Direct line to Stephen Hawking
November 2, 2012 1:57 PM   Subscribe

Where can I find someone willing to help me do relatively simple (ha!) relativistic time dilation calculations?

MeFi answers is awesome, but I don't quite expect someone to be able to just write up the answer for me here... what I need is some place as awesome as MeFi, but with people who are experts in physics.

For the fun of it, here are the things I want to be able to calculate. (Or a simpler version, if it helps the math.)

1. A ship leaves earth and accelerates up to .5c over the course of 1,000 years. it travels for another 1,000 years, and then decelerates for 1,000 years to 0c. How many years have passed
  1. on the ship, relative to the ship?
  2. on the ship, relative to the earth?
  3. on the earth relative to the earth?
  4. on the earth relative to the ship?
  5. at the arrival spot, relative to the arrival spot?
  6. in the radio frequencies transmitted by earth that are currently being received by the ship (e.g. if they were listening to a radio station from earth, what would be the year of it's publication?)
2. The ship in question 1 was moving in a straight line away from earth. If it then started a second journey in a random direction (say, perhaps, at a 90 degree angle) how would you calculate everything from question 1 again?
posted by brenton to Science & Nature (14 answers total) 2 users marked this as a favorite
You could try Physics.SE.
posted by katrielalex at 1:59 PM on November 2, 2012 [2 favorites]

Response by poster: Thanks. Question created. I've found that stack exchange sites tend to be generally irritable when answering questions and prone to putting more energy into disqualifying the question than to actually answering it. (Unlike MeFi, one of the reasons I love it.) Hopefully I'll get some good answers tehre.
posted by brenton at 2:13 PM on November 2, 2012 [2 favorites]

Response by poster: Yep, it's already been flagged as "homework." Hopefully some moderator will advocate for the question to keep it alive.
posted by brenton at 2:23 PM on November 2, 2012 [1 favorite]

Did you happen to google for "relativistic time dilation calculator" ? There's a bunch of them online which looks like they may help.
The 3,000 year trip time - is that measured from the ship or from earth, or is it 2 different questions?
posted by anon4now at 2:27 PM on November 2, 2012

Can you comment or edit to make it understood it's not homework? Are you just learning for fun? Put a bit more context, otherwise, yes, it looks like homework.
posted by barnone at 2:27 PM on November 2, 2012

Homework is a contentious issue on the SE sites. You should add a comment or edit the question to explain that you are not a physicist but you would like to know the answer, or you will get people trying to explain it as if you were taking a course in relativity.
posted by katrielalex at 2:34 PM on November 2, 2012 [1 favorite]

This is fairly simple high school physics. The time dilation formula just uses roots and squares.

Actually, the time dilation at .5c is not going to be very noticeable. Time in the spaceship will only be going 15% slower than passes in an outside frame of reference. There isn't going to be much of a "wow" effect like in the classic twin paradox thought experiment. But, 1,000 years of Earth time would be 870 years of ship time during the time when it is cruising. Ship time during the acceleration and deceleration is going to be a lot closer to Earth time because so much of the acceleration is spent at non-relativistic speeds. And, again, even once you get to .5c, not very much is going on that is interesting in relativistic terms.

Accelerating to .5c over the course of 1,000 years is phenomenally slow. That's 0.00478 m/s^2. By comparison, a car that goes 0 to 60 mph in six seconds will have an acceleration of 4.5 m/s^2. Your ship goes 0 to 60 mph in 17 minutes. If you had your ship accelerate at the rate of 9.8 m/s^2, which would simulate the pull of gravity on Earth, the ship could get to .5c in 177 days, just under six months.

From the point of view of people on Earth, the trip would take 3,000 years. From the point of view of the people on the ship, it would be a bit less than 2,870 years. For the folks at the arrival spot, they would have waited 3,000 years for the ship to arrive.

If people on the ship were at .5c and listening to a radio broadcast from Earth (and we assume that somehow the broadcast signal is strong enough. We will also assume that red-shifting is not a problem), the music would sound a bit slow, but not grossly so. We can't get much more specific than that for specific time instances because of issues defining simultaneity. We can't say how the radio broadcast might sound at a particular point in the journey.

Changing the direction of the ship's travel to a right angle doesn't change the time dilation.

I hope this was helpful for you. While I try to keep my physics muscles in shape, it's been years since I have been in the classroom so I am sure someone will correct me if I got it wrong.
posted by Tanizaki at 3:06 PM on November 2, 2012

I asked (almost) this on the NaNoWriMo site and got some good answers. The best thing to do is plug the numbers into a relativity calculator.
posted by SMPA at 3:11 PM on November 2, 2012

The answer would depend on how the ship accelerates to 0.5c. Assuming a constant rate of acceleration relative to Earth is easiest, but not necessarily realistic if the ship is a rocket. Anyway, making that assumption:

v(t) = 0.5c * t/1000

Then, the tool you require is the equation for the relativistic proper time ("displacement", "line element", "metric"):

ds^2 = dt^2 - dx^2

(x, t) are space and time in some fixed reference frame (Earth's, say).
ds^2 is the square of the proper time experienced by an observer who moves a small distance dx and small time dt relative to the fixed reference frame.

To use this equation, you first take the square root and then "divide" everything by dt:

ds/dt = sqrt(1 - (dx/dt)^2)

This gives the rate at which time goes by for the observer relative to time on earth (ds/dt) in terms of the velocity (dx/dt).

Note that dx/dt = v(t), which you already know/assume. So you need to integrate the above from t = 0 to t=1000.

s = integral(t = 0 ... 1000, sqrt(1 - v(t)^2))

Tanizaki, I think I agree with your answer, except I get "a bit less than 2780" rather than 2870. Do you really get 2870?
posted by hAndrew at 6:54 PM on November 2, 2012

Oh, and be a little bit careful with units, just like I wasn't.
posted by hAndrew at 6:59 PM on November 2, 2012

Tanizaki, I think I agree with your answer, except I get "a bit less than 2780" rather than 2870. Do you really get 2870?

0.5c has a time dilation factor of 1.1547, so 1,000/1.1547=866. I just used 1.15 when I did the first calculation, which gives 869.56, which I rounded up to 870.

I was about to leave work and was therefore too lazy to calculate the change in time dilation for each instant over the 1,000 years that it takes the ship to go from 0 to 0.5c, and the next 1,000 years for it to slow down (and didn't know if this question required such precision). So, I just assumed a little time dilation over each 1,000-year acceleration period because relativistic effects are going to be pretty negligible for most of the acceleration periods. We don't get to a dilation factor of 1.1 until 0.42c, and at 0.3c it's just a factor of 1.05. If you did the differential for the two acceleration legs, than your answer will be more precise than mine.

I was thinking about the ship on the drive home from work. 1,000 years to accelerate to cruising speed. If it had been launched during the Battle of Hastings, it would still be accelerating.
posted by Tanizaki at 7:37 PM on November 2, 2012

There were a couple of details had forgotten/couldn't understand that were nagging at me about this. It took me a couple of days to get to it, but I went ahead and calculated each answer. I took your phrasing, "on X relative to Y" to mean "on X in Y's frame." The answers basically require applying gamma = 1/sqrt( 1- v(t)^2 ) and integrating for the trip. I posted a plot of the times measured.

1. 2779 years: S on the plot. In the ship's frame, the distance to Planet X is foreshortened by a factor gamma. In the ship's frame, the total trip is only 900 light-years (of course in the ship's frame, the planet is travelling toward it).

2. 2779 years: S on the plot. In the Earth's frame, clocks on the ship move slowly by a factor gamma. So this ends up being the same as above, though the reasoning is a bit different.

3. 3000 years: E on the plot. This is the frame of the question, so no transformation applies.

4. 2583 years: E as calc by S on the plot. In the ship's frame, clocks on the Earth move slowly by a factor gamma. So this is Earth time adjusted by gamma^2.

5. 3000 years. This is the same frame as the Earth's.

6. yr 2000: radio time on the plot. This is E as calc by S, but with an additional factor of (1+v(t)) to account for the time delay as the ship moves. At the end of the trip, the Earth and ship are in the same frame, so the year of the signal broadcast has to match up with the Earth's. Of course, the astronauts would say their "Welcome to Planet X" message was broadcast 779 years earlier, while Earth would say 1000.

This is pretty cool, applying travel time delay to the incorrect calculation of Earth time in the ships's frame gets you back to the correct radio time in the Earth's frame and everything matches up. Hey, physics works!
posted by gimletbiggles at 2:54 PM on November 4, 2012

Response by poster: The answers here are mind boggling, both in the sense of being really helpful and way above-and-beyond the call of duty for an answer, and in the sense of blowing my mind with crazy time warping. I am going to have to take some time to digest these answers!

I was thinking about the comment regarding the ships acceleration to 60mph taking 17 minutes, which has got me to thinking what the zero-to-sixty on an aircraft carrier is. I'll have to wait a week to ask on mefi.
posted by brenton at 4:19 PM on November 6, 2012

« Older My pen needs ink   |   Who else dances like Nathan Barnatt? Newer »
This thread is closed to new comments.