In the book of Goldstein, at page 200, the author argues while deriving the Euler's equation(s) that

$$\left.\frac{d\vec L }{dt }\right|_s = \vec \tau,$$ where subscript $s$ denotes the space frame.

and since $\left.\frac{d\vec L }{dt }\right|_s = \left.\frac{d\vec L }{dt }\right|_b + \vec{\omega} \times \vec L,$ we have

$$\frac{d\vec L }{dt }_b + \vec{\omega}\times \vec L = \vec \tau, \tag{*}$$

Then he goes on and calculates $\vec L$ wrt body frame, and derives the rest of the equation.

However, in the equations $(*)$, $\vec L$ is measured wrt space frame, but whose derivative is taken wrt body frame, whereas, the author measures $\vec L$ wrt body frame so that $I$ is a constant tensor, and derives the equation like that.

Clearly, what he derives, namely $(*)$, and how he uses it are different, so how can he do this and that can be a valid argument ?


Let assume that we have a body with a fixed point in space, and let define two reference frames with origin on that fixed point, $O, O'$, the former is an inertial frame whose coordinates are fixed in space, and the latter one is a non-inertial frame whose coordinates are fixed on the body (hence rotates with the body); the former will be called the space, and the latter will be called the body frame.

Let $w$ denote the angular velocity of the coordinates of $O'$ wrt $O$ frame. It can easily be proven that as long as the body frame's origin is on the body and coordinates are fixed on the body, $w$ is independent of body frame we choose to measure, so that we can talk about the angular velocity of the body itself.

Now, w.l.o.g, choose $O'$ as our body frame, and measure $\vec L$; it can be derived that $\vec L = I w$. In here, we calculated $I$ wrt the coordinates of $O$, but since the only difference between $O$ and $O'$ are rotation of coordinates, $I' = IR$, where $R$ is some a rotation matrix. The important thing is to note that since the coordinates of $O'$ are fixed on the body, $I'$ is not a function of time, whereas because of the motion of the body around the origin $I$ is a function of time.

The problem was if we provided $w$ to $O'$, calculating $\vec L$ within $O'$ would not be the same as it is calculated within the space frame; however I can see that the only difference is that the moment of inertial matrix changes with a rotation, i.e with a change of basis; the important point is to note that $I$ is actually the "same" matrix as long as the origin is fixed.

Lastly, for my concerns provided in the comments, $\vec L$ is a frame dependent vector, but it is the same vector as a vector in physical space; you just look at from different angles and measure different things, but get the same result at the end.

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  • $\begingroup$ The vector $\mathbf L$ is the same. The components in each frame are not. The LHS has the basis vectors fixed in time and RHS has the basis vectors rotating in time, that's why a $\boldsymbol \omega\times$ appears $\endgroup$ – ErickShock Apr 12 '19 at 18:00
  • $\begingroup$ @ErickShock How is $L$ the same in both frames ? We do know that $\omega$ is independent of the frame as long as the frames are on the body. Since $I$ is dependent on the frame, $\vec L$ has to be frame dependent. $\endgroup$ – onurcanbektas Apr 12 '19 at 18:15
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    $\begingroup$ My bad, both sides are in the same frame. Just the basis are different: you write $\mathbf L = \sum_i L_i(t) \mathbf e_i = \sum_i L'_i(t) \mathbf e'_i(t)$, where the LHS uses a fixed basis and the RHS uses a rotating basis (we are still in an inertial system in both sides). In the body frame $L$ has the components $L'_i$ and a fixed basis. $\endgroup$ – ErickShock Apr 12 '19 at 19:18
  • $\begingroup$ @ErickShock if the unit vectors are rotating, we are not in an inertial frame. $\endgroup$ – onurcanbektas Apr 12 '19 at 19:33
  • $\begingroup$ In the non-inertial frame the rotating basis will be constant in time. The derivative you wrote, $(d\mathbf L/dt)_b$ is the derivative taken in the non-inertial frame (explicitely: $\dot {\mathbf L}_b = \sum_i \dot{L}'_i \mathbf{e}'_i$). Equation $(*)$ connects the non-inertial derivative (where $\mathbf{e}'_i$ can be take as constant in time) with the inertial derivative (where $\mathbf{e}'_i$ varies with time). $\endgroup$ – ErickShock Apr 13 '19 at 1:29

The equation (*) which you've given above is special in the sense it uses somewhat "unmathematical, physicist's" notation $\frac{d\vec{L}}{dt}_b$. What this is supposed to mean is: let there be components of vector $\vec{L}$ with respect to the body coordinate frame, then calculate time derivative of each of these three components and use the resulting three numbers to define a vector. It could be denoted, say, as $\vec{m}$. But it is customary to denote it

$$ \frac{d\vec{L}}{dt}_b. $$

So, this vector is defined using possibly non-inertial rotating body frame, but the vector itself is actual vector, and we may find its components in the lab frame (which is assumed inertial) or in any frame we like.

However, in the equations (*), $\vec{L}$ is measured wrt space frame, but whose derivative is taken wrt body frame, whereas, the author measures wrt body frame so that is a constant tensor, and derives the equation like that.

First, let me clarify a misconception: there is only one vector $\vec L$, it does not make sense to say "$\vec{L}$ is measured with respect to lab frame" or "$\vec{L'}$ is measured with respect to body frame". There is no separate vector "$\vec{L'}$". There are only separate coordinate triplets of the vector $\vec{L}$, definable for every possible spatial coordinate frame.

For example, in the lab frame we may use components $[L_1,L_2,L_3]$, and in the body frame components $[L_1',L_2',L_3']$.

The equation (*) is written for vectors, thus it does not refer to any particular spatial coordinate frame. We can choose any. It is customary to express everything in the body frame, partially for the reason that you mentioned: tensor of inertia is constant there.

The equation (*) implies these three equations for components $[L_1',L_2',L_3']$ (components of vector $\vec{L}$ in the body frame):

$$ \frac{d[L'_1,L_2',L_3']}{dt} + [\omega_1',\omega_2',\omega_3'] \times [L'_1,L_2',L_3'] = [\tau_1',\tau_2',\tau_3']. $$

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  • $\begingroup$ Then you are basically claiming that, regardless of which frame - inertial or non inertial - measures (or calculates) the angular momentum, s/he will measure the same. This is not something that I'm aware of, so I'm expecting an explanation about why it is true. $\endgroup$ – onurcanbektas Apr 13 '19 at 5:48
  • $\begingroup$ From my point of view, if $\vec L = ı \vec \omega,$ if we take choose our body frame as a frame on the body whose coordinates are fixed on the body, since $\vec \omega$ is independent of frame (as long as we choose as I described) and $I$ depends on the frame, I would expect $\vec L$ to be frame dependent. $\endgroup$ – onurcanbektas Apr 13 '19 at 5:50
  • $\begingroup$ Components of angular momentum will not be the same across different spatial coordinate frames, because lab frame $x$ is not body frame $x$. But the underlying physical quantity - angular momentum $\vec{L}$ and tensor of inertia $I$ - are the same. This property is why these things are called tensors, as opposed to just triplets or matrices or arrays of numbers. The $I$ in the equation $\vec{L} = I\vec{w}$ is independent of coordinate frame as well, it is the tensor of inertia. $\endgroup$ – Ján Lalinský Apr 14 '19 at 12:49
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    $\begingroup$ Now, we can express this frame-independent relation using components, in any frame. In the lab frame, that would be $ L_i = I_{ik}w_k$. In the body frame, it would be $ L_i' = I'_{ik}w_k'$. $\endgroup$ – Ján Lalinský Apr 14 '19 at 12:50
  • $\begingroup$ "This property is why these things are called tensors": is this something generally done in physics ? Modelling frame independent quantities with tensors ? or being a tensor is a result of being frame independent ? Mathematically I know the group theoretical tensors, but I've almost no idea how physicists use them, or what do they mean when they talk about tensor. $\endgroup$ – onurcanbektas Apr 14 '19 at 15:04

I decided to add an answer to your question. The answer, however, became long, so I'll include a TL;DR:

when we say the vector is the same we mean it still has the same magnitude and still points to the same direction in physical space. Axis rotations change the components of a vector, but don't change it's magnitude nor change it's original direction in physical space.

You seem to be confused between what a reference system is and what basis vectors one should use in such a system. We usually choose the basis vectors in any reference system to be orthonormal, but we can still choose between a fixed basis and a moving basis. Let's look at an example: in an inertial frame I can choose as a basis the vectors (I'll stick to 2D for simplicity): $$ \mathbf e_x, \mathbf e_y, $$ which are orthonormal by construction and fixed in place (each are parallel to the axis labeled by their subscript). It may be that my system is not so simple when I use this basis - like when I have a central force - and it's much more convenient to use one that makes the algebra simpler. Fortunately we know from linear algebra that the vectors that describe physical quantities are basis-independent, so I can safely choose a new basis and the final vector will be the same. For a central force it's much better to use the basis $$ \mathbf e_r, \mathbf e_\theta $$ which is orthonormal, but has different directions for every $\theta$. My reference system is still inertial, I just decided to use a "weird" basis that moves around. Indeed, while for the first basis the derivative of an arbitrary vector $\mathbf A$ is $$ \left (\frac{d\mathbf A}{dt} \right )_\text{inertial} = \frac{d}{dt} (A_x \mathbf e_x + A_y \mathbf e_y) = \frac{dA_x}{dt} \mathbf e_x + \frac{dA_y}{dt}\mathbf e_y $$ for a moving basis we have to be careful because the direction of the basis vectors can change in time too, so $$ \left (\frac{d\mathbf A}{dt}\right )_\text{inertial} = \frac{d}{dt} ( A_r \mathbf e_r + A_\theta \mathbf e_\theta) = \frac{dA_r}{dt}\mathbf e_r + A_r \frac{d\mathbf e_r}{dt} + \frac{dA_\theta}{dt}\mathbf e_\theta + A_\theta \frac{d\mathbf e_\theta}{dt} \neq \frac{dA_r}{dt} \mathbf e_r + \frac{dA_\theta}{dt}\mathbf e_\theta. $$

The derivative of the basis of course creates some geometric forces, like for this case there's a centrifugal force that points to positive $\mathbf e_r$, but that's a basis effect, I never changed my reference system!

Now, in rigid-body motion it's way more convenient to express everything in terms of a basis that rotates with the body. So first of all I find such a basis in an inertial system, because Newton's 2nd law is only defined there. This is allows me to write (writing explicitely the time dependence) $$ \mathbf A = A_1(t) \mathbf e_1 + A_2(t) \mathbf e_2 = A'_1(t)\mathbf e'_1(t) + A'_2(t)\mathbf e'_2(t). $$ The laws of physics are the same because I'm in an inertial system still, but now I must also take the derivative of the $\mathbf e'_i$ whenever a derivative of $\mathbf A$ appears in my calculations (this is actually what makes the laws be the same in both basis).

Now suppose I change the reference system to one that rotates with the basis $\mathbf e'_i$, i.e., the body system. There the $\mathbf e'_i$ are obviously fixed in time, so a derivative would be $$ \left ( \frac{d\mathbf A}{dt} \right )_\text{body} = \frac{dA'_1}{dt} \mathbf e'_1 + \frac{dA'_2}{dt} \mathbf e'_2 $$ but unfortunately the laws of physics are not the same anymore, we have to find a way to bring our known physics to this non-inertial frame. We do this by noting that $$ \left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{body} + \boldsymbol \omega \times \mathbf A, $$ now we can substitute the inertial derivative by the non-inertial one in the physical laws and get the corresponding equation of motion in the non-inertial frame.

The thing is: when we say the vector is the same we mean it still has the same magnitude and still points to the same direction in physical space. Axis rotations change the components of a vector, but don't change it's magnitude nor change it's original direction in physical space (this is in sharp contrast to a galilean transformation, which changes vectors from one inertial system to the other, both in magnitude and in direction).

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  • $\begingroup$ Thanks a lot for this clear explanation; let me also try to explain the problem now in a more clear way... $\endgroup$ – onurcanbektas Apr 14 '19 at 4:53

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