Robittybob1 Posted March 6, 2016 Posted March 6, 2016 Try to explain why nothing can go faster than speed of light without using the term relativistic mass. Can it be done?
swansont Posted March 6, 2016 Posted March 6, 2016 The total energy of a massive object tends towards infinity as you approach c. At c it's undefined. At speeds above c, the energy is an imaginary value.
Robittybob1 Posted March 6, 2016 Author Posted March 6, 2016 (edited) The total energy of a massive object tends towards infinity as you approach c. At c it's undefined. At speeds above c, the energy is an imaginary value. That is a start I suppose. Something is moving and you add some more energy it moves faster why can't this go on forever? Are you having to push the additional energy as well as the rest mass energy as you go faster? Is that why "the total energy of a massive object tends towards infinity as you approach c" is any reason to be a restriction? Do we say that kinetic energy has no mass, the kinetic energy of an object does not add to its gravitational mass, but it appears to affect its inertia. But wasn't it impossible to separate inertial mass from gravitational mass? Edited March 6, 2016 by Robittybob1
swansont Posted March 6, 2016 Posted March 6, 2016 That is a start I suppose. Something is moving and you add some more energy it moves faster why can't this go on forever? Are you having to push the additional energy as well as the rest mass energy as you go faster? Is that why "the total energy of a massive object tends towards infinity as you approach c" is any reason to be a restriction?Do we say that kinetic energy has no mass, the kinetic energy of an object does not add to its gravitational mass, but it appears to affect its inertia. But wasn't it impossible to separate inertial mass from gravitational mass? It's not an issue of mass. Even for Galilean kinematics (KE = 1/2 mv^2) you have diminishing returns in increasing speed for each increment of energy increase. For relativity, it's a different dependence, so the speed increase is incrementally even smaller. another consequence of c being invariant.
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 It's not an issue of mass. Even for Galilean kinematics (KE = 1/2 mv^2) you have diminishing returns in increasing speed for each increment of energy increase. For relativity, it's a different dependence, so the speed increase is incrementally even smaller. another consequence of c being invariant. "diminishing returns in increasing speed for each increment of energy increase" and "The speed increase is incrementally even smaller". Hit the nail on the head there. So can we explain it all just with mass and velocity?
Strange Posted March 7, 2016 Posted March 7, 2016 You don't even need mass. One result of special relativity is that we find speed does not add linearly in the way we are used to (at our normal humdrum pace). For example, if you are travelling past the Earth at 0.5c and you fire a rocket ahead of you at 0.5c then, from my perspective on Earth, I will see the rocket moving at 0.8c (rather than at c). http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html This just comes from applying the Lorentz transform.
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 (edited) You don't even need mass. One result of special relativity is that we find speed does not add linearly in the way we are used to (at our normal humdrum pace). For example, if you are travelling past the Earth at 0.5c and you fire a rocket ahead of you at 0.5c then, from my perspective on Earth, I will see the rocket moving at 0.8c (rather than at c). http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html This just comes from applying the Lorentz transform. I've been a bit amused lately where they add speeds recorded by different people. Imaging if the other craft is an alien and they don't speak our language, so we can't communicate with them, how do you describe what is happening then? "For example, if you are travelling past the Earth at 0.5c and you fire a rocket ahead of you at 0.5c then, from my perspective on Earth, I will see the rocket moving at 0.8c (rather than at c). becomes: For example, if you see an alien mothership travelling past the Earth at 0.5c and it fires a rocket ahead at 0.5c then, from my perspective on Earth, I will see the rocket moving at 0.8c (rather than at c)." Is that still true? I think the velocity of the rocket will be c therefore the rocket must be made from photons. Will it look like the light rocket moves ahead of the mothership at 0.5 c? Edited March 7, 2016 by Robittybob1
Strange Posted March 7, 2016 Posted March 7, 2016 Is that still true? Of course. (I struggle to see why the idea of aliens is relevant. Also, not that the speed of the rocket is specified relative to the craft firing it [sorry if I didn't make that clear].) I think the velocity of the rocket will be c And you are wrong.
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 (edited) Of course. (I struggle to see why the idea of aliens is relevant. Also, not that the speed of the rocket is specified relative to the craft firing it [sorry if I didn't make that clear].) And you are wrong. In your example who is measuring the speeds? Are they being measured by two different people? Are you sure the rocket can't go at c, it is made of light after all in my case. If the rocket is made from light it will travel at c (the speed of light) and if the mothership is doing 0.5 c the light front of the rocket must be 0.5 c ahead of the rocket. All my measurements are from the friendly position, for I don't trust the aliens. On board the mothership the aliens think their light rocket is traveling at the speed of light, from their perspective the position I am at is passing them at 1/2 the speed of light. Their conventional rocket that does 1/2 the speed of light is moving at half the speed of the light rocket. From my perspective the conventional rocket is traveling 0.8 c as you say (I'll take your word on that). Edited March 7, 2016 by Robittybob1
swansont Posted March 7, 2016 Posted March 7, 2016 In your example who is measuring the speeds? Are they being measured by two different people? Are you sure the rocket can't go at c, it is made of light after all in my case. Then it's not traveling at 0.5c. It's a non-sequitur. A "rocket made of photons" travels at c in all frames, and you would not observe it unless it collided with your eyes or whatever you used to view it. So it's a complete non-starter for a discussion. We're discussing relativity here, not science fiction.
Eise Posted March 7, 2016 Posted March 7, 2016 The idea of Strange, but a bit simplified: The speed of light is the same for all observers. A rocket flies by, and sends a lightbeam in the direction of its flight. So the astronaut measures a speed of c in the direction of his flight. I see the lightbeam also with a speed of c. If the rocket would be faster than c, I would see the astronaut measuring the light beam behind him, but for the astronaut it would be in front of him. That is logically impossible, so I will see the rocket always with a velocity lower than c.
Strange Posted March 7, 2016 Posted March 7, 2016 In your example who is measuring the speeds? Are they being measured by two different people? Sorry if that wasn't clear. The scenario is: You (or an alien) are on a spaceship which is moving relative to Earth / me at 0.5c. (Note that the implausible numbers have been chosen to make the difference clear and the arithmetic simple). You / alien fire a rocket (torpedo, rescue shuttle, landing craft, whatever) ahead of you at a speed of 0.5c relative to your spaceship. I see your spaceship moving at 0.5c and the rocket travelling at 0.8c. If we were talking old-fashioned physics with trains and baseballs, then if you were on a train going past at 40 MPH and tossed a ball forwards at 10 MPH then I (standing on the platform) would see the ball moving at 50MPH. But if I could measure incredibly accurately, I would see it was a really tiny bit slower than that because velocities do not add linearly. Is that better? If the rocket is made from light it will travel at c (the speed of light) and if the mothership is doing 0.5 c the light front of the rocket must be 0.5 c ahead of the rocket. Your "rocket" is a beam of light. Different rules apply here. You would measure the light moving away from you at c. I would see the light moving at c as well. (I'll take your word on that) You don't have to take my word. The arithmetic is very simple, nothing more complicated than division and square roots.
Mordred Posted March 7, 2016 Posted March 7, 2016 if ship a travelling at 0.5 c fires a rocket at 0.5 c as seen from the ship the rocket will be seen to move at 0.8c. from an external observer. use your relativistic velocity addition formula https://en.wikipedia.org/wiki/Velocity-addition_formula
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 Then it's not traveling at 0.5c. It's a non-sequitur. A "rocket made of photons" travels at c in all frames, and you would not observe it unless it collided with your eyes or whatever you used to view it. So it's a complete non-starter for a discussion. We're discussing relativity here, not science fiction. That 0.5c was 0.5c ahead of the mothership which already doing 0.5c. So the light rocket is definitely going at c in all frames (I'm sure I made that clear). A light rocket could be like laser gun. Sorry if that wasn't clear. The scenario is: You (or an alien) are on a spaceship which is moving relative to Earth / me at 0.5c. (Note that the implausible numbers have been chosen to make the difference clear and the arithmetic simple). You / alien fire a rocket (torpedo, rescue shuttle, landing craft, whatever) ahead of you at a speed of 0.5c relative to your spaceship. I see your spaceship moving at 0.5c and the rocket travelling at 0.8c. If we were talking old-fashioned physics with trains and baseballs, then if you were on a train going past at 40 MPH and tossed a ball forwards at 10 MPH then I (standing on the platform) would see the ball moving at 50MPH. But if I could measure incredibly accurately, I would see it was a really tiny bit slower than that because velocities do not add linearly. Is that better? Your "rocket" is a beam of light. Different rules apply here. You would measure the light moving away from you at c. I would see the light moving at c as well. You don't have to take my word. The arithmetic is very simple, nothing more complicated than division and square roots. Yes that is clear, and I agree. But in my case when the mothership fires the light rocket, how fast relative to the mothership do I see the light front moving? The mothership is doing 0.5c and the light is doing c, so it seems that I would see the craft going half the speed of the light front. Surely?
swansont Posted March 7, 2016 Posted March 7, 2016 I'm sure I made that clear That you are sure is the problem in a nutshell.
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 if ship a travelling at 0.5 c fires a rocket at 0.5 c as seen from the ship the rocket will be seen to move at 0.8c. from an external observer. use your relativistic velocity addition formula https://en.wikipedia.org/wiki/Velocity-addition_formula Yes but if it fires a laser type weapon how fast is the light front moving as seen by the external observer? So is it obvious to the external observer that the craft is going at half the speed of the light front? That is what 0.5c must mean surely?
Mordred Posted March 7, 2016 Posted March 7, 2016 (edited) No, you probably didnt see my post above if ship a travelling at 0.5 c fires a rocket at 0.5 c as seen from the ship the rocket will be seen to move at 0.8c. from an external observer. use your relativistic velocity addition formula https://en.wikipedia.org/wiki/Velocity-addition_formula lets take another example spacecraft is moving at 0.9 c fires a rocket at 0.9c as seen from spacecraft. An external observer will measure 0.9999494975001263 c no matter how close the rocket and craft is to c the addition will always be less than c Yes but if it fires a laser type weapon how fast is the light front moving as seen by the external observer? So is it obvious to the external observer that the craft is going at half the speed of the light front? That is what 0.5c must mean surely? a laser always travels at c in a vacuum to all observers, observer in the craft will see the laser move at c and so will an external observer. yes an external observer will still measure the craft at 0.5c Edited March 7, 2016 by Mordred
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 That you are sure is the problem in a nutshell. in #7 I said "Will it look like the light rocket moves ahead of the mothership at 0.5 c?" That is not from the perspective of the mothership but the perspective of the external observer. .... a laser always travels at c in a vacuum to all observers observer in the craft will see the laser move at c and so will an external observer Yes I know that, but how fast will the laser wavefront move compared to the mothership from the perspective of the external observer?
swansont Posted March 7, 2016 Posted March 7, 2016 in #7 I said "Will it look like the light rocket moves ahead of the mothership at 0.5 c?" That is not from the perspective of the mothership but the perspective of the external observer. That was after your erroneous conclusion that the rocket moving 0.5c relative to the mothership (moving at 0.5c relative to earth) must be moving at c. Your example is anything but clear, for it is not based on a proper application of physics.
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 (edited) That was after your erroneous conclusion that the rocket moving 0.5c relative to the mothership (moving at 0.5c relative to earth) must be moving at c. Your example is anything but clear, for it is not based on a proper application of physics. Sorry about that. In #7 I was varying Strange's statement and I deliberately gave the wrong answer (erroneous conclusion) , but immediately asked if it was right. I was hoping by doing that it would stimulate a degree of discussion, which it has, but not along the lines I was hoping for. Edited March 7, 2016 by Robittybob1
Mordred Posted March 7, 2016 Posted March 7, 2016 see my edit last post Yes I know that, but how fast will the laser wavefront move compared to the mothership from the perspective of the external observer? I caught that question while you were posting
Robittybob1 Posted March 7, 2016 Author Posted March 7, 2016 (edited) ... a laser always travels at c in a vacuum to all observers, observer in the craft will see the laser move at c and so will an external observer. yes an external observer will still measure the craft at 0.5c Yes sounds right. So the craft is 0.5c slower than the light front. Yes that is what I was getting at in #7. For I was trying to see if these problems can be worked out without having to communicate with the other craft. Edited March 7, 2016 by Robittybob1
Mordred Posted March 7, 2016 Posted March 7, 2016 When you describe a problem in relativity it's a needed step to be clear on the observer. Just an fyi
Strange Posted March 7, 2016 Posted March 7, 2016 For I was trying to see if these problems can be worked out without having to communicate with the other craft. Why do you need to communicate? Yes I know that, but how fast will the laser wavefront move compared to the mothership from the perspective of the external observer? The question doesn't really make sense. The speed of light isn't relative to anything. But I suppose you can consider the difference in speed between the light (i.e. c) and the craft (0.5 c) which is (rather painfully obviously) 0.5c.
geordief Posted March 7, 2016 Posted March 7, 2016 (edited) The speed of light isn't relative to anything. .Semantics just or can you say rather that the speed of light is relative to everything equally ? It seems that the speed of light is just one example of the speed limit of everything,isn't it? Do we need a speed limit of some kind just so as not to break causality and does light (and massless particles )just fill that slot? Edited March 7, 2016 by geordief
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