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傷腦筋的相對論
2014/07/15 15:31:36瀏覽2537|回應28|推薦33

Gracie 以光速的 12/13 往右等速移動 156 光分(一光分等於 17,987,547 公里),George 留在原地。終點有一位裁判,他的碼表事先調校到和 George 的完全同步然後前去終點等候 Gracie。

George 計算出當裁判的碼表走到 169 分時 Gracie 應會抵達終點,「同時」他手上的碼表也會顯示 169 分整。他的以爲是對的。

George 也知道以 Gracie 運動的高速,她的時間會慢下來(叫 time dilation),屆時她自己的碼表將不會顯示 169 分鐘,而是 65 分鐘。Gracie 的感覺也不像 169 分鐘,而是 65 分鐘。George 的想法是對的。

之於 Gracie,她認為自己並沒有移動,是 George 往左平移遠去,右端的裁判自己平移來到她處,而整個跑道也自己以 12/13 光速高速往左移動。

由於高速左移的跑道會縮短(根據 Gracie 實際測量,叫 Lorentz contraction),Gracie 算出整條縮短的跑道在她眼前通過後,她的碼表將顯示 65 分,和 George 爲她預估的數值一模一樣。Gracie 也是對的。

由於 George 相對於 Gracie 也是以 12/13 的光速快速移動,所她也知道 George 的碼表將會前進到只 25 分鐘處,而非 169 分(time dilation 照常適用)。裁判的碼表才會顯示 169 分。同理開跑時裁判的碼表早已先行 144 分鐘,而非如 George 所說的歸零,George 的碼表才歸零。但這不是由於 George 說謊,而是由於兩人有相對高速運動,看法不一致。

爲什麼會這樣?因爲之於 George,裁判的碼表和他的碼表同時同步,但之於移動中的 Gracie,裁判的碼表一開始就已經「偷跑」,領先 George 144 分鐘,不同步。這個「領先」的值和 Gracie 的速度以及跑道的長度成正比。Gracie 的認知是對的。

所以 Gracie 到底花幾分鐘完成全程?

Well 根據 Gracie 的碼表是 65 分鐘,根據 George 和裁判的碼表是 169 分鐘。George 人雖不在終點,但 George 會辯稱是 169 分鐘,因為裁判的碼表和他「同步」,雖然他人看不到但可以想見。但 Gracie 知道 George 和裁判都應該說 25 分鐘而非 169 分,Gracie 是對的。

Gracie 叫「裁判不公」!因爲他的碼表先行 144 分。裁判說可是我的碼表和 George 事前調校到同步啊?妳也在場不是?Gracie 說,之於你們可能同步,之於高速運動的我,你應該調到 George - 144 分才算同步。

也就是說如果裁判事先有這樣調校,他的碼表會顯示 169 - 144 = 25 分鐘而不是 169 分鐘。

或 George 如果事先有調校成裁判 + 144 分鐘的話,他的碼表就會顯示正確的 25 分鐘而不是 169 分鐘。

Gracie 還是對的。

也就是說,George 在碼表歸零時應該向裁判發出一道電訊說:「Now」!裁判應該在電訊抵達他時將碼表歸零,這樣兩個人才叫「同步」—— 之於 Gracie 同步,畢竟碼表是要用來測量 Gracie 的不是嗎?

之於 George,他根據自己測量的長度除以 Gracie 的速度所得後宣稱的 169 分是對的,他可以「藉由」裁判和他同步的碼表做此宣告。

之於 Gracie,她根據自己測量的長度除以跑道左移的速度所得後宣稱的 65 分也是對的,George 不會得到這個結論,但他可以「站在 Gracie 的立場」想,接受這個「Gracie 的事實」。

之於事前若有「正確調校」碼表的裁判,他宣稱的 25 分還是對的,George 碼表上的 25 分可以作證,一如前述 George 用裁判碼表的 169 分以為證那般。Gracie 不會得到這個結論,但她可以「站在裁判的立場」想,接受這個「裁判的事實」—— 或不事先正確調校裁判的碼表,就用開跑前 George 和裁判一起歸零的碼表,接受 George 169 分鐘這個「George 的事實」。

所以 Gracie 所花時間的真相是怎樣?

George:169 分鐘

Gracie:65 分鐘

裁判:25 分鐘(假設碼表調校正確)

相對論把「同時」、「現在」的概念整個顛覆掉,任意兩點間沒有什麼叫做「同時」這回事,講「現在」也要在前面加一個主詞 —— 「誰」的現在?

而 Lorentz contraction 爲光子把宇宙縮成零距離,不論多遠它都一開始就已經「在那裏」;time dilation 把光子的「時鐘」無限延緩,不論多久它都是「零歲」,永不 age,所以說 「光子在瞬間遍訪宇宙」。由於光速的這種極限性與絕對性,宇宙才不可能有超光速的現象。光速之爲宇宙絕對極速事所必然,連比如說太陽現在忽然無端消失,地球也要八分二十秒後才會頓失光明,同時循一直線往銀河系某處飛去,因為光的「啟動」和「停止」都以光速傳遞,引力一樣,二者傳到地球都需要八分二十秒,但那只是我們的時間。之於光子或重力子,它們的看法是「地球在太陽消失的同時同分同秒頓失光明並停止繞日」,因為它們瞬間遍在任何處所,不需要時間;它們沒有「距離」的問題,也沒有「時間」這回事。

( 興趣嗜好其他 )
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不傷腦筋不叫相對論 —— 離拿到 certificate 約百分之三還答錯
2014/07/20 21:46

兩隻一模一樣的掃把,一紅一藍,長度皆爲 255 呎。藍的平躺在地面上,紅的以 8/17 光速從左往右飛馳而來。

騎紅掃把的一群人經過時同時出手一舉將藍掃把奪去,藍紅掃把併飛時他們做一個測量。問題是:他們會發現藍掃把比紅掃把長、短、還是一樣 255 呎?

如果紅掃把騎士經過藍掃把時只是測量而不奪取,他們將發現藍掃把被「羅倫茲收縮」爲 225 呎。方程式如下:

255.(1 - ((8c / 17) ^ 2) / (c ^ 2) ^ (1 / 2)) = 225

這是高速相對運動造成的「對方長度縮短,時間走緩」必然現象。但一旦藍掃把被奪後與紅掃把並馳,顯然紅掃把騎士量度後應得到藍掃把與紅掃把一樣長的結論,不是嗎?

這題我答錯了。答案是不管他們怎麼量,藍掃把都不會是 255 呎,而是 225 呎。原因尚不明朗。Dr. Greene promise 以後會解釋其中緣由 —— 這是一個 paradox,叫 pole and barn paradox 或 ladder paradox。

這種課的問題沒有一定是教材絕對可以提供解答的,就是要你絞盡腦汁去想,想到山窮水盡,想到地老天荒。顯然我想得不夠深。

Paradox 在維基百科有解,數學上的細節我拭目以待。


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追到底!
2014/07/19 15:33
問題如下

GolfNut — 無心的邂逅(GolfNut) 於 2014-07-19 15:34 回覆:
I’ve got this haunting if not burning question of thinking that Gracie eventually flies back to George in “How can each observer say the other’s clock runs slow” video (module 22), even though in that video Gracie didn’t actually do so.  The solution of the “Gracie flies away from George” part is now crystal clear, all those 169, 65, 25, 144 minutes make perfect sense to me, after applying principles of time dilation, Lorentz contraction, and asynchronous clocks.  What I still don’t understand is this: why Gracie turns out age less when they meet up again?

I imagined Gracie goes back to George and find George ages more than she does (this is true based on my finding, but not understanding).  Massimo B. answered it in two parts.  I myself did some research and got to know that it’s the famous “Twin Paradox” which Einstein claimed not to be one.  According to Wikipedia, Minkowski provided a solution with a chart explaining that at the juncture of Gracie’s turning-around, George leaps a great deal of time that contributes to his fast aging, mostly.

I am not yet totally clear about Massimo’s second part explanation.  Wikipedia also mentioned “relativistic Doppler shift” that is supposed to explain or quantize Minkowski’s solution.  But my main hurdle is the following.

According to all of the above, it is Gracie who breaks “the symmetry” by involving herself in two reference frames, whereas George involves just one.

Why isn’t it that George involves two and Gracie involves just one reference frame from Gracie’s perspective?

Why is it necessarily true that whoever involves two reference frames ages less?

What is symmetry breaking, what symmetry and how broken? And why in so doing makes one age less?

Will we have a chance to understand all these in this course? After all it seems to be a Special Relativity issue?

I’d be much appreciated if someone can save me from this nightmare.

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大惑初解
2014/07/19 14:55
如下

GolfNut — 無心的邂逅(GolfNut) 於 2014-07-19 14:55 回覆:
有人用 Minkowski’s solution 和此文課題回答我問 Dr. Greene 的問題。


GolfNut:


I am still puzzled by the possibility that if I leave earth straight up in a straight line doing high speed travel for some time then return to earth (straight down), my son could be the same age as I am. Early on I went ahead with the course and am now at Lorentz transformation, so I understand time dilation and how to “design” a flight in order to make myself the same age as my son when I return to earth. What I don’t understand is this: although I “fly up” and “drop back” down to earth, Dr. Greene mentioned something about the returning breaks some symmetry so my scheme may not work, but that was brief and not really clear to me. Anyhow, can’t I (in fact I should) think of the earth as “dropping off” below me and then “rise up” again to greet me? If so why it should be the case that my son aged faster than me? From module 23 video “How can each observer say other’s clock runs slow?” I understand how Gracie is getting three kinds of time: 169 min, 65 min, 25 min., but my question is about actually returning to earth — as if in that video, Gracie finally returns to George. In this case, what’s going to happen? Who will have aged more? My guts feeling is “I leave” first, “earth comes back to me” later, completely mutual, so each part has its aging faster and aging slower trip section, end up nobody is getting any benefit of making herself look younger than her counterpart.


Massimo B.:


In the twin paradox modules it's shown and derived that is the very act of turning around that breaks the simmetry, As Gracie turn around her speed relative to George changes in sign therefore her now slices "rotates" from George's past to George's future (asynchronous nature of clocks) and according to her the most of George's aging happens right there. Here's how she can still claim that George ages less than her in both legs of the journey (time dilation), but adding how much he ages (in her perspective) in the turnaround she agrees that he aged more than her as they meet again at the end of her return journey. I don't know if it's clear in the way I tried to explain it, but the way the professor illustrates it in the course lectures is absolutely clear and indeed revealing. An eye opener. Taking both time dilation and asynchronus nature of clocks into account everything fits perfectly.


Massimo B.:


Maybe I can try to express it better by adding the return part of the journey to the scenario you mentioned. From Gracie's perspective: when she passes by the start line her clock reads 0 just as George clock read 0. But if in that moment Gracie looks at Team George clock at the finish line she reads 144 minutes in that clock. Then, according to her, only 25 minutes elapses on all George's reference clock during the first part of her journey while 65 elapses in her own clock. Hence when Gracie passes by the finish line in her perspective George's clock at the start line will read 0+25=25 minutes and Team George clock at the finish line will read 144+25=169. That's where the video ends. If we add the return part here's what happens: Gracie turns around, that doesn't affect what she reads on the Team George clock at the finish line (169 minutes) cause it's right next to her but she now changed the direction of her motion relative to George, so, given the asynchronous nature of clocks, it's now George's clock at the start line which is be ahead by 144 minuts relative to the one of team George at the finish line right next to her. So, immediately after turning back, she reads 169+144 = 313 minutes on George's clock at the start line. Then again she will claim that only 25 minutes will elapse in team George's clock while 65 minutes elapses on her own. So when she meets George again she reads 65+65 minutes on her clock and 25+144+144+25= 338 minuts in George's clock and she'll agree that George aged more. In her perspective George's clock ticked slowly compared to her own during the whole journey, 50 minuts versus 130, but he aged by 288 minuts just as she turned around, that's why she agrees that he aged more than her in the whole process.
GolfNut — 無心的邂逅(GolfNut) 於 2014-07-19 14:56 回覆:

謝謝人家...


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當助教了
2014/07/18 13:53

Kevin 的疑惑、我的解答:


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慣性參考系的定義
2014/07/16 20:43
Today an inertial frame in the field of classical mechanics is defined as:

An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed.

Hence, with respect to an inertial frame, an object or body accelerates only when a physical force is applied, and (following Newton's first law of motion), in the absence of a net force, a body at rest will remain at rest and a body in motion will continue to move uniformly—that is, in a straight line and at constant speed. 
GolfNut — 無心的邂逅(GolfNut) 於 2014-07-17 13:33 回覆:

Fixed Stars are long gone:

In Newton's time the fixed stars were invoked as a reference frame supposedly at rest relative toabsolute space. In other reference frames either at rest with respect to the fixed stars or in uniform translation relative to these stars, Newton's laws of motion were supposed to hold. In contrast, in frames accelerating with respect to the fixed stars, in particular frames rotating relative to the fixed stars, the laws of motion did not hold in their simplest form, but had to be supplemented by the addition of fictitious forces, for example, the Coriolis force and the centrifugal force.

As we now know, the fixed stars are not fixed. The concept of inertial frames of reference is no longer tied to either the fixed stars or to absolute space. Rather, the identification of an inertial frame is based upon the simplicity of the laws of physics in the frame, in particular, the absence of fictitious forces.


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Asynchronous clock and Lorentz contraction at work
2014/07/16 14:32
一列長度 10 公尺的火車往東速飛馳。經過某站時車頭和車尾裝置的漆罐同時向鐵軌噴漆。列車通過後站務人員前去測量,發現兩點間的距離是 11 公尺。

解釋:從車站的立場看去車頭和車尾的時間並不同步。列車人員所謂的同時其實並不同時。車尾的時間走在車頭的時間之前,所以從車站的立場看去車尾先噴,然後車頭才噴。其間列車已然前進了一公尺,所以鐵軌上的兩漆點間的距離才會是 11 公尺。

列車隨後以同速度倒回,行經車站時列車人員觀測兩漆點間的距離,發現一前一後恰好符合車頭、車尾的位置。

解釋:高速從列車底下往東飛馳的鐵軌(列車自以為靜止,是鐵軌在移動)長度會「羅倫斯收縮」,原長 11 公尺的漆點間軌道會收縮成 10 公尺,與(自以為靜止不動的)列車長度相當,所以漆點完全符合列車頭尾的位置。
GolfNut — 無心的邂逅(GolfNut) 於 2014-07-16 14:42 回覆:

可見時空本是一體,相輔相成。距離(一度空間)除以時間等於速度,速度絕對,因此時空必須互補「成就」速度的絕對性。

前半段是時間的放緩造成誤差(對站務人員而言憑空多出可解釋的一公尺),後半段是空間的收縮造成無誤(對列車人員而言自然消失可解釋的一公尺)。

完美。


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Twin Paradox solution per Minkowski
2014/07/16 13:58

Relativity of simultaneity

For a moment-by-moment understanding of how the time difference between the two twins unfolds, one must understand that in special relativity there is no concept of absolute present. For different inertial frames there are different sets of events that are simultaneous in that frame. This relativity of simultaneity means that switching from one inertial frame to another requries an adjustment in what slice through spacetime counts as the "present". In the spacetime diagram on the right, drawn for the reference frame of the Earth-based twin, that twin's world line coincides with the vertical axis (his position is constant in space, moving only in time). On the first leg of the trip, the second twin moves to the right (black sloped line); and on the second leg, back to the left. Blue lines show the planes of simultaneity for the traveling twin during the first leg of the journey; red lines, during the second leg. Just before turnaround, the traveling twin calculates the age of the Earth-based twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnaround, if he recalculates, he will measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the Earth-based twin. When one transfers from the outgoing inertial frame to the incoming inertial frame there is a jump discontinuity in the age of the Earth-based twin.

GolfNut — 無心的邂逅(GolfNut) 於 2014-07-16 14:00 回覆:
So Gracie will age less when she meet up with George again.

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Twin Paradox isn't
2014/07/16 13:36
In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more. This result appears puzzling because each twin sees the other twin as moving, and so, according to an incorrect naive application of time dilation and the principle of relativity, each should paradoxically find the other to have aged more slowly. However, this scenario can be resolved within the standard framework of special relativity: the travelling twin's trajectory involves two different inertial frames, one for the outbound journey and one for the inbound journey, and so there is no symmetry between the spacetime paths of the two twins. Therefore the twin paradox is not a paradox in the sense of a logical contradiction.
GolfNut — 無心的邂逅(GolfNut) 於 2014-07-16 13:40 回覆:
The paradox centers around the contention that, in relativity, either twin could regard the other as the traveler, in which case each should find the other younger—a logical contradiction. This contention assumes that the twins' situations are symmetrical and interchangeable, an assumption that is not correct. 
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