length contraction & time dilation

monp

Using the concept of length contraction and time dilation explain how is it possible to travel a distance of 1mln light years during the life of a single person?

thanks for any help

jambaugh

Is this a homework problem or are you just curious?

There are two approaches.

Hint 1: The life-time they're talking about is the life-time of the person doing the traveling i.e. going really really fast in the context of special relativity!

Hint 2:The distance they're talking about is the distance as measured by someone not making the trip and so not moving really really fast.

nrqed

As per the rules of the forum, you must show some of your thoughts or work before being helped.


To get you started: consider the situation from two different point of views.

Let's say from Earth, looking at a spaceship travellling at a huge speed toward a star one million light years away. How could the people make it to that star in their own lifetime even though the spaceship is not travelling faster than c? (describe things as seen from Earth, without changing frame!)

Then look at how things look from the point of view of the people in the spaceship.

monp

i' ve calculated that the light need 3,15*1010^13seconds (which is actually about 998858 and nearly six months)to travel a distance of 1mln light years( which is 9,46*10^21meters). The lifetime of single person is about 70 years, which is 2,207*10^9seconds. The person to travle that distance in 70 years should have a velocity much bigger than speed of light which is actually impossible for me, also when this person reach the speed of light he or she will have the gravity of black hole, so I really can't imagine how is possible to travel such a big distance. please give me some explanation.
greatings

Doc Al

Realize that for a moving observer, distances will appear contracted. Say the distance from Earth to Planet X is 1 million light-years (as measured in the earth-planet frame). If you are zooming along in your rocket at a speed of 0.99c (with respect to Earth), what would you measure that distance to be? How long would it take you to cover it, according to your clocks?

What if your speed was 0.999c? Or 0.9999c?

And so on....

malawi_glenn

Light need one milion years to travel one milion light years.

monp

which is 3,15*10^15seconds.

according to Doc Al's if i use the equation for the length contraction 'my road' will really shorter but it'll also be very long distance:/([9,46*10^21*(1-0,9999^2)^1/2]=1,33*10^20)don't have any new ideas

malawi_glenn

yes of course, but why convert to seconds? Then convert 70years to seconds.. :)

Doc Al

For a "stationary" observer who measures the distance to be 1 million light-years, the time for light to travel will be 1 million years. But not for a moving observer, who sees that distance "shortened" due to Lorentz contraction.

malawi_glenn

yes of course.

jambaugh

monp,
Again is this a homework problem or are you just curious?

Keep in mind that the word "relativity" in Special Relativity means that distances and durations you measure or define are relative to your velocity. You see two stars as 1 million ly apart. I --when in motion relative to you-- will see their distances as different. You seeing me traveling between these stars note that it takes me > 1 million years. I see time passing differently.

How different will depend on my speed relative to you.

You need to find a book or webpage on SR and work this out for yourself or you will not gain any understanding. Basically you look at the two events, Me leaving Earth, and Me arriving at the distant star.

You having stayed on Earth see these events as being [itex]\Delta x=1000000 ly[/itex] apart in distance and [itex]\Delta t = 1000000/V[/itex] years in duration where [itex]V[/itex] is my velocity in light-years per year (< 1 and so the speed of light c= 1).

Now work out what my [itex]\Delta x'[/itex] and [itex]\Delta t'[/itex] are using the Lorentz transformation formulas. Especially work out [itex]\Delta t'[/itex] in terms of [itex]\Delta x = 10^6 ly[/itex] and [itex] V[/itex].

Or better yet, work out the "effective velocity" (how far I travel as you see it divided by how much time I see passing) in terms of the actual velocity:

I give it here in an obscured form in terms of hyperbolic trig. You should work out the explicit formula and see if via calculator we get the same numbers:

[tex] V_{eff} = \cosh(\tanh^{-1}(V))[/tex]
[tex] V = \tanh(\cosh^{-1}(V_{eff}))[/tex]

But given I want to make the trip in say 10 years (my time) I need an "effective velocity" of 100,000 light-years per year. This then gives

[itex] V = \tanh(\cosh^{-1}(10^5)) \simeq 0.99999999994999999999874999999994 [/itex] light-years/year i.e. % of the speed of light.

Note that the actual velocity is always less than 1=speed of light in these units.

BTW, Here is a much harder related problem. Assuming someone can only stand say 4 g's of sustained acceleration, how far could they travel in their lifetime of say 100 years assuming they don't need to slow down at their destination? (Note that [itex]1g \simeq 1.03 \frac{ly}{y^2}[/itex])
Assume they start from rest in the frame where we are defining the distance traveled.