Saturday, February 11, 2012

Relatively Faster

Space shuttle moves pretty slowly and stays close to Earth.
Image credit: ISS Expedition 28 Crew, NASA
Following on from my last blog, in which I talked about travelling at an appreciable fraction of the speed of light, today I'm going to add acceleration into the mix. But first, a few other funky consequences and transformations that apply when travelling close to the speed of light.

To recap last week's post, when travelling close to the speed of light, time dilates and length contracts. That means time moves more slowly and distance shrinks. The factor which dictates how much is called the Lorentz factor and is denoted by the Greek letter gamma:

Here v is the speed of your rocket or whatever and c = 3 x 108 m/s is the speed of light.
The rate at which time appears to pass (to an outside observer) in a rocket travelling at v is given by the time that passes for the observer multiplied by gamma (which is always greater than or equal to 1). This is also called the proper time for the people inside the rocket. The apparent distance between A and B for a moving observer is given by the distance between A and B as seen by an observer at rest with respect to the two points (so that A and B don't seem to be moving) divided by gamma.

Moving on

Another funky thing that changes with speed is mass -- it increases proportionally with gamma. Well, it's sort of more accurate to say momentum, and it doesn't mean that you'll feel heavier when you're in a fast-moving rocket, but that it will take more energy or force to accelerate you further. Basically, what this boils down to is the faster you're going, the harder it is to go faster.

What about if we have two rockets, travelling in opposite directions at 0.75c (that is, three quarters of the speed of light)? The apparent speed of one rocket as seen from the other must be less than the speed of light (all speeds of massive objects are less than the speed of light in all inertial reference frames). So we need to use another transformation to work it out. Without going into too much mathematical detail, the equation we need is:

See below for slightly complicated explanation of values.
The tricky part is that now we're talking about three frames of reference, not two. There's a frame of reference for each of our moving spaceships, and the third frame which is dictating how quickly the two spaceships are travelling (in their own frame, of course, each spaceship is stationary and we don't have a problem to work out). This third frame we're going to call the rest frame. We want to work out u, which is how fast spaceship A appears to be travelling from spaceship B's point of view. U is the velocity of spaceship A from the rest frame, v is the velocity of spaceship B from the rest frame. For the equation to make sense, one of U or v has to be negative (to account for the opposite directions part) c remains the speed of light (3x108 m/s).

Whew, OK, bit complicated to keep track of things there.

Getting faster

Next up, let's talk about acceleration when travelling at relativistic speeds. So far, we've only considered things moving at a constant speed. Accelerating frames are, by definition, not inertial (since an inertial frame is defined as not accelerating), so we can't quite apply all the same assumptions to them. When you are accelerating, you are moving into a different inertial frame for each instant that your speed is changing. (Of course, when you stop accelerating, you'll stay in your last frame unless you decelerate.)

If you're in an inertial frame and a relativistic spaceship accelerates past, what acceleration does it appear to have? Well, the following formula will tell us and, it's interesting to note, the apparent acceleration changes with the ship's velocity, even though, from on board the ship, the acceleration feels constant.

I was going to include the equation, but upon further consideration, it's not terribly useful or relevant. Moving on to more practical relativistic space travel, I'd like to point you in the direction of this excellent website. The set of rocket equations as explained on that site are as follows:

Equations taken (and re-typeset) from this excellent website.

So these equations assume that the ship is travelling at a constant acceleration, a. The velocity, v, is the speed it reaches, as measured from a "stationary" reference frame -- which for the sake of brevity I'll call Earth* -- after a t-long period of acceleration. The distance over which the acceleration takes place is d and τ (pronounced tau) is the time that passes for the rocket and the people inside it (generally speaking, less time will pass inside the rocket than for people on Earth).

That inverse cosh function (also called arccosh) in the last equation is a bit of an odd one. It's short for inverse hyperbolic cosine. A good scientific calculator should have the appropriate function (you'd probably have to use the shift key to get to it) and failing that, there's always

There are a few ways to use these equations, depending on the circumstances of your spaceship.

Accelerate constantly until the half way mark, then decelerate until destination

  • So, acceleration is good for us. It maintains things like muscle mass and bone density. It's broadly a good idea to maintain Earth acceleration (9.8 m/s2 is the acceleration due to gravity).
  • If we accelerate the whole way, we'll go splat at our destination. The sensible thing, if we want to accelerate the whole time, is to accelerate constantly to the half way mark, flip the ship and then decelerate the rest of the way. (Flipping the ship is so that the floor doesn't turn into the ceiling. Obviously you'd have to stop the accelerating to do the flipping.)
  • So we use the time taken equation (the first one) but put in half the distance (because we're only accelerating 'til the halfway mark), then double the resultant time to include the time taken to decelerate (conveniently, these things are symmetric).
In general, it's easier to deal with light years (rather than meters) and years (rather than seconds) when we're talking about interstellar distances. However, to get a sensible answer out, we need to put acceleration into years and light years as well. Skipping the maths, 1g = 9.8 m/s2 = 1.03 ly/yr2 so you can use that value for a. You can also just multiply by a factor if you decide you'd like to save fuel by accelerating at only 0.5g or 0.75g. Or save time by going at 1.5g (which humans might be able to adapt to). Also, remember that the speed of light in these units is 1 ly/yr.

To work out the time taken, follow the same procedure as above, but using the time equation (whichever one you're interested in). Again, put in half the distance then double the result.

Accelerate up to a set velocity

Because accelerating for an indefinite period of time might get a bit silly and use up an unrealistic amount of fuel. Also, you'd be smashing into atoms pretty hard and starlight (and the cosmic background radiation if you end up going fast enough) would get blueshifted to higher frequencies. Both of these phenomena would require extra radiation shielding, which adds extra mass and requires extra fuel. So, let's accelerate just up to a set velocity, travel at that velocity for the bulk of the journey and then decelerate again.
  • First we need to know the distance required to reach our desired velocity (and potentially also the time). The procedure isn't too different to the first case. We do need to rearrange the velocity equation a little first. Using c = 1 ly/yr we get:
  • Throw in our final velocity and the desired acceleration and we get the time taken (from an Earthly reference frame). Throw the time into the distance equation and we get out how far we've come when we stop accelerating. Double this to account for the distance and/or time taken decelerating again, subtract that from the total distance and we're left with the distance spent travelling at a constant velocity.
  • You can work out the time that section of travel takes from last time's blog (here).

And there you have it, journey times at relativistic speeds with accelerations. Huzzah!

* Technically not inertial, but it'll do if you ignore the gravity and the motion around the sun. Theoretically we should take the sun as our standard rest frame, so you can pretend I really mean the sun when I say Earth if that makes you feel better. 

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