LOG#021. Journey to the West (East).

I have chosen a very literary title for my article today. I hope you will forget my weakness by exotic arts. The Journey to the West is one of the masterpieces of Chinese literature. His main character is the Monkey King, Sun Wukong. He is a fun character, see him at these two artistical portraits:

For geeks only: Sun Wukong was the inspiring character for the manga/anime masterpiece Dragonbal/Dragonball Z, by A. Toriyama. Son Goku is indeed based on the legendary Sun Wukong.

The Journey to the West is Sun Wukong’s epical adventure. It is one of the Four Great Classical Novels of Chinese literature. These four novels are: Romance of the Three Kingdoms, Water Margin, Journey to the West, and The Plum in the Golden Vase (also called Golden Lotus). In the last times, The Dream of the Red Chamber replaced The Plum of the Golden Lotus as “great classical novel” in China.

By the other hand, Journey to the West was written by Wu Cheng’en in the 16th century during the Ming Dynasty. It is about the voyage of the Monkey King, Sun Wukong, to India and his fantastic and heroic acts. Wikipedia says it clearly:

“(…)The novel is a fictionalised account of the legendary pilgrimage to India of the Buddhist monk Xuanzang, and loosely based its source from the historic text Great Tang Records on the Western Regionsand traditional folk tales. The monk travelled to the “Western Regions” during the Tang Dynasty, to obtain sacred texts (sutras). The bodhisattva Avalokiteśvara (Guanyin) on instruction from the Buddha, gives this task to the monk and his three protectors in the form of disciples — namely Sun Wukong, Zhu Bajie and Sha Wujing— together with a dragon prince who acts as Xuanzang’s steed, a white horse. These four characters have agreed to help Xuanzang as an atonement for past sins(…).”

Western people had the analogue story in the historical (legendary?) voyages of Marco Polo to China…Travelling to the West and East is indeed a nice topic for the time dilation in special relativity and beyond (general relativity and non-inertial frames are also important). We will focus on the SR effect, although we will discuss further some remarks concerning important additional effects from rotating systems and general relativity.

In 1971, a very beautiful an inexpensive experiment was carried out to check the time dilation predicted by special relativity. The experiment is nowadays called Hafele-Keating experiment, and the idea was pretty simple: you can use scheduled airplane flights to carry two very precise atomic clocks around the Earth planet, close to the equator, one in eastbound flights and the other one in westbound flights. This can be sketched in the following picture:

The calculations for the relativistic dilation factor requires two velocity variables:

1st. The velocity of the Earth’s planet rotation at the equatior (or very close to it). We denote it by v_E. Moreover, we know it is equal to v_E=40000km/24 h= 463 m/s.

2nd. The velocity of the ariplanes carrying the atomic clocks, measured relative to the Earth’s surface, our S-frame.

In addition to these velocities, we have also to distinguish 4 different frames:

Frame A. Hypothetical observer moving due west at velocity v_E relative the Earth’s surface, i.e., against the sense of the Earth’s rotation. For him, the sun is always at its zenith and the who Earth planet is rotating underneath him. We can consider he is resting in a good inertial reference frame because he does not participate in the circular motion around the Earth’s axis. Thus, Earth’s rotation arount the sun can be neglected for this frame.

Frame B. This is an observer on the ground. He is moving with v_E relative to the frame A.

Frame C. This observer rests in a plane heading east, he is moving at velocity v_E+v relative to A.

Frame D. This observer is placed in a plane heading west. He is moving at velocity v_E-v relative to A.

Obviously, the B, C, and D frames are not resting in inertial frames! Therefore, we have to perform the relativist calculations in the A’s reference frame. Let us simplify the physical situation with some additional assumption. Suppose that the two airplanes are starting simultaneously at the same point, for instance the point B above, and suppose that they are back simultaneously to B after they have traveled the round-trip around the Earth planet. In the A’s frame, the flight time is equal to \Delta t. The proper times measured by B, C, and D are defined to be \tau_B, \,\tau_C,\,\tau_D and they are related to the flight time in the following way:

\Delta t= \dfrac{\tau_B}{\sqrt{1-\left(\dfrac{v_E}{c}\right)^2}}=\dfrac{\tau_C}{\sqrt{1-\left(\dfrac{v_E+v}{c}\right)^2}}=\dfrac{\tau_D}{\sqrt{1-\left(\dfrac{v_E-v}{c}\right)^2}}

Since the velocities are small compared with the speed of light, we can make a Taylor expansion of the roots. After that expansion, we get the next formulae for the proper times:

\tau_B \approx \Delta t\left[1-\dfrac{1}{2}\left( \dfrac{v_E}{c}\right)^2\right]

\tau_C \approx \Delta t\left[1-\dfrac{1}{2}\left( \dfrac{v_E+v}{c}\right)^2\right]

\tau_D \approx \Delta t\left[1-\dfrac{1}{2}\left( \dfrac{v_E-v}{c}\right)^2\right]

These equations are well enough to calculate the difference between the proper time elapsed in the east-bound airplane and the proper time on the ground:

\Delta \tau^{\mbox{east}}=\tau_C-\tau_B\approx -\dfrac{2v_Ev+v^2}{2c^2}\Delta t

\Delta \tau^{\mbox{west}}=\tau_D-\tau_B\approx +\dfrac{2v_Ev-v^2}{2c^2}\Delta t

If we impose a reasonable velocity for the airplane, say $late v=800km/h=222m/s$, the flight in the A frame takes a time:

\Delta t\approx \tau_B=\dfrac{40000km}{800km/h}=50h=1.8\cdot 10^5s

This number provides, inserted into the previous expressions, the next time delays:

\Delta \tau^{\mbox{east}}=-255ns

\Delta \tau^{\mbox{west}}=+156ns

Some conclusions are straightforward now:

1. On the eastbound flight, time does not elapse as fast as on the ground, as expected. Equivalently, tic-tacs are slower in the eastbound flight compared with the ground tic-tacs.

2. On the westbound flight, time elapses faster. It is logical, since from the A frame viewpoint the airplane heading west has a lower velocity than the observer on the ground B!

I would like to add some additional cool stuff. There are two additional non negligible corrections to be accounted by this experiment:

1st. General relativistic corrections, i.e., the so called gravitational time dilation. Although we have not studied General Relativity, we can understand it from the viewpoint of the non-inertial frames. Gravity itself introduces an extra purely gravitational time delay when you move in the gravitational field. This gravitational time delay reads:

\Delta t_g=\dfrac{\tau}{\sqrt{1-\dfrac{2G_NM}{rc^2}}}

where \tau is the proper time, G_N is the gravitational constant, M is the Earth mass and r=R_T+h is the distance to the centre of the Earth. In the case the root is close to the unity (weak gravitational fields), we can Taylor expand in order to get:

\Delta t_g=\dfrac{g}{c^2}(h-h_0) \Delta \tau

where g is the surface gravity.

2nd. The Sagnac effect. A rotating non-inertial frame suffers an extra time dilation correction:

t_1=\dfrac{\tau}{1-\dfrac{\omega \cdot \mathbf{r}}{c}}=\dfrac{2\pi r}{c}\dfrac{1}{1-\dfrac{\omega \cdot \mathbf{r}}{c}}

t_2=\dfrac{\tau}{1+\dfrac{\omega \cdot \mathbf{r}}{c}}=\dfrac{2\pi r}{c}\dfrac{1}{1+\dfrac{\omega \cdot \mathbf{r}}{c}}

Thus, we have

\Delta t=t_2-t_1=\Delta t_{Sagnac}=-\dfrac{4\pi r\omega\cdot \mathbf{r}}{c^2}\dfrac{1}{1-\dfrac{(\omega \cdot \mathbf{r})^2}{c^2}}

i.e. if \dfrac{\omega\cdot \mathbf{r}}{c}\rightarrow 0, expanding the scalar “dot” product, we finally get

\Delta t_{Sagnac}\approx -\dfrac{4\pi r^2\omega \cos\phi}{c^2}= -\dfrac{4\pi R^2\omega \cos^2\phi}{c^2}

Thus, General Relativity predicts an extra time delay due to the gravitational potential: the deeper a clock is positioned in the gravitational source of this field, the slower is its elapsed time. Supposing a typical cruising altitude about 10000 m, we get and additional 196ns of elapsed time in the airplanes in the course of 50h, in relation to the fixed ground. Therefore:

\Delta \tau^{east}=-255nx+196ns=-59ns

\Delta \tau^{west}=+156ns+196ns=+352ns

From the exact flight data Hafele and Keating theirselves obtained:

\Delta \tau^{east}_{exp}=(-42\pm 23)ns

\Delta \tau^{west}_{exp}=(+275\pm 21)ns

The four atomic clocks measured

\Delta \tau^{east}_{clock}=(-59\pm 10)nx

\Delta \tau^{west}_{clock}=(+273\pm 7)ns

and hence, theory and experiment are in good agreement.

A last remark is also important. A full treatment of the Hafele-Keating experiment, including several paths (or sections) of flight, requires the use of the totally correct time delay (including all the relativistic and non-inertial corrections):

\Delta t_{HK}=\Delta \tau_{SR}+\Delta \tau_{g}+\Delta \tau_{Sagnac}


\displaystyle{\Delta \tau_{SR}=-\dfrac{1}{c^2}\sum_{i=1}^{k}v_i^2\Delta \tau_i}

\displaystyle{\Delta \tau_{g}=\dfrac{g}{c^2}\sum_{i=1}^{k}(h_i-h_o)\Delta \tau_i}

\displaystyle{\Delta \tau_{Sagnac}=-\dfrac{\omega}{c^2}\sum_{i=1}^{k}r_i^2\cos^2 \phi_i\Delta \lambda_i}


One Comment on “LOG#021. Journey to the West (East).”

  1. Let me add that the Journey to the West has a great debt (perhaps in the same sense that the Quixote to chivary books) to chemistry books. It is obvious in some passages, as the one when Lao Den tryes to distill the essence out of the Monkey King. But the point is that most of the Chemistry is about Redox, Acid-Alcali, and other polarities and matches of contraries. The preferred metaphor to describe such reactions was fighting, and so the Journey is a neverending series of matches.

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s