# What would earth look like if traveling at 77% the speed of light?

I've been having a bit of trouble wrapping my mind around this.

If I were in a ship going 77% the speed of light (enough to experience reasonable time dilation) would I see earth going in fast forward? Would people on earth see my ship moving very slowly?

I have this question because I was reading Physics of the Impossible (Micho Kaku) and he stated that if we were to travel this fast, we would get to a nearby star in about 10 years, but on earth it would have been many more years. This leaves me to suspect that if many years passed on earth even though I would experience time as normal, that I would see people in fast forward.

Or would they be moving slowly because relative to me they are moving the speed of light?

• Aug 8, 2013 at 2:16
• @MikeFlynn Maybe summarise this as an answer in your own words: it gets to the heart of relative simultaneity and would be a great thing for the OP to understand. Aug 8, 2013 at 2:53
• Stumbled across this link some time ago. Not exactly what you're asking, but it helps build some intuition (and it's brilliant, imo). gamelab.mit.edu/games/a-slower-speed-of-light Sep 7, 2013 at 3:26
• Although this is about time dilation, not length contraction, as I'd thought from the title, there is an interesting fact to point out about length contraction. Length contraction doesn't tell you what relativistic objects would really look like optically. An interesting theorem in relativistic optics by Roger Penrose (?) states that a sphere still optically appears to be a sphere in all frames of reference.
– user4552
Sep 7, 2013 at 4:03
• when I imagine myself sitting inside a ship that travels at 77% of the speed of light, I definitely see everything fast forward outside the window. and when I imagine myself standing outside and looking at a ship traveling at 77% of the speed of light, I see a ship that's flickering not slow moving or still but something the image of which flickers! :)
– user30685
Oct 7, 2013 at 6:05

## 2 Answers

Your last statement got it right. One of the postulates of special relativity is the principle of relativity, stating that the laws of physics take the same form in all inertial frames. This statement includes the Lorrentz transformations, and so observers in your frame would measure that earth clocks would be slow by a factor of $\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$. Similarly, observers on earth would assert the same about your clocks.

What's happening here is the relativity of phase. Two events judged to be simultaneous in one frame will not be so in another inertial frame. So, say, you observe that earth 's clocks have elapsed $t$ seconds, when yours have elapsed $t'$ seconds. In your frame, these events are simultaneous, but in the earth frame, they aren't. As long as you remain in this frame of reference, that is, you do not turn around, this state will continue.

More specifically, describing what an observer "sees" is not the same as what they "measure.". Instead, one must consider the relativistic Doppler effect. However, in the case where our rocket is simple receding from Earth, the result is simple, and the same as our above analysis - the clocks on Earth appear slow

Mark M's point is mostly right, but I should add that the key point where this weirdness is resolved is that for this experiment to work, the moving observer has to turn around and come back. It is during this period that our moving observer will observe the "rapid aging" of the Earth. This acceleration period is also when the references frames cease to be equivalent.