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                  關于狹義相對論鐘慢效應的研究
                  2021-03-22 16:28:56   來源:互聯網
                  內容摘要
                  作者:王行洪(原文英文版論文發表于【國際科研與現代教育】期刊,DOI為: 10.5281/zenodo.4618575)摘要Abstract:This article discusses time dilation based on special relativity. An experiment…

                  作者:王行洪

                  (原文英文版論文發表于【國際科研與現代教育】期刊,DOI為: 10.5281/zenodo.4618575)

                  摘要Abstract:

                  This article discusses time dilation based on special relativity. An experiment is designed to examine time dilation effects deduced from special relativity. This experiment is similar to but different from the famous “twin paradox” because this experiment doesn’t involve any accelerating process, or decelerating process, or change of travelling directions. Impossible results of the experiment are deduced by using special relativity. Thus it can be concluded that the theory of relativity (special relativity) is incorrect.

                  本文提出一個思想實驗,用來研究狹義相對論引起的鐘慢效應。

                  本實驗與“雙生子佯謬”有所類似但又不同。因為,在本實驗中,實驗對象僅以慣性作勻速直線運動,并不涉及加速過程,也不涉及減速過程,也不涉及轉向過程。實驗對象為太空中的1、2、3三艘飛船,其中飛船1與飛船2相距離若干光年遠并保持相互靜止狀態,其上面的時鐘互相同步。飛船3相對于飛船1和飛船2勻速直線運動,先后與飛船2和飛船1在太空中擦肩而過。本實驗根據狹義相對論,考察此過程中三艘飛船中的時間流逝情況,進行對比后,發現相互矛盾的結果。由此證明,狹義相對論的鐘慢效應是錯誤的。這也說明,狹義相對論是不正確的。

                  Main Text正文:

                  In my previous article [A DISCUSSION ABOUT SPECIAL RELATIVITY], an experiment is designed, by using theory of relativity to deduce the experiment results, to prove that the theory of relativity (special relativity) is incorrect. This article aims to further prove it by looking into the time dilation effect based on special relativity.

                  在本人的上一篇文章【狹義相對論研究】中,設計了一個實驗,運用狹義相對論的理論,分析實驗結果,得出了矛盾的結論,由此證明了狹義相對論的錯誤。本文則著眼于狹義相對論引起的鐘慢效應,進一步對狹義相對論進行證偽。

                  Special relativity?indicates that, for an observer in an?inertial frame of reference, a clock that is moving relative to them will be measured to tick slower than a clock that is at rest in their frame of reference. This case is sometimes called special relativistic time dilation. The faster the?relative velocity, the greater the time dilation between one another, as described by equation in shown by Figure 1:

                  狹義相對論表明,對于任一慣性參考系中的觀察者來說,相對于他們移動的時鐘要比其自身參考系中靜止的時鐘慢。這種情況有時被稱為相對論時間膨脹。相對速度越快,彼此之間的時間膨脹就越大,具體關系如圖1所示的方程式所示:

                  IMG_256

                  圖1

                  According to this theory, time dilation would make it possible for passengers in a fast-moving vehicle to advance further into the future in a short period of their own time. For sufficiently high speeds, the effect is dramatic. For example, one year of travel might correspond to ten years on Earth. Indeed, a constant 1?g?acceleration would permit humans to travel through?the entire known Universe?in one human lifetime.

                  根據該理論,時間膨脹將使快速行駛的飛船中的乘客有可能在其自己的短時間內更快的駛向未來。而且如果相對速度足夠高,其效果是驚人的。例如,飛船中的人旅行一年可能相當于地球上的十年。根據計算,如果使飛船保持恒定的1 g加速度,將使人類在一個人的一生中就能夠穿越整個已知的宇宙。

                  But is it really so?

                  Now we design an experiment:

                  事實真的是這樣嗎?讓我們設計如下這樣一個實驗來進行分析檢驗。

                  At point A, there are 2 spacecrafts (spacecraft 1 and spacecraft 2), both equipped with accurate and synchronized atomic clocks. As shown by Figure 2.

                  在A點,有2個航天器(航天器1和航天器2),均裝有同步的精確原子鐘。如圖2所示。

                  IMG_257

                  圖2

                  And the two spacecrafts (spacecraft 1 and spacecraft 2) start to travel at the same speeds to point B and point C, whereas point B and point C are several light years apart. The distance between point A and point C is the same as the distance between point A and point B. After they arrive at point B and point C. They stop there, static to each other, as shown by Figure 3. According to theory of relativity, we can know that the clocks on the spacecraft 1 and spacecraft 2 are still synchronized and will always be well synchronized as long as spacecraft 1 and spacecraft 2 stay static to each other.

                  這兩個航天器(航天器1和航天器2)以相同的速度行進到B點和C點,而B點和C點相距數光年。 A點和C點之間的距離與A點和B點之間的距離相同。到達B點和C點后,它們停在那里,彼此保持相對靜止,如圖3所示。根據相對論,我們可以知道,航天器1和航天器2上的時鐘仍然是同步的,只要航天器1和航天器2彼此保持靜態,這兩個航天器上的時鐘就會一直保持很好地同步。

                  IMG_258

                  圖3

                  Then, there is a third spacecraft (spacecraft 3) first passing by spacecraft 2(point C) and then passing by spacecraft 1(point B) at a constant high speed, as shown by Figure 4.

                  然后,有第三個航天器(航天器3)以恒定的速度首先經過航天器2(C點),然后經過航天器1(B點),如圖4所示。

                  IMG_259

                  圖4

                  Theory of relativity (special relativity) tells us that, according to the astronaut on spacecraft 3, spacecraft 3 is static while spacecraft 2 and spacecraft 1 are passing by at a constant high speed, as shown by Figure 5. Because we consider that the location of spacecraft 1 is point B and the location of spacecraft 2 is point C, so we can as well say that spacecraft 3 is static while point C and point B are passing by at a constant high speed.

                  相對論(狹義相對論)告訴我們,對于航天器3的宇航員所在的慣性參照系來說,航天器3是靜止的,而航天器2和航天器1則進行勻速直線運動,以恒定的速度經過航天器3,如圖5所示。因為我們認為航天器1的位置是B點,航天器2的位置是C點,所以我們還可以說航天器3是靜止的,而點C和點B進行勻速直線運動,以恒定的速度經過航天器3。

                  IMG_260

                  圖5

                  When spacecraft 3 and spacecraft 2 meet at point C, as shown by Figure 6, astronaut on spacecraft 2 sends a signal to astronaut on spacecraft 3 to tell him the exact time. Then the astronaut on spacecraft 3 will use this information to set the atomic clock on spacecraft 3 to be synchronized with atomic clock on spacecraft 2. At the same time, the astronaut on spacecraft 2 will also send a signal to astronaut on spacecraft 1 to tell him when spacecraft 2 and spacecraft 3 meet. During the whole process, the relative speed of the spacecrafts doesn’t change, i.e., there is neither accelerating process nor decelerating process. So when spacecrafts meet, they just pass by each other without slowing down.

                  For spacecraft 3, the time when spacecraft 2 and spacecraft 3 meet at point C, the time is t3C.

                  For spacecraft 2, the time when spacecraft 2 and spacecraft 3 meet at point C, the time is t2C.

                  For spacecraft 1, the time when spacecraft 2 and spacecraft 3 meet at point C, the time is t1C.

                  Apparently, ?t1C=t2C=t3C.

                  如圖6所示,當航天器3和航天器2在C點擦肩而過時,航天器2上的宇航員向航天器3上的宇航員發送信號,以告知他確切的時間。然后,航天器3上的宇航員將使用此信息將航天器3上的原子鐘設置為與航天器2上的原子鐘同步。同時,航天器2上的宇航員還將向航天器1上的宇航員發送信號,以告知其航天器2和航天器3相遇時的準確時間。在整個過程中,航天器的相對速度不改變,即:既沒有加速過程也沒有減速過程,仍然保持勻速直線運動。

                  對于航天器3,航天器2和航天器3在C點相遇的時間為t3C

                  對于航天器2,航天器2和航天器3在C點相遇的時間為t2C

                  對于航天器1,航天器2和航天器3在C點相遇的時間為t1C

                  顯然,t1C = t2C = t3C

                  IMG_261

                  圖6

                  Then, when spacecraft 3 finally meet spacecraft 1 located at point B, as shown by Figure 7:

                  For spacecraft 3, the time when spacecraft 1 and spacecraft 3 meet, the time is t3B.

                  For spacecraft 2, the time when spacecraft 1 and spacecraft 3 meet, the time is t2B.

                  For spacecraft 1, the time when spacecraft 1 and spacecraft 3 meet, the time is t1B.

                  然后,當航天器3最終遇到位于點B的航天器1時,如圖7所示:

                  對于航天器3,航天器1和航天器3相遇的時間為t3B

                  對于航天器2,航天器1和航天器3相遇的時間為t2B

                  對于航天器1,航天器1和航天器3相遇的時間為t1B

                  IMG_262

                  圖7

                  Theory of relativity (special relativity) tells us:

                  According to astronaut on spacecraft 3, spacecraft 3 is static while spacecraft 2 and spacecraft 1 is moving, so the time passes much slower on spacecraft 1 and spacecraft 2 than on spacecraft 3. That is, ?t1B-t1C=t2B-t2C3B-t3C.

                  Because t1C=t2C=t3C, it is deduced that:

                  t1B=t2B3B

                  According to astronaut on spacecraft 1 and spacecraft 2, spacecraft 1 and spacecraft 2 are static while spacecraft 3 is moving, so the time passes much slower on spacecraft 3 than on spacecraft 1 and spacecraft 2. That is, ?t1B-t1C=t2B-t2C>t3B-t3C

                  Because t1C=t2C=t3C, it is deduced that:

                  t1B=t2B>t3B

                  Thus, there is now a contradictory result: ?t3B1B=t2B3B (t3B-t3C1B-t1C=t2B-t2C3B-t3C)

                  Because t1B, t2B, and t3B are all time readings. And t3B-t3C, t1B-t1C, and t2B-t2C are each a number(For example, they can be 1000 minutes, or 21 years, etc). ?So it is apparently impossible that t3B1B=t2B3B (t3B-t3C1B-t1C=t2B-t2C3B-t3C).

                  相對論(狹義相對論)告訴我們:

                  ?從航天器3上的宇航員的角度來看,航天器2和航天器1在移動,航天器3是靜止的,因此航天器1和航天器2上的時間要比航天器3上的時間慢。也就是說,

                  t1B-t1C=t2B-t2C3B-t3C.

                  由于t1C=t2C=t3C, 因此可以得出:

                  t1B=t2B3B

                  ?從航天器1和航天器2上的宇航員的角度來看,航天器1和航天器2是靜止的,航天器3是移動的。因此航天器3上的時間比航天器1和航天器2上的時間慢。也就是說,

                  t1B-t1C=t2B-t2C>t3B-t3C

                  由于t1C=t2C=t3C, 因此可以得出:

                  t1B=t2B>t3B

                  因此,出現了一個矛盾的結果: ?t3B1B=t2B3B (t3B-t3C1B-t1C=t2B-t2C3B-t3C)

                  因為t1B,t2B和t3B都是時間讀數。 t3B-t3C,t1B-t1C和t2B-t2C分別是一個數字(例如,可以是1000分鐘或21年,等等)。因此,t3B 1B = t2B 3B(t3B-t3C 1B-t1C = t2B-t2C 3B-t3C)顯然是不可能的。

                  Practically, when spacecraft 3 and spacecraft 1 meet, if the astronaut on spacecraft 3 sends a picture of his own clock reading to the astronaut on the spacecraft 1 and the astronaut on spacecraft 1 sends a picture of his own clock reading to the astronaut on the spacecraft 3, both of the astronaut expect to receive a picture showing time reading smaller than his own time reading. This cannot happen in reality.

                  實際上,當航天器3和航天器1相遇時,如果航天器3上的宇航員拍攝一張自己的原子鐘讀數的照片并立即發給航天器1上的宇航員,同時,航天器1上的宇航員也拍攝一張自己的原子鐘讀數的照片并立即發給航天器3上的宇航員,則兩位宇航員都將會各收到一張照片,(只要飛船速度較快)收到的照片顯示的時間讀數遠小于自己的原子鐘的時間讀數。這種情況,顯然不可能在現實中發生。

                  To make the experiment more interesting, we can modify the experiment as follows:

                  Based on the above experiment, when spacecraft 1 and spacecraft 2 are both in location A, they make a schedule. According to this schedule, each year, a new baby will be born in each spacecraft. And the babies born in spacecraft 1 and spacecraft 2 born in the same year will have the same name. When spacecraft 3 meet spacecraft 2, a baby is also born in spacecraft 3. When astronaut on spacecraft 2 tells astronaut on spacecraft 3 the time reading for the setting of clock, the name of the baby is also sent so that the baby born in spacecraft 3 is also named the same name. For example, according to the schedule, the babies born in the year when spacecraft 3 and spacecraft 2 meet is named as Adam. Then there is one person named Adam on each of the 3 spacecrafts.

                  為了使實驗更加有趣,我們可以按以下方式修改實驗內容:

                  基于以上實驗,當航天器1和航天器2都位于點A時(如圖2所示),他們共同制定一份計劃。按照這個計劃,每年,每個航天器上都會出生一個新的嬰兒。并且,根據這個計劃,同一年在航天器1上出生的嬰兒和航天器2上出生的嬰兒將被命名為相同的名字。當航天器3與航天器2相遇時,在航天器3中也出生了一個嬰兒。當航天器2上的宇航員告訴航天器3上的宇航員其原子鐘的時間讀數時,還向他發送了嬰兒的名字,以便在航天器3中出生的嬰兒也被命名為相同的名字。例如,根據該計劃,在航天器3和航天器2相遇的那年出生的嬰兒應該被命名為王健。這樣的話,這三個航天器中將各有一個名叫王健的人。

                  And as said before, when spacecraft 3 and spacecraft 2 meet, the astronaut on spacecraft 2 will also send a signal to spacecraft 1 to tell him when spacecraft 2 and spacecraft 3 meet. So astronaut on spacecraft 1 knows when did spacecraft 3 and spacecraft 2 meet and that the name of the newborn boy in spacecraft 3 is Adam.

                  如前所述,當航天器3和航天器2相遇時,航天器2上的宇航員還將向航天器1發送信號,以告知他航天器2和航天器3相遇的時間,在此同時,航天器2中的宇航員也告知航天器1中的宇航員:在航天器3和航天器2相會時,航天器3中新出生一名嬰兒,名字叫王健。

                  Now we suppose that the relative speed of the spacecrafts is 90% of speed of light in vacuum and the distance between point B and point C is 18 light years. Based on theory of relativity, at the speed of 90% speed of light, the time dilation rate is approximately 2.3. Then:

                  現在我們假設,航天器的相對速度是真空中光速的90%,而B點和C點之間的距離是18光年。根據相對論,在速度為光速90%的情況下,時間膨脹率約為2.3。那么:

                  According to astronaut on spacecraft 3, spacecraft 3 is static while spacecraft 2 and spacecraft 1, with a distance of 18 light years, is moving at the speed of 90% speed of light towards spacecraft 3.

                  so when spacecraft 3 meets spacecraft 1, Adam on spacecraft 3 is 20 years old. Because of time dilation effect, the Adam on spacecraft 1 will be 20/2.3=8.7 years old.

                  According to astronaut on spacecraft 1, spacecraft 1 and spacecraft 2 are static while spacecraft 3 is moving at the speed of 90% speed of light towards spacecraft 1. And the distance between spacecraft 1 and spacecraft 2 is 18 light years.

                  so when spacecraft 3 meets spacecraft 1, Adam on spacecraft 1 is 20 years old. Because of time dilation effect, the Adam on spacecraft 3 will be 20/2.3=8.7 years old.

                  ?從航天器3上的宇航員的角度來看,航天器3是靜止的,而航天器2和航天器1的距離為18光年,它們以真空中光速90%的速度勻速向航天器3移動。

                  因此,當航天器3與航天器1相遇時,航天器3上的王健已經20歲了。由于時間膨脹效應,航天器1上的王健將是20 / 2.3 = 8.7歲。

                  ?從航天器1上的宇航員的角度來看,航天器1和航天器2都是靜止的,航天器2和航天器1的距離為18光年,航天器3以真空中光速90%的速度勻速向航天器1移動。

                  因此,當航天器3與航天器1相遇時,航天器1上的王健已經20歲了。由于時間膨脹效應,航天器3上的王健將是20 / 2.3 = 8.7歲。

                  When spacecraft 3 and spacecraft 1 meet, if the Adam on spacecraft 3 sends his own picture to the Adam on spacecraft 1, and the Adam on spacecraft 1 also sends his own picture to the Adam on spacecraft 3. What will happen?

                  According to above analysis based on theory of relativity (special relativity), what will happen is that, on each of the 2 spacecraft, a 20-year-old young man named Adam holds the picture of a 8.7-year-old boy named Adam.

                  This is apparently impossible.

                  當航天器3和航天器1相遇時,如果航天器3上的王健將自己的最新照片發送給航天器1上的王健,航天器1上的王健也將其自己的最新照片發送給航天器3上的王健,會發生什么情況呢?

                  根據相對論(狹義相對論)進行分析,將會發生的情況是:在這兩個航天器中,各有一個名叫王健的20歲年輕人拿著自己才收到的一個名叫王健的8.7歲男孩的照片。

                  這顯然是不可能的。

                  Or we can also suppose, when spacecraft 3 and spacecraft 1 meet, we let both the Adam on spacecraft 1 and the Adam on spacecraft 3 wave through the window to each other. What will happen?

                  20-year-old young man named Adam on spacecraft 1 waves to a 8.7-year-old boy named Adam on spacecraft 3?

                  20-year-old young man named Adam on spacecraft 3 waves to a 8.7-year-old boy named Adam on spacecraft 1?

                  20-year-old young man named Adam on spacecraft 1 waves to a 20-year-old boy named Adam on spacecraft 3?

                  8.7-year-old young man named Adam on spacecraft 1 waves to a 8.7-year-old boy named Adam on spacecraft 3?

                  ......

                  Well, at least we can conclude that the theory of relativity(special relativity) is incorrect.

                  或者我們也可以設定,當航天器3和航天器1相遇時,我們讓航天器1上的王健和航天器3上的王健彼此通過航天器的舷窗互相揮手致意。會發生什么?

                  一個在航天器1上名為王健的20歲年輕人與一個在航天器3上名為王健的8.7歲男孩互相揮手致意?

                  一個在航天器1上名為王健的8.7歲男孩與一個在航天器3上名為王健的20歲年輕人互相揮手致意?

                  一個在航天器1上名為王健的20歲年輕人與一個在航天器3上名為王健的20歲年輕人互相揮手致意?

                  一個在航天器1上名為王健的8.7歲男孩與一個在航天器3上名為王健的8.7歲男孩互相揮手致意?

                  ......

                  至少,我們現在可以得出結論,根據相對論(狹義相對論)推導出來的鐘慢(時間膨脹)效應是不正確的。那么,狹義相對論的理論也就是不正確的。

                  Conclusion結論:

                  Because the time dilation effect results in impossible situation, the theory of relativity (special relativity) is incorrect.

                  基于狹義相對論的鐘慢(時間膨脹)效應導致出現不可能的推論結果,因此,狹義相對論是不正確的。

                  References:

                  Einstein A. (1916), Relativity: The Special and General Theory?(Translation 1920), New York: H. Holt and Company.

                  Xinghong Wang, “A Discussion about Special Relativity”, International Journal of Scientific Research and Modern Education, Volume 6, Issue 1, Page Number 4-7, 2021.

                  Calder, Nigel (2006).?Magic Universe: A grand tour of modern science.?Oxford University Press. p.?378.?ISBN?978-0-19-280669-7.









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