Different coordinate system as opposed to different reference frame

Question

I'm having a hard time getting the difference between the two. In Euler's equations of rotating bodies for example, we have:

$$ \mathbf{\dot{L}}+\mathbf{\omega} \times \mathbf{L} = \mathbf{\Gamma},$$

where the torque $\mathbf{\Gamma}$ is (please correct me if I'm mistaken) in the space (inertial) reference frame, but expressed in the coordinate system fixed with the rotating body (because the other terms in the equation are expressed in this coordinate system—the inertia tensor is diagonal with respect to this coordinate system).

My question is then the following: what is the difference between talking about objects in a certain reference frame as opposed to talking about them represented in a certain coordinate system which is fixed with the said reference frame? Both cases seem pretty similar to me.

Edit: if it's still not clear what I'm struggling with: It is the statement that Euler's equation, for example, is with respect to an inertial frame of reference, but with respect to coordinate axes fixed with a rotating body. How is it not a contradiction? Aren't the axes variant with time as rotation, making the reference point from there non-inertial?

Answer 1

This is a problem I've struggled with before. The issue is that most textbooks are not clear enough in their definitions of the fixed and rotating reference frames. I've tried to be as comprehensive as possible in my answer, I hope it helps!

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To describe the motion of a rigid body through space we must make use of two separate reference frames:

• Firstly, a fixed (i.e. inertial) reference frame with a set of orthonormal basis vectors $\mathbf{e}_i$. These vectors are constant in time to all inertial observers. I will call the inertial reference frame $K$, and this set of vectors $S = \{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$.

• Secondly, the body frame - a non-inertial reference frame which moves with the rigid body. A second set of orthonormal basis vectors $\mathbf{e}'_j$ move with this frame. I will call this frame of reference $K'$, and this set of vectors $S' = \{\mathbf{e}_1', \mathbf{e}_2', \mathbf{e}_3'\}$.

The diagram below illustrates this picture (the weird blob is the rigid body!).

Any vector $ \mathbf{A}$ in $n$-dimensional space can be written as a linear combination of either set of basis vectors: we can write (using $\sum_j \equiv \sum_{j=1}^3$) $$ \mathbf{A} = \sum_i A_i\mathbf{e}_i = \sum_j A'_j\mathbf{e}'_j. $$

In particular, if the vector $\mathbf{A}$ is fixed to the rigid body, then the the projection of $\mathbf{A}$ onto each of the $\mathbf{e}_i$ varies with time and thus, the components $A_i$ are functions of time. However, since the projection of $\mathbf{A}$ onto each of the $\mathbf{e}'_j$ is constant with time, the components $A'_j$ are constant with time too. This argument holds if the vector $\mathbf{A}$ is fixed to inertial frame, $K$, or if the vector $\mathbf{A}$ is fixed to the body frame, $K'$.

Importantly, an observer from either reference frame can use either set of basis vectors. The difference is in how they view the behavior of each set of vectors: from the perspective of $K$, the basis vectors $\mathbf{e}_i$ are fixed, but the basis vectors $\mathbf{e}'_j$ are functions of time; however, to an observer in $K'$, i.e. from the perspective of the rigid body, the $\mathbf{e}'_j$ are fixed, and it is instead the vectors $\mathbf{e}_i$ that are time dependent.

Thus, when observed from $K'$, $$ \left(\frac{\text{d}}{\text{d}t}\right)_{K'}\mathbf{e}'_j = \mathbf{0}. $$ So the rate of change of a vector $\mathbf{A} = \sum_j A'_j\mathbf{e}'_j$ appears to be $$ \left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K'} = \sum_j \dot{A'}_j\mathbf{e}'_j. $$ However, when written in terms of the orthonormal set $S$, the $\mathbf{e}'_j$ basis vectors are functions of time. I show below that, as observed from $K$, $$ \left(\frac{\text{d}}{\text{d}t}\right)_{K}\mathbf{e}'_j = \boldsymbol{\omega}\times\mathbf{e}'_j\, \tag{1} $$ for some time-dependent vector $\boldsymbol{\omega}= \boldsymbol{\omega}(t)$. So, \begin{align} \left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K} &= \left(\frac{\text{d}\sum_j {A'}_j\mathbf{e}'_j }{\text{d}t}\right)_{K} &&\text{by construction} \\ &= \sum_j \dot{A'}_j\mathbf{e}'_j +A'_j\dot{\mathbf{e}}'_j &&\text{product rule} \\ & = \sum_j \dot{A'}_j\mathbf{e}'_j + A'_j \boldsymbol{\omega}\times\mathbf{e}'_j &&\text{by Eq.~1} \\ & =\sum_j \dot{A'}_j\mathbf{e}'_j+ \boldsymbol{\omega}\times(A'_j \mathbf{e}'_j) &&\text{homogeneity} \\ & = \sum_j \left(\dot{A'}_j\mathbf{e}'_j\right) +\boldsymbol{\omega}\times\mathbf{A}. &&\text{linearity} \end{align} Thus, we write that $$ \left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K'} = \left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K} - \boldsymbol{\omega}\times\mathbf{A}. \tag{2} $$ This is the crux of the problem: to an observer in the moving frame, every vector appears to have an additional term in its time derivative, when, in fact, it is the frame itself that is changing.

With this, Euler's equations follow quite naturally. Let $\mathbf{T}$ be the torque on the body, and let $\underline{\mathbf{I}}\boldsymbol{\omega}$ be the angular momentum of the body, where $\underline{\mathbf{I}}$ is the moment of inertia tensor. The key point is that as measured in $K'$, the moment of inertia tensor is constant. (This is not true in $K$, as the distribution of the mass of the body with respect to the $\mathbf{e}_i$ basis changes with time.) Using equation (2), this means that \begin{align} \left(\frac{\text{d}}{\text{d}t}\right)_{K'}\underline{\mathbf{I}}\boldsymbol{\omega} &= \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\boldsymbol{\omega} &&\underline{\mathbf{I}}~\text{constant in}~ K' \\ &= \underline{\mathbf{I}}\left[\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega}-\underbrace{\boldsymbol{\omega}\times\boldsymbol{\omega}}_{\mathbf{0}}\right] &&\text{by Eq.~2} \\ &= \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega}.&& \tag{$\star$} \end{align} Finally, the rate of change of angular momentum in the body is equal to the applied torque: \begin{align} \mathbf{T} &= \left(\frac{\text{d}\underline{\mathbf{I}}\boldsymbol{\omega} }{\text{d}t}\right)_{K} \\ &= \left(\frac{\text{d}\underline{\mathbf{I}}\boldsymbol{\omega}}{\text{d}t}\right)_{K'}+\boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}) &&\text{by Eq.~2} \\ &= \underline{\mathbf{I}}\left(\frac{\text{d}\boldsymbol{\omega} }{\text{d}t}\right)_{K}+\boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}). &&\underline{\mathbf{I}}~\text{constant in} ~K' \end{align} This is Euler's equation! It holds from the perspective of the inertial frame, but is derived using the basis vectors of the non-inertial frame.

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All that remains is to prove Equation (1). To do so, it is best to forget about the body frame $K'$, and to think of the $\mathbf{e}'_j$ as simply a set of vectors which move with respect to the $\mathbf{e}_i$, as seen from $K$. The moving basis vectors $\mathbf{e}'_j(t)$ are related to the fixed basis vectors $\mathbf{e}_i$ by a time dependent rigid rotation, which can be described by a matrix $\underline{\mathbf{R}}(t)$. Using Einstein notation, we have $$ \mathbf{e}'_j(t) = R_{ji}(t) \mathbf{e}_i, \qquad\text{and}\qquad \mathbf{e}_i = R_{ji}^\top(t) \mathbf{e}'_j(t). \tag{3} $$ Since $\underline{\mathbf{R}}$ is a rotation matrix, it is orthogonal. So, for all times $t$ it satisfies $$ \underline{\mathbf{R}}(t)\underline{\mathbf{R}}^\top(t) = \underline{\mathbf{I}}, $$ where $\underline{\mathbf{I}}$ is the identity matrix. Differentiating this with respect to time we find \begin{align} \underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^\top(t)+\underline{\mathbf{R}}(t)\underline{\dot{\mathbf{R}}}^\top(t) &= \underline{\mathbf{0}} \\ \underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^\top(t) + \left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^\top(t)\right)^T &= \underline{\mathbf{0}}. \end{align} So, the matrix $\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^\top(t)$ must be antisymmetric. This implies that there exists some vector $\boldsymbol{\omega} = (\omega_1, \omega_2, \omega_3)$ such that $$ \underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^\top(t) = [\boldsymbol{\omega}\times]\,. $$ Here, the cross-product matrix $$ [\boldsymbol{\omega}\times] = \begin{pmatrix}0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 &0 \end{pmatrix} $$ acts on vectors like a cross product as $[\boldsymbol{\omega}\times]\mathbf{A} = \boldsymbol{\omega}\times\mathbf{A}$. If you have not seen this before look here 2. For our purposes, it will be most useful to know the components of this matrix in the moving basis $\mathbf{e}'_j$. Using $\boldsymbol{\omega} = \omega'_j\mathbf{e}'_j$, we have $$ \left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^\top(t)\right)'_{ij} = \mathbf{e}'^T_i\left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)\mathbf{e}'_j = \epsilon_{ijk}\,\omega'_k. $$ With this, we can calculate the time derivatives of the primed basis vectors: \begin{align} \dot{\mathbf{e}}'_j(t) &= \dot{R}_{ji}(t)\, \mathbf{e}_j &&\text{by Eq. 3} \\ &= \dot{R}_{ji} \sum_{k=1}^n R^\top_{kj}\,\mathbf{e}'_k &&\text{by Eq. 3} \\ &= \epsilon_{ikm}\,\omega'_m \mathbf{e}'_k \\ &= \epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_m)\mathbf{e}'_k\\ &= \frac{1}{2} \epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_m)\mathbf{e}'_k - \frac{1}{2}\epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_k)\mathbf{e}'_m\\ &=\frac{1}{2}\epsilon_{ikm}\boldsymbol{\omega}\times(\mathbf{e}'_k\times\mathbf{e}'_m)\\ &=\frac{1}{2}\epsilon_{ikm}\boldsymbol{\omega}\times(\epsilon_{kml}\mathbf{e}'_l)\\ &=\delta_il\boldsymbol{\omega}\times\mathbf{e}'_l\\ &=\boldsymbol{\omega}\times\mathbf{e}'_i(t).\tag{2} \end{align} Thus, as seen from the inertial frame, there exists an instantaneous angular velocity vector $\boldsymbol{\omega}$ around which the primed basis vectors $\mathbf{e}'_i(t)$ rotate.

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Finally, here is an alternative derivation of Euler's equation that I am particularly fond of. We proceed by directly by differentiating the moment of inertia tensor. To do this, we align the $\mathbf{e}'_i$ with the principal axes of the body. We denote the principal moment of the body along the $\mathbf{e}'_i$ axis as $I_i$. Since the body is rigid, the pricipal moments are constants, and so we can explicitly write the moment of inertial tensor as $$ \underline{\mathbf{I}} = \sum_i I_i \mathbf{e}'_i \otimes \mathbf{e}'_i $$ (summation convention not used here). The time derivative of $\underline{\mathbf{I}}$ (in $K$) is then $$ \left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}} = \sum_i I_i\left[\dot{\mathbf{e}}'_i\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes\dot{\mathbf{e}}'_i\right] = \sum_i I_i\left[(\boldsymbol{\omega}\times\mathbf{e}'_i)\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes(\boldsymbol{\omega}\times\mathbf{e}'_i)\right]. $$ Now, let $\mathbf{b}$ and $\mathbf{c}$ be arbitrary vectors, and $\underline{\mathbf{A}}$ be a linear operator. Using the definition of the tensor product, we have $$ \left[(\underline{\mathbf{A}}\mathbf{b})\otimes\mathbf{c}\right]_{ij}=\left(\underline{\mathbf{A}}\mathbf{b}\right)_i c_j = A_{ik}b_k c_j = \left(\underline{\mathbf{A}}\right)_{ik}\left(\mathbf{b}\otimes\mathbf{c}\right)_{kj} = \left[\underline{\mathbf{A}}(\mathbf{b}\otimes\mathbf{c})\right]_{ij} $$ and $$ \left[\mathbf{b}\otimes(\underline{\mathbf{A}}\mathbf{c})\right]_{ij}=b_i\left(\underline{\mathbf{A}}\mathbf{c}\right)_j = b_i A_{jk} c_k = \left(\mathbf{b}\otimes\mathbf{c}\right)_{ik}\left(\underline{\mathbf{A}}^T\right)_{kj} = \left[(\mathbf{b}\otimes\mathbf{c})\underline{\mathbf{A}}^T\right]_{ij}. $$ Therefore, \begin{align} \left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}}&= \sum_i I_i\left[(\boldsymbol{\omega}\times\mathbf{e}'_i)\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes(\boldsymbol{\omega}\times\mathbf{e}'_i)\right] \\ &= \sum_i I_i\left[[\boldsymbol{\omega}\times](\mathbf{e}'_i\otimes\mathbf{e}'_i)+(\mathbf{e}'_i\otimes\mathbf{e}'_i)[\boldsymbol{\omega}\times]^T\right]\\ &=[\boldsymbol{\omega}\times]\underline{\mathbf{I}}-\underline{\mathbf{I}}[\boldsymbol{\omega}\times] \end{align} where $[\boldsymbol{\omega}\times]$ is the operator defined in the previous section, and in the last line I have used that this operator is anti-symmetric. Thus, the time derivative of the angular momentum is \begin{align} \mathbf{T} &= \left(\frac{\text{d}}{\text{d}t}\right)_{K} (\underline{\mathbf{I}}\boldsymbol{\omega}) \\ &= \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega}+\boldsymbol{\omega}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}}\\ &=\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega} + [\boldsymbol{\omega}\times]\underline{\mathbf{I}}\boldsymbol{\omega}-\underline{\mathbf{I}}\underbrace{[\boldsymbol{\omega}\times]\boldsymbol{\omega}}_{\boldsymbol{\omega}\times\boldsymbol{\omega}=\mathbf{0}}\\ &=\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega} + \boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}) \end{align} as before.

Answer 2

The word frame is intended to refer to physical matter. For example we talk of the Earth frame, and if you are travelling by car you will naturally use the car as your reference frame when considering objects in the car. A reference frame is the matter relative to which coordinates are defined. I have defined these in The Large and the Small:

• A reference frame consists of the reference matter, the apparatus, and the procedures, required to determine a spacetime coordinate system.
• A coordinate system is a mapping from physical events to coordinates with the form $(t, x, y, z)$ where $t$ is the time of the event and $(x, y, z)$ describes the position of the event (for example, in terms of distances, north, east and up, from some chosen fixed point, or origin).

(this is more strictly a spacetime coordinate system. In Newtonian mechanics you can consider space coordinates and take time as a parameter).

Thus the frame and the coordinate system are closely related, and in many cases (such as you describe) it is not necessary to distinguish between them. We may talk of motion in a reference frame, meaning that coordinates are implicit (or known) or we may talk of motion in coordinates, knowing that coordinates have been derived from an implicit frame.

if it's still not clear what I'm struggling with, it is the statement that Euler's equation for example, is with respect to an inertial frame of reference, but with respect to coordinate axes fixed with a rotating body. How is it not a contradiction? Aren't the axes variant with time as rotation, making the reference point from there non-inertial?

You are right, accounts of Euler's equation are confusing. It would not make much sense to work in the frame of the body (which is what many accounts seem to suggest) because then you would simply have $\mathbf\omega = \mathbf 0$. Euler's equation does apply relative to an inertial frame of reference, in which we can write $$ \mathbf T = \dot {\mathbf L}.$$ But this is a difficult equation to use, because ${\mathbf L} = I\omega$. Hence

$$\dot {\mathbf L} = \dot {\mathbf I}\omega + {\mathbf I}\dot\omega$$

and $\dot {\mathbf I}$ is not easy to work with. Instead we rewrite $ \mathbf I$ in the body frame, determined in general from the eigenvectors of $ \mathbf I$. Then $\dot {\mathbf I} =0 $, and $ \mathbf T = \dot {\mathbf L}$ is rewritten as $$ \mathbf T = \dot {\mathbf L} + \mathbf \omega \times \mathbf L.$$

i.e. Euler's equation. IOW we do not write the equation in the rotating frame, but only adapt the equation so as to use the components of the Moment of Inertia Tensor, $\mathbf I$, in the rotating frame, because in this frame the components are constant.

In other words, we are not really describing the motion in the rotating frame, but simply writing vectors in terms of vectors parallel to the principle axes, while actually remaining in the inertial frame.

To make this clearer, consider the simplest case of circular motion in a plane about the origin. In the inertial frame we can describe any vector in terms the unit vectors $\mathbf i$ and $\mathbf j$, but it is often more convenient to use the unit vectors $\mathbf {\hat r}$ and $\mathbf {\hat {\theta}}$ (this is directly applicable in the case where the principle axes are aligned with $\mathbf {\hat r}$ and $\mathbf {\hat {\theta}}$). We have, for any vector $\mathbf a$,

$$ \mathbf a = a_i \mathbf i + a_j \mathbf j = a_r \mathbf {\hat r} + a_{\theta} \mathbf {\hat {\theta}} $$

Then when we differentiate $\mathbf a$ with respect to time, terms appear in the inertial frame from the derivatives of $\mathbf {\hat r}$ and $\mathbf {\hat {\theta}}$ because $\mathbf {\hat r}$ and $\mathbf {\hat {\theta}}$ are varying in the inertial frame.

This is made confusing in rigid body treatments, because $a_r$ and $a_{\theta}$ are also the components of $\mathbf a$ in the rotating frame. Imv, it is not actually helpful to think of them like that. In the rotating frame $\mathbf {\hat r}$ and $\mathbf {\hat {\theta}}$) are constant vectors, and we would need to introduce inertial (fictitious) forces to describe the motion.

Answer 3

Consider a set of point particles with momenta $\mathbf{p}_i$, acted on by forces $\mathbf{f}_i$. If we wanted to formulate the equations of motion of these particles in the frame fixed to the center of mass of the system, we could write:

$$\dot{\mathbf{p}}'_i+\mathbf{F}=\mathbf{f}_i,\tag1$$

where $\mathbf{F}$ is a fictitious force, and $\mathbf{p}'_i$ are the momenta of the particles in the moving (non-inertial) frame.

By analogy you can view $-\mathbf{\omega} \times \mathbf{L}$ as an "inertial torque". In the same way as we could reduce $(1)$ to the simple equation

$$\dot{\mathbf{p}}'_i=\mathbf{f}_i'\tag2$$

by introducing the total force $\mathbf{f}_i'=\mathbf{f}_i-\mathbf F$, the Euler's equation can be reduced to the form of the equations of motion in inertial frame:

$$\mathbf{\dot{L}} = \mathbf{M},\tag3$$

with $\mathbf{M}=\mathbf{\Gamma}-\mathbf{\omega} \times \mathbf{L}$ being the torque in the rotating frame.

So the reason why right-hand side of the Euler's equation contains a strange term of "dual nature" is because the left-hand side also contains such a term. Remember that $\mathbf\omega$ is also a quantity defined with respect to the inertial frame, despite being described in rotating coordinates—the same way as $\mathbf\Gamma$.

Answer 4

Euler's equation is time derivative of the definition of angular momentum expressed at the center of mass of a body (notice the G subscript).

$$ \boldsymbol{H}_{\rm G} = \mathbf{I}_{\rm G} \boldsymbol{\omega} \tag{1} $$

_{$\boldsymbol{H}_{\rm G}$ is the angular momentum vector summed at the center of mass, along the inertial reference frame, $\boldsymbol{\omega}$ is the rotational vector along the inertial frame, and $\mathbf{I}_{\rm G}$ is the 3×3 mass moment of inertia tensor summed up at the center of mass, along the inertial reference frame .}

But the orientation of the coordinate system does not matter as long as the quantities involved use the same orientation, and the coordinate system is inertial (co-moving with the center of mass) at every instant. In the above scenario $\mathbf{I}_{\rm G}$ changes with time because the body is rotating. This is why the time derivative of (1) is

$$ \boldsymbol{\tau}_{\rm G} = \mathbf{I}_{\rm G} \boldsymbol{\alpha} + \boldsymbol{\omega}\times \boldsymbol{H}_{\rm G} \tag{2} $$

The second part comes from the product rule of differentiation as it is equal to $( \tfrac{\rm d}{{\rm d}t} \mathbf{I}_{\rm G}) \boldsymbol{\omega} = \boldsymbol{\omega}\times \mathbf{I}_{\rm G} \boldsymbol{\omega}$.

In order to properly use (1) and (2), the mass moment of inertia tensor needs to be re-oriented from the body coordinate directions to the inertia reference frame. Consider the 3×3 rotation matrix $\mathbf{R}$ that transforms from the body reference frame to the inertial reference frame (in orientation only). The mass moment of inertia on the inertial frame is found from

$$ \mathbf{I}_{\rm G} = \mathbf{R}\, \mathbf{I}_{\rm G}^\text{body} \mathbf{R}^\top \tag{3}$$

_{Where $\mathbf{I}_{\rm G}^\text{body}$ is the mass moment of inertia, summed at the center of mass, along the body orientation.}

In the above equation $\mathbf{I}_{\rm G}^\text{body}$ is fixed in time, and only the rotation matrix $\mathbf{R}$ changes with time.

Now we can take advantage of the fact that (1) and (2) may be expressed in any coordinate orientation as long as the reference orientation is still an inertial frame (non-rotating).

So at some instant, we define this reference frame to coincide with the body frame, transforming (1) and (2) into the following (which is only valid for this moment in time).

$$ \boldsymbol{H}_{\rm G}^\text{body} = \mathbf{I}_{\rm G}^\text{body} \boldsymbol{\omega}^\text{body} \tag{4} $$

$$ \boldsymbol{\tau}_{\rm G}^\text{body} = \mathbf{I}_{\rm G}^\text{body} \boldsymbol{\alpha}^\text{body} + \boldsymbol{\omega}^\text{body}\times \boldsymbol{H}_{\rm G}^\text{body} \tag{5} $$

The advantage here is that $\mathbf{I}_{\rm G}^\text{body}$ is fixed in time and thus no need to invoke (3) prior to using the equations above. This simplifies that calculations as any variability above are solely due to variability in $\boldsymbol{\omega}^\text{body}$.

This is the true form of Euler's equations in body coordinate, but it leads to a seeming contradiction as you pointed your in your question.

The apparent contradiction is that if (5) is expressed on body coordinates (and thus inertia is fixed) then it is not on an inertial reference frame, and thus invalid, and additionally, since inertia is fixed, the second term of (5) should be zero since it comes from the time derivative of inertia.

The answer is (as I eluded above) that (5) is expressed on an co-moving coordinate frame that happens to coincide with the rotating frame at one instant only. So all the calculations are done the same as in (2), except oriented at a different direction.

But the result is that you cannot integrate $\boldsymbol{\alpha}^\text{body}$ over time to find the next $\boldsymbol{\omega}^\text{body}$ since at a future time the body coordinate system is oriented differently. You can however integrate $\boldsymbol{\alpha}$ to find a future $\boldsymbol{\omega}$ since both are in the same coordinate orientation.

So what use is of (5)? The answer is when $\boldsymbol{\alpha}^\text{body}$ is expressed in terms of Euler angles, or some other parameterization of rotation, then the angle acceleration can be integrated over time to yield future configurations of the mechanism. NASA uses body frame oriented coordinate systems since that is what sensors on spaceship measures, but they still require inertial gyroscopes in order to transfer everything into an inertial reference frame for purposes of time integration. Lose a gyro, and you lose control of the spacecraft. This is how important the use of inertial reference frames is in 3D dynamics.

Answer 5

A coordinate system has an ordered set of real numbers associated with every point in the manifold. A reference frame is a manifold itself. So a manifold can be described by different coordinate systems. A reference frame has no notion of coordinates.

Answer 6

I will answer the question by writing in logical order the fundamental theoretical relations and emphasizing the clauses of their applicability. The whole discussion takes place in a same reference system $\Sigma $.

The angular momentum of a mechanical system (relative to a pole $O$) is the sum of the angular moments of all mass-points: $$ \boldsymbol{L_o} = \sum \boldsymbol{r_i}\times m_i\boldsymbol{v_i} $$

Let $O$ be a point settled to a rigid body. The moment of inertia related to $O$ is the following linear operator $\boldsymbol{J_o} $ acting on the angular velocity $ \boldsymbol {\omega} $ of the body: $$ \boldsymbol{J_o} \boldsymbol{\omega} = \sum m_i\boldsymbol{r_i}\times(\boldsymbol{\omega}\times\boldsymbol{r_i}) \qquad \text{pole O fixed to a rigid body} $$

It is easy to see that (G center of mass) $$ \boldsymbol{L_o} = m \boldsymbol{r_G}\times\boldsymbol{v_o} + \boldsymbol{J_o} \boldsymbol{\omega} \qquad \text{pole O fixed to a rigid body} $$

Deriving $\boldsymbol{J_o} \boldsymbol{\omega}$ respect to time we have $$ \frac{d}{dt}(\boldsymbol{J_o} \boldsymbol{\omega}) = \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega}\times \boldsymbol{J_o} \boldsymbol{\omega} \qquad\text{pole O fixed to a rigid body} $$

Proof:

$\frac{d}{dt}(\boldsymbol{J_o} \boldsymbol{\omega}) = \frac{d}{dt}\sum_i m_i \boldsymbol{r_i}\times(\boldsymbol{\omega}\times\boldsymbol{r_i}) = \\ = \sum_i m_i(\boldsymbol{v_i}-\boldsymbol{v_o}) \times (\boldsymbol{\omega}\times\boldsymbol{r_i}) + \sum_i m_i \boldsymbol{r_i}\times (\frac{d\boldsymbol{\omega}}{dt}\times \boldsymbol{r_i})+ \sum_i m_i \boldsymbol{r_i}\times [\boldsymbol{\omega} \times (\boldsymbol{v_i}-\boldsymbol{v_o})] =\\ = \sum_i m_i(\boldsymbol{\omega}\times\boldsymbol{r_i})\times(\boldsymbol{\omega}\times\boldsymbol{r_i}) + \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} + \sum_i m_i \boldsymbol{r_i}\times [\boldsymbol{\omega} \times (\boldsymbol{\omega}\times\boldsymbol{r_i})]= $\

and thanks to the Jacoby-identity we have further:

$= 0 + \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} - \boldsymbol{\omega}\times \sum_i [(\boldsymbol{\omega}\times\boldsymbol{r_i})\times m_i \boldsymbol{r_i}] - \sum_i (\boldsymbol{\omega}\times\boldsymbol{r_i})\times(m_i\boldsymbol{r_i} \times \boldsymbol{\omega}) = \\ % = 0 + \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} - \sum_i \boldsymbol{\omega}\times [(\boldsymbol{\omega}\times\boldsymbol{r_i})\times m_i \boldsymbol{r_i}] - 0 \\ % = \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega}\times \sum_i[ m_i \boldsymbol{r_i}\times(\boldsymbol{\omega}\times\boldsymbol{r_i})]\\ % = \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega}\times \boldsymbol{J_o} \boldsymbol{\omega} \qquad QED$

If the point $O$ is fixed in $\Sigma$ or if $O\equiv G$ (center of gravity) we have more simply: $$ \boldsymbol{L_o} = \boldsymbol{J_o} \boldsymbol{\omega} \qquad\qquad\qquad\text{point O of the rigid body fixed in $\Sigma$ or $O\equiv G$} $$ $$ \frac{d \boldsymbol{L_o}}{dt} = \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega}\times \boldsymbol{L_o} \qquad\text{point O of the rigid body fixed in $\Sigma$ or $O\equiv G$} $$

So far dynamics had nothing to do with it, but we must now suppose that the $\Sigma $ reference system is inertial.

If the pole $O$ is fixed or coincident with the center of gravity $O\equiv G $ the derivative with respect to the time of the angular momentum is equal to the resultant torque $\boldsymbol{\Gamma_o}$ of the external forces applied to the system:

$$ \frac{d\boldsymbol{L_o}}{dt} = \boldsymbol{\Gamma_o} \qquad \text{pole O fixed in $\Sigma$ or $O\equiv G$} $$

Under the condition that the pole $O$ (settled to the body) is a fixed point in the inertial frame $ \Sigma$ or that $ O \equiv G $ (center of gravity of the body) the Euler equation for the motion of a rigid body has therefore the form

$$ \boldsymbol{\Gamma_o} = \boldsymbol{J_o} \frac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega}\times \boldsymbol{L_o} $$

The notation $ (\frac{d\boldsymbol {L}}{dt})_{rot} $ used in place of $ \boldsymbol {J_o} \frac{d\boldsymbol{\omega}} {dt} $ is ambiguous and confusing!

The correct vector form of Euler's equation is independent of the choice of a particular coordinate system and was here obtained by working only in an inertial frame of reference $\Sigma $ (that is without use of any rotating reference system).

Answer 7

If you want to give 300 rep. points, while that's all you have, it seems to me that you don't give a **** about reputation and are truly interested in understanding. That's the way, ahu, ahu,...!!!

Let me begin to mention two things you wrote:

1)

If it's still not clear what I'm struggling with, it is the statement that Euler's equation, for example, is with respect to an inertial frame of reference, but with respect to coordinate axes fixed with a rotating body.

I'm having a hard time getting the difference between the two: A different coordinate-system as opposed to a different reference-frame. I meant the other terms are expressed in a coordinate system fixed to the rotating body, and it isn't the same, is it? That's the essence of my question.

In the first citation, there seems to be a contradiction (so it seems to me at least; maybe you made a mistake while typing; I don't know). You say that Euler's equation is wrt an inertial (non-force/acceleration) frame and at the same time wrt the co-rotating axes in a non-inertial frame. I don't quite get what you are saying there, but nevertheless I know what your question is about.

Considering the second citation.
The essence of your question. Let's consider a Euclidean space (in the case of special relativity, a Minkowski-space). One can "put" on this 3-d space different coordinate systems: the Cartesian coordinates (most used), polar coordinates, cylinrical coordinates, or (in a 2-d reference frame or Euclidean plane) elliptical coordinates. The frame itself doesn't change. The frame can have a velocity, but you can always perform a Galilean transformation to make the frame be at rest wrt to yourself). In the case of a moving 4-d reference frame (inertial frame) in special relativity the transformation becomes a Lorenz tranformation.
So the essence is that a coordinate-system can be changed to suit the problem, while the reference frame stays the same.

When you change to a non-inertial frame (say from inertial to non-inertial in the case of Euler's equation), you change the frame itself (from inertial to non-inertial). In the non-inertial frame, again, different coordinate-systems can be used, as is indeed done in generel relativity. In a rotating frame general relativity comes into play (though the Euler equations where already there before GR), as the spacetime frame (the non-inertial frame) is curved. Don't confuse this with curved coordinate systems which can be put on a flat space.
I'll don't bother you with the math. You can look that up for yourself.

So, one more time: one and the same inertial, or non-inertial frame (or reference frame), can be equiped with different coordinate systems, while one and the same coordinate system doesn't imply that there is just one associated reference frame.

Answer 8

A reference frame does not imply a coordinate system. A coordinate system tells you how you label points in space (or spacetime). A reference frame describes your motion with respect to the bodies in the problem.

For instance, consider throwing a ball directly upwards under gravity. If you are standing on the ground (i.e. your reference frame is defined by being at rest with respect to objects on the surface of the earth), the ball takes a nice parabolic trajectory. This trajectory can be described in different coordinate systems. For instance, if you set your origin to the initial location of the ball just as it leaves your hand, and you use $y$ to label an axis that is normal to the ground with the sky being the positive direction, then the equation that describes the motion of the ball is:

$y(t) = vt - \frac{1}{2} g t^2 \\ x(t) = 0$

However, you could define a coordinate system where $x$ is the upwards direction, you could describe the problem in spherical coordinates, or you could move your origin such that $y=0$ is the point $100$ meters above where the ball was thrown

$y(t) = -100 + vt - \frac{1}{2} g t^2 \\ x(t) = 0$.

These are all descriptions in the same reference frame, but with different sets of coordinates.

You could also describe the same situation in a car that is moving at $2m/s$ with respect to the rest frame we described earlier, but with the origin at the starting location of the ball. In this case the equations take the form

$y(t) = vt - \frac{1}{2} g t^2 \\ x(t) = -2t$.

As you can see, you can use many different coordinate systems to describe the same situation within a particular reference frame.

Answer 9

Involving a second reference frame, induces an origin translation from one reference frame to another,- despite any coordinate system used. So, in all of them origin translation happens. For example, given these RF's :

Origin translation vector in cartesian coordinate system is : $$ T_{_{OO~^\prime}} = \begin{bmatrix}x_{_{O~^\prime}} \\ y_{_{O~^\prime}}\end{bmatrix} $$

Same translation vector in polar coordinate system looks like :

$$ T_{_{OO~^\prime}} = \begin{bmatrix}r_{_{O~^\prime}} \\ \varphi_{_{O~^\prime}}\end{bmatrix} $$

Just changing coordinate system alone,- does not involve an origin transfer vector.