# Frame Independance of acceleration ?

1. Jun 30, 2010

### Austin0

It seems clear that all inertial frames would agree that a frame was accelerating based on instantaneous (short interval) relative velocity measurements.
That all frames would agree that the velocity was changing over time relative to their own frame but it is not so clear if they would all agree on the quanttitative rate of change.
So assuming a common rest metric in all frames would they calculate the same coordinate acceleration I.e. the same N (m/s /s ) ???
Logically it seems that they would not ,,but thats just a guess so I am hoping for the real answer. Thanks

2. Jun 30, 2010

### Ich

I don't know what you mean by "a common rest metric in all frames", but coordinate acceleration is frame-dependent - contrary to http://en.wikipedia.org/wiki/Proper_acceleration" [Broken].

Last edited by a moderator: May 4, 2017
3. Jun 30, 2010

### Austin0

I meant that all frames would share the same metric if they accelerated to a state of rest wrt each other.
SO it sounds like you are saying that inertial frames with differing relative velocities would calculate quantitatively different coordinate acceleration rates for the same accelerating frame ,correct ?

How is proper acceleration frame independant? Simply based on internal accelerometer readings.?
But not as measured by inertial frames or by internal calculations of acceleration based on measurements of instantaneous velocity relative inertial frames , right ?
Thanks.

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4. Jun 30, 2010

### my_wan

Yes, all observer, unlike velocity, will agree when something is accelerating or not, but will disagree on how fast it is accelerating. Proper acceleration is the acceleration as measured by the object being accelerated, such as an accelerometer.

5. Jun 30, 2010

### Ich

Again, I'm not sure I understand what you mean. Proper acceleration is a http://en.wikipedia.org/wiki/Four-acceleration" [Broken], which is a frame-independent ("covariant") mathematical object. You can calculate it, given the momentary velocity and coordinate acceleration in an inertial frame.

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6. Jun 30, 2010

### Austin0

OK I will try again. Given an accelerating system .
Observing inertial frames would calculate different quantitative coordinate accelerations but would all agree on calculated four-acceleration. Yes maybe??

Given constant internal proper acceleration would the 4-acceleration as calculated in internal frames be constant?

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7. Jun 30, 2010

### Ich

Yes. (disclaimer: just like it is the case with any vectors, they would agree on the magnitude of that vector, but not its direction. That's obvious if you rotate the coordinate system, and it's true for Lorentz transformations also)
Yes.

8. Jun 30, 2010

### Meir Achuz

"Given constant internal proper acceleration would the 4-acceleration as calculated in internal frames be constant?"
No, its components would change like a four vector, and the three-vector acceleration we measure have more complicated transformation.

9. Jul 1, 2010

### Ich

Right, only the magnitude is constant.

10. Jul 1, 2010

### Austin0

If I am understanding you correctly (???) then:
Corrdinate acceleration is based on raw data. The normal acceleration math applied to the direct instantaneous relative velocity measurements; no transformation involved.
If a gamma is derived from an instantaneous v and the coordinate acceleration is transformed you get the four -acceleration vector. The magnitude of which is the only relevant component and is invariant.

This raises another question:
Based on the assumption that the coordinate acceleration of an accelerating frame (AF) would diminish over time in all relative inertial frames wouldn't this mean that the time component of the 4-acceleration would increase relative to the spatial components?
Wouldn't it follow that the magnitude of the derived 4-vector would also diminish??

Given an AF initially at rest wrt inertial frame IF(R) :
Wouldn't the magnitude of the 4-vector be a function of where along the acceleration course the measurements were taken?

Looking at AF after initiation of acceleration in IF( R) the coordinate acceleration profile would be essentially flat with increasing negative slope as v--->c.
At the instant of initiation all other IF's would agree on the onset of acceleration [although disagreeing on the location and time]
But relative to their frames the point on the acceleration curve would be different.
In their coordinates the acceleration would begin not from rest but from some initial relative velocity.
It would be equivalent to [effectively indistinguishable from] a frame that had started at rest in their frame and was at a different point in the acceleration curve.

If this is at all accurate wouldn't it mean that if you compare measurements in IF(R) with with measurements in another frame with signiificant relative velocity IF(2)
at initiation (PI) and at a later point (P2) when the profile is still relatively flat in IF(R)
is it unreasonable to assume that :
1) The coordinate acceleration measured in IF(R) would have diminished less than the coordinate acceleration measured in IF(2)

2) The IF(R) 4-vector magnitude would have also diminished less than the magnitude of IF(2) 4-vector.

Have I gone astray somewhere here?
Thanks to all . The responces have been very helpful

11. Jul 1, 2010

### Ich

In a way, yes.
The magnitude is not a component.
It diminishes only in frames where v is increasing. Independent of the value of (coordinate or proper) acceleration, the time component increases faster than the space component.
No.
No.
the profile wrt velocity? Essentially yes, but getting flat again as v->c. $a=\gamma^3 A$, with A=proper acceleration.
You lost me here. Maybe all this is based on the wrong assumption that proper acceleration is a function of time or velocity?

12. Jul 2, 2010

### Austin0

Based on the assumption that the coordinate acceleration of an accelerating frame (AF) would diminish over time in all relative inertial frames wouldn't this mean that the time component of the 4-acceleration would increase relative to the spatial components?

If A is constant and the magnitude of the 4-vector is constant I dont understand how the time component could increase relative to the spatial component without changing the magnitude???

I have given this all a lot of thought and some reading up on proper A and I think I understand the basic concept and where I went wrong .
I started with the premise that at some point as v---->c all frames would measure coordinate acceleration as 0. Reach a practical limit of measuremt of velocity change.
That any frame observing after that point would consider the AF as an IF.
At that point i inferred that the measured and calculated (A) would neccessarily also be 0 Working back from this point I assumed that at some point (a) would be detectable but vanishingly small and increase infinitesimally 0---> and that the calculated A would also increase infinitesimally 0---> . If I am now understanding correctly as soon as (a) was detectable at all (A) would immediately go from 0 to the full initial value.
Or am I still missing it?
In any case thanks for your input it helped
.

You lost me here. Maybe all this is based on the wrong assumption that proper acceleration is a function of time or velocity

13. Jul 2, 2010

### Ich

The magnitude is $\sqrt{A_x^2-A_t^2}$. At increases faster than Ax, but the (eventually infinitesimal) difference still gives a=const. IOW, as v->c, the vector becomes (almost) lightlike.
Yes.

14. Jul 2, 2010

### brainstorm

Acceleration due to gravity does not seem to be frame-independent to me. An object accelerating due to gravity has to be moving relative to the source of the object endowing it with gravitational force, no? If it is moving in orbit at constant altitude, its acceleration is expressed as distance from the object it is orbiting. If it is losing altitude, its acceleration is expressed in decreasing distance to the center of the attracting body, no? How is that frame-independent?

15. Jul 2, 2010

### djy

In the framework of GR, gravity is not a force; it is spacetime curvature. Absent some other force (like an electromagnetic field), objects move along geodesics. By definition, if an object is moving along a geodesic, then the magnitude of its proper acceleration is zero, and this is a frame-invariant scalar.

16. Jul 2, 2010

### brainstorm

I can accept the idea that objects traveling through curved spacetime are independent of the object responsible for the gravity they traverse since those objects' gravity and the spacetime between them can be interpolated as the spacetime itself. However, I don't see how this changes the fact that any frame's duration is limited by whether it is converging toward another mass and the duration to impact.

I'm not familiar enough with the idea of a "geodesic" to know if it applies to all curved trajectories through spacetime or only those that don't converge toward gravitational fulcrums. If the term refers to paths that escape convergence with other bodies by curving between them, then I agree that these objects are in a sense "still" and therefore frame-independent.

17. Jul 2, 2010

### djy

I'm not sure what the 'duration to impact' has to do with the instantaneous acceleration experienced at any given time.

Assuming no forces besides gravity are involved, then all particles are always moving along geodesics. I'm not sure what you mean by 'escape convergence', but I'll suppose that you mean that the trajectory is an escape trajectory. Again, regardless of the type of orbit, the path through spacetime is a geodesic and the magnitude of proper acceleration is zero.

I'm also not sure what you mean by 'objects are in a sense "still"'.

18. Jul 2, 2010

### brainstorm

True, but what I mean is that an object's frame may be described as independent of the object its about to crash into, but that frame will be rather irrelevant once the crash occurs. The frame of an object orbiting at constant altitude of other object(s), on the other hand, is more or less permanently immune from collision and therefore more frame-independent, I would say, except for the fact that its relative isolation is due to its relationship with the object it's orbiting and that object's relationship with other objects it's not crashing into.

By "convergence," I mean when two or more objects accelerate into each other due to gravity, or are on a convergent course. Objects whose momentum causes them to maintain a constant distance from one another could be called "balanced" and those moving away from each other could be called "divergent," I think, except to the extent they could be moving in elliptical orbit in which case their divergence is a build-up of potential energy to be expressed as convergent motion at a later moment. Do you find these terms problematic?

I have the sense that objects that have achieved constant altitude from other object(s) can be described as "still" insofar as they are not converging or diverging from other objects. I say this because I cannot determine a better situation to describe as "still" or "at rest." An object on Earth "at rest" is in fact inhibited from expressing its gravity as motion by friction, which causes it to push against the ground in an attempt to further proceed in the direction of the Earth's center.

An object in a geodesic (if I'm using this term correctly), on the other hand, does not have any innate potential to decrease its distance from any other bodies, so I feel it can be described as "still" or "at rest." Does that make sense?

19. Jul 2, 2010

### djy

Based on reading your posts I'm not sure you understand what a reference frame is. It is simply a coordinate system -- an assignment of labels (coordinates) to points in spacetime.

Frame-independent just means that a quantity is the same in all reference frames. It has nothing to with whether objects associated with the reference frames "crash" or otherwise cease to exist.

An object in a perfectly circular orbit is moving along a geodesic and experiences no acceleration. An object sitting on the Earth, however, also maintains a constant distance from the Earth's center, but it is (as you said) primarily due to electromagnetic forces between the molecules of the object and the molecules of the Earth. It does experience an acceleration of 9.8 m/s^2, which is manifested as a curved (as opposed to geodesic) path in spacetime.

That acceleration, like all proper acceleration, is frame-independent: the proper acceleration of the object is measured as 9.8 m/s^2 in all reference frames.

That's not correct. Any free-falling object is moving along a geodesic. A free-falling meteor, for example, moves along a geodesic toward the earth until it reaches the atmosphere, when electromagnetic forces (friction) accelerate it, deviating it from its former geodesic path.

20. Jul 2, 2010

### brainstorm

System of points/coordinates relative to what? How can you label points in spacetime abstractly when spacetime itself is the product of gravitation and dynamism among objects?

If the Earth and moon were assigned as two points in a spacetime system, and the moon was subsequently decelerated relative to the Earth, the two coordinates would begin accelerating toward each other until they converged, would they not? In that case, would you still measure spacetime in reference to the Earth and a point estimated to approximate the previous orbital altitude of the moon?

Here I follow you until you contrast "curved" with "geodesic." "Geodesic" seems to refer to any path of least resistance through gravitationally-contoured spacetime. Does "curved" refer to the constant deceleration of the object on Earth relative to its gravitational acceleration? Why is that called "curved?"

Ok, so geodesic refers to any path through spacetime caused by an object's unimpeded momentum.

What I don't get is if you view a "frame" as an arbitrarily assigned set of coordinates unrelated to actual objects moving in relation to one another, why wouldn't you be 100% free to define a frame however you please? To me, a frame seems like it should refer to an actual configuration of objects moving relative to each other.

In that sense, an object converging with another object can only be doing so within an unadjusted frame to the extent that the frame is based on other objects that are not converging. For example, a meteor could crash into the Earth within the frame of the Earth and moon without the frame becoming nonsense. However, if the frame included only the meteor and the Earth, the frame would become meaningless at the moment the meteor converged with the Earth. At that point, the line connecting two points would become a single point. How does the frame retain dimensionality without including other, non-convergent points?

21. Jul 3, 2010

### djy

Time is a coordinate in spacetime; it is not exterior to it. You can think of spacetime as a static surface that happens to have four dimensions, and everywhere the surface is locally Minkowskian. This is analogous to the surface of a sphere, which has two dimensions and everywhere the surface is locally Euclidean.

One can imagine assigning all manner of wacky, arbitrary coordinate systems to the surface of a sphere. Similarly, one can also do this with spacetime.

Basically, yes.

It's up to you to choose whatever reference frame you want.

A geodesic is simply the generalization of a "straight" line to a curved manifold, for example, a great circle on the surface of a sphere. I used the word "curved" informally to describe a line that isn't "straight".

Any object undergoing acceleration is not following a geodesic.

You are.

It's certainly possible to construct reference frames that contain singularities, or in general don't assign coordinates to every point in spacetime. That doesn't mean those points cease to exist. It just means your reference frame isn't complete.