Traveling near the speed of light

1. Jul 27, 2015

S_David

• Thread title edited to reflect the topic better, travel "near" c rather than "at" c
Hello all,

Suppose we can travel on a spacecraft at the speed of light, how long it would take for the person on the spacecraft to travel one light year, not to a person observing him/her from Earth, if there is any difference?

Thanks

2. Jul 27, 2015

Staff: Mentor

We can't. Nothing with nonzero rest mass can move at the speed of light.

If you mean, suppose we can travel at some speed close to the speed of light, but not at it, then you can use the relativistic time dilation formula to calculate how much elapsed time the person on the spacecraft will experience. Any basic relativity textbook will discuss this.

3. Jul 27, 2015

S_David

I am an Electrical engineer, so I am no familiar with the theories of relativity. I am just curious if we want to travel to some distant planets at the speed of light, or as you corrected me, at a speed close to the speed of light, how the people on the spacecraft will experience the trip. For example, if someone of age 30 traveled on the spacecraft, how old will he/she be when one light year distance is crossed by that person on that speed.

Excuse me if my question is still not legitimate.

4. Jul 27, 2015

Staff: Mentor

If you're going to ask questions about objects moving at speeds close to the speed of light, you're asking a question that involves relativity, so it will probably be helpful to you to at least take a look at the theory.

It depends on the speed. Let's pick a speed of 87% of the speed of light (more precisely, $\sqrt{3} /2$ times the speed of light), since at that speed, the relativistic time dilation factor is 2, which is a nice round number. That means that, while one year would elapse on Earth clocks (or clocks at the destination) during the trip, the person would only age six months (and would only experience six months' worth of subjective time during the trip).

Of course, you can make the elapsed time the person experiences during the trip as short as you like by pushing their speed closer to the speed of light. Some representative values:

At a speed of $0.968$ c, the time dilation factor is 4, so the person would only age three months.

At a speed of $0.997$ c, the time dilation factor is 12, so the person would only age one month.

At a speed of $0.9998$ c, the time dilation factor is 52, so the person would only age one week.

At a speed of $0.999996$ c, the time dilation factor is 365, so the person would only age one day.

And so on. The general formula is that the time dilation factor $\gamma$ is given by

$$\gamma = \frac{1}{\sqrt{1 - v^2}}$$

Note that we are also assuming that the ship's destination is at rest relative to Earth, and that clocks on Earth and clocks at the destination are synchronized. That probably seems innocuous at this point, but many apparent "paradoxes" in relativity textbooks are based on not being careful enough about considering such assumptions.

5. Jul 27, 2015

Kyle.Nemeth

One light year is defined as the distance that light travels in one year. So if we perhaps were to assume that one could travel on a spaceship at c, the speed of light, then it would seem to the person on the spaceship that they had traveled 1 LY (~5,879,055,313,800 miles) in one year, and the observer on this ship would seem convinced that they are one year older than when they started the journey.

This, of course, disregards the fact that it is impossible for any massive object to accelerate to c and we are also not considering any reference frame other than the observer on the ship; an observer in any other reference frame than the ship's frame would make different measurements.

6. Jul 27, 2015

S_David

Thank you for taking the time to reply. I am not sure how you inferred the aging from the time dilation factor, but assuming there is a clock on the spaceship synchronized with a clock on Earth, does this mean that the spaceship will reach one light year distance in less than one year on the synchronized clock on the spaceship? It would take so for the clock on the Earth, but what about for the clock on the spaceship?

7. Jul 27, 2015

Staff: Mentor

Yes, but distance and time are frame-dependent in relativity. See below.

No. First, once again, the ship can't move at c, and you can't just cavalierly disregard that. (See further comment below.) I shouldn't have disregarded it myself in the numbers I gave before; I'll give revised numbers in a forthcoming reply to the OP.

Second, to the person on the ship, the distance between Earth and the destination is length contracted, by the same factor that their time is dilated. So if the ship is moving at 0.866c relative to Earth, they will only age half a year if the trip takes a year according to Earth clocks. (At that speed, the ship would only cover 0.866 light year in a year, not 1 light year; again, see revised numbers in my forthcoming reply to the OP.)

Don't do that. Moving exactly at c is fundamentally different from moving at any speed short of c. Failing to distinguish the two in relativity is only going to cause confusion and problems.

8. Jul 27, 2015

Staff: Mentor

Because that's what the time dilation factor means, physically. See below.

There can't be. In relativity, you can't synchronize clocks that are in relative motion.

It means that the spaceship will reach one light-year distance, according to Earth observers, in less time, according to clocks on the spaceship, than it does according to clocks on Earth. The distance, according to the spaceship, will also be less than the distance according to Earth observers.

However, if the distance is one light-year according to Earth observers, it will take the spaceship longer than one year, according to Earth clocks, to cover that distance, because the spaceship is moving slower than c. Or, alternatively, if the trip takes one year according to Earth clocks, the ship will cover less than one light-year of distance, according to Earth observers. The numbers I gave before didn't take that into account, so they were incorrect. Here are corrected numbers, for both possible corrections just described:

For a trip of one light-year according to Earth observers, at a speed of $\sqrt{3} / 2 \approx 0.866$ c relative to Earth, the trip will take $2 / \sqrt{3} \approx 1.155$ years according to Earth clocks. Since the time dilation factor is 2, the trip will take only $0.577$ years according to clocks on the spaceship. Also, the distance between Earth and the destination will be length contracted, according to spaceship observers, to 1/2 light-year.

For a trip that takes one year according to Earth clocks, at a speed of $\sqrt{3} / 2 \approx 0.866$ c relative to Earth, the trip will cover $\sqrt{3} / 2 \approx 0.866$ light-years according to Earth observers. Again, since the time dilation factor is 2, the trip will take only 1/2 year according to clocks on the spaceship, and the distance between Earth and the destination wil be length contracted, according to spaceship observers, to $0.433$ light-years.

For higher speeds, the time dilation/length contraction factor will get larger, so the times and distances according to the spaceship will change; but the times and distances according to Earth will be the same.

Once again, you really should take some time to learn about relativity before considering these questions; as you can see, it makes a difference.

9. Jul 27, 2015

Kyle.Nemeth

Yes, hence why I stated that I was only considering the frame of the observer on the ship.

It is understood that the ship can not move at c. You're right, I agree that one should not "cavalierly" disregard that no massive object could move at c. To do so would incorrectly assume that the special theory of relativity is valid for observers moving at c.

Wouldn't this observer measure his/her "own time" or "proper time" from a clock on their spaceship? Wouldn't this observer's time be dilated relative to observers on Earth and not to the ship observer him/herself?

Aren't these the measurements made by an Earth observer? Wouldn't the observer on the ship seem to have aged ½ a year relative to an observer on Earth and vice versa for the observer on the ship (the Earth observer aging ½ a year relative to the ship observer), hence where the Twin Paradox is recognized?

Edit: Pardon my inadequate use of the website's accessories, I'm still learning how to use them.

(Moderator's note: No problem, I used magic admin powers to edit your post to format the quotes properly.)

Last edited by a moderator: Jul 27, 2015
10. Jul 27, 2015

Staff: Mentor

Then you are considering a different scenario from the OP of this thread. In the OP, the distance was one light-year in the Earth frame. That means the distance is shorter than one light-year in the ship frame. In your scenario, as I understand it now, the distance is one light-year in the ship frame, which means it's longer than one light-year in the Earth frame.

No, it would be to incorrectly assume that moving at c works the same as moving slower than c. It doesn't. But SR applies just as well to things moving at c (like light) as to things moving slower than c.

Yes. I was referring to the Earth frame, since that's the scenario that is under discussion in the thread generally. I didn't realize you were talking about a different scenario (see above).

The distance of 0.866 light-year is in the Earth frame. The elapsed time of half a year is the proper time elapsed on the ship's clock. See further comments below.

According to the Earth frame, yes.

According to the ship frame, yes; but comparing this to the previous observation is comparing apples and oranges. See below.

This isn't a twin paradox scenario because the ship does not return to Earth (at least, not in the scenario proposed by the OP to this thread). That means that relativity of simultaneity enters into any attempt to compare elapsed time on Earth with elapsed time on the ship. In the standard twin paradox, where the ship returns to Earth, the two clocks (Earth and ship) can be compared directly at both the start and end of the trip. They can't in this scenario because the ship doesn't end up on Earth.

To expand on this a bit more, let me first describe the invariants--the direct observations, which are fixed and independent of the choice of reference frame--and then describe how each frame (Earth and ship) would describe what happens. I'll use the numbers for the OP's scenario, where the distance in the Earth frame is one light-year.

The invariants: The ship departs Earth; at the instant it departs, Earth clocks and ship clocks both read zero. The ship and Earth (and the beacon) have a relative velocity of $0.866$ c. The ship travels to a beacon out in deep space; at the instant it arrives, the ship's clock reads $0.5774$ years. The beacon's clock at that same instant reads $1.155$ years.

According to the Earth frame: The distance from Earth to the beacon is one light-year. At the instant the ship departs Earth, the beacon's clock reads zero (the beacon and Earth clocks have been previously synchronized to ensure this is true in the Earth frame). At the instant the ship arrives at the beacon, Earth clocks read $1.155$ years. So the ship, moving at $0.866$ c, took $1.155$ years to travel one light-year.

According to the ship frame: The distance from Earth to the beacon is 1/2 light-year. At the instant the ship departs Earth, the beacon's clock reads $0.866$ years (because the beacon and Earth clocks are not synchronized in the ship frame; the beacon's clock runs ahead of Earth's by this amount). The beacon takes $0.5774$ years, in the ship frame, to cover half a light-year (in this frame, the ship is at rest and the Earth and beacon move) at a speed of $0.866$ c. But the beacon's clock is time dilated in this frame, so only $0.2887$ years elapse on the beacon's clock during the trip. So when the beacon arrives at the ship, the beacon's clock reads $1.155$ years, and the ship's clock reads $0.5774$ years. Since Earth clocks are running behind the beacon's clock in this frame by $0.866$ years, Earth clocks read $0.2887$ years when the beacon arrives at the ship.

So, as you can see, the invariants are the same in both frames (as they must be); but relativity of simultaneity must be taken into account as well as time dilation and length contraction in order to properly describe things in a given frame.

11. Jul 27, 2015

S_David

I knew it was not going to be easy, but I have this unclear idea that traveling large distance at speeds close to the speed of light, doesn't mean it is experienced the same to observers on Earth and to observers on the spaceship. For example to travel to a planet 1400 light years away at speeds close to the speed of light, doesn't mean it would take around 1400 years for the crew on the spaceship to reach the planet. You used a lot of details explaining that.

But, what do mean clocks in relative motion cannot be synchronized? If someone takes a watch on a spaceship traveling at a large fraction of c, it wouldn't tick the same way it were on Earth? That is what I meant by synchronization. You touched on this by using the phrase "according to clocks on the spaceship". Do you mean the same thing?

Do you recommend an article or a simple textbook talking about these things?

Thanks

12. Jul 27, 2015

S_David

Also, I inferred from your explanations that as we approach c, we can travel one light year in almost no time "according to clocks on the spaceship", right?

13. Jul 27, 2015

Staff: Mentor

We can travel one light-year according to Earth observers in almost no time according to clocks on the spaceship. To observers on the ship while the ship is traveling, the distance will be much less than one light-year.

14. Jul 27, 2015

S_David

Yes, we need to make a reference for the distance as well. Thanks, that helped.

15. Jul 27, 2015

Staff: Mentor

It would take at least 1400 years according to observers on Earth and the planet. The elapsed time experienced by the crew of the ship would be much less than 1400 years. There is no absolute time in relativity.

Consider what we mean when we say that two clocks at rest relative to each other are synchronized. We mean that corresponding ticks for each clock happen "at the same time"; for example, both clocks read 12 noon "at the same time". We can verify this by having the clocks send light signals back and forth, each signal containing a "time stamp" giving the time of emission according to the clock that sent the signal. Each clock can then verify that, if they take the time of emission and add the light travel time, the readings match. For example, two clocks one light-second apart could each emit light signals once per second, and each clock would receive the other's signals once per second, with a one-second delay--so clock #2 receives clock #1's signal marked "second #1" when clock #2 reads "second #2", and vice versa.

What we mean when we say that clocks in relative motion cannot be synchronized is that all of the above breaks down; corresponding ticks of the clocks no longer happen "at the same time". If the clocks exchange light signals, they will find that they are out of sync; the readings will no longer match up, after adjusting for light travel time.

It would seem to the person on the ship to be ticking perfectly normally. But clocks on Earth (and on the planet the ship was traveling to) would not. This could be seen by exchanging light signals between the clocks, as above.

Taylor and Wheeler's Spacetime Physics is a good introductory textbook. An online textbook is here:

http://lightandmatter.com/sr/

16. Jul 27, 2015

S_David

Interesting!! So, based on all the presented information, we can say that if there is a clock on the destination location that is synchronized with a clock on Earth (assuming both Earth and the destination are in rest), then when the spaceship reaches the destination, the elapsed time on the clock in the destination will be shorter than the time elapsed on the clock on Earth, right?

Thanks for the book and for your replies.

17. Jul 27, 2015

Staff: Mentor

No. We can say that the elapsed time on the spaceship clock when it reaches the destination will be shorter than the elapsed time on the clock at the destination location (which is synchronized with Earth clocks).

18. Jul 28, 2015

Stephanus

I've read the terms, blue/red shifted, lenght contraction, time dilation, etc...
So, if you see object blue-shifted to you (I don't know how do we realize that it's blue-shifted. Franhover line? Okay, somehow we can) then using doppler factor, you will know it's speed.
You know the speed, you'll know it's distance according to bouncing signal and divide the time trip by 2 (although the object has already moved to some distance when the object bounce our signal back, and even that we can calculate the object position assuming it doesn't change velocity). No matter what clock that we see on the object (the object might have use a very inaccurate grand father clock/wall clock) we can count its time dilation according to v.
And all after that, we CAN'T synchronize clock?
What if somehow we see that there are two comoving object, we consider ourselves at rest. And through some complicated calculation, can't we just send them signals to indicate "Start reset the clock to zero"?
And we can send reset signal to the clocks according to their rest world line. Or we can even send reset signal command to the clocks according to our reset world line
It doesn't mean that we can't synchronize the clock ,right. What about using some algebra I supposed?

19. Jul 28, 2015

harrylin

The short answer which you probably got from the earlier replies, is that while it takes about one year for the person on Earth, it takes almost no time for the person in the spaceship. Thus the velocity of light corresponds to an infinite velocity if we compare the distance to be traveled with the proper elapsed clock time (Einstein: "the velocity of light in our theory plays the part, physically, of an infinitely great velocity").

Last edited: Jul 28, 2015
20. Jul 28, 2015

newjerseyrunner

Wait, how is a light year defined? Wouldn't the definition of how long a light year is depend greatly on your reference frame? Time and space both dilate; isn't the a light year's length inversely proportional to gamma of your reference frame? If you were traveling 0.9998c away from earth, wouldn't a light year to you be 52 times longer than a light year for an observer on earth?