# B Can velocity be determined in flat space?

1. Nov 21, 2016

### Chris Miller

Would there be any way in a universe expanded into flat space, with no celestial objects visible, for a spacecraft to determine its velocity? It could know its acceleration was say 10 M/sec in a certain direction (though it could also be decelerating), and from this determine its minimum velocity. As it approached c, what would change from its frame of reference?

2. Nov 21, 2016

### Staff: Mentor

What does this mean? What cosmological model are you referring to?

Velocity is always relative; if there are no other objects visible, there's nothing to define velocity relative to.

If by "acceleration" you mean proper acceleration--the acceleration felt by the crew of the spacecraft and measured with an accelerometer--then yes, this can be known by local measurements (the accelerometer I just referred to), even if there are no other objects visible.

No, the direction of proper acceleration is definite; you can measure that locally (just measure which end of the spaceship is "down", which end you can stand on the way you would stand on a floor on Earth).

How? There is no such thing as "minimum velocity".

Relative to what?

3. Nov 21, 2016

### Ibix

It's worth noting that you can't just accelerate. You must have an exhaust plume or something, so you'll have an external reference once you start accelerating even if you didn't have before.

4. Nov 21, 2016

### Chris Miller

True, but all the plume would be good for is confirming your rate of acceleration?

5. Nov 21, 2016

### Chris Miller

Thanks Peter.

Flat space? A universe expanding towards darkness and flatness (under Hubble's "constant").

So velocity is meaningless then? You could (measurably) accelerate at 10 M/sec/sec forever?

I just meant that if you assumed a starting velocity of, at minimum, 0, then you could use your acceleration history to determine a minimum velocity.

Your starting frame of reference (before any acceleration).

6. Nov 21, 2016

### pervect

Staff Emeritus
Special relativity would be the obvious example of a flat space (and flat space-time). I believe we've already answered your question in the context of special relativity, there's no way to determine your velocity with respect to "space".

So if you're basically asking the same question again, hoping for a different answer, the answer isn't any different, it's still no. If you are asking a different question, I may not be understanding the intent of your question.

Pre-relaltivity physics did have the concept of an "ether", and according to those pre-relativistic theoreis, it would be possible to find the absolute velocity of an object. relative to this hypothetical ether. However, when people tried to actually perform the experiments that would tell them the absolute velocity of the Earth - they always got zero. Since the Earth is orbiting the sun, this was not expected. Many ingenious efforts were made to try to modify and save the ether theories, but none of them were successful in explaining all experimental results. The conclusion we've drawn from experimental results to date is that special relativity is correct, which implies that the answer to your question is "no" and will remain "no".

Look at our FAQ's on experimental test of special relativity if you want more detail on the extensive set of experiments that have tested the theory.

7. Nov 21, 2016

### Chris Miller

Not at all trying to challenge SR. E.g., my GPS would be useless without it. Just trying to get my head around the whole no universal frame of reference thing. I get that every point in the universe is associated with its own now, but don't they all lie on the same (expanding) 4D "surface"? Still can't figure what happens to this relative velocity-less spaceship after a year of 1G acceleration...

8. Nov 21, 2016

### jbriggs444

A spaceship has a perfectly well defined velocity relative to any inertial frame you specify. In particular, if you accelerate at 1 g for one year, you end up with a well defined velocity relative to the inertial frame in which you started out at rest.

9. Nov 21, 2016

### Chris Miller

Thanks jbriss (aka Schultz), and 1 year at 10 M/sec/sec > 300,000,000 M/sec? This is my problem. It's been made clear that everything in my frame feels "normal," if which, then I'd feel/calculate myself/craft having accelerated past c? Are you saying nothing has to exist in my related frame of reference?

10. Nov 21, 2016

### jbriggs444

Google relativistic velocity addition. Adding 10 meters per second (relative to the moving body) to a velocity that is already near light speed (relative to the initial inertial frame) will not result in an increase of 10 meters per second.

11. Nov 21, 2016

### Staff: Mentor

If you are referring to our best-fit cosmological model, it is not "expanding towards" flatness; it is already spatially flat, as seen by comoving observers (and always has been).

Unless it's relative to something, yes.

You could (if you had sufficient fuel) fire your rocket engine such that you feel this acceleration forever, yes. But, as jbriggs444 has pointed out, that doesn't mean your velocity relative to an object that remains at rest at your initial starting point will increase by that amount forever. Velocity addition doesn't work that way in relativity.

12. Nov 21, 2016

### Staff: Mentor

I'm not sure what you mean by this, but see below.

No. The applicable "surface" in our cosmological models is a 3D "surface", a surface of constant coordinate time in the standard FRW coordinates that are used to describe our models of the universe. Or, to put it in more physical terms, it is a surface on which all comoving observers--all observers who see the universe as homogeneous and isotropic--measure the same "age of the universe", the same proper time along their worldlines since the Big Bang.

However, note that this has nothing to do with the issue you are having about acceleration. Your issue there is, as I and others have said, a simple failure to use the correct relativistic velocity addition formula.

13. Nov 21, 2016

### Mister T

Special relativity doesn't demand there be no preferred frame of reference. It simply tells us that it's not needed, and that there's no way to distinguish it from any other inertial reference frame.

It has a velocity relative to the frame of reference in which it was at rest before you started accelerating it. But it also has a velocity relative to every other inertial reference frame. There's nothing special about that original rest frame, or indeed any other.

Are you still clinging to the notion that the rest frame in which you were born is somehow special? It is for you, but it's not any more special or any less special than any other.

14. Nov 21, 2016

### The Bill

If the universe the spaceship is accelerating in is truly empty but for the ship, then nothing is different at all between any two times under the same acceleration.

Now, if you add an interstellar medium with its own reference frame, you have a different story. Even a very low density medium, say 1 hydrogen atom per 1000 cubic meters in its own rest frame, will become dangerous at a high enough relative velocity.

15. Nov 21, 2016

### Staff: Mentor

And defining a reference from which to measure your speed.

You could, of course, do this virtually since you don't really need an object, just a starting point which is the spaceship's chosen "original" location. After that chosen time and location, you just measure your acceleration and calculate your new speed and direction with respect to that remembered, virtual point. It isn't fundamentally different from how Navy submarines navigate by dead reckoning, just with an arbitrarily chosen starting point and reference frame. It's navigation on a blank map.
Not meaningless, relative -- just like it has been since Galileo defined it the way it is used today. Everything is fine for people when they have a reference frame to fall back on (the surface of the Earth), even if they recognize on an instinctive level that that isn't a necessary rest frame (otherwise riding on an airplane would get weird). But for some strange reason, whenever people learn Einstein's Relativity they add-in the concept of a universal rest frame when they've never needed the concept before.

 Ehh, that may be overly harsh: When people first learn about Einstein's relativity and the universal speed limit, but don't yet know about Relativistic velocity addition, it may create a logical gap that is too far to cross....that's kinda how things were prior to Einstein anyway.

Last edited: Nov 21, 2016
16. Nov 22, 2016

### Battlemage!

For me personally, it comes down to this with regard to velocity addition: it's experimentally verified that the speed of light doesn't depend on the speed of its source. If you think about that carefully, that leads to only one of two conclusions when it comes to velocity addition: either (1) light has its own weird space time laws (which makes no sense because both light and the chemicals that we are made of are governed in large part by electromagnetic laws) or (2) speeds in general don't add the way we think of intuitively (but our intuition has to be at least close for things we observe on a daily basis).

I mean, seriously think about the notion of the speed of light not depending on the speed of its source, which means inertial observers all agree on the speed of light. If you follow that very basic idea to its logical conclusion, it makes a whole lot of sense on an intuitive level that u is not just v + u'. You probably won't be able to figure out the actual formula without some solid algebra, but it should become obvious that simple Galilean velocity addition can't be right (that or there are two different sets of laws of physics, one for light and one for mechanical objects... but again that makes zero sense since light is electromagnetic and so are the atoms we are made of).

I'm not saying special relativity in general is intuitive, but if you really think about the experiments that show the constancy of the speed of light, you eventually have to come to the conclusion that something isn't right with classical velocity addition.

17. Nov 22, 2016

### Chris Miller

By "surface" I was only analogizing our 4D universe to a 3D universe in which a 2D plane has been curved into an expanding 3D shape (e.g., bubble)... and now see what you mean by our universe's surface being 3D (as the bubble's was 2D). So, thanks. But isn't every point on the bubble's surface expanding outwards along a (different?) 3rd D vector, and every point on our universe's 3D surface outwards in a unique 4th D vector?

18. Nov 22, 2016

### Chris Miller

I really appreciate all the effort and expertise my novice questions have garnered here. The notion of colliding with a single hydrogen atom at ~c being dangerous is thought-provoking, as are the Doppler shifting of even radio waves into the gamma end of the EM spectrum. (Will have to plan my trip pretty carefully.)

To my (still confused) thinking, the constancy of c in all inertial frames, waning velocity addition under constant acceleration/thrust (implying increasing mass) and length foreshortening all spring from time dilation. If my clock's running relatively slow then my 300,000,000 M has to be a greater distance than your 300,000,000 M for our measurements of c to be the same. Similarly my 1 second of thrust is going to be a lot more thrust than your 1 second of thrust. But in my ~c frame of reference, my 1 second (e.g., your 100 years) of thrust is still going to feel like 1 second, consume 1 second's worth of fuel, and increase my velocity by the same amount it has every preceding second. Only to you (my "stationary" observer) will it have taken 100 year's worth of time and fuel? Like, if I accelerate a particle to .999999999997 c within a stationary frame (accelerator) using external (magnetic) energy in say a minute, from the particle's "perspective" it may have been accelerated to this velocity in a nanosecond.

19. Nov 22, 2016

### PeroK

In terms of understanding SR, you are still not getting the relativity of motion. The gist of a lot of what you say is false, because in your frame of reference you are (instantaneously) at rest and it's the stationary observer who is moving at $c$ and whose clock is (to you) running slow.

If you imagine your acceleration taking place in stages of: $10ms^{-2}$ for $1s$ then no accleleration for $1s$, then repeat ad infinitum. Each time you stop acclerating, you are back in an inertial reference frame, with nothing absolute to distinguish it from your initial state. You are not absolutely getting any faster; nor are you ever absolutely in a reference frame at "approx $c$".

This is a tricky problem in a way, because there is a natural reference frame of loosely "the rest frame of galaxies". In the same way that high speed motion relative to the Earth causes real effects like air resitance and the sound barrier and the danger of hitting a "stationary" tree, so motion through space at high speed relative to the matter and radiation sources has real effects, such as very high energy radiation and high-energy collisions.

It's a mistake to take that as some sort of absolute motion.

In general, you are far better nailing SR for constant velocity before you think about accelerating frames. It's almost impossible to grasp the scenarios you are trying to understand before you have a firm grasp of SR for constant velocity.

20. Nov 23, 2016

### bahamagreen

I have never liked the velocity addition explanation that says in effect, "The velocities don't add the same way you are used to thinking for everyday movements."

In the everyday world we "choose addition" intuitively understanding that addition as an operation is cumulative, associative, and linear.
Being told that we are using the "wrong velocity addition" because v+v< 2v makes it sound like something is wrong with addition, that real physical addition is not linear, or that it is only linear and applicable for some applications but not for others or all.

21. Nov 23, 2016

### jbriggs444

They don't. Because they are not velocities within the same frame of reference.

22. Nov 23, 2016

### Mister T

Velocities don't combine the same way you are used to thinking for everyday movements.

It's analogous to combining slopes. Suppose you have a roof with a pitch of 5/12, meaning it rises 5 inches for every 12 inches of run, and you set a wedge on top of it that had the same slope of 5/12. To find the slope of the upper surface of the wedge you wouldn't add them to get 10/12. That's not the correct way to combine slopes.

What you can do is add angles. In this case it would be about 22.6° + 22.6°, or 45.2°.

But if the slope is very small compared to 1, then adding slopes will give you very nearly the correct answer. So for someone who considers these small slopes to be "everyday" slopes, adding them works.

But you were an engineer, technician, or scientist working with particle beams, your everyday velocities would be too large to add. You'd have to figure out another way to combine them, analogous to adding angles rather than slopes.

23. Nov 24, 2016

### Chris Miller

Great analogy, though I don't think bahamagreen is questioning the science, but only the vernacular. I have a similar problem with 2 objects having relativistic relative velocities each expecting the other to be time dilated. Maybe it's just that, as a programmer, I can't get my head around the statement t1>t2 && t2>t1 ever returning TRUE.

24. Nov 24, 2016

### Ibix

It won't. But t1>t2 && T2>T1 might. The times being measured in different frames aren't, in general, referring to the same range of events.

Think of two cars travelling along straight diverging roads at equal speeds. Both will see the other falling behind, so both say the other is travelling slower. It's not contradictory, they're just measuring different things.

25. Nov 24, 2016

### Staff: Mentor

It never does, and the fact that you write it that way shows that you are still missing the point.

The correct way to write it would be $t1_a > t2_a \, \&\& \, t2_b > t1_b$ where the subscripts identify the reference frame.

The point is that the time between two events depends not only on the two events but also on the reference frame. They are different quantities. All of the rules of logic and math still apply to these different quantities.