Assuming I had a nice and speedy space ship and i were in deep space,
at what velocity would I have to travel to experience a time dialation of 2:1 compared to deep space?
If my ship’s speedometer, affected by high velocity time dialation as well, was registering the velocity of light I wouldn’t really be traveling at the speed of light from the perspective of an inert observer due to the effects of time dialation. What would my observed velocity be if i were traveling
a: away from the observer?
b: towards the observer?
c: askew the observer?
Time passes the same for each observer. One second = one second on board a spaceship as it does on earth. It is one of the fundamental tenets of Special Relativity (if not THE fundamental tenet) that the speed of light is constant for all observers. No matter where you are or how fast you are going you will always measure the same speed for light in a vacuum.
The trick lies in how one observer views another. From earth if I was watching you in your spaceship going near light speed things would look odd to me. Your clocks would run slower, you look squished a bit and so on. However, from your perspective aboard the spaceship everything seems just fine. Clocks are running normally and so on. What’s more, from your perspective on the spaceship, it would seem as if you were standing still and it was me who was moving. Now you can see where the name Relativity came from. All your measurments are relative to the person making the measurments. There is no absolute time or speed…just reference frames.
Deep space is not a reference frame. There is no absolute reference frame. Velocity can only be measured with respect to a reference frame. If you are moving at sqrt(3)/2 relative to me, then (if I did the calculation correctly) your clock would appear to be running half as fast as mine, as I measure it. Since the situation is symmetric, my clock would be running half as fast as yours, as you measure it.
Your second question is confusing. Your ships speedometer must measure your speed relative to some reference frame not attached to your ship. (If the reference frame were attached to your ship, then your velocity would always be zero.) If it said you were traveling at the speed of light relative to some inertial frame, then it is broken.
I think that the situation mangomerlot is describing is the speed at which time dilqation allows you to travel one light year while one year of time passes in your travelling frame of reference…
if (for instance) a chain of eleven tranceiver stations were placed equidistant from each other in a line between Earth and Tau Ceti, eleven light years away,
then they could be said to be one light year apart in our frame of reference and that of an observer on Tau Ceti;
now it is possible for a ship to accelerate until the crew see one of these stations pass by per year, so that they experience a velocity of one ‘light year’ per year.
This is the only way for a speedometer to display ‘light speed’, but it is pretty arbitary, and has no more significance than a coincidence…
I seem to remember this happens at somwhere around 0.8c, but I might be well off the mark.
Working in the reference frame of those on the ship, they see the distance between the stations contracted:
L’ =L sqrt(1-v^2/c^2)
In their frame, their speed is
v=L’/t’ = (L/t’) sqrt(1-v^2/c^2)
If we want to have (L/t’) = c, then we get
v = c sqrt(1-v^2/c^2)
which gives the answer above.
Interesting - I never thought of this as “traveling one light year per year” (which of course it isn’t…)
If i were traveling at an apparent velocity of one light year per year according to the instruments aboard my supposed spaceship , what would my velocity in a meters per second ratio be from the perspective of an inert observer?
So .866c gives you a time dilation factor of 2, while .707c would only give you a time dilation factor of 1.414 (relative to the same frame you’re measuring your speed against)
You have to be careful here though. No one measures the rocket to be traveling at c.
Earth Frame
The rocket travels 10 light years in 14.14 years at a velocity of .707c
Rocket Frame
The rocket observer measures the distance of the moving Earth frame to be contracted to .707 light years, the time to be 10 years and the relative velocity between him and the earth to also be .707c
Its only when comparing Earth frame distance with rocket frame time that you get one light year per year.
Can someone tell me: if I accelerated away from Earth at 10ms[sup]-2[/sup] (to make for a comfortable trip and hopefully make the sums easier!) for 6 months, and then decelerated at the same rate for 6 months, then repeat back to Earth. 2 years passes for me aboard ship, how long on Earth?
Assume whatever makes this as simple as possible: I’m just trying to understand how this works…
Sorry I was unclear. As Desmostylus explained already, I was giving the speed at which you move “one light year per year”, not the speed which gives a time dilation of 2.
Well my first posting got destroyed - hope this works.
My website has a page that will calculate all the relativistic changes: http://www.1728.com/reltivty.htm
If you want a dilation factor of 2, merely travel at:
0.8660254037844386 the speed of light OR
161,325.2880798581 miles per second OR
259,627.88449097934 kilometers per second.
Hmmm, seems this is not in agreement with previous postings - this could get interesting.
g8rguy
I guess my initial search on this message board overlooked those answers - my mistake for not spotting them.
However, it’s nice to see I got the correct answer too - considering I made such a big deal out of the whole thing (and at a rather late stage in the discussion).
I would have entered the discussion much earlier but the Message Board is getting VERY slow - and it ate my first posting. I guess this is happening to lots of other Dopers too.
