Twin Paradox

UPDATE: Data Does Not Match Special Relativity Time Dilation \| Open Letter On Twin Paradox \| Open Letter On Special Relativity \| Sign Open Letter \| Report \| Survey Questionnaire \| Mainstream Response \| Debate Methodology \| Can't Be Due To Relative Velocity \| Can't Be Due To Turnaround Acceleration \| Time Difference Accumulation Is Not Indeterminate \| Dingle's Question \| Students \| Blog Discussion \| Two Step Argument \| Three Step Argument \| Two Step Argument #2 \| UpdateIndex \| Help

VIEW: Home \| Data Does Not Match Special Relativity Time Dilation \| Open Letter On Twin Paradox \| Open Letter On Special Relativity \| Sign Open Letter \| Report \| Survey Questionnaire \| Mainstream Response \| Debate Methodology \| Can't Be Due To Relative Velocity \| Can't Be Due To Turnaround Acceleration \| Time Difference Accumulation Is Not Indeterminate \| Dingle's Question \| Students \| Blog Discussion \| Two Step Argument \| Three Step Argument \| Two Step Argument #2

Visit the TeacherWeb® videos to learn about the features, modifications, and customizations available on your website. Click here to learn more. Can't Be Due To Relative Velocity <blockquote style="margin-right: 0px;" dir="ltr"> <p><strong><u><font size="5">The Constant Velocity-Symmetric Effects Class</font></u></strong> </p> <p><font size="5">In section 4 of Einstein's seminal 1905 paper defining special relativity (A. Einstein "On the Electrodynamics of Moving Bodies," Annalen der Physik <strong>17</strong>, 891 (1905) - as translated by Saha and Bose), Einstein describes a Twin Paradox scenario and writes:</font></p> <dir> <dir> <p><font size="5">"From this, the following peculiar consequence follows. Suppose at two points A and B of the stationary system two clocks are given, which are synchronous in the sense explained in 3 when viewed from the stationary system. Suppose the clock at A to be set in motion in the line joining it with B, then after the arrival of the clock at B, they will no longer be found synchronous, but the clock which was set in motion from A will lag behind the clock which had been all along at B by an amount 1/2t(v/c)<sup>2</sup>, where t is the time required for the journey." </font></p> </dir></dir> <p><font size="5">Hardly a rigorous proof – it’s just a stated conclusion. What was the "From this" that Einstein was basing his "peculiar consequence"? Einstein had just derived the time dilation equation applicable for all inertial observers. It seems that, at best, Einstein left out some steps in going from "all inertial observers <em>observe</em> time dilation" to "a net time difference in <em>proper</em> time for one inertial observer versus another inertial observer".</font></p> <p><font size="5">Einstein's "explaining" the net proper time difference as a function of relative velocity for the constant velocity segments is the primary member of this class of reconciliation argument. However, the explanation leads to a rather glaring contradiction. Many proponents of the "there is no paradox" group recognized this flaw and defined a whole array of alternate explanations that were logically equivalent but just more obscure. These arguments include all explanations where the net proper time difference is alleged to ALWAYS accumulate evenly during the constant velocity segments with half accumulating in the outbound segment and half in the inbound segment and to accumulate smoothly during these constant velocity segments - such arguments include some employing relative simultaneity. All these reconciliation arguments can be refuted using the Methodology and logic described below.</font></p> <p><font size="5"><u><strong>Key Questions/Points For This Class of Arguments</strong></u></font></p> <p><font size="5">By asking questions of the proponent of this class of reconciliation argument, we establish that they hold or agree that:</font></p> <p><font size="5">1a) All clocks in a given inertial frame (e.g., the R-frame) tick at the same rate - i.e., accumulate proper time at the same rate.</font></p> <p><font size="5">1b) As a corollary to 1a) above, if a clock in an inertial frame accumulates a net proper time difference versus another clock in another frame, then it also accumulates the same net proper time difference versus all other clocks in that other inertial frame. For example, if S's clock during the constant velocity outbound segment accumulates a net time difference of 50 billion ticks versus the stay-at-home twin's clock at rest in R, then it must also have accumulated a net time difference of 50 billion ticks versus all clocks at rest in R as all those clocks are ticking in unison.   </font></p> <p><font size="5">2) For a Twin Paradox scenario with the outbound relative velocity equal to the inbound relative velocity, half the net proper time difference is claimed to accumulate in the outbound constant velocity segment and half the net proper time difference is claimed to accumulate in the inbound constant velocity segment in this class of reconciliation argument.</font></p> <p><font size="5">3) The net proper time difference is claimed to accumulate smoothly during both constant velocity segments. In other words, if the outbound constant velocity segment is held to account for the accumulation of 50 billion ticks of net proper time difference, any sub-segment that's <strong>1/10th</strong> the whole segment would have to account for the accumulation of 5 billion (i.e., 1/10th of 50 billion) ticks of net proper time difference.</font></p> <p><font size="5"><strong><u>The Numbers</u></strong></font></p> <p><font size="5">Given the base parameters for the specific scenario defined in the <a href="http://worknotes.com/Physics/SpecialRelativity/TwinParadox/apt2.aspx" target="_blank">Methodology</a> section, the proponent of this class of reconciliation argument would fill in the proper time intervals for the various segments of S's round trip as follows:</font></p> <p><strong><u><font size="5"> <table style="border: 1px dotted #7f7c75;" border="0" cellspacing="0" cellpadding="2" width="425"> <tbody> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><strong><u><font size="2" face="Arial"> </font></u></strong> <p align="left"><strong><u><font size="2" face="Arial">Segment on S's Worldline </font></u></strong><u><font size="2" face="Arial"></font></u><font size="2" face="Arial"></font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><strong><u><font size="2" face="Arial"> </font></u></strong> <p align="right"><strong><u><font size="2" face="Arial"># of Proper Time Ticks</font></u></strong><u><font size="2" face="Arial"></font></u><font size="2" face="Arial"></font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E1-E1a (Initial Acceleration)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">500</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E1a-E2 (Constant velocity outbound)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">(50,000,000,000 - 1,000)</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E2-E2a (Turnaround Acceleration - Part 1)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">500</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E2a-E3 (Turnaround Acceleration - Part 2)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">500</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E3-E4a (Constant velocity inbound)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">(50,000,000,000 - 1,000)</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E4a-E4 (Final Deceleration)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">500</font></p> </td> </tr> </tbody> </table> </font></u></strong></p> <p><font size="5"> <table style="border: 1px dotted #7f7c75;" border="0" cellspacing="0" cellpadding="2" width="425"> <tbody> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><strong><u><font size="2" face="Arial"> </font></u></strong> <p align="left"><strong><u><font size="2" face="Arial">R's Worldline </font></u></strong><u><font size="2" face="Arial"></font></u><font size="2" face="Arial"></font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><strong><u><font size="2" face="Arial"> </font></u></strong> <p align="right"><strong><u><font size="2" face="Arial"># of Proper Time Ticks</font></u></strong><u><font size="2" face="Arial"></font></u><font size="2" face="Arial"></font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" height="17" width="61%"><font size="2" face="Arial"> </font> <p align="left"><font size="2" face="Arial">E1-E4 (Whole scenario)</font></p> </td> <td style="border: 1px dotted #7f7c75;" height="17" width="39%"><font size="2" face="Arial"> </font> <p align="right"><font size="2" face="Arial">200,000,000,000</font></p> </td> </tr> </tbody> </table> </font></p> <p><font size="5">The 4 acceleration segments for S,  E1 - E1a,  E2 - E2a, E3 - E4a and E4a - E4, are equal and are defined as 500 proper time ticks for S. For this class of reconciliation argument, the focus is on the constant velocity segments so we will use round numbers and talk of those segments having 50 billion ticks instead of the more precise "50 billion minus 1,000 ticks".</font></p> <p><font size="5">Below we only deal with the proper time associated with the above worldline segments for S and with the whole worldline for R as described/agreed to by proponents of the argument themselves. </font></p> <p><font size="5"><strong><u>The Problems</u></strong></font></p> <p><font size="5"><strong style="font-weight: bold;">Problem 1):</strong> Einstein's original claim that the net proper time difference was due to time dilation was immediately met by the critics' question, "<em>How can a symmetric cause, time dilation as a function of relative velocity, produce an asymmetric effect?</em>" Hence, the critics claimed that from the traveling twin's point of view it should have been the stay-at-home twin who lost proper time and aged less and, hence, that claim and similar claims lead to a contradiction and were invalid. <br /><br />In turn, the relativists argued that because the traveling twin accelerated and Special Relativity didn't cover  acceleration, the traveling twin's view was totally invalid. The fact that this argument was accepted is very revealing and shows a tendency to accept any remotely plausible argument for Special Relativity while not seriously considering any argument that questioned any aspect of Special Relativity. A little thought would have shown that accepting the relativists' argument and applying it consistently would rule out using Special Relativity for any purpose. For example, all earth labs are rotating/accelerating with the earth and most/all instruments and/or observers have accelerated or will accelerate in the future. Hence, claiming that the traveling twin cannot use Special Relativity even for the constant velocity segments because he has or will accelerate does not resolve the quoted problem above that was raised by the critics, but rather it creates a worse problem.<br /><br /><strong>Problem 2):</strong> If one ignores Problem 1 above and just looks at a single Twin Paradox scenario, everything may seem fine with this reconciliation argument. However, if one adds a second Twin Paradox scenario where S plays the role of the stay-at-home twin and the whole 2nd scenario begins and ends during S's constant velocity outbound segment and the traveling twin, in this 2nd scenario, is at rest in R in his outbound segment, then this reconciliation argument leads to contradictory conclusions. On the one hand, when this class of reconciliation argument is applied to the original Twin Paradox scenario, it claims that S <strong><font color="#ff0000">loses</font></strong> net proper time versus all clocks at rest in R for the segment E1a-E2 including all its sub-segments. On the other hand, when this class of reconciliation argument is applied to the 2nd, nested Twin Paradox scenario, it claims that S <strong><font color="#0000ff">gains</font></strong> net proper time for at least one sub-segment of E1a-E2 versus all the clocks at rest in R.  A more detailed discussion of this point is given below. </font></p> <p><font size="5">In the specific scenario defined in the <a href="http://worknotes.com/Physics/SpecialRelativity/TwinParadox/apt2.aspx" target="_blank">Methodology</a> section, R, the stay-at-home twin accumulates 200 billion ticks between E1 and E4, the start and the end of S's round trip. And S accumulates 100 billion ticks between E1 and E4. According to the proponents of this class of reconciliation argument, half of that total net proper time difference, i.e., 50 billion ticks, accumulated during the constant velocity segment of the outbound journey (E1a-E2). </font></p> <p><font size="5">Furthermore, by "Key Point 1b)" above, this implies that S accumulates a net proper time difference of 50 billion ticks versus all clocks at rest in R during the constant velocity outbound segment and further, this accumulation occurs smoothly and evenly for each sub-segment of that constant velocity outbound segment.</font></p> <p><font size="5">Now we'll construct a 2nd, nested Twin Paradox scenario that begins at E1a and ends with E2. We'll have a new observer, R1, who starts next to S at rest in the S-frame as S ends his initial acceleration (E1a). R1 starts off on his own (R1's own) Twin Paradox scenario round trip. R1's constant velocity segments are done at 0.866c with respect to the S-frame and done so that for R1's outbound constant velocity segment, R1 is at rest in the R-frame. R1 completes his round trip and comes to rest in S again just as S starts his turnaround acceleration (E2) for the original scenario.</font></p> <p><font size="5">S accumulates 50 billion ticks between E1a and E2. Hence, consistently applying the reconciliation argument, R1 accumulates 25 billion ticks for his round trip. From the previous paragraph, we know that S is smoothly and evenly losing proper time ticks for the whole E1a-E2 segment for all clocks at rest in R, including R1 during R1's constant velocity outbound segment since for that segment, R1's at rest in R. Also, consistently applying this class of reconciliation argument, we must conclude that for the nested scenario, while R1 is at rest in R during his constant velocity outbound segment, R1 is smoothly and evenly losing proper time ticks versus S. Thus, this class of reconciliation argument leads to two contradictory conclusions. </font></p> <p><font size="5">This class of reconciliation argument cannot be the general solution that explains how the net proper time difference accumulates.</font></p> <p><font size="5"><strong>Problem 3): </strong>The above should be sufficient. Further, we know from experience that giving multiple arguments can lead the proponent to think, for example, "<em>I can't address problem 1, however, I don't find argument 2 convincing, therefore, I'll dismiss both.</em>" Despite this point, some other problems with this class of reconciliation argument might be thought provoking for some and even help show what's fundamentally wrong with this class of reconciliation argument and point the reader to the correct solution. So here goes.</font></p> <p><font size="5">We'll reference the Twin Paradox scenario defined in the <a href="http://worknotes.com/Physics/SpecialRelativity/TwinParadox/apt2.aspx" target="_blank">Methodology</a>. However, we'll modify this classic scenario such that S does not start at rest in R, but rather the scenario begins with S, already traveling at relative velocity v with respect to R, having an arbitrarily near miss with R at what we had called E1a. The rest of the scenario continues as defined.</font></p> <p><font size="5">However, let's say that when S gets to the space-time point previously marked by E2, a decision is randomly made as to whether S or R will be the traveling twin. In this modified scenario, at the space-time point previously marked by E2, there's complete relativistic symmetry between S and R as both will have been at rest in an inertial frame the whole time.</font></p> <p><font size="5"> If it's decided that S is the traveling twin, then, at E2, S commences his turnaround acceleration. The constant velocity class of reconciliation argument holds that 1/2 the net time difference has accumulated by E2 - in other words, by E2, S has accumulated a net proper time difference of <strong><font color="#ff0000">minus</font></strong> 50 billion ticks versus R.</font></p> <p><font size="5">However, if it's decided that R will be the traveling twin, then, the constant velocity class of reconciliation argument holds that a very different net time difference has accumulated at the space-time point previously marked by E2 - in other words, at that point, S has accumulated a net proper time difference of <strong><font color="#0000ff">plus</font></strong> 50 billion ticks versus R. This is a contradiction.<br /><br />[Note that all of the above arguments not only show that Special Relativity's time dilation cannot be used to explain the net proper time difference, but also show that Special Relativity's time dilation cannot be  describing proper time or what's happening physically. (For further discussion on that topic, see <a href="htmlpage7.aspx" target="_self">Dingle's Question page</a>.) The above also shows that no symmetric construct can be describing asymmetric physical changes.]  </font></p> </blockquote> <p><font size="3">© 2004, 2012 </font></p> Automatically clean input to optimize for the web. Or, would you like to view and optionally restore one of these previous versions? Password: Forgot your password?

Twin Paradox

Can't Be Due To Relative Velocity

©2013 TeacherWeb, Inc.