- #1
Ignorantsmith12
- 13
- 4
- TL;DR
- To test my understanding of special relativity, I imagined a scenario in which a spaceship entered an impossible reference frame and then tried to calculate what the universe would look like to said ship. I chose an impossible reference frame to be absolutely sure what is and is not possible in special relativity. Please critique. By the way, I'm not actually in high school. I graduated a long time ago; I had to pick a prefix, and I am just that bad at math.
I’ve always been bad at math, and therefore physics, and therefore I'm ashamed. It makes me feel like I don’t belong in this universe, but I can't leave, so I keep trying to learn math, and this scenario I came up with is one example of that. I imagined a scenario involving an impossible reference frame, and then used some equations in special relativity to calculate what it would look like. I would appreciate it if people explained what I got wrong and what, if anything, I got right. Thank you for your input.
In case some of the base assumptions I am using are wrong, I listed three of them here.
In case some of the base assumptions I am using are wrong, I listed three of them here.
- The speed of light is the same in all reference frames.
- Any reference frame where the above is true is allowed by special relativity, but…
- Just because the rules of physics allow something doesn’t, on its own, mean it’s possible, as is the case with my impossible reference frame.
Let’s pretend that we have a ship that somehow achieved a path in spacetime that is so incredibly long that the CMB appears to be moving .99C in the "x" direction. So far as I know, this is not possible, since decelerating relative to the CMB just means accelerating in the opposite direction; even so, I think this is a good self-test. Let’s then assume the ship moves .6c in the "y" direction. After an Earth year of traveling in the "y" direction, it stops moving in that direction, and it speeds back up to the speed of the CMB in the "x" direction, and encounters a ship that had been moving along with the CMB from the beginning. How much wider was the universe for the ship in the "x" direction when it found this very long path through spacetime, and how much time elapsed for both ships, and do both ships have the same spacetime metric?
Calculating the “full width” sounds straightforward. First, find a random piece of empty space in the universe. The distance between our Sun and Alpha Centauri is about 3.8 light-years, I think. I’ll just say 4. Next, we take the equation for length contraction, which is $$L’=L{\sqrt{1-v^2}}$$. I already set v as a percentage of c, so that’s why you don’t see it divided by c^2. We know the coordinate length in light-years. So we divide both sides of the equation to find the proper length. So $$4\gamma$$. According to my math, if I input .99c into this equation, I get that the distance between the sun and Alpha Centauri is now 28.55 light-years, so basically the distance between objects is over seven times what it was when at rest relative to the cosmic background. Objects within the universe, on the other hand, will seem contracted by the same amount.
Calculating the time both ships experience is next. Let's start with the ship going .99c, aka ship A. $$ T=t\gamma$$. So we said one year, so that’s $$dt=1$$; square it and it’s 1. Then $$00.99c^2=98$$ $$ 1-.98=0.02$$; square root that, and you get 0.141. Ship B, the one in a super rest frame in the x direction, but moving .6c in the y direction, experiences .8 years. Let's take a moment to talk about this result, because it is odd: it apparently, but does not actually, violate the speed-of-light barrier. According to ship A, ship B moved .6 light years in .141 years. That means Ship B moved faster than light from Ship A’s perspective. Indeed, since everything in the universe but Ship B has a path through spacetime that is as short, or shorter than the CMB, the whole universe saw Ship B move faster than light. I don’t think relativity is broken, however. What happened is simply that ship B avoided traveling a lot of distance that ship A did by making its path long in the x direction, but short in the y direction. In other words, it found a shortcut. If there is any doubt, just imagine ship B with headlights, then trying to race its own light to the finish line. The light will win every time. The light speed barrier hasn’t been broken.
Now for the hard part. The check. So let's say as the two ships are about to meet, two supernovae go off, one at x=0 and the other at x=10, and at time t=1 from the perspective of the super rest x-direction frame of ship B. Let's start with ship A. So we have these two equations $$t'=\gamma(t-vx)$$ and $$x’=\gamma(x-vt)$$ So that’s $$7.09(1-.99*10)=-63.09=t$$ and for our x we get $$7.09(10-.99*1)=63.872.$$ Now, let's do this with ship B too because of that Y direction movement. $$1.25(1-.6*10)=-6.25$$ and $$1.25(10-.6*1)=11.75.$$Ok now for the check. $$ds^2=-dt^2+dx^2+dy^2+dy^2$$. For ship B we get $$–(6.25^2)+11.75^2=99$$ and for Ship A, doing the exact same thing I get 99. How did I do?