Non-inertial Reference Frames

In summary, the conversation discusses the issue of deriving the rotation matrix and its time derivative in non-inertial reference frames. The conversation includes the use of nomenclature and notation to explain how the frame-dependent representations of a vector are related and how to calculate the derivative in a rotating frame. The solution to the problem is provided, which involves taking the derivative of the operator.
  • #1
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[SOLVED] Non-inertial Reference Frames

Homework Statement


State Newton's Second Law of motion in a non-inertial reference frame.


Homework Equations


In an inertial frame F=ma
If S and S' are two reference frames, the same point is related by
r=A(t)r'+b(t) where A is an orthogonal matrix. (Galilean Relativity).


The Attempt at a Solution


Let r be a point in an inertial frame S, and r' be the same point in the frame S'.
r=A(t) r' + b(t), A orthogonal.
Differentiate twice with respect to time
[tex]\ddot{r}=\ddot{A}r'+2\dot{A}\dot{r'}+A\ddot{r'}+\ddot{b}[/tex]
Where a dot represents a time derivative. Now it would be handy to get this in a more familiar form, in terms of angular velocity.

For any vector Q in an infinitesimal circular rotation
[tex]Q=Q+w\times Q dt[/tex]
Where x is the cross product, and w is the angular velocity. (This is easier to see with a diagram).
So this gives us a rotation from identity: for a general rotation we can think of it as lots of infinitesimal rotations one after another
[tex]A=(I+\frac{t\vec{w}\times}{N})^N[/tex] in the limit N approaches infinity.
[tex]A=exp(t \vec{w} \times)[/tex]
Ok, so let's look at the special case when the axes of S and S' are aligned at t=0:
[tex]A(t)=I+t \vec{w}\times + \frac{t^2}{2} \vec{w}\times(\vec{w}\times)[/tex] + higher order terms
Now we in general allow w to vary
[tex]\dot{A(t)}=\vec{w}+t\dot{\vec{w}}\times+t \vec{w}\times(\vec{w}\times)[/tex] + higher order terms
[tex]\ddot{A(0)}=2\dot{\vec{w}}\times+\vec{w}\times(\vec{w}\times)[/tex]

And here lies the problem, plugging into my original equation I get a term of the form:
[tex]2\dot{\vec{w}}\times r'[/tex]
Which is TWICE the angular velocity term I should have.

Possible solutions:
If I ignored the second term in the first time derivative of A this problem would vanish, but it would be inconsistent.
I'm wondering if perhaps in deriving my rotation matrix ignoring the variation in w, even though it is second order, leads to an issue.

If someone could give me a hint where I'm going wrong it would be very much appreciated.
Cheers.
 
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  • #2
fantispug said:
I'm wondering if perhaps in deriving my rotation matrix ignoring the variation in w, even though it is second order, leads to an issue.
You're derivation of the rotation matrix and its time derivative is the problem.

Here is how I prefer to look at things.

Some nomenclature first:
  • [tex]\mathbf q[/tex] A vector, sans any concept of representation.

  • [tex]\mathbf q_F[/tex] The vector [tex]\mathbf q[/tex] as expressed in reference frame F.

  • [tex]\frac{d\mathbf q}{dt_F}[/tex] The derivative of some vector quantity q as seen by an observer attached to (i.e., not moving in) reference frame F.
  • [tex]\frac{d}{dt_F}[/tex] The above frame-dependent derivative written in operator form.

With this notation, the derivative of some vector quantity q from the perspective of an inertial frame I and a rotating frame R are related via

[tex]\frac{d\mathbf q}{dt_I} = \frac{d\mathbf q}{dt_R} + \mathbf {\omega}\times \mathbf q[/tex]

where [itex]\mathbf {\omega}[/itex] is the rotation rate of frame R with respect to frame I. In operator form, this becomes

[tex]\frac{d}{dt_I} = \frac{d}{dt_R} + \mathbf {\omega}\times[/tex]

Back to this vector quantity q. The frame-dependent representations of the vector are related via

[tex]\mathbf q_I = \mathbf A_{R\to I} \mathbf q_R[/tex]

Differentiating with respect to time,

[tex]{\boldmath {\dot q}}_I =
{\boldmath {\dot A}}_{R\to I} \mathbf q_R +
\mathbf A_{R\to I} {\boldmath {\dot q}}_R[/tex]

The first form of the vector derivative addressed the frame of the observer but not the frame in which the vector is expressed. The second form simply differentiated a bunch of scalars (the individual components of a matrix and a vector are just scalars.) We need to reconcile the two forms. Differentiating the frame-dependent components of some vector yields the frame-dependent derivative of that vector. Thus the left-hand side of the first form results from calculating the derivative of q as expressed inertially while the right-hand side results from calculating the derivative in the rotating frame:

[tex]{\boldmath {\dot q}}_I =
\mathbf A_{R\to I}(
{\boldmath {\dot q}}_R + \mathbf {\omega}_R\times \mathbf q_R)[/tex]

Equating this form with the second form,

[tex]\mathbf A_{R\to I}(
{\boldmath {\dot q}}_R + \mathbf {\omega}_R\times \mathbf q_R) =
{\boldmath {\dot A}}_{R\to I} \mathbf q_R +
\mathbf A_{R\to I} {\boldmath {\dot q}}_R[/tex]

and thus

[tex]{\boldmath {\dot A}}_{R\to I} \mathbf q_R =
\mathbf A_{R\to I}(\mathbf {\omega}_R\times \mathbf q_R)[/tex]

In operator form,

[tex]{\boldmath {\dot A}}_{R\to I} =
\mathbf A_{R\to I}\mathbf {\omega}_R\times[/tex]

Taking the derivative of this operator,

[tex]{\boldmath {\ddot A}}_{R\to I} =
\mathbf A_{R\to I}(
\mathbf {\omega}_R\times \mathbf {\omega}_R\times +
{\boldmath {\dot \omega}}_R)[/tex]
 
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  • #3
Awesome answer, thanks.

I think I need to spend a bit of time getting my ideas straight - I think the changing unit vectors, and how to view them in the non-inertial reference frame are confusing me.

Thanks again for your clear exposition.
 

1. What is a non-inertial reference frame?

A non-inertial reference frame is a coordinate system that is accelerating or rotating relative to an inertial reference frame. In this type of frame, objects appear to experience fictitious forces that do not actually exist in reality.

2. How does a non-inertial reference frame differ from an inertial reference frame?

In an inertial reference frame, objects follow Newton's laws of motion and experience no acceleration unless acted upon by an external force. However, in a non-inertial reference frame, objects appear to experience forces due to the acceleration of the frame itself.

3. What are some examples of non-inertial reference frames?

Examples of non-inertial reference frames include a car taking a sharp turn, a rotating amusement park ride, and a person on a spinning playground carousel. In each of these cases, the frame itself is accelerating or rotating, causing objects within it to appear to experience fictitious forces.

4. How is the Coriolis force related to non-inertial reference frames?

The Coriolis force is a fictitious force that appears to act on objects moving in a non-inertial reference frame due to the rotation of the frame. This force plays a significant role in weather patterns and the motion of objects on a spinning planet.

5. Why is it important to consider non-inertial reference frames in scientific calculations?

Non-inertial reference frames play a crucial role in understanding the behavior of objects in the real world. Many natural phenomena, such as the motion of planets and the Earth's weather patterns, can only be accurately described by taking into account the effects of non-inertial reference frames. Additionally, considering non-inertial frames is necessary for making accurate predictions and calculations in fields such as engineering and physics.

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