2.27 Times the Speed of Light.. ?

Of?

James: maybe I can explain it using some different numbers. You’re near Earth in your spaceship, and I’m out near Alpha Centauri in mine. We’re pretty much at rest, and we agree that we’re separated by 3.96 light years. I now start heading towards you at .99c as measured by you. You calculate that it will be four years before I reach earth. I of course now measure distance and time rather differently. As far as I’m concerned, the distance from me to Earth looks length-contracted by a factor of 7 down to .57 light years. ( 1/√(1-v²/c²) → 1/√(1-.99²) = 1/√(1-.98) = √(.02) = 1/.1414 = 7 ). I’m also time-dilated by a factor of 7, so as far as I’m concerned it’ll be 4/7 or .57 years before I reach earth.

You now start heading towards me at .99c as measured by observers on earth. The velocity addition formula says I measure our closing speed s = (v + u) / (1 + vu/c²). This works out to be (.99 + .99) / ( 1 + .99 * .99) = 1.98 / 1.9801 = .99995c. That’s how fast I would say you were travelling if I considered myself to be at rest. But if I did, I’d also have to assert that the distance between the Earth and me is shortening. We crash into each other after .285 years subjective time, which is 2 years as measured by observers on Earth. At that instant the distance from me to Earth has shortened to .285 light years, and you’ve come across it at .99995c. This “shortening” is what’s missing from your scenario.

There is no length contraction except by the measurements of someone while they are traveling.

In my scenario, no measurement except time is being taken while any motion is taking place. Length contraction doesn’t apply.

And before anyone mentions it, relativity of simultaneity doesn’t apply either. :sunglasses:

No problem re the length contraction James. The distance between here and Alpha Centauri doesn’t actually change when you travel at .99c. Your measurements of distances in the universe are altered by your motion through the universe, that’s all. Your measurements of the distance diminish as your speed increases. Some might say that this is “real”, but it’s no more real than the way measurements of size are diminished by distance. You can speed up and slow down and see a distance concertina-ing, just as you can move back and forth and see a size concertina-ing.

As for measuring time, a clock clocks up motion. When you measure time, what you’re actually measuring is cumulative local motion. Your measurements of this motion are altered by your motion through the universe.

Note that this doesn’t mean that special relativity is wrong. Just that the way people describe it is wrong, because they rely on a “mystical” non-explanation for the constant speed of light. See The Other Meaning of Special Relativity for the right way to describe it. We measure the speed of light to be the same because in simple terms, we’re made of it.

Right, so back to the question; “What happened such as to cause a measurement of 2.27c?” {{without introducing an absolute frame}}

The answer seems to be: “You wouldn’t get a measurement of 2.27c.”

Trust me enough to accept that there really is an answer that shows how easy those numbers would be real. It isn’t that complicated of a scenario. The challenge isn’t really how it could happen, but rather how it could happen without an absolute frame.

No I mean… you wouldn’t do the calculation by saying 2/.88

You are ignoring the fact that your clock would have been running slower the faster you moved… for one.

Those are the only measurements they have.

No. I am asking for you to “do the math” and show me how you do it without introducing an absolute frame.

The astronauts don’t believe the figure either.

But they USED an absolute frame in their figures… namely the speed of the clock.

The source suggesting otherwise is what I am seeking. MMP, as usual, gets it.

I should add that I don’t know the actual numbers of how motion and time are related to “do the math” on this… but in principle (ignoring other factors like contraction ect) , I imagine it would be relatively easy.

We have the distence and the numbers on the clock… Assuming the numbers on the clock are a function of parts moving at the speed of light that have to not only run their normal course, but now also move in the direction you are moving at the speed that you are moving,(effectively increasing the distence, and thus time, they need to travel in order to advance the numbers on the clock) we could calculate the speed at which you would need to travel in order to have your clock show .88 after a distence of 2Ls.

All reference frames are independent of all observers. That is what special relativity is all about. Any observer can choose to do whatever calculations they wish in whatever (inertial) reference frames they wish and everything will work out fine.

This is a fine thing for you to stipulate, but it isn’t special relativity. In special relativity, the Lorentz transformations are required to translate from one description using reference frames to another such description. It doesn’t matter whether in a particular cooked up scenario anyone is actually doing measurements. If you aren’t using reference frames, you aren’t doing physics; certainly not with special relativity.

We can do the actual numbers fairly easily, assuming that the one spaceship that “moved” did so with instant acceleration at the beginning and end of its trip.

Let’s say that, in the initial reference frame A, the position of the moving ship is 0 on the x axis and the position of the target ship is 2 on the x axis. In this reference frame, the moving ship went from t,x coordinates 0,0 to a,2 and the target ship went from 0,2 to a,2, where a is unknown.

The moving ship uses its magic engine and moves, according to its clock, 0.88 seconds.

That means that in another reference frame, A’, the moving ship went from t’, x’ coordinates 0,0 to 0.88,0 and the target ship when from t’, x’ coordinates of b,c to 0.88,0, where b and c are unknown.

We have three unknowns in this scenario. Fortunately, as this scenario involves a translation from one system to another, the unknowns are coordinated through the relative velocity of the reference frames and the frames are chosen based on the relative motion of the ships. So we really have four unknowns: a, b, c, v.

We know ds^2 = - dt^2 + dx^2 = - dt’^2 + dx’^2 . (NB: We are using a “d” here to represent a delta.)

So - dt^2 + 2^2 = - (0.88)^2 + dx’^2 .

Therefore dt^2 + dx’^2 = 4 + 0.7744 = 4.7744

dt^2 = (a - 0)2^2 = a^2
dx’^2 = (0 - 0)^2 = 0

So, a^2 = 4.7744 and a ~ 2.185.

The standard Lorentz transformation in our coordinate systems (where the speed of light = 1) to t’ is:

t’ = [1/(1 - v^2 )^1/2 ](t - vx)

This gives us

b = t’ = [1/(1 - v^2 )^1/2 ](0 - v2) = -2v/(1 - v^2 )^1/2

for the t’ coordinate of the target ship at the start of movement in the second frame.

Also useful to us is the inverse transformation:

t = (t’ +vx’)[1/(1 - v^2 )^1/2 ]

We can get from that,

a = t = (0.88 + v(0))[1/(1 - v^2 )^1/2 ], so t = 0.88[1/(1 - v^2 )^1/2 ]

for the coordinate in the initial reference frame when the two ships collide.

And the translation for the x’ coordinate is:

x’ = [1/(1 - v^2 )^1/2 ](-vt + x)

This gives us

c = x’ = [1/(1 - v^2 )^1/2 ](-v0 + 2) = 2/(1 - v^2 )^1/2

If we shorthand g = [1/(1 - v^2 )^1/2 ], then a = 0.88g, b = -2vg, c = 2g.

We know right off the bat (ignoring the approximate value for a) that g = 2.185/0.88 = 2.483.

Some algebra gets us that v = 0.917 (approximately) and that (approximate values) b = -4.554, c = 4.966.

So we know that according to the reference frame in with the two ships were originally (and finally) stationary, the moving ship travelled for 2.185 seconds at a velocity of 0.917 light seconds/second. We also know that in our chosen second reference frame, the traget ship moved from time index -4.554 to time index 0.88 (5.434 seconds) to move the distance from 4.966 to 0, a speed of approximately -0.914. Given the rounding of the figures used throughout, we can see that these cases are symmetrical, i.e., that the motion for one frame is the opposite for the motion of the other.

In no case does this reasoning appeal to a reference frame in which there is an absolute speed. We chose reference frames that were easy to work with, but we can use the Lorentz transformations to work with any inertial reference frames.

Bonus question (pretty easy): At time t’=0, how far away are the ships?

Well thank you for at least making an attempt.

But …

In that setup, you have the “moving ship” not moving. The “travel clock” as defined earlier displays time of travel. So you could not have a frame wherein t’ went from 0 to 0.88 and also x’ went from 0 to 0, unless you are proposing that perhaps he traveled in a circle?
Unfortunately,…” the rest of your math no longer applies.

Still not seeing a source.

If this were real, one ought have been forthcoming. But I’ll wait a pinch longer.

I still don’t know what you are asking. You could at least be nice enough to make your “Source?” into a complete sentence.

Your comment makes me infer that you do not know the way that special relativity (and, indeed, Galilean relativity) is used. We are free to use a reference frame in which any given body is at rest. Since the clock of the problem is the one at rest on the moving ship, this seemed like the appropriate reference frame for the clock.

If there are some other details that you left out, then you are free to actually provide the details to make this a proper physics problem if you wish. Doing so would actually highlight the substantive issues, if any, that you have with special relativity.

What Xunzian is asking is for your source for the many claims that you have made, probably specifically that one can construct a physical system as you describe in which the two astronauts meet after one moves for 0.88 seconds, that this physical system is one accepted by contemporary physicists, and that this acceptance requires the use of some absolute reference frame. Given that this example lacks enough details to be a well-posed problem, it is probable that you have copied this scenario from another source. Regardless of the origin of the scenario, your claim that it is accepted by contemporary physicists suggests that you either have a source for this claim.

I think that you would do yourself and everyone else a favor if you would leave out your self-aggrandizing insinuations.

Can you explain what “at rest on the moving ship” means?

If you (or he) professes that the situation could not occur, simply prove it by eliminating ALL possible options, the same way anything is proven. But realize that your attempt, despite the error, was at least proclaiming that such a situation could indeed occur without adding an absolute frame. Fix the error and maybe try again.

I don’t actually care if one can construct it – I’m pretty sure that one is out the window! What I need is some legitimate outside source making the same theoretical claim as exists in the OP. If that position is widely accepted, that oughtn’t be a problem. If it isn’t, well, then we have our answer!

Well, either you know how to do it or you don’t. If you do know how to use reference frames, then why put up a front of ignorance?

Well, like Galileo explained centuries ago, we can set up many different coordinate systems as reference frames. We can identify a moving ship and use it as a reference for objects that are at rest with respect to the ship. In the case of this scenario, we identify a ship that moved and we pick out a particular coordinate system with that ship for the period that it is moving in another reference frame. That in that picked out coordinate system, the “moving” ship is at rest. If you have indeed studied relativity theory, I am surprised that I have to explain this to you.

Do you deny that you took this from a source? If you did, could you please provide the source? In any case, could you please provide the details relevant to fleshing out this example and could you provide your evidence that this scenario is something accepted by contemporary physicists?

To what error do you refer?