Calculating an impossible reference frame as my own homework problem

[Ignorantsmith12]

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.

• 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?

 

[Dale]

So, if this is supposed to be standard special relativity then the size of the universe is infinite. So any fraction or multiple of infinity will also be infinite.

 

[PeterDonis]

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.

I don't understand what you mean by this. What does the length in spacetime of the ship's path have to do with the apparent speed of the CMB relative to it?

 

[PeterDonis]

a ship that had been moving along with the CMB from the beginning.

It's impossible for any ship to move along with the CMB. The CMB moves at the speed of light. And CMB radiation is moving at the speed of light in all directions, so it's not even clear what "moving along with the CMB" would mean.

 

[PeterDonis]

Ship B has a path through spacetime that is as short, or shorter than the CMB

It's impossible for any ship to have a path through spacetime that's shorter than the path of a light ray. The length of a light ray's path through spacetime is zero.

 

[PeterDonis]

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.

I don't understand what this means either.

 

[PeterDonis]

How did I do?

I'm not sure your scenario can even be analyzed, since it appears to contain internal contradictions. I've pointed a few of them out in my previous posts.

Here's a suggestion: try to draw a spacetime diagram of your scenario. I'm not sure that's even possible with the scenario as you've specified it in the OP of this thread--which would mean that the scenario does indeed contain internal contradictions.

 

[pervect]

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.

Assuming impossible things, in general, leads to contradictions. In fact, given inconsistent assumptions, it's fairly well known that one can prove anything one desires.

The classic example is "Given that 2+2 =5, show that I am the King of England".

I will append the details of the proof shortly, but I want to make my point as clearly as I can first. Making incorrect assumptions leads to nonsense. It's not a tool for learning. In fact, it's important to make every effort to avoid making incorrect assumptions, because, using perfectly valid logic, one can derive ANYTHING from inconsistent premises.

That said, here's a version of the well-known proof. (It's sometimes attributed to Bertrand Russel, but a quick online check suggested his version was slightly different, and the basic idea predates Russel).

Given that 2+2 = 5 , we can conclude that 4=5. Subtract 3 from both sides, then we have 1=2. The set of me and the King has two members,, but, we've shown that two is equal to one. Therefore the set of me and the King has one member. Thus, I am the King of England. QED. (Note that one can use the same argument to show that I am the Queen of England, the Pope, or anyone else).

So - to expand you understanding, you need to find another way, one that does not involve impossiblities/ inconsistent assumptions.

One example of how you might do this is the idea of "null coordinates". For a simple example, assign cooridnates (t,x) to a two dimensional space-time. (You can generalize the argument to a 4d spaccetime by assigning coordintes (t,x,y,z) if you prefer, but I'll do the simpler case.

The details about how to go about this are rather complex, at least the ways I know of doing it are. For instance, there is something called the Newman-Penrose formulation of general relativity. But that advanced treatment probably won't help you in your goal of understanding special relativity better. My personal opinion is that while there are lessons to be learned from this approach by borrowing some ideas from affine geometry, it's not useful to someone who is struggling to learn with special relativity.

If you are the curious sort, it might be interesting anyway to know that you can mark regular intervals along a null wordline (such as a worldine where u=constant or v=constant), but that while you can mark regular intervals along such a worldine, the numerical "length" of any such intervals will always be dependent on the observer. In the usual treatment of special relativity, we focus on things, called invariants, that are NOT dependent on the observer. If you are not familiar with this treatment, Taylor and Whereler's "Space-time Physics" is a good source for a treatment of special relativity that does focus on invariants.

But interestingly, one does not actually need to be able to assign numerical lengths to do geometry - there's a branch of geometry called affine geometry that treats this.. See for instance the wiki https://en.wikipedia.org/wiki/Affine_geometry

In mathematics, affine geometry is what remains of Euclidean geometry when ignoring (mathematicians often say "forgetting the metric notions of distance and angle.

While I find this fascinating, I can't say that it will particularly help you if your main goal is to learn special relativity for the first time. Taylor and Wheeler's geometric treatment (in their book Space Time Physics) focussing on invariants is probably much more helpful if you're doing special relativity for the first time. I don't actually know your background, apologies if I've made some incorrect assumptions about your background and/or interests.