# Travel time according to both traveling and resting observers

First, my understanding of time dilation, which is typically derived based on a clock that ticks every time a beam of light emitted from the floor is bounced back by a mirror on the ceiling. Such a clock on a spaceship traveling at velocity v ticks more slowly according to an observer A at rest, from whose perspective the light needs to travel a longer distance, in comparison to how the clock ticks according to another observer B on the spaceship. In a similar manner, the length of any object in the spaceship along the direction of v appears shorter according to A, in comparison to the length measured by B. In other words, time dilation and length contraction exist only to A at rest, but never experienced by B traveling at v.

Based on such understanding, I have the following questions:

(1) How long does it take for the spaceship to go from Earth to the nearest star (assumed to be 4 light years away), according to both A on Earth and B on the spaceship? Here we assume the velocity of the spaceship is $$v=c/2$$ and $$\gamma=1/\sqrt{1-v^2/c^2}=2/\sqrt{3}=1.15$$.

I got these answers from chatGPT: the travel time is $$t=8$$ years according to A, but $$t'=\gamma t=1.15\times 8=9.24$$ years according to B. But I think the second part of the answer is wrong. Since B's clock on the spaceship is $$\gamma$$ times slower than A's clock according to A, shouldn't B's travel time be $$\gamma$$ times shorter instead of longer? If so, B's travel time would be $$t/\gamma=8/1.15=6.96$$ years according to A, which is the time needed to travel the contracted distance $$4/\gamma$$ light years at velocity $$v=c/2$$. However, as B does not experience any slowing down of time or contraction of length (as mentioned above), why does he experience a shorter (or longer) travel time? After all, what is the travel time according to the clock on the spaceship?

(2) If we assume B does experience a shortened travel time (by a factor $$\gamma$$), then can the travel time be arbitrarily shortened by letting v approach c asymptotically (in theory)? If so, can we take this to the limit to claim that a photon at velocity c can travel through any distance instantly from the photon's perspective (while it takes the photon a year to travel one lightyear from human observer's perspective)?

• Yes, one could, theoretically, travel to the farthest ends of the observable universe in a human lifetime. See e.g. math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html Practically we will never even get close to that because of the required energy expenditure. In reality we will probably "travel" at the speed of light, but not in form of baryonic matter but simply as an "upload" over an electromagnetic channel (radio waves or light). A roundtrip to Pluto would take less than a day that way. It's no more than a slightly longer commute. Commented Feb 1 at 5:36
• Your question keeps confusing A and B. You probably should choose descriptive names instead. And not to trust ChatGPT. Lest your correct ideas be polluted. Commented Feb 1 at 5:52
• Thanks for reminding me of the confusion caused by A and B. I made some correction and hopefully now it is clear that A is the observer at rest and B is the observer traveling at velocity v.
– Ruye
Commented Feb 1 at 6:14

Forget ChatGPT, it's Relativity answers are often nonsense.

The way to analyse these problems is to recognise that the spacetime interval between two events is invariant and agreed on by observers in all inertial frames of reference. Thus $$c^2 (\Delta \tau)^2 = c^2 (\Delta t)^2 - (\Delta x)^2 = c^2(\Delta t')^2 - (\Delta x')^2\ ,$$ where $$\Delta \tau$$ is a proper time interval between two events on the worldline of an observer, measured on their own watch, $$\Delta t$$ and $$\Delta t'$$ are time intervals between the two events measured in two frames that are moving with respect to each other along the x-axis and $$\Delta x$$ and $$\Delta x'$$ are the spatial separations between the events measured in those two frames.

Recognising this, then the problem is easy. We can make $$\Delta x = 4$$ light years in our "stationary" frame of reference (just a label). If the spaceship travels at $$c/2$$ in this frame, then $$\Delta t = (\Delta x)/(c/2) = 8$$ years. Using the spacetime interval equation above we see then that the proper time experienced by the spaceship in getting from A to B isjust $$\sqrt{8^2 - 4^2} = 6.93$$ years, which is indeed $$\Delta t/\gamma$$, and is also $$\Delta t'$$, since the spaceship is present at both events with $$\Delta x'= 0$$.

How can this time be shorter than 8 years? Because the distance between A and B, as measured in the stationary frame (4 light years), wouuld appear shortened by a factor of $$\gamma$$ when inferred by an observer on the spaceship.

The second part of your question is a correct summary. The proper time interval "experienced by a photon" is always zero between any two events, so it can be "everywhere all at once", to borrow a phrase. Nothing with mass can travel at $$c$$, but it can get arbitrarily close and thus its proper time interval to travel between two spacetime events can get arbitrarily small.

• Thanks for the answer! But just try to understand it a little more intiutively regarding the reason why B experiences shortened travel time. Is the reduction of B's travel time caused by (1) time dilation, or (2) distance contraction? Isn't the case that B does not experience either of them? Could you elaborate a little more about the "length is moving with respect to the spaceship"? Is the "length" here the distance between A and B?
– Ruye
Commented Feb 1 at 8:18
• @Ruye there is nothing intuitive about time dilation or length contraction. If a 1 metre ruler moves passed you at $c/2$ and you timed when the ends passed you, then you would think it had a length 1/1.15 m. Commented Feb 1 at 8:30
• Got it, thanks!
– Ruye
Commented Feb 1 at 9:41