Traveling near the speed of light

[EngWiPy]

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

 

[PeterDonis]

Suppose we can travel on a spacecraft at the speed of light,

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.

 

[EngWiPy]

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.

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.

 

[PeterDonis]

I am no familiar with the theories of relativity.

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.

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.

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.

 

[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.

 

[EngWiPy]

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.

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?

 

[PeterDonis]

One light year is defined as the distance that light travels in one year.

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

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.

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.)

This, of course, disregards the fact that it is impossible for any massive object to accelerate to c

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.

 

[PeterDonis]

I am not sure how you inferred the aging from the time dilation factor

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

assuming there is a clock on the spaceship synchronized with a clock on Earth,

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

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 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.

 

[Kyle.Nemeth]

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

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

No. First, once again, the ship can't move at c, and you can't just cavalierly disregard that.

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.

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.

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?

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;

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.)

 

[PeterDonis]

I was only considering the frame of the observer on the ship.

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.

To do so would incorrectly assume that the special theory of relativity is valid for observers moving at c.

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.

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?

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).

Aren't these the measurements made by an Earth observer?

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.

Wouldn't the observer on the ship seem to have aged ½ a year relative to an observer on Earth

According to the Earth frame, yes.

and vice versa for the observer on the ship (the Earth observer aging ½ a year relative to the ship observer

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

hence where the Twin Paradox is recognized?

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.

 

[EngWiPy]

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.

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

 

[EngWiPy]

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?

 

[PeterDonis]

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?

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.

 

[EngWiPy]

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.

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

 

[PeterDonis]

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.

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.

what do mean clocks in relative motion cannot be synchronized?

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.

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?

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.

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

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

http://lightandmatter.com/sr/

 

[EngWiPy]

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.

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?

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

http://lightandmatter.com/sr/

Thanks for the book and for your replies.

 

[PeterDonis]

f 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?

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).

 

[Stephanus]

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

I've read the terms, blue/red shifted, length 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?

 

[harrylin]

Hello all,

Suppose we can travel on a spacecraft at [nearly] 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

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").

 

[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?

 

[Stephanus]

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?

Wait, how is ten trillions KM defined? Wouldn't the definition of how long far ten trillions KM is depend greatly on your reference frame? Time and space both dilate; isn't the ten trillion KM's length inversely proportional to gamma of your reference frame?

If you were traveling 0.9998c away from earth, wouldn't ten trillions KM to you be 52 times longer farther than ten trilions KM for an observer on earth?
Not 52, but 50. And the answer is yes. But this is also true.
"a light year to an observer on Earth be 50 times longer farther for you, too"
A light year is not a time unit, it's a length unit.
1 light year is ten trillions KM
1 light second is 300 000 KM
1 KM is 1000 m
1 feet is 12 inch, etc...

 

[EngWiPy]

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).

And the crew's aging process would follow their own clock? This basically means if we could travel at a speed that approaches the speed of light, we almost would not age any older!

 

[Stephanus]

And the crew's aging process would follow their own clock? This basically means if we travel at speed the approaches the speed of light, we almost would not age any older?

We DON'T travel!. It's the universe that travel approaches the speed of light. It's the universe that won't age. We still feel 1 second is 1 second.
I remember an Einstein joke (or anectode?)
Once on a train in US (Einstein had already been in US), a student recognized him. And the student ask, "I'm sorry professor, when will New York stop at the train?"
How do you beat gravity force? Jump out of the window, and the floor (and the earth, the moon, everything) will move toward you accelerated at 9.8ms. As long as you stay on the room, you are actually travel and accelerated at 9.8 m/s^2
BUT DON'T PROVE IT!

 

[CookieSalesman]

Guys, everyone knows you can't travel at the speed of light. However, traveling at 99.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999% of the speed of light is much longer to say. We get it, E=mc2+pc2

 

[PeterDonis]

I don't know how do we realize that it's blue-shifted. Franhover line?

Or something similar, yes; we have to have some way of knowing the frequency of the light when it was emitted, so we can compare it with the frequency that we receive. Spectral lines are one way of doing this; they are like "markers" that are emitted at known frequencies.

And all after that, we CAN'T synchronize clock?

Yes. Remember that the problem isn't just time dilation; it's relativity of simultaneity. There is no absolute definition of what "at the same time" means, and clock synchronization requires defining what "at the same time" means.

This doesn't mean that one observer can't calculate what a clock, moving relative to him, reads at any event on that clock's worldline; of course he can (given that he knows the motion of the other clock relative to him). But that's not the same as clock synchronization.

 

[harrylin]

And the crew's aging process would follow their own clock? This basically means if we could travel at a speed that approaches the speed of light, we almost would not age any older!

Once more: #19
Indeed we almost would not age at all between the start and finish. Of course, according to a co-moving reference system this is because the distance is almost zero.

 

[Stephanus]

Guys, everyone knows you can't travel at the speed of light. However, traveling at 99.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999% of the speed of light is much longer to say. We get it, E=mc2+pc2

Is that 107 nines?

 

[PeterDonis]

Wouldn't the definition of how long a light year is depend greatly on your reference frame?

No. The distance between two objects is frame-dependent, but the definition of a light-year is not.

 

[PeterDonis]

the crew's aging process would follow their own clock?

Yes.

 

[PeterDonis]

If you were traveling 0.9998c away from earth, wouldn't ten trillions KM to you be 52 times longer farther than ten trilions KM for an observer on earth?
Not 52, but 50. And the answer is yes.

No, it isn't. Length contraction is not the same as changing the definition of the unit of length.

 

[Stephanus]

Or something similar, yes; we have to have some way of knowing the frequency of the light when it was emitted, so we can compare it with the frequency that we receive. Spectral lines are one way of doing this; they are like "markers" that are emitted at known frequencies.

And what if we don't know the original frequency? A celestial body approaching (traveling away) us, we will see the process in that celestial body slower/look faster, after adjusting with doppler effect, it will be slower, of course. Surely I hope nature won't fool us.

Yes. Remember that the problem isn't just time dilation; it's relativity of simultaneity. There is no absolute definition of what "at the same time" means, and clock synchronization requires defining what "at the same time" means.

This doesn't mean that one observer can't calculate what a clock, moving relative to him, reads at any event on that clock's worldline; of course he can (given that he knows the motion of the other clock relative to him). But that's not the same as clock synchronization.

Seriously. The ONLY way we can synchronize our clock with other object if we comoving with it?

 

[PeterDonis]

what if we don't know the original frequency?

Then we can't use frequency to obtain a Doppler shift.

A celestial body approaching (traveling away) us, we will see the process in that celestial body slower/look faster, after adjusting with doppler effect, it will be slower, of course.

It will be slower if the object is moving away from us, or faster if the object is moving towards us. If we know the "natural" rate of the process (the rate at which an observer at rest with respect to the body would observe it), then yes, we can use the rate we observe to derive a Doppler shift. Here the rate of the process is the "frequency" we are using to measure the Doppler shift.

The ONLY way we can synchronize our clock with other object if we comoving with it?

Yes. Remember that "synchronize" here has a very specific technical meaning: it means "synchronize" in the sense of constructing an inertial frame such that both clocks tick coordinate time in that frame. That requires the clocks to be at rest relative to each other.

You appear to be using "synchronize" in a more general sense, to mean something like "clocks exchanging information about their times and rates". Of course any two clocks can do this using light signals. But that in itself doesn't count as "synchronization" in the technical sense I described above, which is the sense we need if we're talking about inertial frames in SR.

 

[Stephanus]

No, it isn't. Length contraction is not the same as changing the definition of the unit of length.

But if we move away/toward earth. Earth would see that 1 ly for us is 50 ls for earth, right.
And we also see, that 1 ly for us is 1 week for earth.
That's why I answer yes.

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?

 

[Stephanus]

But if we move away/toward earth. Earth would see that 1 ly for us is 50 ls for earth, right.
And we also see, that 1 ly for us is 1 week for earth.
That's why I answer yes.

Argghh, I know the answer, but I can't rephrase my sentence. The length is $\sqrt\frac{1}{1-v^2}$

 

[PeterDonis]

if we move away/toward earth. Earth would see that 1 ly for us is 50 ls for earth, right.

If we properly take into account relativity of simultaneity as well as length contraction, yes. I strongly suggest that you draw a spacetime diagram of this scenario. If you do so correctly, you will see that the two "lengths" (one light-year according to Earth, vs. one light-year according to us) are at different "angles" in spacetime, so you can't just compare them directly and expect to learn anything useful.

In other words, the two lengths are different in Earth's frame, not because one is shorter in the $x$ direction ($x$ is the space coordinate according to Earth) than the other, but because one (the length according to us) has a component in the $t$ direction ($t$ is the time coordinate according to Earth). Spacelike intervals are given by $\sqrt{\Delta x^2 - \Delta t^2}$; the length according to us has the same $\Delta x^2$ as the length according to Earth, but the length according to us also has a $\Delta t^2$, whereas the length according to Earth does not.

 

[Stephanus]

Yes. Remember that "synchronize" here has a very specific technical meaning: it means "synchronize" in the sense of constructing an inertial frame such that both clocks tick coordinate time in that frame. That requires the clocks to be at rest relative to each other.

You appear to be using "synchronize" in a more general sense, to mean something like "clocks exchanging information about their times and rates". Of course any two clocks can do this using light signals. But that in itself doesn't count as "synchronization" in the technical sense I described above, which is the sense we need if we're talking about inertial frames in SR.

Thanks! It helps me clear the way to understand SR.

 

[Stephanus]

Yes. Remember that "synchronize" here has a very specific technical meaning: it means "synchronize" in the sense of constructing an inertial frame such that both clocks tick coordinate time in that frame. That requires the clocks to be at rest relative to each other.

I'm sorry. I have this scenario in mind.

At event A, Green collect information from Blue. Green knows Blue left/right speed, velocity, distance, etc..
Can't Green at event B after make some calculation send command to Blue left and Blue right.
Left: "Set your clock at so and so"
Right: "Set your clock at so and so"
Green could have done that with pen/paper and watch.

 

[PeterDonis]

Can't Green at event B after make some calculation send command to Blue left and Blue right.

Yes, but the clock readings the Blue clocks are commanded to set will not be consistent with coordinate time in the inertial frame in which the Blue clocks are at rest. (Actually, I'm assuming this would be the case, since otherwise I don't see why you'd be raising the point. But you haven't described the calculation Green does, so we don't know what commands Green would actually send in the scenario you've described.) So this won't be "clock synchronization" in the technical sense I defined.

 

[Stephanus]

Yes, but the clock readings the Blue clocks are commanded to set will not be consistent with coordinate time in the inertial frame in which the Blue clocks are at rest. (Actually, I'm assuming this would be the case, since otherwise I don't see why you'd be raising the point. But you haven't described the calculation Green does, so we don't know what commands Green would actually send in the scenario you've described.) So this won't be "clock synchronization" in the technical sense I defined.

So, at A. Red collects information, doppler, speed, distance.
At B, Red sends command to event "Set:00" -> "Set your clock at 00:00"
At B, Red sends command to event "Set:2.5" -> "Set your clock at 00:02.50"
I think this is how to resolve the problem. Should have calculated it of course. But I just boost the world line.

 

[PeterDonis]

I think this is how to resolve the problem.

What "problem" are you trying to resolve? Why does Red choose those particular values to send to Blue and Green? What is Red trying to accomplish?

 

[Stephanus]

What "problem" are you trying to resolve? Why does Red choose those particular values to send to Blue and Green? What is Red trying to accomplish?

Ahhh, of course. Red should choose the right picture. Thanks!
[It doesn't make sense for Red to consider Blue/Green is moving. Red should convert to Blue/Green at rest]

 

[PeterDonis]

Red should choose the right picture.

"Should" for what purpose? Again, what is Red trying to accomplish?

 

[Stephanus]

"Should" for what purpose? Again, what is Red trying to accomplish?

Purpose? I don't know . I think we are too deep in math, that I forgot it's about phsyics. I'm just doing math. Okay, I'll concentrate on twins paradox.

 

[Nugatory]

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?

Yes, different observers will disagree about both the distance covered and the time between departure and arrival for a given journey. Thus, you can never speak of a distance of one light-year (or any other distance, for that matter) without specifying a frame.

In this thread, we've specified in the early posts that the Earth and the endpoint of the journey are at rest relative to each other and separated by one light-year (that is, a light signal sent from Earth and reflected back from a mirror at the destination will come back two years after it is sent). We're then figuring the distances and times using a frame in which the spaceship is at rest, and comparing these with the values we have in the frame in which the Earth is at rest.

That's a reasonable enough thing to do and long as we remember that there's nothing special about any of these frames, not even the one in which the Earth is at rest. It just so happens that that frame is particularly convenient to use if you're trying to answer the question "How long will I have to wait for the ship to complete a round trip?".

 

[PeterDonis]

you can never speak of a distance of one light-year (or any other distance, for that matter) without specifying a frame.

Just to clarify what I was saying earlier, this is true, but it's also true that we can define a "light-year" as a unit of distance without specifying a frame. It's the distance light travels in one year. The frame dependence comes in when we realize that "the distance light travels in one year" refers to different sets of events in spacetime for different frames.

 

[Nugatory]

Just to clarify what I was saying earlier, this is true, but it's also true that we can define a "light-year" as a unit of distance without specifying a frame. It's the distance light travels in one year. The frame dependence comes in when we realize that "the distance light travels in one year" refers to different sets of events in spacetime for different frames.

Ah - right. Yes, that's an important clarification.

 

[Kyle.Nemeth]

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.

In the original post, S_David asks how long it would take for a person on a spaceship to travel one light year, not to an observer on the Earth. This led me to believe that we want to know how long the trip takes relative to the ship observer. The original post does not explicitly state that the distance is one light year in the Earth frame, so is it implied by this portion of the question,

how long it would take for the person on the spacecraft to travel one light year,

that the distance must be 1 light year in the Earth frame since S_David is on Earth "measuring" the light year?

No, it would be to incorrectly assume that moving at c works the same as moving slower than c. It doesn't.

So the theory does apply, since if I were to suppose an observer were moving at c, the theory will tell me what this means? For example, as v approaches c, γ approaches ∞, so if we were to consider the total energy of an observer of mass m with speed c,

E = γmc² = ∞

Thus, in order to be moving at c, the observer must have a total amount of energy that is infinite, but it is impossible to transfer an infinite amount of energy to any massive object since this would contradict the Law of Conservation of Energy.

Also, if we consider the reference frame of a light beam moving parallel to the length L₀ (L₀ is the length measured by an observer on a spaceship at rest) between two points in outer space, then in the light beams frame,

L = (√1-β²) L₀ = 0

Is this why the theory applies to light moving at c?

But SR applies just as well to things moving at c (like light) as to things moving slower than c.

Isn't light the only thing that moves at c?

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.

Is it also due to relativity of simultaneity that, for the ship's frame, the Earth and the beacon clocks are not synchronized?

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).

Is this because the ship is moving toward the clock at v = 0.866c, so that light from the beacon's clock reaches the ship 0.866 years earlier than light from the Earth's clock?

 

[Nugatory]

So the theory does apply, since if I were to suppose an observer were moving at c, the theory will tell me what this means?

The theory does not apply. The rules for travel at speeds less than the speed of light are derived starting from an assumption that is equivalent to denying the possibility of an observer moving at the speed of light (there are more mathematically rigorous ways of stating this, but that's a digression here) so they cannot apply in such a situation. If you try applying them and get a result that makes sense, that's just an accident. (If I try applying the theory that 1=0 to various math problems, every once in a while I'll find that I've lucked into the right answer, but that doesn't make it right to apply that "theory").

Isn't light the only thing that moves at c?

No. Gravitational waves also do, and it's a historical accident that we call the universe's invariant speed "the speed of light". We measured the speed of light before we discovered that the universe had to have an invariant speed and that several things including light would have to move at that speed.

Is it also due to relativity of simultaneity that, for the ship's frame, the Earth and the beacon clocks are not synchronized?

You could say that, and you wouldn't be wrong, but it's something of an uninteresting tautology. Consider what it means to say that two clocks are synchronized. You're basically saying that they both read the same at the same time - and relativity of simultaneity says that "at the same time" means different things on different frames so clocks synchronized in one frame will not be synchronized in another.

Is this because the ship is moving toward the clock at v = 0.866c, so that light from the beacon's clock reaches the ship 0.866 years earlier than light from the Earth's clock?

No. Time dilation and relativity of simultaneity are what's left after you've allowed for light travel time.

 

[PeterDonis]

In the original post, S_David asks how long it would take for a person on a spaceship to travel one light year, not to an observer on the Earth.

Yes, but the OP wasn't clear about whose frame the one light-year distance was supposed to be relative to. In subsequent discussion, it seems clear that he meant one light-year relative to Earth, not the ship. However, either case can be analyzed; the underlying principles are the same either way.

This led me to believe that we want to know how long the trip takes relative to the ship observer.

Yes; at least that was the original question in the OP. There have been others as the thread has gone on.

The original post does not explicitly state that the distance is one light year in the Earth frame, so is it implied by this portion of the question...

Yes, agreed.

So the theory does apply, since if I were to suppose an observer were moving at c, the theory will tell me what this means?

No. The theory says that only things with zero rest mass (like light) can move at c; an "observer" (meaning something like you or me or a spaceship or a measuring device) can't. That's because any "observer" must have nonzero rest mass.

Have you looked at any basic SR textbooks? If you haven't, you should. This is all very basic SR.

if we consider the reference frame of a light beam

There is no such thing; a light beam moves at c in every reference frame, so there can never be a frame in which it at rest, and "the reference frame of a light beam" assumes that the light beam is at rest in the frame.

Again, if you haven't looked at an SR textbook, you should. This is very basic SR. (There is also a Physics Forums FAQ explaining that there is no such thing as the rest frame of a photon.)

Is this why the theory applies to light moving at c?

No. The theory applies to light moving at c because it correctly predicts the behavior of light moving at c. It just doesn't do it the way you are thinking it does. An SR textbook will explain how it actually does.

Isn't light the only thing that moves at c?

No. Anything with zero rest mass moves at c. Light is the only commonly encountered thing that does, but there are other particles in physics that also have zero rest mass (gluons and gravitons).

Is it also due to relativity of simultaneity that, for the ship's frame, the Earth and the beacon clocks are not synchronized?

Yes. But also see Nugatory's reply.

Is this because the ship is moving toward the clock at v = 0.866c, so that light from the beacon's clock reaches the ship 0.866 years earlier than light from the Earth's clock?

No. The ship is moving relative to Earth and the beacon, so the difference in light travel times from Earth to the ship, vs. from the beacon to the ship, is not constant. The difference in clock readings between the beacon and Earth, in the ship frame, is due to relativity of simultaneity.

 

[PeterDonis]

The theory does not apply.

Careful. The theory applies perfectly well to light and other things with zero rest mass. It just doesn't model them the way Kyle.Nemeth is thinking it does.