quick question on time dilation

[theANTIcraig]

if i were to travel to a star...say 5 light years away (10 LR round trip) at 99.99999999 percent the speed of light and then come back to earth...which of the following would occur:

 

1) 10 years would pass for me (the traveler) where thousands of years would pass for those back on earth or,

 

2) a few hours would pass for me due to time and space dilation, and 10 years would pass for those back on earth

 

thanks

[JohnB]

2. Depending on how fast you accelerated.

[swansont]

The spaceship can't go faster than c, so it can't beat a light pulse that will take 10 years to make the trip. So something like option b.

[djeinstine]

A traveler can travel to the center of the galaxy and back which is 60 light years accelerating at 1g (earth's gravity) toward the speed of light and when he came back to earth 10,000 years would have passed. This is considered time travel to the future using time dilation. I don't remember the exact wording but (this came from this month's sciam - asymmetry of time)

[thedarkshade]

I'd go with option be, though it seems like a pretty rough approximation to me, but yet a more rational option.

 

You can very easily find out the Lorentz factor and using that you could quite accurately calculate how much time would take you to do the whole trip.

 

i.e. when you travel at 0.99c Lorentz factor has a value of 7 (7.1). So that means that if you want to travel to a star thats 7 light years away and you're traveling at 0.99c, then you'd actually get there in a year!

[Cap'n Refsmmat]

A year from whose reference frame?

[Kyrisch]

It would be a combination of the two. If your trip is 10 LY, and you travel at 99.99999999% c, then your trip will take, from your frame of reference, 10 years and change. When you return, at that close to the speed of light, [math]\frac{10}{0.00000000002}[/math] years will have passed.

[pioneer]

Look at it this way. If we sent a light beacon to a star that is 10 light years away it will take 10 years in our reference to reach it. A light year is how far light travels in a year. If you were sitting on the photon, it would appear to happen almost instantaneously, due to maximum time dilation. If we travel less than C, the result is the earth would see 10+ years while the moving reference would see 0+ years.

 

One of the conceptual problems has to due with relative reference. Only the reference with energy added to created velocity will have real time dilation. If you don't add energy you don't get any.

[thedarkshade]

A year from whose reference frame?

Traveler's frame of reference. It his his clock that is moving 7 times slower then ours here on Earth.

[Radical Edward]

if i were to travel to a star...say 5 light years away (10 LR round trip) at 99.99999999 percent the speed of light and then come back to earth...which of the following would occur:

 

1) 10 years would pass for me (the traveler) where thousands of years would pass for those back on earth or,

 

2) a few hours would pass for me due to time and space dilation, and 10 years would pass for those back on earth

 

thanks

 

look at it this way. At high velocity, you get length contraction, so what looks like 10 light years here on earth would not look like 10 light years if you were flying from earth to a distant star at 0.999999x the speed of light.

[freezy]

There seems to be a bit of confusion as to what the final outcome would be. I recommend looking at http://this link on the Twin Paradox as it deals precisely with this issue.

 

Using the numbers from the OP and the reciprocal of the Lorentz factor, we get:

 

From Earth's inertial frame of reference:

[math]t=2d/v

=2*5/.9999999999

=10.000000001

[/math]

So everyone on Earth would note that the trip took about 10 years, which would be expected, traveling that close to the speed of light.

 

From the pilot's rest frame:

[math]t'=\sqrt{1-v^2/c^2}[/math]

[math]t'=\sqrt{1-(.9999999999c^2/300000}[/math]

[math]t'=0.0000141421...[/math]

 

[math]0.0000141421*5=0.0000070710[/math] (time for one way trip with length contraction)

[math]0.0000070710/.9999999999c=0.00000000002[/math] (the amount of aging the pilot undergoes, one way)

[math]0.00000000002*2=0.00000000004[/math] (round trip aging)

 

So the pilot aged about 0 years. The pilot would essentially be the same age as when he/she left whereas everyone on Earth would have aged 10 years. Everyone on Earth would claim that the pilot was gone for 10 years. The pilot would claim that it took him/her no time at all to make the trip. Checking the clocks, it would show that neither is lying, thus the paradox.

 

(I'd appreciate it if someone would check my calculations )

[Motor Daddy]

So if it took almost zero time for the pilot, and he traveled the distance of ten light years, what was his velocity?

[iNow]

Relative to whom/what?

[Motor Daddy]

Relative to whom/what?

 

What was the ships velocity (d/t)?

[iNow]

How are you still not getting this? The ships velocity relative to whom/what?

[Motor Daddy]

How are you still not getting this? The ships velocity relative to whom/what?

 

I am asking what the ship's velocity was. Velocity is d/t. Distance divided by time.

Edited June 20, 2008 by Motor Daddy

[Cap'n Refsmmat]

Right. But who are we measuring the velocity compared to? For example, I could say that I'm stationary right now, compared to the Earth, but I'm moving at a ludicrously high speed compared to the sun, and the sun's moving at a high speed compared to the center of the galaxy, which is moving at some speed through the universe...

 

For calculations like this you need to define your reference point, so to speak.

[Motor Daddy]

Didn't we just establish that the ship traveled a distance of 10 light years during the trip?

[Cap'n Refsmmat]

According to the people on Earth, yes. However, length contraction takes effect at such ludicrously high speeds, changing the distance he travels, and I don't know the math to figure out what that would do to his velocity according to his spaceship's speedometer.

[Motor Daddy]

According to the people on Earth, yes. However, length contraction takes effect at such ludicrously high speeds, changing the distance he travels, and I don't know the math to figure out what that would do to his velocity according to his spaceship's speedometer.

 

So the faster he travels the less distance he travels?

[iNow]

So the faster he travels the less distance he travels?

 

Relative to whom?

[Edtharan]

I am asking what the ship's velocity was. Velocity is d/t. Distance divided by time.

You can only determine what the velocity of something is by comparing it to another object. So compared to the pilot, the ship is travelling at 0 km/s, but compared to the Earth the ship would be travelling at some fraction of the speed of light.

 

According to the pilot (in this example), the time taken will be less. This means that he must (according to his frame of reference), the distance must also be less.

 

Remember Velocity is Distance divided by Time. 300 / 100 gives the same result as 30 / 10.

[Motor Daddy]

Relative to whom?

 

Relative to the ship's speedometer and odometer, which measures distance traveled.

 

Does the odometer rack up less miles the faster the ship travels?

 

If the odometer was at zero at the beginning of the trip, what does it read after traveling 10 light years?

 

You can only determine what the velocity of something is by comparing it to another object. So compared to the pilot, the ship is travelling at 0 km/s, but compared to the Earth the ship would be travelling at some fraction of the speed of light.

 

According to the pilot (in this example), the time taken will be less. This means that he must (according to his frame of reference), the distance must also be less.

 

Remember Velocity is Distance divided by Time. 300 / 100 gives the same result as 30 / 10.

 

Correct, traveling at the velocity of 3 ft/sec, you will travel 30 ft in 10 seconds, or you will travel 300 ft in 100 seconds.

 

So what was the ships velocity if the ship traveled 10 light years at .99c (as stated in the OP?)

 

The distance traveled is known. The velocity of travel is known.

Edited June 20, 2008 by Motor Daddy
multiple post merged

[iNow]

So what was the ships velocity if the ship traveled 10 light years at .99c (as stated in the OP?)

 

Relative to what?

 

 

Hopefully, you'll start getting annoyed soon and realize why I keep asking.

[Motor Daddy]

Relative to what?

 

 

Hopefully, you'll start getting annoyed soon and realize why I keep asking.

 

How was the velocity of .99c determined?

How was the distance of 10 light years determined?