# Faster than light conundrum

1. Apr 11, 2014

### lowemack

In theory with a powerfull enough ship a space man can travel to a distant star, say Proxima Centura (4.2LY away) in 2.1LY by his own clock, yet on Earth and Proxima Centura more than 4.2 years will have passed.

My question is why will he not measure Proxima Centura coming towards him at twice the speed of light?

Surely speed is distance over time and in his reference frame he as travelled at twice the speed of light.

2. Apr 11, 2014

### georgir

The distance will seem shorter to him as well.

3. Apr 11, 2014

### lowemack

How would you know that it is the distance that is shorter and not the speed faster?

4. Apr 11, 2014

### Staff: Mentor

By measuring it.

5. Apr 11, 2014

### abitslow

"in his reference frame he as travelled at twice the speed of light. "
You have just stated that Einstein and all the physicists after him are wrong. You either do not know what you are writing, or fail to understand the first thing about relativty. To wit: no massive object can attain the speed of light, which is the same for ALL (inertial) observers.
It is ALSO nonsense to claim that theoretically, a person could travel to alpha centuri gamma in 2.1 Light years.
Are you seriously confusing distance with time?
Perhaps you meant 2.1 years? There is NO known technology which could theoretically allow such a trip to be made. I guess if you invoke magic pixie dust, then you might as well abandon the rest of the Laws of Physics, too, huh?
Speed is distance divided by time. Light, in order to be traveling the same speed in any inertial reference frame, must 'warp' BOTH time and space so that their ratio remains constant. These individual effects are called time dilation and length contraction. For the traveling observer, the duration of the trip will be shorter and so will its length, RELATIVE TO THE EARTH'S FRAME OF REFERENCE. That is, a clock on the ship will tick fewer times, and the ship will log fewer miles, than what a telescope observing it from Earth will measure.
There are no contradictions in special relativity, there are no paradoxes. If you encounter one, it is because you are failing to understand the theory. To be honest the best way to understand the effects of velocity on time and space, a space-time diagram is the way to go. Explanation of how to use one can be found on the web. For instance in the Stanford Univ. lecture series on special relativity, even if the math is a bit too advanced for you, Susskind demonstrates how to use the diagrams pretty clearly.

6. Apr 11, 2014

### ChrisVer

@abitslow:
There is no actual reason to say that length and time have different dimensions/units... As you can say a mass has units of energy, the same thing you can say about distance and time (you can always set c=1 and get that [length]=[time])

7. Apr 11, 2014

### lowemack

How do you measure a distance in a reference frame that is moving close to C?

8. Apr 11, 2014

### ZapperZ

Staff Emeritus
Why would this be a problem? For all you know, according to another reference frame, we are moving VERY close to c. Yet, we have no problem measuring the speed of light.

Zz.

9. Apr 11, 2014

### Staff: Mentor

Because you are at rest and the star is rushing towards you, this is the same problem as measuring the distance to any fast-moving object.

Because it's moving, what you're doing is measuring the distance to the point in space where it was at a particular time. In practice we'd use radar, in a thought experiment we'd note where we are at the time that we take the distance measurement, where the star is at that time, and then measure the distance between these two points (which aren't moving) at our leisure with a meter stock.

10. Apr 11, 2014

### lowemack

Apologies, I did mean 2.1 years, not light years, and I am pretty sure that from the spaceships clock the journey could be done in 2.1 years, relativity does allow this.

11. Apr 11, 2014

### lowemack

How does a spaceship log fewer miles, how can you tell it is the distance that got shorter and not you arriving quicker than faster than light travel would allow.

How would you be able to tell when you arrive at Alpha Centura in 2.1 years that the distance you travelled was less and you didn't travel faster than light, because at the end of the day you arrived at something that was 4.2LY away in 2.1 years, from your point of view.

12. Apr 11, 2014

### Staff: Mentor

Realize that both distance and time are measured differently in different frames. The 4.2 LY is the distance measured by earth observers, not spaceship observers. The 2.1 years is the travel time measured by spaceship observers, not earth observers. (You'll get all sorts of nonsense by mixing and matching measurements from different frames.)

13. Apr 11, 2014

### lowemack

How would the spaceman be able to tell that the distance he travelled was less than 4.2LY.

If he had a tape measure on the spaceship, and he fastened the end to earth, would it not measure 4.2LY when he got to proxima centura?

If it measures less because the distance travelled is shorter, what would happen to the tape measure when he stopped at the destination?

14. Apr 11, 2014

### PAllen

Since the tape is at rest relative to earth (once unrolled), this is a measurement in earth frame rather than rocket frame.

The practical way to measure distance to an object speeding towards you is radar, as has been suggested. Starting long before earth passes you at high speed, you are sending out signals to the far away planet (encoded with their transmission times, for simplicity). You note when the earth passes you, then you note when you receive a signal (from the distant planet) whose transmission time compared to reception time is twice as far back as when earth passed. Half this time difference times c is your radar measurement of how far apart earth and the planet were at the time earth passed you.

Another theoretical way to measure such distance in the rocket frame is to imagine a series of space buoys laid out between earth and the planet. You see them going by you at v (local doppler measurement). If this virtual road is going by you at v, for time t (as you measure it), you conclude v*t is the road length that has gone passed you (irrespective of what anyone else claims about the distance between buoys).

Even more impractical, you have a whole bunch of meter sticks at rest relative to your rocket (moving the same velocity as you relative to earth), laid out in front of you. You have to synchronize clocks with other rockets (at rest relative to you) along the rulers. You use the Einstein convention for this. Then, you note when earth passed you, and ask the rockets along the rulers which one saw the other planet go by at that time, and what the ruler reading is. It will take a while to collect this information, obviously.

All of these measurements will lead to the same value, and it be shorter than 4.1 light years by the gamma factor.

15. Apr 11, 2014

### ZapperZ

Staff Emeritus
Maybe it is time for you to go back to the basics and learn a bit more about "length contraction" and how it is derived:

http://arxiv.org/abs/1203.5333

Zz.

16. Apr 11, 2014

### ghwellsjr

The way to measure distance is with a laser range finder or a radar range finder. In this case, the space man could send out a radar signal when he leaves Earth and measure with his on-board clock how long it takes for him to receive the echo. He assumes that the signal took the same amount of time to get to the star as it took to get back to him and that it traveled at the speed of light in both directions. This allows him to establish how far away the star was at a particular time in his trip. When he gets to the star, he can establish the speed of the star and he can calculate how far away it was when he started his trip.

A spacetime diagram will help. Here is one showing the Earth in blue, the star in red and the space man in black, along with his radar signal and its echo (you have to look very carefully just before the second black dot). The dots represent one-year increments of elapsed time on each clock. This is for the mutual rest frame of the Earth and the star. The space man is traveling at 0.8945c. I have shown his path as extending past the star so that you can easily estimate his travel time, according to his on-board clock at 2.1 years:

As you stated, in the Earth/star frame, it takes him a little over 4.2 years to get there, namely 4.7 years.

The echo of the radar signal that he sent when he left Earth arrives back at him at his time of just under 2 years, we'll call it 1.98 years. He divides that by two and establishes that the star was 0.99 light-years away from him at his time 0.99 years after he left Earth. When he arrives at the star at his time of 2.1 years, he calculates that it took 2.1-0.99=1.11 years for the star to travel 0.99 light-years toward him during the final part of the trip. This calculates to a speed of 0.892c (close enough for our eye-balling off the graph).

Now he can calculate where the star was when he left Earth by multiplying his time interval between Earth and star by the star's speed, 2.1*0.892=1.87 light-years. But to see this, we have to transform all the coordinates of the events in the above diagram to his rest frame during the trip:

Does this all make perfect sense to you? Any questions?

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17. Apr 12, 2014

### ghwellsjr

We can illustrate this in a spacetime diagram, in fact a couple. This scenario assumes that the space man did not actually leave Earth but was approaching Earth from a long distance away and passing by Earth on his way to the star (or planet). We want to show how he establishes how far away the star was at the moment he passed Earth according to the space man's inertial rest frame. In order to do this correctly, we have to make sure the space man is also inertial, that is, he is not changing speed or direction. As PAllen said, the space man is continually sending out coded radar signals (so that he can match them up with their echoes), keeping track of when they were sent, then detecting their echoes and keeping track of when they were received. Some time later, he goes through his log of data and averages each sent time with its matched echo time. Then he looks for the average time that corresponds to when he passed Earth and takes the difference between the echo time and the sent time and divides that by two and calculates how far light would travel in that time.

So here is a diagram showing the one radar signal sent with its matched echo whose average time equals the time he passed Earth:

Since he passed Earth at time 0, we can see that the average of -1.87 and 1.87 is also 0 and the distance to the star is 1.87 light-years in his inertial rest frame. It doesn't look like it in the above diagram because that is the Earth/star inertial rest frame.

But we can transform all the coordinates of the above diagram to the rest frame of the space man and we get:

Now we can easily see that the star is 1.87 light-years away from Earth when the space man passed the Earth according to the Earth/star inertial rest frame.

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18. Apr 12, 2014

### ghwellsjr

In the previous post I showed how an inertial space man would establish the distance to the star at the moment he passed the Earth.

But what about a space man who was on the Earth and then took off for the star? Does this mean that the star was 4.2 light-years away from him the moment before he left and then 1.87 light-years the moment after he left? Well now we have to consider the rest frame of a non-inertial observer and there is not a standard way to do this so we are free to pick whatever scheme we want to answer the question. I personally like to use the same method that the inertial observers use, which is the radar method. The non-inertial space man does exactly what the inertial space man does but he will get a different set of data and will draw different conclusions.

To begin with, here is a spacetime diagram showing some of the radar signals (the thin black lines) sent by the space man and their echoes (the thin red lines):

Here is a log of the data the spaceman collects:

Code (Text):

Sent    Rcvd    Star's
Time    Time    Proper Time
-10.2   -1.8    -6
-9.2    -0.8    -5
-8.2    0.04    -4
-7.2    0.28    -3
-6.2    0.52    -2
-5.2    0.76    -1
-4.2    1.0 0
-3.2    1.24    1
-2.2    1.48    2
-1.2    1.72    3
-0.2    1.96    4
2.4 2.4 5
3.4 3.4 6
4.4 4.4 7
After the space man arrives at the star, he can calculate the distance to the star as a function of his own Proper Time and he will get this diagram of his non-inertial rest frame or as you said, his point of view:

Note that when the trip starts at his Proper Time of 0, the star has already started moving towards him so that it is much closer (actually 1.6 light-years away) than its starting distance of 4.2 light-years according to this particular way of establishing the non-inertial rest frame of the space man. Note that this gives a different distance than the inertial Earth/star rest frame in the previous post (which was 1.87 light-years).

Hope this helps. Let me know if you need more explanation or have any questions.

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19. Apr 15, 2014

### lowemack

Thanks ghwellsjr. Great explanations that help a lot.
Would the spaceman still get the same distance (2.1LY) if he tried to measure the distance he travelled away from Earth by using this method of radar/laser echo?

20. Apr 18, 2014

### ghwellsjr

The scenario is symmetrical so whatever distance he measures to the star as he is traveling toward it will be the same distance he measures to the Earth as he travels away from it. However, the distances are changing so they both are functions of his time and although there are moments when those distances are 2.1LY, that particular distance is not significant.

I think you probably meant to say one of the shorter distances that I calculated for you in post #17 (1.87LY) or in post #18 (1.6LY). Why are there two different answers? Because they are measuring two different things. The first one is the distance between the Earth and the star in the inertial rest frame of the spaceman during the trip. This distance is a constant so that when he is colocated with the Earth, the star is 1.87LY in front of him. Later, when he is colocated with the star, the Earth is 1.87LY behind him. This distance also corresponds to the simple formula, distance equals speed multiplied by time. Since he calculated his speed at 0.892c and his travel time is 2.1 years, he gets 1.87LY.

However, you are trying to reconcile the fact that before the spaceman started his trip and after he finished it, the distance between the Earth and the star is a different constant, 4.2LY according to their mutual rest frame so it seems that he is traveling double the speed of light since it took him only 2.1Y of travel time. But if we actually carry out the radar measurements that he performs in establishing his own non-inertial rest frame, we see that the star begins moving toward him long before the Earth starts leaving him and the Earth continues moving away from him long after the star reaches him. That is what I want to show you now.

In order to do this more clearly than I did in my previous diagrams, I'm adding tick marks every tenth of a year and labeling the dots that are spaced one year apart when they don't coincide with the Coordinate Time. I'm also labeling the tick marks corresponding to the sending of a radar signal, the point of reflection and the receiving of the radar echo. And I'm showing a minimal set of radar signals and their reflections at strategic points along the scenario. These points are whenever there is a change in the spaceman's trajectory for either sending the radar signals or receiving their echoes. And then I need another pair of radar signals at the beginning and at the end.

This spacetime diagram is an extension of the first one in post #18:

There are two sets of logs that the black spaceman keeps, one for radar signals sent to the red star and one for radar signals sent to the blue Earth. Here is the one for the star:

Code (Text):

Sent    Rcvd    Star's  Calc    Calc
Time    Time    Time    Time    Dist
-10 -1.6    -5.8    -5.8    4.2
-8.4    0.0 -5.2    -4.2    4.2
0.0 1.98    4.2 0.99    0.99
2.1 2.1 4.7 2.1 0.0
12.4    12.4    4.2 15  0.0

And here is the log for the blue Earth:

Code (Text):

Sent    Rcvd    Earth's Calc    Calc
Time    Time    Time    Time    Dist
-10 -10 -10 -10 0.0
0.0 0.0 0.0 0.0 0.0
0.12    2.1 0.5 1.11    0.99
2.1 10.5    8.9 6.3 4.2
4   12.4    10.8    8.2 4.2

Note that I have added into the above two logs the calculations that the spaceman makes from his radar data to establish the Time and Distance to the star and the Earth. Recall that the Calculated Time is the average of the Sent Time and the Received Time and the Calculated Distance is one half of the difference between the Sent Time and the Received Time multiplied by the speed of light which in our case is 1.

From the above calculations and the observed time on the clock of the star and the Earth, the spaceman can construct his non-inertial rest frame. Note that he has to linearly interpolate the worldlines between the calculated events and add in the appropriate number of dots and tick marks based on the deltas between the observed Proper Times from the star or from the Earth:

I didn't draw in the radar signals but if you want, you can print the diagram and draw them in yourself, just remember to draw them at 45 degree angles. You will see that the radar signals, their echoes and the Proper Times of reflection follow the same patterns as they did in the first diagram in this post.

You can see from the above diagram that the star starts moving at just over 0.6c toward the spaceman 4.2Y before the Earth starts to leave the spaceman so that it happens to be 1.6LY away when the Earth starts to move. The Earth moves away at 0.8945c for 1.1 years according to the spaceman but it ages only 0.5 years during this time. Meanwhile, the star is moving just over 0.6c and continues until the spaceman has reached one year. At that point, the speed of the star goes up from 0.6c to 0.8945c so that in the next 1.1 years for the spaceman, the star has aged 0.5 years. There is an overlap only of 0.1 years from the spaceman's time of 1.0 years to 1.1 years when the speeds of the star and Earth are identical and the distance between them is a constant, 1.87LY, the same distance that they are separated in the inertial rest frame of the spaceman during the trip (see post #17). Here is a spacetime diagram zoomed in on this portion of the overlap so you can see the measurements more clearly:

After this period of time, the spaceman becomes colocated with the star while the Earth continues to move away at just over 0.6c. Finally, at the spaceman's time of 6.3 years, the Earth comes to rest 4.2LY away from the spaceman who has been colocated with the star for 4.2 years. Note in the diagram two above that the Proper Times for the star and the Earth have once again become synchronized with each other and ticking at the same rate as the spaceman's clock but he has lost 2.6 years because in the star/Earth rest frame (first diagram above) his trip took 4.7 years but his clock only advanced 2.1 years. For example, if you look at the second diagram in this post at the red dot labeled 10 (years), the Coordinate Time is 7.4 years which is the same at the spaceman's Proper Time and that gives us a difference of 2.6 years, the amount of time the spaceman lost during his trip according to the clocks synchronized in the star/Earth rest frame.

Now I have one more diagram to show you, the distance between the star and the Earth according to the spaceman's non-inertial rest frame:

The important point to notice about this diagram is that it shows that the distance between the star and the Earth is 1.6LY at the beginning (0 years) and ending (2.1 years) of the trip but in between it is 1.87LY for 0.1 years (from 1.0 years to 1.1 years) according to the radar method of establishing a non-inertial rest frame.

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