D H Posted January 27, 2012 Share Posted January 27, 2012 Thank you IM for your reply. But if the speed is unchaged because the source of the light, the glass rod, and the device which measures the speed of that light are all on the Earth at rest with respect to each other -- i.e. in a single frame of reference, then it will be the same if all that stuff will be placed in a rocket and launch to space with the speed eg. 0,5 c. Am I right? Huh? You really should learn the subject you are trying to criticize before you criticize it. As a starter, suppose you see two space ships each moving directly away from you, but in opposite directions, with each space ship going at 3/4 the speed of light as perceived by you. The people on each space ship will perceive that they are moving away from you at 3/4 the speed of light. They will also perceive that they are moving away from the other space ship at 24/25 the speed of light -- not 1.5 times the speed of light as Newtonian mechanics would suggest. Link to comment Share on other sites More sharing options...
Bart Posted January 27, 2012 Author Share Posted January 27, 2012 (edited) Huh? You really should learn the subject you are trying to criticize before you criticize it. As a starter, suppose you see two space ships each moving directly away from you, but in opposite directions, with each space ship going at 3/4 the speed of light as perceived by you. The people on each space ship will perceive that they are moving away from you at 3/4 the speed of light. They will also perceive that they are moving away from the other space ship at 24/25 the speed of light -- not 1.5 times the speed of light as Newtonian mechanics would suggest. Sorry DH, I am not criticize the subject only I want to explain my doubts. I think we do not understand each other, because your answer do not explain my question. I have learned SR, may be as deep as you but I am not so happy follower of that theory as you might expect. Can you be so kind to answer my simple question putted in the post #25 to IM? Edited January 27, 2012 by Bart Link to comment Share on other sites More sharing options...
D H Posted January 27, 2012 Share Posted January 27, 2012 Try rephrasing that question. As it stands it's pretty much gobbledygook. You aren't communicating. Link to comment Share on other sites More sharing options...
Bart Posted January 28, 2012 Author Share Posted January 28, 2012 (edited) Try rephrasing that question. As it stands it's pretty much gobbledygook. You aren't communicating. The answer from IM: " The speed of light in a glass rod will show no difference, whether the rod is parallel to the Earth or perpendicular to the Earth. This result does not violate special relativity. Why? Because it is being performed in a single frame of reference. The source of the light, the glassrod, and the device which measures the speed of that light are all on the Earth at rest with respect to each other -- i.e. in a single frame of reference. The motion of the Earth has no effect because the entire experiment and the Earth are moving together. " So my question is very clear: If the above explanation is OK, whether it is also true the following statement: The speed of light in a glass rod will show no difference, whether the rod is parallel to the rocket or perpendicular to the rocket, because it is being performed in a single frame of reference. The researcher, the source of the light, the glass rod, and the device which measures the speed of that light are all on the rocket at rest with respect to each other -- i.e. in a single frame of reference. The motion of the rocket has no effect because the entire experiment and the rocket are moving together. " DH, if this question is still unclear to you, then I would think that you play with me in the "cat and mouse", and you do not want to explain anything. Edited January 28, 2012 by Bart Link to comment Share on other sites More sharing options...
swansont Posted January 28, 2012 Share Posted January 28, 2012 As long as the rocket is not accelerating, there is no difference in performing the experiment — any inertial frame of reference will give the same result. Link to comment Share on other sites More sharing options...
D H Posted January 28, 2012 Share Posted January 28, 2012 The motion of the rocket has no effect because the entire experiment and the rocket are moving together. Correct. You could even have the rocket moving at 99.9999% the speed of light with respect to some reference, and the motion of the rocket will still have no effect. There is no absolute reference frame by which to judge something's speed. Everything is relative. In the case of this experiment, that the rocket is moving with respect to something else has zero impact on the outcome of the experiment. Link to comment Share on other sites More sharing options...
Bart Posted January 28, 2012 Author Share Posted January 28, 2012 (edited) Correct. You could even have the rocket moving at 99.9999% the speed of light with respect to some reference, and the motion of the rocket will still have no effect. There is no absolute reference frame by which to judge something's speed. Everything is relative. In the case of this experiment, that the rocket is moving with respect to something else has zero impact on the outcome of the experiment. Thanks Swansont and DH for your explanations. So, does it mean that regardless of whether the rocket is standing on the Earth or already is flying in space, the speed of light in the glass rods ( parallel to the rocket or perpendicular to the rocket), which will be observed by the researcher in the rocket, will always show no difference and remains the same as on Earth? Edited January 28, 2012 by Bart Link to comment Share on other sites More sharing options...
D H Posted January 28, 2012 Share Posted January 28, 2012 Yes. The laws of physics are the same in all inertial frames of reference. Link to comment Share on other sites More sharing options...
Bart Posted January 28, 2012 Author Share Posted January 28, 2012 (edited) Yes. The laws of physics are the same in all inertial frames of reference. Thank you DH. So in that case, the light-clocks made of glass that were used in the experiment presented earlier in this forum on the link: http://dl.dropbox.com/u/26262175/TimeDilation.pdf should always indicate the same time as the clocks on the Earth. Is my understanding correct? Edited January 28, 2012 by Bart Link to comment Share on other sites More sharing options...
D H Posted January 28, 2012 Share Posted January 28, 2012 So in that case, [/size] the light-clocks made of glass that were used in the experiment presented earlier in this forum on the link: http://dl.dropbox.com/u/26262175/TimeDilation.pdf should always indicate the same time as the clocks on the Earth. Is my understanding correct? Your understanding is incorrect. A request: Stop with the font/size games. Just use the defaults. Your understanding is incorrect. That was a bit terse. The people on the Earth will see the rocket's clock as running slow. The people on the rocket with see the Earth-based clocks as running slow. Link to comment Share on other sites More sharing options...
Bart Posted January 28, 2012 Author Share Posted January 28, 2012 (edited) Your understanding is incorrect. A request: Stop with the font/size games. Just use the defaults. That was a bit terse. The people on the Earth will see the rocket's clock as running slow. The people on the rocket with see the Earth-based clocks as running slow. Sorry but I did not use any font/size game. May be in my notebook was something wrong defaulted. My question was about some other matter. It not concerned on how the observers see each other clocks on Earth and in the rocket, but on whether the clock in the rocket, due to the fact that the speed of light in its glass is the same as on Earth, is ticking at the same local frequency in the rocket as the ticking of clocks on the Earth? My understanding is that the clocks in the rocket must have identical frequency (measured locally), as the frequency measured localy for the clocks on the Earth. It means that the local time idications in the rocket and the Earth will be the same Edited January 28, 2012 by Bart Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted January 28, 2012 Share Posted January 28, 2012 Yes, an observer sitting on the rocket will see that the clock is ticking at exactly the frequency he'd expect if he were sitting on Earth and watching a stationary clock. Link to comment Share on other sites More sharing options...
Bart Posted January 28, 2012 Author Share Posted January 28, 2012 Yes, an observer sitting on the rocket will see that the clock is ticking at exactly the frequency he'd expect if he were sitting on Earth and watching a stationary clock. Thanks. It means that the local time indications in the rocket will be exacly the same as on the Earth. Thus it denies the existence of time dilation, and proves constancy of the time in the universe ? Link to comment Share on other sites More sharing options...
D H Posted January 28, 2012 Share Posted January 28, 2012 Thanks. It means that the local time indications in the rocket will be exacly the same as on the Earth. Correct. Thus it denies the existence of time dilation, and proves constancy of the time in the universe ? Incorrect. Link to comment Share on other sites More sharing options...
swansont Posted January 28, 2012 Share Posted January 28, 2012 Thank you DH. So in that case, the light-clocks made of glass that were used in the experiment presented earlier in this forum on the link: http://dl.dropbox.com/u/26262175/TimeDilation.pdf Please note that this is not a physical experiment, it is a thought experiment. Any inconsistencies presented are going to be a result of incorrectly applying relativity to the scenario. If you want to analyze actual data, use e.g. the Vessot rocket experiment. http://www.astro.ucla.edu/~wright/vessot.htm Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted January 28, 2012 Share Posted January 28, 2012 Thanks. It means that the local time indications in the rocket will be exacly the same as on the Earth. Thus it denies the existence of time dilation, and proves constancy of the time in the universe ? No; an observer on the ground will see the rocket's clock as running slow. You can see this effect on light clocks in this online demonstration: http://www.refsmmat.com/jsphys/relativity/relativity.html#light-clock In the first version (hit Play to start), photons bounce back and forth at a known rate. When you press Next, you switch to the reference frame of the high-velocity observer; to him, the photons must travel on the diagonal, so the light clock takes longer to tick. An observer moving with the light clock will not notice this effect. (Note: the demonstration will work best in recent versions of Firefox and Chrome.) Link to comment Share on other sites More sharing options...
Bart Posted January 29, 2012 Author Share Posted January 29, 2012 (edited) It means that the local time indications in the rocketwill be exacly the same as on the Earth. Correct. Thus it denies the existence of time dilation, and proves constancy of thetime in the universe ? Incorrect. Why incorrect? If I, being in a flying rocket, have on my clock an indication eg. 12.00 and at the same time, you on the Earth have also on your clock an indication 12.00, and after an hour I have an indication 13.00 and you also have an indication 13.00, etc. Thus, in my frame in a rocket and in your frame on the Earth, time passes at an identical rate. Does not it? Edited January 29, 2012 by Bart Link to comment Share on other sites More sharing options...
swansont Posted January 29, 2012 Share Posted January 29, 2012 Why incorrect? If I, being in aflying rocket have on my clock an indication,eg 12.00 and at the same time, you areon Earth have also on your clock an indication 12.00, and after an hour I havean indication 13.00 and you also have anindication 13.00, etc. Thus, in my framein a rocket and in your frame on the Earth, time passes at an identical rate,is not it? Creating two frames where you have identical time dilation does not extrapolate to "constancy of the time in the universe". It's trivial to find two frames where the time won't agree. Link to comment Share on other sites More sharing options...
D H Posted January 29, 2012 Share Posted January 29, 2012 Why incorrect? If I, being in a flying rocket, have on my clock an indication eg. 12.00 and at the same time, ... Your conclusion does not follow from the premises. I cut your post off with the words "and at the same time" because that is the source of your error in understanding. One of the key consequences of relativity is that even simultaneity is relativity. There is no universal "at the same time". You are wasting your time trying to disprove relativity mathematically. You aren't going to be able to that. The mathematics of special relativity is very sound. What you should be doing is trying to understand that mathematics. Just because the mathematics of special relativity is very sound does not mean that it is right. After all, the mathematics of Newtonian mechanics is also very sound. Just because a theory is internally consistent does not mean it is correct. Scientific theories have an extra constraint: They have to describe reality. Experiment upon experiment have shown that it is relativity theory, not Newtonian mechanics, that describes what transpires at very high speeds. Link to comment Share on other sites More sharing options...
rigney Posted January 29, 2012 Share Posted January 29, 2012 (edited) Creating two frames where you have identical time dilation does not extrapolate to "constancy of the time in the universe". It's trivial to find two frames where the time won't agree. If relativity is correct, as to space and time being curved, wouldn't it be impossible to find two time frames that did agree? Edited January 29, 2012 by rigney Link to comment Share on other sites More sharing options...
D H Posted January 29, 2012 Share Posted January 29, 2012 If relativity is correct, as to space and time being curved, wouldn't it be impossible to find two time frames that did agree? No. How did you arrive at that conclusion? Any two co-moving frames will agree on time. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted January 29, 2012 Share Posted January 29, 2012 Why incorrect? If I, being in a flying rocket, have on my clock an indication eg. 12.00 and at the same time, you on the Earth have also on your clock an indication 12.00, and after an hour I have an indication 13.00 and you also have an indication 13.00, etc. Thus, in my frame in a rocket and in your frame on the Earth, time passes at an identical rate. Does not it? After whose hour? The observer on Earth will claim that the rocket's clock is running slowly. An observer on the rocket will claim that it is running at exactly the right speed. Watch the animation I posted. Link to comment Share on other sites More sharing options...
rigney Posted January 29, 2012 Share Posted January 29, 2012 No. How did you arrive at that conclusion? Any two co-moving frames will agree on time. Unless two photons are superimposed, wouldn't they have different time frames? Link to comment Share on other sites More sharing options...
swansont Posted January 29, 2012 Share Posted January 29, 2012 Unless two photons are superimposed, wouldn't they have different time frames? Photons are not in an inertial frame. "Time frame" for a photon makes no sense. Link to comment Share on other sites More sharing options...
rigney Posted January 30, 2012 Share Posted January 30, 2012 (edited) Photons are not in an inertial frame. "Time frame" for a photon makes no sense. There is much in this link that I am totally ignorant of, which you might readily explain? But since this is my first look at it, I'll likely need more than a bit of time trying to figure it out.http://www.circlon-theory.com/HTML/motionofphotons.html Edited January 30, 2012 by rigney Link to comment Share on other sites More sharing options...
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