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TIME DILATION. WHY do clocks that are |
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| Sep8-11, 08:08 PM | #52 |
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TIME DILATION. WHY do clocks that areMy favorite way is the spacetime diagram approach, but you should read them all and pick the one you like the best. |
| Sep9-11, 12:22 AM | #53 |
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At a speed of 0.6c the time dilation factor is 0.8. So let's assume that both twins are going to watch the other one age during the trip. There are many ways we could do this but let's just say that each twin has a blinking light that flashes exactly once per second and they each count the other one's flashes during the trip. As soon as the traveler starts out at 0.6c, they will each observe the other one flashing at exactly one have the rate of their own. This is a combination of time dilation and the delay in the light travel time. Let's say that after many days, the traveler turns around. He will immediately now see the flashes from his earth bound twin come in at double his own rate. But what will the earth bound twin see? Well, he won't see anything different until several days later because the sight of his twin turning around is subject to the delay in the light travel time. Eventually though, he will see the light flashes coming from his twin suddenly go from one half his rate to double his rate. But this will happen near the end of the trip. It is this imbalance in the ratio of the observed rates by each twin of the other twin's flashes that accounts for the difference in the total count that each one makes of his twin's flashes and thus the amount that each one has aged during the trip. Look up Relative Doppler for more details. |
| Sep9-11, 03:33 AM | #54 |
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It makes more sense to say that it is proportional to the light shift, because the light shift represents the distance. I cannot also imagine that the light from a moving in circle object will be red shifted. It is probably an error. Since we do the measurements, the light travels in our frame of reference and since there is no difference in the distance, it MUST not show shift to red or blue, otherwise we will observe a paradox. The red shift claim for circular moving object is very much like the explanation of the ladder paradox; it says that although both doors open simultaneously the ladder will see that the closer door opens first. That's wrong, because the ladder sees the doors as light reflection and since the back door opens simultaneously with the front door, there will not be door at the back, to reflect a ligh which will travel to the lader and show a presence of a door. To imagine it in easy way, we can put light source on the inside of the back door, and the switch to turn the light on we place on the front door when it opens completely. If they open simultaneously, the ladder will never see the light beam from the back door, because new paradox will emerge; we will have two light beams, one directed to the ladder and one perpendicular down from the opened back door. What do you say about that, DaleSpam? Same is with the red shift of the light from a moving in circle clock. If we say that we observe red shift, the clock will still see its own reflection from a mirror on "our side" as not shifted in any direction light. I put "Our side" between quotation marks, because we are actually a mirror in the clock's frame of reference. In that case we will end up with two different wave lengths for same light, measured on same distance in two frames of reference. |
| Sep9-11, 06:27 AM | #55 |
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Or it would be even simpler if we take out the whole turnaround event. Since this is purely theoretical example we can put an event of instant turnaround where the moving brother goes through one portal and comes out another at exactly the same distance but moving in reverse direction (towards Earth). That way the Earth bound brother won't have the turnaround event to mess things up for him. What happens in that situation then? DaleSpam i opened that page you linked, and i still find it difficult to agree with the explanations. I take into consideration the fact that i don't understand most of the math in there, but still... the example that you prefer with the graph still takes into account only the POV of the earth bound and does its math only from there. I opened another explanation which thankfully didn't have any math in it and the way they used to explain it there was with the whole inertial frame missing from the one brother. But even there they write that a possibility of the moving brother not experiencing any acceleration and gravitational forces with an instant turnaround could be considered and the explanation of that was left to the "reader". I take that as a failure to explain the problem fully and explaining it only when certain conditions are met and others aren't. |
| Sep9-11, 07:54 AM | #56 |
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So the bottom line is that the traveler counts the half-rate flashes from the earth twin for the same amount of time that he counts the double-rate flashes, but the earth twin counts the low-rate flashes from the traveler for most of the trip and doesn't start counting the high-rate flashes until near the end so he ends up with a much smaller count than the traveler does. And remember, they are each counting the one-second flashes from the other twin. |
| Sep9-11, 08:16 AM | #57 |
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You'll eventually need to include the acceleration of the flying brother, but I don't see how that would resolve the time difference, because the brother on Earth (or in a motionless space craft) also sees himself as accelerating away. If the G-force during the acceleration helps somehow, then you will be right :) |
| Sep9-11, 11:30 AM | #58 |
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| Sep9-11, 11:35 AM | #59 |
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Earlier in this thread you have been asking about the light travel time but now you seem to believe that it is non-existent. Are you saying that when the traveler is far away from the earth twin, the earth twin will still see him change direction half way through the trip and he won't have to wait for the light from that action to reach him? If you believe this, then you are believing in action-at-a-distance or an infinite speed for light. Are you sure you want to maintain this position? |
| Sep9-11, 02:50 PM | #60 |
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Now try to imagine what happens to both brothers and you'll see no difference for both of them. One of them will do the real traveling, but the other will see himself as traveling the same path with the same acceleration, speed, turning around and arriving back. When the "brother in motion" turns back, he will keep seeing the flashes sent to him before he turned back, and some time latter he will start seeing the flashes sent after he turned around. Same for the other brother; he will see himself as turning around some time later after the "real turn" was made from his brother. I do not imply limitless light speed in this example, for there is no space for such suggestion. |
| Sep9-11, 02:55 PM | #61 |
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| Sep9-11, 03:27 PM | #62 |
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If you don't agree with this can you please explain where you think I'm wrong? |
| Sep9-11, 03:49 PM | #63 |
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I have a question about time too.
Sorry if this is a dumb question: if theres a moving train with a mirror in it, and an observer(guy1) inside the train shines a light into it and measures the time it takes for it to get to the mirror and come back. He will measure a shorter time than an observer(guy2) outside of the train that is standing still. So the observer(guy2) says the time goes slower inside the train because its moving, but can the observer(guy1) in the train not say that time is going slower for the guy2 because in guy1's restframe guy 2 is moving? if so can someone explain this to me? sorry if this is a dumb question, I am just curious |
| Sep9-11, 03:53 PM | #64 |
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You'll probably ask me how this change anything. Few posts earlier I asked you the question "what do we consider as light and how do we measure its speed". One photon is not light and its speed should not represent the speed of light. The speed of light should be represented by the frequency with which we meet every successive wave. The faster we move through the waves the faster we meet every next one thus changing the frequency, respectively the "speed of light". This way we will see that the speed of light depends on the speed of the observer and on the speed of the source. If you accept this, the theory of relativity will start to make no sense. The example with the two garage doors treats overlapping simultaneous events to which you cannot apply simultaneity for the simple reason that you cannot observe them as two events; you see only one of them and you should treat them as one. The fact that you KNOW about the second event does not make it present to work with it and to apply values to it for later use. Ignoring this will create paradoxes like the example I gave you with the two light beams from the back garage door. |
| Sep9-11, 04:16 PM | #65 |
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Sometime I also need to adjust the environment and the events to my way of thinking, and then I know that I'm somewhere wrong. If the departing brother measures x distance, the other will measure it as x as well, and if one of them measure n velocity it is n for the other as well. Both of them observe the other with the same delay since they are symmetrically placed from the point in time where the events started. |
| Sep9-11, 04:52 PM | #66 |
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Please study this page, you seem to be under some severe misconceptions about how things work: http://www.edu-observatory.org/physi...periments.html |
| Sep9-11, 05:04 PM | #67 |
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| Sep9-11, 05:39 PM | #68 |
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Pardon me, but the rest of your answer seams very much like a dodging. The fact that SR takes in to account "many things that are invariant under the Lorentz transform" does not answer how do you see the "overlapping simultaneous events" (the two garage doors) Not mentioning that my note puts in doubt an important part of SR. I guess I have to wait until someone else comes with explanation. I can imagine how annoying a guy like me can be in the midst of a company like you guys. Later these days I'll post a thought experiment with graphics and after your answer I'll stop bothering you :) I wish you a great weekend :) |
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