Implications of the statement Acceleration is not relative

In summary, the statement "Acceleration is not relative" has significant implications in the context of understanding the twin paradox in the theory of relativity. This statement suggests that the rocket twin cannot be considered at rest while accelerating, which is crucial in resolving the paradox. While this idea may seem shocking and goes against the principle of relativity, it is supported by the fact that acceleration can be independently measured or felt, and that an observer in an accelerating frame may consider themselves at rest. This concept is also evident in Einstein's work, where he explores the equivalence of inertial and gravitational mass and considers an observer in an accelerating chest to be at rest.
  • #316


Mentz114 said:
Movement is relative. Acceleration does not always imply movement.
It would help if you would avoid the unqualified term "acceleration". I believe that you mean that "proper acceleration does not always imply movement".
 
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  • #317


GregAshmore said:
I begin with a caveat:
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.

In this thread several methods for constructing the worldline of the rocket have been proposed. One of these had the rocket at the same position throughout, so the acceleration along that path would be zero. Obviously, that cannot be the method that is to be used.

Whatever the method that is to be is to be used, the path along the worldline must have non-zero length; the worldline must show the rocket as having traveled some distance. (I suppose that the worldline of the rocket is to be drawn with reference to an inertial frame, but that is not necessary for the success of my argument.)

The world line can be considered in any coordinates, including one where the rocket always has coordinate position zero, and is a vertical line in said coordinates. In such coordinates, the coordinate acceleration is zero and the proper acceleration is non-zero. The proper acceleration is defined as a covariant (or absolute) derivative of 4-velocity along the world line. Note that 4-velocity has a nonzero time component in such coordinates. This absolute derivative (of 4 velocity by proper time - which is just measured clock time for the rocket) has, as part of its expression, connection coefficients. These can be related to measurements of g forces. Thus the rocket, setting up coordinates themselves, at rest in those coordintes, directly computing proper acceleration according to its defining formula expressed in those coordinates, comes up with a nonzero value.
 
  • #318


PAllen said:
The world line can be considered in any coordinates, including one where the rocket always has coordinate position zero, and is a vertical line in said coordinates. In such coordinates, the coordinate acceleration is zero and the proper acceleration is non-zero. The proper acceleration is defined as a covariant (or absolute) derivative of 4-velocity along the world line. Note that 4-velocity has a nonzero time component in such coordinates. This absolute derivative (of 4 velocity by proper time - which is just measured clock time for the rocket) has, as part of its expression, connection coefficients. These can be related to measurements of g forces. Thus the rocket, setting up coordinates themselves, at rest in those coordintes, directly computing proper acceleration according to its defining formula expressed in those coordinates, comes up with a nonzero value.
Can you provide a link to an explanation of the steps in this calculation?
 
  • #319


DaleSpam said:
It would help if you would avoid the unqualified term "acceleration". I believe that you mean that "proper acceleration does not always imply movement".
Yes, I fell into the same error I was advising against.

Mentz114 said:
You must say whether you mean proper or coordinate acceleration.
:redface:
 
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  • #320


DaleSpam said:
The questions, as asked, are unanswerable. I have modified them as little as possible to make them answerable.

Yes.

Yes

Yes.

Yes.

You have enormous flexibility in choosing coordinate systems. You can choose charts such that each of those statements is true. You can also choose charts such that each is false.
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).

I'll ask two simple questions for now.

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?
 
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  • #321


GregAshmore said:
But there are surely degrees of fiction.
Only two that are of relevance:
- influences quantitative predictions (physics)
- doesn't influence quantitative predictions (not physics)
 
  • #322


GregAshmore said:
Can you provide a link to an explanation of the steps in this calculation?

See:

http://en.wikipedia.org/wiki/Proper_acceleration#In_curved_spacetime

noting that the calculations they give for curved spacetime also apply exactly to non-inertial coordinates in flat spacetime (special relativity). Also, they really apply to inertial coordinates as well, except that the connection coefficient goes to zero.

I can't find a good online link right now for the relation of connection coefficients to physical measurement, but it is well known that in the local 'rest frame' of an accelerating rocket (with no spin), all the connection coefficients are zero except the (x,tt), (y,tt), and (z,tt) components, and that these correspond to the measured g force in the x,y and z directions. For g force felt doesn't change direction, you can define x as the direction in which you feel g force; then only the (x,tt) connection component is nonzero, and it is (within units) equal to the g force you measure.

Then, the formula for proper accleration I linked, in a rocket rest frame, the (ordinary) derivative of the 4-velocity is zero; and the only term of the connection expression that doesn't vanish is the (x,tt) component = g force, times the time component of 4-velocity (which is the only component non-zero for such an accelerated frame) squared. You get, finally, that proper acceleration computed in rocket rest frame is proportional to g-force measured in rocket rest frame.

[One reference that covers all of this in detail, but at a much more advanced level than I assume is appropriate for you, is section 13.6 of Gravitation, by Misner, Thorne, and Wheeler.]
 
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  • #323


GregAshmore said:
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.
Here is the wikipedia page on worldlines: http://en.wikipedia.org/wiki/World_line

It doesn't give a good definition, but should give you the idea. Perhaps someone else with a good textbook can give a formal definition.

GregAshmore said:
In this thread several methods for constructing the worldline of the rocket have been proposed. One of these had the rocket at the same position throughout, so the acceleration along that path would be zero.
The proper acceleration along a worldline can be non-zero even if the object is at the same position throughout (coordinate acceleration is zero). The classical example is that of an object at rest in Schwarzschild coordinates (e.g. sitting in a chair on the surface of the earth).

GregAshmore said:
Whatever the method that is to be is to be used, the path along the worldline must have non-zero length; the worldline must show the rocket as having traveled some distance.
The path does need to have non-zero length, but that length can be purely in the time dimension. For an object "at rest" their worldline does not travel any spatial distance, but instead mantains a constant spatial position and parallels the time axis.

GregAshmore said:
Here is my logic:
1. An absolute quantity cannot be dependent on a non-absolute quantity.
I assume that by "absolute" you mean "frame invariant". However, this statement is wrong.

The prototypical example of an invariant quantity is the spacetime interval [itex]ds^2=-c^2dt^2+dx^2+dy^2+dz^2[/itex]. As you can see, the spacetime interval is dependent on frame variant quantities, both in terms of being logically dependent (definition) and mathematically dependent (derivatives). But those dependencies are such that under a change of frame the spacetime interval remains unchanged.

The property of frame variance or frame invariance cannot be deduced simply by the method of looking to see if it is dependent on a frame variant quantity.

GregAshmore said:
2. Proper acceleration is absolute.
Yes. (again assuming "absolute" means "frame invariant")

GregAshmore said:
3. Proper acceleration is derived from, and therefore dependent on, proper velocity.
OK. Although that isn't the only way to define it.

GregAshmore said:
4. It follows that proper velocity is absolute.
No that doesn't follow. See 1 above. However, proper velocity is the spacelike part of the four-velocity which is absolute.

GregAshmore said:
5. Proper velocity is derived from, and therefore dependent on, the distance traveled along the worldline.
I am not sure what you mean here.

GregAshmore said:
6. It follows that the distance traveled along the worldline is absolute.
The spacetime interval along the worldline is indeed absolute (frame invariant). I don't think that it follows from the above, but it is correct.

GregAshmore said:
7. The distance traveled along the worldine is, by definition, the distance through spacetime traveled by the rocket.
Yes, but it is better to use the term spacetime interval rather than distance. In the case of a massive object like a rocket the interval is timelike so it is a "distance" which is measured by clocks.

GregAshmore said:
8. A "distance traveled" is by definition "movement".
I have no problem with this. It seems to lend itself more to a LET-type interpretation of relativity than a block-universe interpretation, but I think it is OK.

GregAshmore said:
9. It follows that the rocket has experienced absolute movement through spacetime.
No problem with this either, but the movement that you are describing is like the movement of a cursor pointing to different points along a fixed line.

GregAshmore said:
When you tell the resting rocket observer that he had proper acceleration, you are also telling him that he moved some absolute distance through spacetime. That is precisely the charge he intended to deny when he made the claim to be permanently at rest.
No, when the resting rocket observer sees his clock tick he is moving some absolute distance through spacetime. The proper acceleration is not relevant. You cannot stop moving through spacetime simply by being at rest in space.
 
  • #324


GregAshmore said:
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.
 
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  • #325


GregAshmore said:
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 Schwarzschild 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.
 
  • #326


GregAshmore said:
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,[itex]\phi[/itex]), 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:

[itex]A^\mu = \dfrac{d U^\mu}{d \tau} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda[/itex] where [itex]U^\mu[/itex] is proper velocity, and [itex]\Gamma^\mu_{\nu \lambda}[/itex] is the so-called "connection coefficients" that are different for different coordinate systems. The two pieces of the proper acceleration
[itex]\dfrac{d U^\mu}{d \tau}[/itex]
and
[itex]\Gamma^\mu_{\nu \lambda} U^\nu U^\lambda[/itex]
are not physically meaningful by themselves, but the combination is physically meaningful.
 
  • #327


GregAshmore said:
(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).
 
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  • #328


GregAshmore said:
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:
[itex]\dfrac{d^2 x}{dt^2} = 0[/itex]
[itex]\dfrac{d^2 y}{dt^2} = 0[/itex]

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

Since [itex]\dfrac{d^2 r}{dt^2}[/itex] 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 [itex]\dfrac{d^2 r}{dt^2}[/itex], but the combination
[itex]\dfrac{d^2 r}{dt^2} - r (\dfrac{d \theta}{dt})^2[/itex]
 
  • #329


GregAshmore said:
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.
 
  • #330


stevendaryl said:
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."

http://www.relativity.li/en/epstein2/read/c0_en/c1_en/
 
  • #331


1977ub said:
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.
 
  • #332


GregAshmore said:
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.

GregAshmore said:
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.
 
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  • #333


DaleSpam said:
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.
 
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  • #334
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.
 
  • #335
GregAshmore said:
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.

GregAshmore said:
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.

GregAshmore said:
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.
 
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  • #336
GregAshmore said:
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.
 
<h2>What does it mean when it is said that acceleration is not relative?</h2><p>When it is said that acceleration is not relative, it means that the acceleration of an object is independent of the observer's frame of reference. This means that the acceleration of an object will be the same regardless of who is observing it.</p><h2>How is this different from the concept of relative motion?</h2><p>Relative motion refers to the motion of an object in relation to a particular frame of reference. In contrast, the statement that acceleration is not relative means that the acceleration of an object will be the same in all frames of reference, regardless of the relative motion between the observer and the object.</p><h2>What are the implications of this statement in terms of Newton's laws of motion?</h2><p>This statement has significant implications for Newton's laws of motion. It means that the laws of motion are valid in all frames of reference, and the acceleration of an object will be the same regardless of the observer's frame of reference. This helps to explain the universality of these laws and their applicability in various scenarios.</p><h2>How does this concept apply to real-world situations?</h2><p>In real-world situations, the concept that acceleration is not relative means that the acceleration of an object will remain the same regardless of the observer's perspective. This is particularly useful in fields such as physics and engineering, where understanding the behavior of objects in motion is crucial.</p><h2>Are there any exceptions to this statement?</h2><p>Some scientists argue that there may be exceptions to this statement in extreme scenarios, such as near the speed of light or in the presence of strong gravitational fields. However, for most everyday situations, the statement that acceleration is not relative holds true and can be applied successfully.</p>

What does it mean when it is said that acceleration is not relative?

When it is said that acceleration is not relative, it means that the acceleration of an object is independent of the observer's frame of reference. This means that the acceleration of an object will be the same regardless of who is observing it.

How is this different from the concept of relative motion?

Relative motion refers to the motion of an object in relation to a particular frame of reference. In contrast, the statement that acceleration is not relative means that the acceleration of an object will be the same in all frames of reference, regardless of the relative motion between the observer and the object.

What are the implications of this statement in terms of Newton's laws of motion?

This statement has significant implications for Newton's laws of motion. It means that the laws of motion are valid in all frames of reference, and the acceleration of an object will be the same regardless of the observer's frame of reference. This helps to explain the universality of these laws and their applicability in various scenarios.

How does this concept apply to real-world situations?

In real-world situations, the concept that acceleration is not relative means that the acceleration of an object will remain the same regardless of the observer's perspective. This is particularly useful in fields such as physics and engineering, where understanding the behavior of objects in motion is crucial.

Are there any exceptions to this statement?

Some scientists argue that there may be exceptions to this statement in extreme scenarios, such as near the speed of light or in the presence of strong gravitational fields. However, for most everyday situations, the statement that acceleration is not relative holds true and can be applied successfully.

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