2020必威
关于狭义相对论钟慢效应的研究
2021-03-22 16:28:56   来源:互联网
内容摘要
作者:王行洪(原文英文版论文发表于【国际科研与现代教育】期刊,DOI为: 10.5281/zenodo.4618575)摘要Abstract:This article discusses time dilation based on special relativity. An experiment…

2020必威 www.mizhidayu.com 作者:王行洪

(原文英文版论文发表于【国际科研与现代教育】期刊,DOI为: 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.

本文提出一个思想实验,用来研究狭义相对论引起的钟慢效应。

本实验与“双生子佯谬”有所类似但又不同。因为,在本实验中,实验对象仅以惯性作匀速直线运动,并不涉及加速过程,也不涉及减速过程,也不涉及转向过程。实验对象为太空中的1、2、3三艘飞船,其中飞船1与飞船2相距离若干光年远并保持相互静止状态,其上面的时钟互相同步。飞船3相对于飞船1和飞船2匀速直线运动,先后与飞船2和飞船1在太空中擦肩而过。本实验根据狭义相对论,考察此过程中三艘飞船中的时间流逝情况,进行对比后,发现相互矛盾的结果。由此证明,狭义相对论的钟慢效应是错误的。这也说明,狭义相对论是不正确的。

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【狭义相对论研究】中,设计了一个实验,运用狭义相对论的理论,分析实验结果,得出了矛盾的结论,由此证明了狭义相对论的错误。本文则着眼于狭义相对论引起的钟慢效应,进一步对狭义相对论进行证伪。

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:

狭义相对论表明,对于任一惯性参考系中的观察者来说,相对于他们移动的时钟要比其自身参考系中静止的时钟慢。这种情况有时被称为相对论时间膨胀。相对速度越快,彼此之间的时间膨胀就越大,具体关系如图1所示的方程式所示:

IMG_256

图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.

根据该理论,时间膨胀将使快速行驶的飞船中的乘客有可能在其自己的短时间内更快的驶向未来。而且如果相对速度足够高,其效果是惊人的。例如,飞船中的人旅行一年可能相当于地球上的十年。根据计算,如果使飞船保持恒定的1 g加速度,将使人类在一个人的一生中就能够穿越整个已知的宇宙。

But is it really so?

Now we design an experiment:

事实真的是这样吗?让我们设计如下这样一个实验来进行分析检验。

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点,有2个航天器(航天器1和航天器2),均装有同步的精确原子钟。如图2所示。

IMG_257

图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.

这两个航天器(航天器1和航天器2)以相同的速度行进到B点和C点,而B点和C点相距数光年。 A点和C点之间的距离与A点和B点之间的距离相同。到达B点和C点后,它们停在那里,彼此保持相对静止,如图3所示。根据相对论,我们可以知道,航天器1和航天器2上的时钟仍然是同步的,只要航天器1和航天器2彼此保持静态,这两个航天器上的时钟就会一直保持很好地同步。

IMG_258

图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,有第三个航天器(航天器3)以恒定的速度首先经过航天器2(C点),然后经过航天器1(B点),如图4所示。

IMG_259

图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.

相对论(狭义相对论)告诉我们,对于航天器3的宇航员所在的惯性参照系来说,航天器3是静止的,而航天器2和航天器1则进行匀速直线运动,以恒定的速度经过航天器3,如图5所示。因为我们认为航天器1的位置是B点,航天器2的位置是C点,所以我们还可以说航天器3是静止的,而点C和点B进行匀速直线运动,以恒定的速度经过航天器3。

IMG_260

图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.

如图6所示,当航天器3和航天器2在C点擦肩而过时,航天器2上的宇航员向航天器3上的宇航员发送信号,以告知他确切的时间。然后,航天器3上的宇航员将使用此信息将航天器3上的原子钟设置为与航天器2上的原子钟同步。同时,航天器2上的宇航员还将向航天器1上的宇航员发送信号,以告知其航天器2和航天器3相遇时的准确时间。在整个过程中,航天器的相对速度不改变,即:既没有加速过程也没有减速过程,仍然保持匀速直线运动。

对于航天器3,航天器2和航天器3在C点相遇的时间为t3C。

对于航天器2,航天器2和航天器3在C点相遇的时间为t2C。

对于航天器1,航天器2和航天器3在C点相遇的时间为t1C。

显然,t1C = t2C = t3C。

IMG_261

图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,当航天器3最终遇到位于点B的航天器1时,如图7所示:

对于航天器3,航天器1和航天器3相遇的时间为t3B。

对于航天器2,航天器1和航天器3相遇的时间为t2B。

对于航天器1,航天器1和航天器3相遇的时间为t1B。

IMG_262

图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).

相对论(狭义相对论)告诉我们:

?从航天器3上的宇航员的角度来看,航天器2和航天器1在移动,航天器3是静止的,因此航天器1和航天器2上的时间要比航天器3上的时间慢。也就是说,

t1B-t1C=t2B-t2C3B-t3C.

由于t1C=t2C=t3C, 因此可以得出:

t1B=t2B3B

?从航天器1和航天器2上的宇航员的角度来看,航天器1和航天器2是静止的,航天器3是移动的。因此航天器3上的时间比航天器1和航天器2上的时间慢。也就是说,

t1B-t1C=t2B-t2C>t3B-t3C

由于t1C=t2C=t3C, 因此可以得出:

t1B=t2B>t3B

因此,出现了一个矛盾的结果: ?t3B1B=t2B3B (t3B-t3C1B-t1C=t2B-t2C3B-t3C)

因为t1B,t2B和t3B都是时间读数。 t3B-t3C,t1B-t1C和t2B-t2C分别是一个数字(例如,可以是1000分钟或21年,等等)。因此,t3B 1B = t2B 3B(t3B-t3C 1B-t1C = t2B-t2C 3B-t3C)显然是不可能的。

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.

实际上,当航天器3和航天器1相遇时,如果航天器3上的宇航员拍摄一张自己的原子钟读数的照片并立即发给航天器1上的宇航员,同时,航天器1上的宇航员也拍摄一张自己的原子钟读数的照片并立即发给航天器3上的宇航员,则两位宇航员都将会各收到一张照片,(只要飞船速度较快)收到的照片显示的时间读数远小于自己的原子钟的时间读数。这种情况,显然不可能在现实中发生。

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.

为了使实验更加有趣,我们可以按以下方式修改实验内容:

基于以上实验,当航天器1和航天器2都位于点A时(如图2所示),他们共同制定一份计划。按照这个计划,每年,每个航天器上都会出生一个新的婴儿。并且,根据这个计划,同一年在航天器1上出生的婴儿和航天器2上出生的婴儿将被命名为相同的名字。当航天器3与航天器2相遇时,在航天器3中也出生了一个婴儿。当航天器2上的宇航员告诉航天器3上的宇航员其原子钟的时间读数时,还向他发送了婴儿的名字,以便在航天器3中出生的婴儿也被命名为相同的名字。例如,根据该计划,在航天器3和航天器2相遇的那年出生的婴儿应该被命名为王健。这样的话,这三个航天器中将各有一个名叫王健的人。

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,当航天器3和航天器2相遇时,航天器2上的宇航员还将向航天器1发送信号,以告知他航天器2和航天器3相遇的时间,在此同时,航天器2中的宇航员也告知航天器1中的宇航员:在航天器3和航天器2相会时,航天器3中新出生一名婴儿,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:

现在我们假设,航天器的相对速度是真空中光速的90%,而B点和C点之间的距离是18光年。根据相对论,在速度为光速90%的情况下,时间膨胀率约为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.

?从航天器3上的宇航员的角度来看,航天器3是静止的,而航天器2和航天器1的距离为18光年,它们以真空中光速90%的速度匀速向航天器3移动。

因此,当航天器3与航天器1相遇时,航天器3上的王健已经20岁了。由于时间膨胀效应,航天器1上的王健将是20 / 2.3 = 8.7岁。

?从航天器1上的宇航员的角度来看,航天器1和航天器2都是静止的,航天器2和航天器1的距离为18光年,航天器3以真空中光速90%的速度匀速向航天器1移动。

因此,当航天器3与航天器1相遇时,航天器1上的王健已经20岁了。由于时间膨胀效应,航天器3上的王健将是20 / 2.3 = 8.7岁。

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.

当航天器3和航天器1相遇时,如果航天器3上的王健将自己的最新照片发送给航天器1上的王健,航天器1上的王健也将其自己的最新照片发送给航天器3上的王健,会发生什么情况呢?

根据相对论(狭义相对论)进行分析,将会发生的情况是:在这两个航天器中,各有一个名叫王健的20岁年轻人拿着自己才收到的一个名叫王健的8.7岁男孩的照片。

这显然是不可能的。

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.

或者我们也可以设定,当航天器3和航天器1相遇时,我们让航天器1上的王健和航天器3上的王健彼此通过航天器的舷窗互相挥手致意?;岱⑸裁??

一个在航天器1上名为王健的20岁年轻人与一个在航天器3上名为王健的8.7岁男孩互相挥手致意?

一个在航天器1上名为王健的8.7岁男孩与一个在航天器3上名为王健的20岁年轻人互相挥手致意?

一个在航天器1上名为王健的20岁年轻人与一个在航天器3上名为王健的20岁年轻人互相挥手致意?

一个在航天器1上名为王健的8.7岁男孩与一个在航天器3上名为王健的8.7岁男孩互相挥手致意?

......

至少,我们现在可以得出结论,根据相对论(狭义相对论)推导出来的钟慢(时间膨胀)效应是不正确的。那么,狭义相对论的理论也就是不正确的。

Conclusion结论:

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

基于狭义相对论的钟慢(时间膨胀)效应导致出现不可能的推论结果,因此,狭义相对论是不正确的。

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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