Recognitions:

## Implications of the statement "Acceleration is not relative"

 Quote by GregAshmore 1. An absolute quantity cannot be dependent on a non-absolute quantity.
An absolute quantity cannot be defined in terms of a non-absolute quantity, but it is often convenient to use non-absolute quantities for calculating about absolute quantities, and then the mechanics of the calculation may depend on a non-absolute quantity. For example:

The distance between a ship on the surface of the ocean and the location of an iceberg is an absolute quantity; either that distance is zero and the ship is sinking or it's non-zero and the ship isn't sinking. However, when the coast guard broadcasts an iceberg warning, it uses non-absolute coordinates (latitude and longitude, zero longitude is chosen based on an accident of British maritime history) to identify the location of the iceberg; and it's up to the ship's captain to calculate the absolute distance between his ship and the iceberg.

The ship's captain uses a formula involving the (non-absolute) latitude and longitude to calculate the (absolute) distance so, it's easy to make the mistake of thinking that the distance is defined in terms of latitude and longitude. In fact the absolute distance is defined by the two absolute points (location of ship and location of iceberg) and the latitude and longitude values were determined by those points.

Recognitions:
 Quote by GregAshmore The definition of proper acceleration has been given as "calculated along the path of the worldline." This definition is ambiguous because it does not define how the worldline is constructed.
True enough, but of very little practical significance as there is no serious disagreement as to what a worldline is nor how to construct them.

What is confusing is that there are different ways of drawing them, according to the coordinate axes you use. For example: The worldline of a particle hovering just outside a black hole will look like a vertical straight line on a piece of paper if you use the Schwarzchild t coordinate as the vertical axis and the r coordinate as the horizontal axis. Use K-S coordinates for the axes and the worldline will look like a hyperbola on your sheet of paper. But we're talking about the exact same set of points in spacetime either way.

 Quote by GregAshmore True, the physical content does not change in this example.
The physical laws are never changed by changing coordinates. Coordinates are just labels we give to points in space and time. Whether we label points by (x,y,z), or by latitude and longitude and altiude, or by (r,θ,$\phi$), can't make any difference to the physics.

 This is not true in the case of the resting rocket. In that coordinate system, no force acts on the Earth, yet it accelerates.
But coordinate acceleration isn't physical. Or at least, it's only partly physical. An object's coordinate velocity can change because the object is being acted on by a force, but it can also change because your coordinate system is curvilinear or noninertial. The physically meaningful quantity is not coordinate acceleration, but acceleration relative to the inertial paths.

Mathematically, proper acceleration, which is the physically meaningful quantity, is expressed as:

$A^\mu = \dfrac{d U^\mu}{d \tau} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda$ where $U^\mu$ is proper velocity, and $\Gamma^\mu_{\nu \lambda}$ is the so-called "connection coefficients" that are different for different coordinate systems. The two pieces of the proper acceleration
$\dfrac{d U^\mu}{d \tau}$
and
$\Gamma^\mu_{\nu \lambda} U^\nu U^\lambda$
are not physically meaningful by themselves, but the combination is physically meaningful.

Recognitions:
 Quote by GregAshmore (I suppose that the worldline of the rocket is to be drawn with reference to an inertial frame
Not necessarily. Drawing the worldline just requires choosing some convention (aka frame) for assigning coordinates to points on the worldline; then draw coordinate axes on a piece of paper; and start plotting points using these axes. There's no requirement for an inertial frame here.

(It is true that it's generally easier to draw straight lines, and in flat spacetime the worldline of an object that is experiencing no proper acceleration will be a straight line using an inertial frame and Minkowski coordinates, so we tend to use these a lot. But that's just a convenience).

 Quote by GregAshmore True, the physical content does not change in this example.
I should point out that if you buy that the physics is not changed when you go from Cartesian coordinates to Polar coordinates, then it's exactly the same type of change in going from inertial coordinates to noninertial coordinates.

In rectangular coordinates, the path of an object traveling inertially is given by:
$\dfrac{d^2 x}{dt^2} = 0$
$\dfrac{d^2 y}{dt^2} = 0$

In polar coordinates, the same path is given by:
$\dfrac{d^2 r}{dt^2} = r (\dfrac{d \theta}{dt})^2$
$\dfrac{d^2 \theta}{dt^2} = -\dfrac{2}{r} \dfrac{dr}{dt} \dfrac{d \theta}{dt}$

Since $\dfrac{d^2 r}{dt^2}$ can be nonzero even with no physical forces acting, an object will "accelerate" without any physical cause for that acceleration. An object's radial velocity is not constant, in general, even with no forces acting. The physically meaningful acceleration is not $\dfrac{d^2 r}{dt^2}$, but the combination
$\dfrac{d^2 r}{dt^2} - r (\dfrac{d \theta}{dt})^2$

 Quote by GregAshmore Back to the original scenario. 1. Prior to the firing of the rocket, if you select the specific coordinate system, do you make both the Earth and the rocket move, and in such a way as to maintain unchanged the distance between them?
To get into the "spirit" of relativity, you should think in terms of everything moves. For any object whatsoever, if it waits a second, it's at a different spacetime location than it was a second ago. So everything has a nonzero velocity through spacetime. But you can choose coordinates so that the spatial component of velocity is zero for some object.

 Quote by stevendaryl To get into the "spirit" of relativity, you should think in terms of everything moves. For any object whatsoever, if it waits a second, it's at a different spacetime location than it was a second ago. So everything has a nonzero velocity through spacetime. But you can choose coordinates so that the spatial component of velocity is zero for some object.
There's always this visualization from Epstein:

"The reason you can't go faster than the speed of light is that you can't go slower. There is only one speed. Everything, including you, is always moving at the speed of light."

Recognitions:
 Quote by 1977ub There's always this visualization from Epstein: "The reason you can't go faster than the speed of light is that you can't go slower. There is only one speed. Everything, including you, is always moving at the speed of light." http://www.relativity.li/en/epstein2/read/c0_en/c1_en/
Or "A watch is to time as an automobile odometer is to distance; if the time on your watch is changing, you're moving; and the direction is forwards in time". This isn't exactly rigorously scientific, and some people dislike the analogy.... But it is one way of interpreting the constant and non-zero magnitude of the four-velocity.

Mentor
 Quote by GregAshmore You have said that you can cause motion by choosing a specific coordinate system. I am asking questions about what happens when that specific coordinate system is chosen. You can't avoid answering the questions by attempting to use some other coordinate system(s). ... You have chosen a "specific" one coordinate system that you have chosen
OK. If I am the one choosing the specific coordinate system then the one I would choose is the rocket's radar coordinates, as described in the Dolby and Gull paper I linked to earlier.

 Quote by GregAshmore 1. Prior to the firing of the rocket engine, you select the specific coordinate system. Do you make the rocket move? 2. While the engine is firing, you select the specific coordinate system. Do you make the rocket move?
No, the rocket is always at x=0, by definition, and therefore it never moves since dx/dt=0 always.

Also, the radar coordinate system covers the entire spacetime, so I only select it once, I don't make any new selection before during or after firing the engine.

 Quote by DaleSpam OK. If I am the one choosing the specific coordinate system then the one I would choose is the rocket's radar coordinates, as described in the Dolby and Gull paper I linked to earlier. No, the rocket is always at x=0, by definition, and therefore it never moves since dx/dt=0 always. Also, the radar coordinate system covers the entire spacetime, so I only select it once, I don't make any new selection before during or after firing the engine.
Thank you for the further information. This gives a much different impression than you have given so far. Up to now, you have made it sound as though the act of selecting the coordinate system at the appropriate time is what causes the motion of the Earth.

[Edited to remove reference to an earlier discussion on this forum.]

In my opinion, it is wrong to say that a choice made by an analyst is the cause of anything in the system being analyzed. The physical system will behave according to the laws of nature, regardless of how, or whether, the analyst chooses to go about his business. The analyst is a spectator of the scene, not an actor in it. (Unless he happens to also be the one firing the rocket.)

You may disagree as to the use of the term "cause"; that is of course your right. But you might think about stating the case for causation in a way that emphasizes the properties of nature rather than your prerogative to choose how you analyze nature.
 Thank you all for the details on how proper acceleration is calculated. From this moment on, I am by [my] rule not permitted to speak further on the subject until I have learned to do the calculation for myself. This will do it for me on this thread. I learned a lot. Hopefully I will show a bit more competence as I move forward with study and especially working of problems. I owe George a rework of my analysis of the twin paradox. I'll post it when it's done--could be a week or two.

Mentor
 Quote by GregAshmore Thank you for the further information. This gives a much different impression than you have given so far. Up to now, you have made it sound as though the act of selecting the coordinate system at the appropriate time is what causes the motion of the Earth.
It is the selection of the coordinate system which causes the motion of the earth. I don't know what you think that I have said differently now than I have at any time previously.

Perhaps you were simply not aware that coordinate systems on spacetime cover both space and time in a single coordinate system? I don't know how you could be unaware of that fact in a discussion about spacetime, especially given the references I and others have provided. Particularly the Dolby and Gull reference which I have repeatedly recommended and which clearly spells out how to develop such a coordinate system.

 Quote by GregAshmore The physical system will behave according to the laws of nature, regardless of how, or whether, the analyst chooses to go about his business. The analyst is a spectator of the scene, not an actor in it. (Unless he happens to also be the one firing the rocket.)
Agreed.

 Quote by GregAshmore But you might think about stating the case for causation in a way that emphasizes the properties of nature rather than your prerogative to choose how you analyze nature.
The point is that some things which you think belong to nature actually do not belong to nature but to the analysis itself. The choices the analyst makes don't cause any changes in nature, but they do cause changes in the analysis.

Whether or not a given object is moving is not a property of nature, it is a property of the analysis. Therefore, the analysts choices are in fact the cause.

Mentor
 Quote by GregAshmore I am by [my] rule not permitted to speak further on the subject until I have learned to do the calculation for myself.
A very wise rule. If you have trouble with the computations, don't hesitate to ask. I would not consider that "speaking further on the subject".

Also, if you use Mathematica, I can share code as needed, although writing your own is itself quite instructive.