# A to B, faster then c.

1. May 11, 2006

### Enos

I've thought of this last year but I need the opinions of others to see if it is theoretically possible to travel a light year in less then a year. Of course this thought experiment should not violate relativity.

Lets say we create electromagnetic elevator type of technology. Where you have the positive charge on one end and the negative charge on the other and they are off until the timer reaches zero. Each machine repels from the other at a fraction of the speed of light once the timer reaches zero. There are no laws against space receeding faster then light so would this also apply to traveling faster then light using receeding technology or am I missing something in my thought experiment?

Before charge:
{Ground}[][][][][][][][][][]A_________________B

After charge:
{Ground}[]++[]--[]++[]--[]++[]--[]++[]--[]++[]B

Would something happen to the ground or the traveller due to the combined forces?

Last edited: May 11, 2006
2. May 11, 2006

### JesseM

Where did you get this idea? In special relativity, where space is not expanding, it is impossible for anything to move faster than light in any given inertial reference frame. It is possible for an inertial observer to observe two objects moving apart faster than light, even though neither is moving faster than light in his own frame--for example, if I observe one object moving to the left at 0.8c and another moving to the right at 0.7c in another frame, then the distance between them is increasing at 1.5c. However, an observer travelling alongside either object would not observe the other object to be moving away at 1.5c, he would always see the other object moving at less than c.

3. May 11, 2006

### Enos

I meant no laws against space receeding faster then light. I'll edit my post.

4. May 11, 2006

### Tom Mattson

Staff Emeritus
Say you are on Earth and you want to get to some point whose distance from Earth is 1 ly as measured by an Earthbound observer (that's important). You can do that in an arbitrarily short proper time without ever exceeding the speed of light in any frame. That's because the faster you go, the shorter the distance between the Earth and your destination is (thanks to length contraction).

5. May 11, 2006

### Enos

What if the machines were then swtiched to attract rather then repel. There wouldn't be a twin paradox would there? Perhaps slighty but not the whole effect due to the multiple machines?

6. May 11, 2006

### JesseM

Are you talking about the expansion of space in general relativity? But two charges repelling each other wouldn't cause space to expand between them, there is nothing about your example that goes beyond special relativity.

7. May 11, 2006

### Enos

I am talking about the expansion of space. But I am not using the expansion of space but rather that is where I got the idea of traveling faster then light using receeding technology rather then trying to get a vessel to travel faster then light. It is theoretically possible to get an object travel lets say 50,000km/s. I'll use older technology to make it easier to understand what I am trying to say. Ok, we somehow made Hydraulic Lifts to push 50,000km/s and we put 3 lifts over each other.

Before hydraulic push:
}[]{}[]{}[]{A_________B

After hydraulic push:
}--[]--{}--[]--{}--[]--{B

}--[ = 50,000km/s
x6 = 300,000km/s

Last edited: May 11, 2006
8. May 11, 2006

### JesseM

You seem to be saying that although you were inspired by the fact that things can move apart faster than light if the space between them is expanding, you believe this is possible even when the space between them is not expanding, ie in the static flat spacetime of special relativity. But this is wrong--in special relativity it is not possible for one object to move away from another faster than light, as seen in one of the object's inertial rest frame.

9. May 11, 2006

### Enos

Is there an explantion on how this is not possible with this thought experiment? Because technically the object is only moving away from the nearest lift while the other lifts are moving away from each other at 50,000km/s. Even though the total of the seperation is 300,000km/s the seperate machines only travel 50,000km/s away from each other. So this shouldn't violate relativity.

10. May 11, 2006

### JesseM

The reason is that velocities don't add the same way in relativity when you switch from one reference frame to another. Say I am on a ship, and I shoot a missile which moves at 0.4c to my right in my rest frame. You are sitting on earth, and you in turn see my ship to be moving at 0.4c to your right in your rest frame, the same direction as the missile. You will not see the missile as moving at 0.4c + 0.4c = 0.8c; instead you must use the formula for addition of velocities in relativity, (u + v)/(1 + uv/c^2), which in this case would give (0.4c + 0.4c)/(1 + 0.16) = 0.69c. As long as u and v are less than or equal to c, their sum according to this formula will also be less than or equal to c. The reason velocity addition works differently in SR than how it does in classical mechanics has to do with the fact that each observer measures velocity=distance/time using rulers and clocks which are at rest relative to themselves (with the clocks synchronized in their frame), which means each observer will see other observers' rulers shrunk relative to their own, their clocks slowed down relative to their own, and their clocks out-of-sync.

11. May 12, 2006

### Enos

Hmm... Thanks for the input and the link. I was aware that nothing was technically going faster than light and perhaps my choice of words when saying "faster than light" was wrong. But with this type of traveling one can travel across great distances away from earth and return without a costly time dilation right?

12. May 12, 2006

### JesseM

What do you mean by "this type of travel"? Your scheme just involves objects moving apart, not moving apart and later coming back together. Anyway, if something moves away from the earth and later comes back, to figure out how long the trip took all that matters is the object's own speed in the earth's reference frame, the movement of those intermediate lifts is irrelevant. If its speed is some constant v on both legs of the trip, then whatever time t it takes in the earth's frame, the time dilation will mean it will have taken only $$t * \sqrt{1 - v^2/c^2}$$ according to the object's own clocks. And if it travelled to a star whose distance was d light years away in the earth's frame, then the trip there and back will have taken a time of t = 2*d/v years in the earth's frame. For example, if it was going to a star 100 light years away at 0.5c in the earth's frame, it won't get back to earth for 2*100/0.5 = 400 years according to earth-clocks...but according to its own clocks the time would only be $$400*\sqrt{1 - 0.5^2}$$ = 346.4 years because of time dilation. And if it was going to that same star at 0.999c, it would get back to earth 2*100/0.999 = 200.2 years later according to earth-clocks, but only $$200.2*\sqrt{1 - 0.999^2}$$ = 8.95 years later according to its own clocks. The time according to its own clocks can be reduced as much as you like thanks to time dilation, but the time according to earth-clocks for it to travel to a star 100 light years away and back can never be less than 200 years.

Last edited: May 12, 2006
13. May 12, 2006

### Enos

Thanks for the replies, another mistake I made was that the force needed to push this type of machine at that velocity would require more then what is theoretically possible.

14. May 12, 2006

### DaveC426913

The primary factor you need to consider is JesseM's post #10: velocities do not add linearly.

At classical speeds (say, Mach 1) the numbers add the way we expect:
.5Mach + .5Mach +.5Mach = 1.5Mach
.5Mach + .5Mach +.5Mach +.5Mach = 2.0 Mach
.5Mach + .5Mach +.5Mach +.5Mach +.5Mach = 2.5Mach

At relativistic speeds (near speed of light), the numbers do not simpy add:
.5c + .5c +.5c = ~.9c
.5c + .5c +.5c +.5c = ~.99c
.5c + .5c +.5c +.5c +.5c = ~.999c

No matter how many hydraulic lifts you put in your chain, the sum of their speeds will always add up to less than c.

Last edited: May 12, 2006