# Rotating reference frames and acceleration

My question stems from a conversation I had recently with another physics buddy of mine and has to do with rotating reference frames and acceleration. Say, in a non-rotating reference frame you have an object with a known position. For the sake of argument, say it has a position A of 0i + 2j + 0k m from an arbitrary point in the fixed reference frame B. Now, switch to a rotating reference frame with angular velocity of 1 rad/s with a similar setup where point B is at the origin. Point A, in this rotating reference frame, would now appear to have a tangential acceleration.
Now, I've always been told (and read) that acceleration isn't relative, but it would seem here that it is. Can anyone help me with this?

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#### Dale

Mentor
Coordinate acceleration is relative to a given coordinate system. Proper acceleration is frame invariant, so it is not relative to any given coordinate system and all coordinate systems agree on its value.

#### dauto

In your example, the second referential system is non-inertial. You will know that because you will observe the presence of two fictitious forces - the centrifuge force and the Coriolis force. That's what's meant by the statement "acceleration isn't relative". It does not mean that there are no referential frames in which the object will be seen to have an acceleration even though it had no acceleration in another reference frame. It means that if a object is not accelerated in a INERTIAL reference frame than it is not accelerated in all INERTIAL reference frames. And you can tell whether or not a reference frame is inertial based on the presence or absence of fictitious forces.

#### Chestermiller

Mentor
From my understanding, which may or may not be correct, proper acceleration is relative to 4D space-time itself (which I regard as stationary). So, in this sense, I regard proper acceleration as absolute.

Chet

#### Dale

Mentor
You have to be careful with the word "absolute". Some people take "absolute" to be a synonym with "invariant", in which case it is OK. But other people take "absolute" to refer to quantities in the absolute reference frame (aka the aether frame), in which case you can convey an unintended meaning. I prefer to use the word "invariant" when I intend the first meaning, just so that there is no confusion.

#### dauto

From my understanding, which may or may not be correct, proper acceleration is relative to 4D space-time itself (which I regard as stationary). So, in this sense, I regard proper acceleration as absolute.

Chet
No, proper acceleration is the acceleration as measured from the point of view of the referential frame in which the accelerating object is (instantaneously) at rest. There is no such a thing as the reference frame of space-time itself. That would an equivalent to the aether theory

#### D H

Staff Emeritus
I concur with Dale. The term "absolute" can drag in a lot of unnecessary baggage. If "absolute" means "with respect to a local non-rotating, free-falling frame" then yes, proper acceleration and proper rotation are "absolute". On the other hand, if "absolute" means some global, absolute reference frame then no, that beast doesn't exist in general relativity.

Nonetheless, there is something special about local inertial frames. Even though there are no windows in Einstein's windowless elevator car, a passenger can still measure proper acceleration using an ideal accelerometer and measure proper angular velocity using an ideal rate gyro.

This has been really helpful, thanks guys! I think my problem stemmed from coordinate vs. proper acceleration and my definition of absolute. I read up on wikipedia and I think I get it now

#### Chestermiller

Mentor
No, proper acceleration is the acceleration as measured from the point of view of the referential frame in which the accelerating object is (instantaneously) at rest. There is no such a thing as the reference frame of space-time itself. That would an equivalent to the aether theory
What would be fundamentally incorrect about regarding 4D space-time as the relativistic replacement of the pre-relativistic 3D aether? I don't know whether this works in general relativity, it would not lead to any incorrect results in solving special relativity problems (of course, within the framework of the difference between the 3D spatial metric, and the 4D Minkowski metric). I find it easy to conceptualize all objects moving relative to space-time at the speed of light, but in different "(time) directions." I think of the proper acceleration of an object as the rate of change of its 4 velocity with respect to proper time, which is determined solely by the rate of change of the direction of its 4 velocity, since the magnitude of its 4 velocity is fixed. I know that this picture may ruffle the feathers of many physicists, but it how I like to think about it as an engineer. It seems to simplify things for me.

Chet

#### WannabeNewton

(1) Nothing moves in space-time. Particles don't move on their worldlines; the worldline of a particle represents the particle's history so it isn't some kind of physical trajectory.

(2) Proper acceleration is not the rate of change of the 4-velocity with respect to proper time if you are speaking of regular derivatives. The proper acceleration is $a^{\mu} = u^{\nu}\nabla_{\nu}u^{\mu}$ = $\frac{du^{\mu}}{d\tau} + \Gamma^{\mu}_{\nu\gamma}u^{\nu}u^{\gamma}$. In other words it is the absolute derivative of the 4-velocity with respect to proper time. Local experiments can be performed to determine proper acceleration unambiguously.

#### A.T.

(1) Nothing moves in space-time. Particles don't move on their worldlines; the worldline of a particle represents the particle's history so it isn't some kind of physical trajectory.
"Move" is maybe a misleading word. But why shouldn't an observer sort the events on a world line according to his coordinate time, and interpret this as "advancing" along the world line over time? Seems like a purely interpretational issue to me.

#### dauto

The statement at point is "proper acceleration is relative to 4D space-time itself (which I regard as stationary)". That statement makes no sense.

#### TrickyDicky

(2) Proper acceleration is not the rate of change of the 4-velocity with respect to proper time if you are speaking of regular derivatives. The proper acceleration is $a^{\mu} = u^{\nu}\nabla_{\nu}u^{\mu}$ = $\frac{du^{\mu}}{d\tau} + \Gamma^{\mu}_{\nu\gamma}u^{\nu}u^{\gamma}$. In other words it is the absolute derivative of the 4-velocity with respect to proper time. Local experiments can be performed to determine proper acceleration unambiguously.
Just a minor terminology corection (I know you you know this better than me), you are defining 4-acceleration there, not proper acceleration, wich is the derivative of proper velocity wrt coordinate time, thus a 3-vector (with the same magnitude as 4-acceleration).
"Move" is maybe a misleading word. But why shouldn't an observer sort the events on a world line according to his coordinate time, and interpret this as "advancing" along the world line over time? Seems like a purely interpretational issue to me.
In SR and with objects following inertial motion maybe that interpretation is possible, but in GR I think WannabeNewton is right.
The statement at point is "proper acceleration is relative to 4D space-time itself (which I regard as stationary)". That statement makes no sense.
I make no sense of it either, proper acceleration is a 3-vector and therefore relative to 3D space.That is precisely the mistery of its "absoluteness" (or invariance better term as Dale points out) together with 3-rotations.

#### Chestermiller

Mentor
(1) Nothing moves in space-time. Particles don't move on their worldlines; the worldline of a particle represents the particle's history so it isn't some kind of physical trajectory.
I agree with AT on this.
(2) Proper acceleration is not the rate of change of the 4-velocity with respect to proper time if you are speaking of regular derivatives. The proper acceleration is $a^{\mu} = u^{\nu}\nabla_{\nu}u^{\mu}$ = $\frac{du^{\mu}}{d\tau} + \Gamma^{\mu}_{\nu\gamma}u^{\nu}u^{\gamma}$. In other words it is the absolute derivative of the 4-velocity with respect to proper time.
This is what I meant by proper acceleration. To me, physically, this represents the rate of change of the 4 velocity vector with respect to time along a worldline.

Chet

#### Dale

Mentor
This has been really helpful, thanks guys! I think my problem stemmed from coordinate vs. proper acceleration and my definition of absolute. I read up on wikipedia and I think I get it now

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