<p>It doesn't have to be a visible telescope, any electromagnetic radiation will act in the same way.</p>

<p>I'll give you the answer, but if you want the details you need to look at a description of the resolution to the so-called  <a href="https://en.m.wikipedia.org/wiki/Twin_paradox" rel="nofollow noreferrer">twin paradox</a>, for for example <a href="https://www.cpp.edu/~ajm/materials/twinparadox.html" rel="nofollow noreferrer">here</a>.</p>

<p>Assume that the spaceship travels  outwards and back to Earth at the same speed, with a  brief period of acceleration in between. From the point of view of the observer on the spaceship, they will receive signals at at <em>lower</em> frequency on the outward half of the journey. The change in frequency to depends on how fast they go and is given by the relativistic Doppler effect formula.
$$f_{\rm out} = f \left(\frac{c - v}{c + v}\right)^{1/2} = \alpha f,$$
where $\alpha &lt; 1$ and $f$ is the frequency of the light/signal on Earth. So the Earth would appear <em>slowed down</em>.</p>

<p>On the way back the observed frequency 
$$f_{\rm in}= f \left(\frac{c + v}{c - v}\right)^{1/2} = \frac{f}{\alpha}$$ and the Earth would appear <em>speed up</em>.</p>

<p>According to the observer on the spaceship they spend half the travel time in each of these phases. The average frequency they receive is
$$\bar{f} = \frac{(f_{\rm out}+f_{\rm in})}{2} = \frac{f}{2} \left(\alpha + \frac{1}{\alpha}\right),$$ which is always $&gt;f$.</p>

<p>This means that <em>on average</em> things on Earth appear speeded up, such that the Earth is older upon return, but what is seen from the spaceship is divided into distinct outbound and inbound phases.</p>

CC BY-SA 3.0
2017-10-10T21:32:13.990Z
@barrycarter The answer can be provided *either* in terms of time dilation *or* the Doppler effect. I have chosen the latter as simpler.
ProfRob
CC BY-SA 3.0
2017-10-10T06:46:00.963Z
2017-10-10T14:10:11.877Z
2017-10-10T14:10:11.877Z
ProfRob
<p>Suppose you traveled off Earth and back at relativistic speeds, so that when you come back much longer time elapsed on Earth than you experienced. Suppose you had an excellent telescope through which you could observe the details of the surface of the Earth down to human scale, all the time.  </p>

<p>What will you see?  Will you see people moving incredibly fast? Will you see people progressing from birth to old age in a matter of seconds?</p>

<time-dilation>
If you looked at Earth while travelling at relativistic speeds, will you see sped-up activities?
Mark G
# If you looked at Earth while travelling at relativistic speeds, will you see sped-up activities?

(Asked by: username_0 on 2017-10-10)

Suppose you traveled off Earth and back at relativistic speeds, so that when you come back much longer time elapsed on Earth than you experienced. Suppose you had an excellent telescope through which you could observe the details of the surface of the Earth down to human scale, all the time.

What will you see? Will you see people moving incredibly fast? Will you see people progressing from birth to old age in a matter of seconds?

**Tags:** time-dilation


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## 1 answers

### Answer by username_1 on 2017-10-10. Score: 5

It doesn't have to be a visible telescope, any electromagnetic radiation will act in the same way.

I'll give you the answer, but if you want the details you need to look at a description of the resolution to the so-called [twin paradox](https://en.m.wikipedia.org/wiki/Twin_paradox), for for example [here](https://www.cpp.edu/~ajm/materials/twinparadox.html).

Assume that the spaceship travels outwards and back to Earth at the same speed, with a brief period of acceleration in between. From the point of view of the observer on the spaceship, they will receive signals at at *lower* frequency on the outward half of the journey. The change in frequency to depends on how fast they go and is given by the relativistic Doppler effect formula.
$$f\_{\rm out} = f \left(\frac{c - v}{c + v}\right)^{1/2} = \alpha f,$$
where $\alpha < 1$ and $f$ is the frequency of the light/signal on Earth. So the Earth would appear *slowed down*.

On the way back the observed frequency
$$f\_{\rm in}= f \left(\frac{c + v}{c - v}\right)^{1/2} = \frac{f}{\alpha}$$ and the Earth would appear *speed up*.

According to the observer on the spaceship they spend half the travel time in each of these phases. The average frequency they receive is
$$\bar{f} = \frac{(f\_{\rm out}+f\_{\rm in})}{2} = \frac{f}{2} \left(\alpha + \frac{1}{\alpha}\right),$$ which is always $>f$.

This means that *on average* things on Earth appear speeded up, such that the Earth is older upon return, but what is seen from the spaceship is divided into distinct outbound and inbound phases.

> **Comments:**
> **username_1**: @barrycarter The answer can be provided *either* in terms of time dilation *or* the Doppler effect. I have chosen the latter as simpler.  

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