- #1
fantispug
- 105
- 0
[SOLVED] Non-inertial Reference Frames
State Newton's Second Law of motion in a non-inertial reference frame.
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).
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.
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.