2.27 Times the Speed of Light.. ?

This “problem” is entirely flawed from the beginning.

First, however, a note. Special relativity doesn’t simply say that there is no absolute reference frame. It assumes that changing the reference frame one uses does not change the fundamentals of physics and that the speed of light is the same in all reference frames. This means that for a large range of reference frames, we can choose whatever we want and we will get the same events happening according to the laws of physics. It also means that there is no absolute reference frame that is picked out by the events in physics; i.e., there is not one special frame where everything works out or where all the laws of physics have their true form.

OK, so now to the problems.

One thing that is a prerequisite to any discussion of physics is a reference frame for measurement. In non relativistic scenarios we can assume a vague reference frame, but in relativistic scenarios we need to understand what reference frame we are using because it matters to the calculations. In this case, we are not told the reference frames in which the “2 light-seconds (2Ls) apart” is given.

In this case, I could presume that both ships are at rest relative to one another and that the distance given is in a reference frame currently at rest with respect to the ships. However, this is information that should be supplied at the start.

We have to know what a “travel clock” is. Does this measure the travel against a given reference frame? The reference frame that should have been given above but wasn’t? And what was the speed relative to the given reference frame? Is it the measurement of how long the engine was firing?

The distance between each ship is not a constant thing, it is something that depends on the reference frame that we use to determine the distance. Through choice of reference frame, we can make the distance between the ships arbitrarily small.

I fear that in the above question there is a confusion between an absolute reference frame and a reference frame as an intellectual artifact. An absolute reference frame is a particular frame that is picked out by peculiar physical laws or phenomena; this is something that is effectively ruled out by special relativity. A reference frame is an intellectual artifact that we use to describe physical situations and always has been; this is something used by special relativity. In order to describe distances and use physics, we need a reference frame.

I am afraid that I simple do not understand what this is supposed to mean.

I’m glad I’m not the only one. Nice expansion Physbang.

Nice explanation.

The question itself is paradoxical though, therein lies the problem, if you understood special relativity you would never ask that particular question in that particular way. you’d maybe say 1.98x the speed of light, but 2.27 would break the laws of physics as we know them in the first place and would be out of kilter with all measurable scientific fact.

Right, what I’m saying is that the 0.88s thing doesn’t happen.

The second postulate of special relativity is: As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

All entities involved in the example are inertial frames of reference, since two objects traveling at c towards each other are at a constant velocity, and any stationary observer would also be in an inertial frame. So, we are good to go. No violations here. It is that pesky bit in the denominator that normally gets ignored because, well, how often do we calculate velocity for things approaching c? It’s all about Einstein’s flashlight.

Now there is a surprising first statement from you. {{NOT!}}

You are given all of the information necessary to answer the question of what happened. The challenge is to do it without introducing an absolute frame which is by definition independent of the observers.

Exactly correct.

But the 0.88secs does happen. So explain how. :sunglasses:

Source?

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.