (not likely, in my case) come here and ask for clarification. This does happen on this forum, I grant you, but not often enough to my taste. Rather, those with a poor understanding of what they have read come and read the riot act to the professionals here. This is OK, I believe, as far as it goes, provided ONLY they are willing to acknowledge there are those here who are professionals in the subject they are pontificating about and thus are willing to become "students" (as far as this makes sense on an internet chat room) BUT, what my gets my goat, and it seems that of some others, is when these people pretend to knowledge, skills and education that they manifestly don't posses. I cannot for the life of me see why anyone would want to do this. I can think of no member here "of good standing" (whatever that might mean) who has not had to ask a question and been grateful for the responses. I know that I have. Our Jack seems to be a case in point, but I can think of several others in the same category, of those who are quite clearly a little confused about the mathematical niceties of, say, coordinate transformations in flat (or otherwise) spacetime, and refuse to acknowledge that relativity has a very great deal more to it than trains and embankments. Alpha showed us this quite beautifully, and yet I doubt our Jack even read it, and if he did, has not the skills to critique it. And yet again he persists on what he terms "calling Alpha to put up". Likewise, I recall there was a thread entitled something like "I question string theory", but no real grounds were given, certainly no math was offered as I remember. So what's the beef? Just don't like it? Tough: I don't "like" QM, but there it is. /end rant Ben, you may delete or trash this as off-topic, at your discretion Post ReplyCreate New Thread