apologies if this has been posted before. There are three triplets, Dick, Mick and their idiot brother Nick. Dick and Mick get into identical rocketships. In Nick's "rest" frame, they both jet off at the same time, travel the same distance in opposite directions, then return at the same time. the paths taken are linear. So to Nick, both would have experienced equal time dilation. However, wouldnt Dick and Mick see their other two siblings as having experienced time dilation instead? How do we answer this paradox?
Yes, if each sees theirself as at rest, they will measure the others to be time dilated differently. The solution is the same as for the more traditional hypothetical, both Dick and Mick experience accelerations and changes in their inertial frames of reference. Though in SR it is the difference in velocities that time dilates clocks, it is the accelerations that sets the clocks apart from oneanother, and establishes which were actually moving relative to their common initial frame of reference. Which of the three experiences acceleration during the hypothetical tells us which were also changing frames, and whose clocks would ultimately be time dilated, relative to their initially synchronized frame of reference.
If Tom's clock shows total elapsed proper time of \(T\) when Dick and Mick are reunited then Dick and Mick's clocks will show total elapsed proper time of: \(\tau_D=\tau_M=\int_0^T {\sqrt{1-(v(t)/c)^2}dt}\)
This Non-paradox is answered in the same way as one answers the original twin paradox: By taking into account the entirety of SR, including length contraction and the relativity of simultaneity, and by realizing that Dick and Mick do not remain in the same inertial frames throughout the scenario.
okay, let's say i replace Tom with their triplet, Nick. Whats the end result when all 3 are reunited? Also, how does acceleration affect time dilation?
Let's not. Did you get the explanation I gave you? What did you understand? Do you see the \(v(t)\) in the formula? What does it tell you?
Acceleration is only a part of it. Length contraction is the part you want to focus on and why length contraction will clarify to you how there is no paradox.
Emil, Chan Rasjid, Herbert Dingle, and now you, eram, have all made the same mistake of trying to use a toy, mathless, philosophically inspired distortion of special relativity instead of actually using special relativity. As such, your predictions not only sometimes disagree with the predictions of special relativity, they sometimes have blatant logical inconsistencies -- Chan Rasjid and Herbert Dingle compounded their error by blaming special relativity when in fact the problem was with their failure to learn the subject that they were criticizing.
If Nick takes off from the start point and returns the same way his brothers did, they will all be the same age when reunited. One way to look at this is the relativity of simultaneity approach: Imagine you are traveling from Clock A to Clock B. In their own rest frame Clocks A&B are synchronized. In your frame, they are not, Clock B reads a later time than clock A. As you travel from one to the other, both clocks run slow compared to your own. So both clocks accumulate less time than yours during the the trip. For example. we'll assume you left Clock A when it read 12:00 and it takes an hour by your clock to reach Clock B, while Clocks A and B advance by 1/2 hr. However, As stated above, in your frame, Clock B reads later than Clock A in your frame (1 1/2 hr later in this example). So when you leave clock A when both it and your clock read 12:00, Clock B reads 1:30. 1/2 hr accumulates on A and B by the time you reach Clock B, so now clock A reads 12:30 and Clock B reads 2:00 while your clock reads 1:00. You accelerate to reverse direction and head back to clock A. (we'll assume that this happen in a short enough time that you remain in the vicinity of Clock B.) You have now switched to a new inertial frame. In this new frame, it is Clock A that reads later than Clock B. Thus, when you leave clock B with it still reading 2:00, Clock A now reads 3:30. CLock A runs half as fast as your clock during the return trip and accumulates 1/2 hr, reading 4:00 upon your return while your clock reads 2:00. You have aged 2 hrs during the trip while 4 hrs have passed on Clock A. Of course, in the rest frame of A and B, it just took 4 hrs for you to make the round trip while your clock ran 1/2 as fast as theirs and accumulated 2 hrs.
Why would switching Tom with someone else change the result? He's just a clock. Replace all three twins with clocks if you wish. Tom's clock will progress the most because he remains inertial. Generally, the clock which experiences the 'most' acceleration progresses the least. (Some may have a problem with this characterization but it's a great starting point if you're trying to develop an SR intuition)
Alice makes a round-trip over a distance L in a certain time T at constant speed and with zero time spent at the turnaround. (Both L and T being measured in a certain inertial frame). Bob, who is on a trip of constant proper acceleration, passes Alice at both that start and end of her round-trip. By chance or fate, Alice's and Bob's measurement of proper time for their respective round-trips is exactly the same. Question 6: What is the magnitude of Bob's proper acceleration as a function of L and T? Question 7: At what speed (measured in the "certain inertial frame") does Bob pass Alice's start and end point? Question 8: How far (measured in the "certain inertial frame") is Bob's turnaround point relative to Alice's start and end point? Question 9: How does this help settle the dispute between RJ Beery and Tach?
Off-the-cuff, your scenario is problematic because Alice experiences infinite acceleration on the turnaround. I like the idea, though, we should pursue it. Note I put the word most in quotes to account for this sort of analysis...
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heh only Dingle didn't visit SciForums. The way you bring it up is as though they're all my friends lol. Hmmm, and I didn't make any predictions. Just asking what the result would be. *sees wall of math* no wonder they failed to learn it properly. Nevermind, I'll try my best.
So according to Tach's equation I know how much dilated time has elapsed, and it is also due to the changing of inertial frames. Does this mean that Dick (in his rocket ship) doesn't always see his two other triplets as undergoing time dilation? Will there be some form of time compression during the turnaround? does Bob make a turnaround as well? question 9 is a tricky one. Edit: What do I mean by time compression? Let's say for Mick, he was the same age as Rick and Dick at the start. At their reunion, he finds that Dick is as old as him and Rick is older. So if Rick was the same age and now became older, he must have aged faster, some form of time compression; which is the reverse of time dilation, sort of like fast-forwarding a video. Apologies if I happen to be testing you guys' patience, I'm taking quite a while to absorb this.
thanks. Neverfly told me that there a lot of old men on these forums, many of whom have attained a high level of education, so I should show some respect.
