# Time Dilation. I don't get it!

## Main Question or Discussion Point

Suppose, you live on a planet in orbit around Alpha Centaury which is approximately 4 light years away from our own Sun. You board a spaceship bound for Earth, and fire up the engines at the exact moment you observe the Earth to be located at the position of the Vernal Equinox. Your observation of the equinox does of course trail its occurrence by 4 years due to your distance from the actual event. Your spaceship accelerates almost instantly from zero to about 0.999 c. For the most part of the journey you travel at constant speed.

Now, here is the question: How many Vernal Equinoxes are you going to observe during your transit to Earth?

According to Special Relativity, you, the passenger of the spaceship, are at rest since the reference system of the ship is a valid inertial system as long as the ship moves at a steady rate. You see surrounding space, Sun and Alpha Centaury with it, move past your position at a constant speed of 0.999 c.

The solar system is in steady linear motion in respect to the ship's frame of reference hence you observe terrestrial clocks running slower then your own. One Earth-hour is dilated to 22 of your Ship-hours. Simultaneously, the distance between Sun and Alpha Centaury appears contracted to 1/22 of the initial distance measuring now a bit less then 0.2 light-years in the ship's reference frame.

If special relativity were correct, then according to the ship's clock the trip would last less then 0.2 years. Since for the ship-bound observer 22 ship-years correspond to one Earth-year, the passenger won't see the Earth move much past the position of the Vernal Equinox during the journey.

But what happened to the 4 Vernal Equinoxes which did already occur by the time the start signal, the information of the first equinox, reached the ship at Alpha Centaury? How can the ship miss the visual information of those events?

If, on the other hand, the passenger does indeed observe those four Vernal Equinoxes, then he does so within 0.2 ship-years. That would mean that he witnessed the passage of at least 4 Earth-years within 0.2 ship-years. This observation contradicts the time-dilation effect predicted by special relativity. Remember! In the reference frame of the ship the Earth is in motion and must therefore "Age" slower, not faster, then the ship.

An Earth-bound observer sees things slightly different. In his frame of reference, 4 Vernal Equinoxes occur while the ship is in transit, a total of 8 occurred since the one (Vernal Equinox) that served as the start signal. The ship arrives 0.004 years after the information of its departure from Alpha Centaury. The Earth-bound observer expects the passenger to count 8 Vernal Equinoxes during his 4 Earth-years in transit.

There is a difference between counting 0, 4 or 8 occurrences of the Vernal Equinox. The observation of the passenger, assuming he can count and remains focused on the Earth during the journey, can impossibly confirm all those predictions.

What am I doing wrong?

Related Special and General Relativity News on Phys.org
JesseM
Keep in mind the difference between when events happen in a frame and when observers at rest in that frame see the events, as discussed on this thread and this one. If you want to figure out how often the ship-observer sees the equinoxes as opposed to how often they occur in his frame, you have to remember to take into account the Doppler Effect. In the inertial frame where the ship-observer is at rest after the acceleration, the Earth only has a vernal equinox once every 22.366272 years. However, the Earth is also moving towards the ship at 0.999c, so every 22.366272 years the Earth will have moved 22.366272*0.999 = 22.343906 light-years closer to the ship in this frame (of course the ship accelerated so it wasn't actually at rest in this frame for that long, but we can imagine a third observer who has always been traveling at 0.999c relative to Earth, and whose trip is timed right so that the guy on the ship is moving right alongside this third observer after accelerating away from Alpha Centauri--obviously every time the light from an equinox reaches one of their eyes after that, it will reach the other one's eyes at the same moment). If each successive light signal has 22.343906 less light years to travel to reach the resting ship than the previous light signal, then the total time for each successive signal to get from the Earth to the ship will be reduced (as compared to the time for the previous signal) by 22.343906 years in this frame. This means that even though there are 22.366272 years between each equinox in this frame, the time between the ship-observer seeing each successive equinox is only 22.366272 - 22.343906 = 0.022366 years. So if the distance from Earth to Alpha Centauri is only 0.17884 light-years in this frame, and Earth is moving towards the ship at 0.999c, it will take 0.17884/0.999 = 0.179 years for Earth to reach the ship, which means that if the person on the ship is seeing the light from a new equinox every 0.022366 years, he'll see 0.179/0.022366 = 8 equinoxes between leaving Alpha Centauri and passing Earth.

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Dale
Mentor
Hi Xeinstein, JesseM is correct. There is a http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop3.html#c3" that may interest you. Just consider the Equinox to be a signal with a frequency of 3.17E-8 Hz (once every year). According to the calculator on that page, at .999c that signal is blueshifted to a frequency of 1.42E-6 Hz (once every 8.2 days).

As you already mentioned the answer of 8 equinoxes is obvious if you consider the situation from the earth's perspective. There are 4 equinoxes already "on the way" and 4 more that occur while the ship is arriving. The images from all of those must be observed as the ship travels.

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jtbell
Mentor
Just consider the Equinox to be a signal with a frequency of 3.17E-8 Hz (once every year).
Or use "cycles per year" as your unit of frequency, which doesn't change the equation but simplifies the arithmetic a lot!

Dale
Mentor
D'oh!

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