anything and everything can and is moving faster than c

SpeedFreek · Oct 16, 2007

I am pondering (or researching) and will bet back to you on this one vanDivX.. but I'm thinking that if length contraction is positive in your direction of travel, then length contraction would be negative in the opposite direction (i.e. it would make distances look larger). There doesn't need to be a specific term for that, the contraction would be either positive or negative. (This would be the same as acceleration, which in physics is referred to as acceleration whether you are speeding up or slowing down. What we might call deceleration is just negative acceleration, or acceleration in the opposite direction.) It would mirror what light does, as it is stretched (redshifted) when moving away from you, and is compressed (blueshifted) when moving towards you. We have to remember that when looking forwards from a relativistic ship, light-travel time or aberration actually make the distance in front look larger, and it is only when you subtract these factors from your on-board calculations that you could work out how much length has contracted in your direction of travel. This next bit is definitely correct (I have checked!) - if you travel 5 light years at 86.6% of light speed, the relativistic change factor is 2. A journey of 5 ly at 86.6% of c would look to an observer to take 5.77 years, just as you would expect. But on the ship, time runs half as fast and length contraction in your direction of travel causes the distance to your destination to look halved too. So you would travel 5 light years in 2.88 years, and your destination would (when you start, moving at 0.866c) look only 2.5 light years away, so there is no paradox in your observations from that ship. I am currently finding out what the view "out the back" would be, and will get back to you. <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

SpeedFreek · Oct 16, 2007

Ok, I did my research and found that my analysis of the problem is indeed incorrect. Both time-dilation and length contraction are symmetrical in both directions, for the traveller. Light-travel time causes a distorted view of things, but the underlying principle is that observers in different frames cannot agree on the simultaneity of events. The basic example still stands, but I was mistaken in my view of why the ship sees what it does. But all observers do see the light moving at c, relative to them and that is the important bit! <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

vandivx · Oct 17, 2007

<blockquote>In reply to:<hr />I'm thinking that if length contraction is positive in your direction of travel, then length contraction would be negative in the opposite direction (i.e. it would make distances look larger). There doesn't need to be a specific term for that, the contraction would be either positive or negative. (This would be the same as acceleration, which in physics is referred to as acceleration whether you are speeding up or slowing down. What we might call deceleration is just negative acceleration, or acceleration in the opposite direction.)<hr /></blockquote> <img src="/images/icons/smile.gif" /> nice try BUT assuming that we talk about the official interpretation of SR I point out to you that the formula for length contraction does not allow for length dilation but only for contraction and sign has nothing to do with it (still I got what you said there, read below what is my take on it) L=L'(sqrt(1-(v/c)sq)) when velocity is zero then L=L' and when it approaches c, L approaches zero, that is the observed length L can only get shorter for astronaut on board the ship the v=0 naturally and so for him the length doesn't change, observer on Earth relative to which the rocket is traveling near the speed of light sees the rocket as shortened and that includes the measuring stick the astronaut is using to measure the speed of light in his own ref frame and it would be absurd if the observer on Earth saw the stick shoter if the astronaut measured light going in the direction of travel and longer if he measured light speed as it went backwards, length is in exactly the same category in this respect that time is, you can't have it both ways for either one <img src="/images/icons/wink.gif" /> I am perfectly aware that in physics we typically talk only about acceleration and the term deceleration is rarely if ever used (witness for example all the talk about Pioneer probe suffering anoma <div class="Discussion_UserSignature"> </div>

SpeedFreek · Oct 17, 2007

I already told you I was incorrect, and length contraction is symmetrical in both directions, so there is the same amount of contraction in front as behind. My interpretation of events in my previous 2 posts was wrong, but the principle (all observers measure light moving at c) does stand up to scrutiny. The two observers, in relative motion, can't agree about what distant events are simultaneous. In particular, the observers on the ship might judge a pair of events, positioned fore and aft, to be simultaneous for them, while you would measure the "aft" event preceding the "fore" event. (Note that this is not an issue relating to delays in signals transmitted by light; this is an actual shift in what is, or is not, simultaneous.) So when the shipboard observers find that a pair of lightbeams, travelling fore and aft, are making journeys across equal distances in equal times, you (the "stationary" observer) tell them that they've messed up the synchronization of their distant clocks, and the journey times are not equal. (They, meanwhile, observe that you have messed up your synchronization in a precisely symmetrical way. So when one observer is happy with a light-travel time experiment, the other says it's messed up by faulty synchronization.) But I tend to agree with you that there seems to be something lacking in all this, some missing part of the puzzle. I will have to look deeper into the relativity of simultaneity, which seems to be the basis of SR. <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

vandivx · Oct 18, 2007

ok, I didn't quite get what you meant by symmetrical, my fault what I am getting at doesn't require consideration of simultaneity of events at all I believe, you can separate the fore and aft events by having them performed separately traveler shines a beam of light up ahead when traveling near speed c and he finds the beam covers given path in a given time in his reference frame and he finds it travels ahead at speed c, meantime observers on Earth watching the rocket fly past at near speed of light see the guy shining the beam of light up ahead and by some measurement determine that the beam's speed is c in their reference frame of Earth as expected and their interpretation why the traveler also measured speed c is that his clock slowed down and his measuring stick grew shorter and therefore he measured the speed of the light beam leaving the rocket in the aft direction at speed c Earth observers see the traveler apply his shortened measuring stick to the distance the light beam made as it was advancing ahead of the rocket and understand why he believes the distance he measured to be longer than it was from point of view of the Earth observes and the same with his clock when he calculates the speed - his clock ticks slow and that also affects his calculations of the speed of that beam of light and both effects conspire against him in that they bump up the speed of that beam to c for him while the Earth observers see that the beam of light is not advancing on the rocket that much because the rocket itself already travles almost at the speed of light later on the traveler on the ship decides to perform the same experiment but in opposite direction, that is in the aft direction and lo and behold he finds the light speed as it leaves him also moving at speed c (at least that's what SR theory says he should measure), meanwhile the Earth observers measure by some means the speed of the beam of that aft light and find it also moving at <div class="Discussion_UserSignature"> </div>

SpeedFreek · Oct 18, 2007

No need to apologise for your lengthy "dummies" description, it is the same method I use to explain concepts like the metric expansion of space or manifold spaces (my specialities!), and usually aids understanding, but I tend to agree that with SR, physical descriptions are less straightforward than they might at first seem. I'm now dipping my toes into SR and finding things more difficult to visualise! <img src="/images/icons/smile.gif" /> One thought that struck me though, was how would the traveller be able to measure the speed of their emitted lights? It seems relatively easy for an external observer to see a ship moving across their view to find a means to measure it's speed. And then it also seems easy for that observer to measure the speed of the lights emitted fore and aft from that ship. But how, when travelling on a relativistic ship, might one measure the speed of ones fore and aft lights? How could one work out how far a photon has receded in the direction or the opposite direction of travel? Could this be the key to the problem? Perhaps the time-dilation and length contraction are red herrings in this case, and the aberration of light should be the focus of our analysis? <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

origin · Oct 18, 2007

I think it is straight forward that for the observer on a speeding ship, a light on the back of the ship would have a messured speed of c. In nonreltivistic terms if you are in a car traveling at 80 mph and you throw a baseball out the back at 60 mph the observer in the car will see the ball recede from the car at 60 mph. An observer outside of the car will see the ball traveling in the same direction as the car at 20 mph. With relativity if a ship is going 99% of c then it makes perfect sense that a light beam shone out the back of the ship will recede from an observer on the SHIP at c. The problem is that an observer outside of the ship will also see the light moving at c, instead of .01c. So the distance between the ship and the light beam will increase at a rate of 199% c. How can both of these observations be going on at the same time? Good question - I obviously need to go back and study my physics book some more <div class="Discussion_UserSignature"> </div>

larper · Oct 18, 2007

<blockquote>In reply to:<hr />One thought that struck me though, was how would the traveller be able to measure the speed of their emitted lights? <hr /></blockquote> Easy. Send a probe out in front of the ship, traveling 1mph relative to the ship for 1 hour. After 1 hour, shine your headlights on the probe and measure the round trip time. You will get the amount of time it takes light to travel 2 miles. You will also get a doppler reading that says the probe is receding at 1mph. <div class="Discussion_UserSignature"> Vote Libertarian </div>

SpeedFreek · Oct 18, 2007

<blockquote>In reply to:<hr />I think it is straight forward that for the observer on a speeding ship, a light on the back of the ship would have a messured speed of c. In nonreltivistic terms if you are in a car traveling at 80 mph and you throw a baseball out the back at 60 mph the observer in the car will see the ball recede from the car at 60 mph. An observer outside of the car will see the ball traveling in the same direction as the car at 20 mph. <hr /></blockquote> I'm not sure what you are implying here, or did I misunderstand you? In classical mechanics (non relativistic terms), if you are in a car travelling at 80mph and you throw a baseball out the back at 60mph you will see the ball recede from the car at 140mph!. The ball moves aft at 60mph relative to the ground and the car moves forward at 80mph relative to the ground. *EDIT: I am totally wrong here, of course... I blame the new puppy I have, for interfering with my concentration. <blockquote>In reply to:<hr />With relativity if a ship is going 99% of c then it makes perfect sense that a light beam shone out the back of the ship will recede from an observer on the SHIP at c. The problem is that an observer outside of the ship will also see the light moving at c, instead of .01c. So the distance between the ship and the light beam will increase at a rate of 199% c. How can both of these observations be going on at the same time? Good question - I obviously need to go back and study my physics book some more <hr /></blockquote> Yup, that's the problem we are mulling over, here. <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

larper · Oct 18, 2007

<blockquote>In reply to:<hr />I'm not sure what you are implying here, or did I misunderstand you? In classical mechanics (non relativistic terms), if you are in a car travelling at 80mph and you throw a baseball out the back at 60mph you will see the ball recede from the car at 140mph!. The ball moves aft at 60mph relative to the ground and the car moves forward at 80mph relative to the ground. <hr /></blockquote> No, even in classical mechanics, its all relative. The ball indeed strikes the ground moving in the same direction as the car, traveling 20mph relative to the road. <div class="Discussion_UserSignature"> Vote Libertarian </div>

larper · Oct 18, 2007

<blockquote>In reply to:<hr />With relativity if a ship is going 99% of c then it makes perfect sense that a light beam shone out the back of the ship will recede from an observer on the SHIP at c. The problem is that an observer outside of the ship will also see the light moving at c, instead of .01c. So the distance between the ship and the light beam will increase at a rate of 199% c. <hr /></blockquote> So what? Who cares if an observer in a 3rd reference frame measures the rate of separation between two things as being faster than c? So what? That is not the problem that relativity solves. Heck, it's not a problem at all. The "problem" that SR solves is that both the ship and the outside observer both measure the speed of the light beam as c, relative to themselves. <div class="Discussion_UserSignature"> Vote Libertarian </div>

SpeedFreek · Oct 18, 2007

No, even in classical mechanics, its all relative. The ball indeed strikes the ground moving in the same direction as the car, traveling 20mph relative to the road. Dowp! Of course, you are right.. I'm not thinking properly and have now totally confused myself.. I will take a rest from this subject for a while. <img src="/images/icons/frown.gif" /> I feel such a fool, making such a basic mistake! <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

origin · Oct 18, 2007

So what? Who cares if an observer in a 3rd reference frame measures the rate of separation between two things as being faster than c? So what? That is not the problem that relativity solves. Heck, it's not a problem at all. I wasn't implying that it was a problem. I was simply trying to describe what an observer outside the ship would see. <div class="Discussion_UserSignature"> </div>

origin · Oct 18, 2007

I will take a rest from this subject for a while. I feel such a fool, making such a basic mistake! Trying to describe SR when traveling at near light speeds is so confusing, especially when you are switching from obsevers in different inertial frames, that it is not surprising to make a mistake now and then. No big deal... <div class="Discussion_UserSignature"> </div>

vandivx · Oct 19, 2007

"The "problem" that SR solves is that both the ship and the outside observer both measure the speed of the light beam as c, relative to themselves." --- that's correct, just what we talk about here, the point is this: if we assume SR to be correct in its results, how can the traveler arive at his measure in terms of shrinking ruler and slowing down clock when he measures speed of light emited backwards from ship... those factors work fine for light sent forward from ship but they don't seem to square up with opposite direction what is needed is not just shrinking rulers and slowing down clocks but also expanding rulers and speeding up clocks up to twice their normal rate when observed from outside reference frame taken to be at rest but standard interpretation of SR knows nothing about such things as expanding rulers and speeding up clocks if changed length and time are used to explain light sent forward as they typically are, they should be used to explain the light sent backward too I would think, otherwise those changes are not valid inferences from SR and shouldn't be used at all boy I wish Einstein didn't get sidetracked by GUT and continued working on SR and GR for the rest of his life integrating the two and ironing the paradoxes of SR and all that vanDivX <div class="Discussion_UserSignature"> </div>

larper · Oct 19, 2007

<blockquote>In reply to:<hr />what is needed is not just shrinking rulers and slowing down clocks but also expanding rulers and speeding up clocks up to twice their normal rate when observed from outside reference frame taken to be at rest but standard interpretation of SR knows nothing about such things as expanding rulers and speeding up clocks <hr /></blockquote> Nope. I am in a ship. I shine a line in one direction versus another. What is the difference? What does "in front vs behind" mean? Now, I am outside the ship. He whizzes by me at near c velocity. He shines a light in his direction of travel just before he gets to me. I see a blue light. He shines the same light out the back after he has passed me. I see a red light. Both light beems are moving at exactly c. What problem requires expanding rulers and fast clocks? <div class="Discussion_UserSignature"> Vote Libertarian </div>

SpeedFreek · Oct 19, 2007

I am in a ship. I shine a line in one direction versus another. What is the difference? What does "in front vs behind" mean? Ahh! And there is the crux of the matter, the thing we have forgotten - the ship can consider itself at rest. Suddenly things are becoming clearer. <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

xmo1 · Oct 20, 2007

Ok folks, the biggest problem in this forum is over thinking the problem. If you do that you miss the point of the subject. There are two trains (ha^3) moving past each other at 10mph each. They are moving away from each other at 20mph, but each train is still moving at 10mph. That is what is meant by frame of reference (your frame, or the frame of reference of each train). Trains traveling at 1200 feet per second are called bullet trains. <div class="Discussion_UserSignature"> DenniSys.com </div>

origin · Oct 22, 2007

What problem requires expanding rulers and fast clocks? Here is the problem. If I am in a Jet traveling at 330 m/s and I have a multi directional speaker send out a sound, the sound will move along direction of travel at ~10 m/s and behind me it will recede at ~670 m/s relative to the jet. An outside observer will clearly hear the doppler shift and would measure the wave front speed at ~340 m/s. Obviously, the speed of sound will be different relative to the observer. With light however if a ship is traveling at 95% of c and there is a light on the ship the light will move along the direction of travel of the ship at c and will recede from the ship at c. Of course an outside observer would measure the light as traveling at c, but the light move away from the ship at 5% of c and receded from the back of the ship at 195% the speed of c. See the problem? Yes the light would be doppler shifted, but all of the observers independant of their different velocities would measure the velocity of the light at c. This is clearly counter intuitive based on how sound waves behave, which seems 'normal'. <div class="Discussion_UserSignature"> </div>

larper · Oct 22, 2007

<blockquote>In reply to:<hr />If I am in a Jet traveling at 330 m/s and I have a multi directional speaker send out a sound, the sound will move along direction of travel at ~10 m/s and behind me it will recede at ~670 m/s relative to the jet. <hr /></blockquote> No, it won't. <div class="Discussion_UserSignature"> Vote Libertarian </div>

vandivx · Oct 22, 2007

<blockquote>In reply to:<hr />Nope. I am in a ship. I shine a line in one direction versus another. What is the difference? What does "in front vs behind" mean? Now, I am outside the ship. He whizzes by me at near c velocity. He shines a light in his direction of travel just before he gets to me. I see a blue light. He shines the same light out the back after he has passed me. I see a red light. Both light beems are moving at exactly c. What problem requires expanding rulers and fast clocks?<hr /></blockquote> I suppose you are aware that length contraction and time dilation are being used all the time to explain how one observer can understand why the the other one is measuring what he does (that he measures speed of light being c even though he is moving relative to you) I am not disputing that both of them measure speed c in all directions and that each can consider himself at rest 'rear direction' or 'front vs behind' means opposite directions along the line of relative motion btw the two observers so that if they are receeding from each other then front direction is when you point your beam of light away from your fellow observer and behind direction is when you shine it at him while the distance between you two is increasing preferably at near the speed of light each observer can consider himself to be perfectly at rest while looking at what the other does and of course each measures speed of light being c in his 'rest frame', nobody is disputing that, what I am after is that invocation of modified rulers and clocks to explain why the other observer you see as moving relative to you measures also the speed of light being c as you do when he emits beam of light in the direction of his motion all I am saying is that that kind of explanation (shrinking ruler & slowing down clock) doesn't hold when he instead shines the light back at his fellow observer in eff <div class="Discussion_UserSignature"> </div>

vandivx · Oct 22, 2007

<blockquote>In reply to:<hr />Ahh! And there is the crux of the matter, the thing we have forgotten - the ship can consider itself at rest. Suddenly things are becoming clearer.<hr /></blockquote> <img src="/images/icons/smile.gif" /> I wish I too would like some simple insight that would give me absolution and make me see the light as the unbelieving Thomas saw it when he put his hand on the wounds of christ (just making fun point, it is not any alusion to you or anybody) fact is I am bothered by this insight into SR for like a decade now (at least) and I used to think I am just deluded and that when I examine it critically later on I will see where I made a mistake (you know that line about 'you sleep on it'... but I slept on it many times and so far got no results except being rested <img src="/images/icons/smile.gif" /> ) , I know one can make a logical slip and be deluded on such simple point through some simple oversight of something obvious like this idea that "the ship can consider itself at rest", alas so far the absolution evades me and I have no peace of soul take this problem being posed as devil's advocate kind of thing and I hope it will sow doubts in your mind as it did in mine LOL yes the ship can consider itself at rest but what we are talking about here is what you see when you look at the ship as outside observer and see it moving near the speed of light I know it all dovetails nicely in calculations, what I am disputing is that the results of those calculations have any pictorial explanation, i.e., that they can be understood in visual or graspable terms, as I said if I am true about this problem then SR will join QM as being mathematically correct but nobody will understand how it all works vanDivX <div class="Discussion_UserSignature"> </div>

larper · Oct 22, 2007

Question it all you want, but it is indeed how the universe works. The classic model of a "shrinking ruler" is to take a meter stick. Make a box that is slightly less than a meter long. The box has two doors. The exit door is initially closed, and the entry door is initially open. Shoot the stick at the box at near the speed of light. The entry door is triggered to close as soon as the end of the stick has passed by, and the exit door is triggered to open just before the front of the stick hits it. There is a frame of reference in which an observer will see the stick inside the box with both doors closed. This is really how the universe works. <div class="Discussion_UserSignature"> Vote Libertarian </div>

SpeedFreek · Oct 22, 2007

"yes the ship can consider itself at rest but what we are talking about here is what you see when you look at the ship as outside observer and see it moving near the speed of light" But we dealt with what the outside observer would see earlier on. The problem we had was reconciling their view with the view from the ship. Remember, if the ship were travelling at 90% of c, we would as an outside observer see the light moving away in front at c (10% of c faster than the ship) and light moving away behind from the point of emission at c, with the ship moving away from the point of emmission at 90% of c, and therefore moving away from the emitted rear light at 190% of c. The problem occurred when trying to explain these numbers from the view outside the ship, if the ship sees its fore and aft light both moving away at c. (edited to correct last paragraph) <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

larper · Oct 22, 2007

And the problem is made even worse when you use a photon and then state "sees the light emitted." To see light, it has to be aimed at you, usually by reflecting off something. So, you really need to state that the ship shines a light at an object in front of the ship, and measures the round trip time. <div class="Discussion_UserSignature"> Vote Libertarian </div>

anything and everything can and is moving faster than c

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