Who Can Solve the Mystery of the True Torque?

[paradisePhysicist]

Alright how do I explain this. I am looking for the true and absolute equation of torque, specifically, the ratio of angular acceleration to torque.

Basically, I got the equations from online, but then found out from another website that most online equations are bogus, my estimate is 99%. The 99% of equations are basically a=I^-1*trq. The website says that real torque equation is not: a=I^-1*trq, but actually: a=I^-1*(tqr-w*L).

website: https://cwzx.wordpress.com/2014/08/30/from-torque-to-angular-acceleration-correctly/

At the point I am not sure who's correct, the website, 99% of the online tutorials, or neither. This is because something even weirder happened also.

The weirder thing that happened, is something that happened when testing the equations to Box 2D. After some trial and error I got the equations to produce the exact same motions as Box 2D. The weird part though, was that none of my equations actually require the usage of local rotations to match up with torque. Somehow, despite my equations not using any local rotation variables, it matches with Box 2D.

So in summary, I am wondering who's got the real torque, the website, the 99% of online tutorials, or Box 2D, or none of them. And maybe there should be a video proof of torque equations, like putting a rectangular shaped object on an air hockey table, slapping it with a steady impulse force, perhaps a something like a gravity propelled swinging contraption, and measuring the rotations via an aerial view camera.

 

[vanhees71]

I don't know what the page is talking about, because the guy didn't tell us to which reference frame he refers to. Obviously he's discussing the spinning top. Then usually the Euler equations are given, and they refer to the body-fixed (non-inertial, rotating) frame of reference. The Euler equation is nothing else than the equation of motion for the angular momentum. It's time derivative is the torque, and you have to take the "covariant" time derivative for components wrt. a rotating reference frame,
$$\mathrm{D}_t \vec{J}=\dot{\vec{J}}+\vec{\omega} \times \vec{J}=\vec{\tau}.$$
On the other hand you have the relation between $\vec{J}$ and $\vec{\omega}$, given by the tensor of inertia (whose components wrt. the body-fixed frame is constant in time, and you can always choose it to be diagonal by choosing the principle vectors of the tensor as the basis of your body-fixed frame):
$$\vec{J}=\hat{\Theta} \vec{\omega}.$$
So your equation reads [EDIT: corrected in view of #13]
$$\hat{\Theta} \dot{\vec{\omega}} + \vec{\omega} \times \hat{\Theta} \vec{\omega}=\hat{\tau}.$$
This coincides with what the guy on the website writes, if he really refers to the body-fixed frame, and it's in any textbook on classical mechanics that's worth the paper it is printed on ;-).

 

[paradisePhysicist]

I don't know what the page is talking about, because the guy didn't tell us to which reference frame he refers to. Obviously he's discussing the spinning top. Then usually the Euler equations are given, and they refer to the body-fixed (non-inertial, rotating) frame of reference. The Euler equation is nothing else than the equation of motion for the angular momentum. It's time derivative is the torque, and you have to take the "covariant" time derivative for components wrt. a rotating reference frame,
$$\mathrm{D}_t \vec{J}=\dot{\vec{J}}+\vec{\omega} \times \vec{J}=\vec{\tau}.$$

Forgive me if I'm misunderstanding, I am a bit tired at the moment. But wouldn't an applied angular acceleration be the same result regardless of if the reference frame is inertial or non inertial? An angular acceleration simply changes the omega ω, which is the angular velocity.

I must confess I am not fully understanding the difference between an inertial and non-inertial reference frame. This website: https://forceinphysics.com/non-inertial-frame-of-reference/ seems to say that someone walking down the street is a non-inertial reference frame, which doesn't make sense to me, because if someone is walking at a steady pace then no acceleration is going on (abstractly, but IRL of course there is acceleration due to friction).

Also I was wondering, how is it possible that my physics code does not need to know the local orientation of an object, yet produces the same behavior as Box 2D? Is that somehow possible or is Box 2D not based on physical behavoir? I have double checked my code but I suppose I could triple check it if there is no reason it should work.

 

[jbriggs444]

This website: https://forceinphysics.com/non-inertial-frame-of-reference/ seems to say that someone walking down the street is a non-inertial reference frame,

Ignore that article. The author may be trying to say something correct, but he hides it well with nonsense. A person walking down the street with constant velocity is at rest in an inertial frame.

Intuitively, a non-inertial frame is one that is accelerating or rotating. More formally, it is one in which objects free from external forces fail to move in straight lines at constant speeds -- i.e. in frames in which Newton's first law does not hold good.

For instance, a spinning carousel defines a non-inertial reference frame because thrown balls trace out curved arcs relative to the carousel.

It is pretty clear that if you measure angular velocity relative to a spinning carousel and then again relative to a stationary Earth, the two measurements will differ.

 

[vanhees71]

If you have an inertial frame of reference you can define an origin and a righthanded Cartesian basis and then describe position and momentum (and with them all other relevant quantities) in terms of the corresponding components. This I write as $\vec{x}$ and $\vec{p}$ (considering one particle, but that's enough to get the concept). Since the basis vectors are time-independent the components of the time derivative of these vectors are just the time derivatives of the components, and the Newtonian laws of motion are written always in reference to an inertial frame, particularly the equation of motion for the components of $\vec{x}$ simply reads
$$m \ddot{\vec{x}}=\dot{\vec{p}}=\vec{F}(\vec{x}).$$
Now we refer to an arbitrary reference frame which is accelerated and rotating relative to the inertial frame. The non-inertial frame is uniquely determined by an origin $\vec{R}(t)$ (components with respect to the inertial frame) and a right-handed basis $\vec{e}_j'$ fixed in the non-rotating frame. Thus there is a time-dependent rotation matrix such that
$$\vec{e}_j' = D_{ij}(t) \vec{e}_i.$$
To describe the motion in terms of components wrt. the non-inertial frame, we still have to use Newton's Laws in the inertial frame. So we write (Einstein summation for Cartesian components applies)
$$\vec{x}(t)= x_j'(t) \vec{e}_j'(t) = D_{ij}(t) x_j'(t) \vec{e}_i.$$
Now consider an arbitrary vector with components $V_i$ wrt. the inertial basis and $V_j'$ wrt. the non-inertial basis. Then
$$\dot{V}_i=(D_{ij} \dot{V}_j'+\dot{D}_{ij} V_j') \vec{e}_i.$$
Now since $\hat{D} \in \mathrm{SO}(3)$ we have $D_{ij} D_{ik}=\delta_{jk}$ and thus $\vec{e}_i=D_{ik} \vec{e}_k'$. So we get
$$\dot{V}_i= (D_{ij} \dot{V}_j' + \dot{D}_{ij} V_j') D_{ik}) \vec{e}_k'.$$
So the components of the time derivative wrt. the non-inertial basis is
$$\mathrm{D}_t V_k'=\dot{V}_k' + D_{ik} \dot{D}_{ij} V_J'.$$
So we get in addition to the normal time derivative of the component $V_k'$ an extra term from the rotation of the non-inertial basis relative to the inertial basis. To analyse this extra term we note that
$$D_{ik} D_{ij}=\delta_{kj} \; \Rightarrow \; \dot{D}_{ik} D_{ij}+D_{ik} \dot{D}_{ij}=0$$
and thus
$$\Omega_{kj}'=D_{ik} \dot{D}_{ij}=-\dot{D}_{ik} D_{ij}=-\Omega_{jk}'.$$
So $\Omega_{kj}'$ is antisymmetric and thus can be written in terms of three vector components $\omega_i'$ as
$$\Omega_{kj}'=\epsilon_{k i j} \omega_i',$$
from which our "covariant time derivative" gets
$$\mathrm{D}_t V_k' = \dot{V}_k' + \epsilon_{kij} \omega_i' V_j'.$$
This can be finally written in vector notation
$$\mathrm{D}_t \vec{V}'=\dot{\vec{V}}' + \vec{\omega}' \times \vec{V}'.$$
So our Newtonian equation of motion now reads
$$m \mathrm{D}_t^2 \vec{x}'=\vec{F}'(\vec{x}').$$
Now
$$\vec{v}'=\mathrm{D}_t \vec{x}' = \dot{\vec{x}}' + \vec{\omega}' \times \vec{x}'$$
and then
$$\mathrm{D}_t^2 \vec{x}' = \mathrm{D}_t \vec{v}' = \ddot{\vec{x}}' + 2 \vec{\omega}' \times \dot{\vec{x}'} + \vec{\omega} ' \times (\vec{\omega}' \times \vec{x}') + \dot{\vec{\omega}}' \times \vec{x}'.$$
So the acceleration written in terms of the non-inertial components gets pretty involved. In the equation of motion one often brings all the terms from the acceleration on the left-hand side execept the first one to the right-hand side, i.e., one writes an equation which looks like Newton's equation in non-inertial frames with additional "force terms" (the socalled "inertial forces") on the right-hand side:
$$m \ddot{\vec{x}}'=\vec{F}'(\vec{x}') - 2m \vec{\omega}' \times \dot{\vec{x}}' -m \vec{\omega}' \times (\vec{\omega}' \times \vec{x}') - m \dot{\vec{\omega}}' \times \vec{x}'.$$
The inertial forces are named as Coriolis force, centrifugal force, and angular acceleration force.

Sometimes in addition you have to separate the position vector in terms of the position vector of the origin of the non-inertial frame, $\vec{R}$ and the position vector wrt. this origin, i.e.,
$\vec{x}=\vec{R}+\vec{r}$. Then the eom. for the non-inertial position components becomes even more complicated.

 

[paradisePhysicist]

Ignore that article. The author may be trying to say something correct, but he hides it well with nonsense. A person walking down the street with constant velocity is at rest in an inertial frame.

Oh ok lol.

Intuitively, a non-inertial frame is one that is accelerating or rotating. More formally, it is one in which objects free from external forces fail to move in straight lines at constant speeds -- i.e. in frames in which Newton's first law does not hold good.

For instance, a spinning carousel defines a non-inertial reference frame because thrown balls trace out curved arcs relative to the carousel.It is pretty clear that if you measure angular velocity relative to a spinning carousel and then again relative to a stationary Earth, the two measurements will differ.

I guess what confuses me is the measurement of the angular velocity of the carousel will be the same regardless of if someone is on the carousel or on the world. If someone is on the world they simply take a point particle of the carousel and see how long it takes for the point particle to arrive back at its original location, in order to determine the frequency. If someone is on the carousel they simply take a point particle of world and see how long till they see that point particle again in their 2d vision space, to determine the frequency of the carousel.

 

[vanhees71]

All vectors are completely independent of the choice of the reference frame but their components are not! In the notation of my previous posting from a moment ago the components transform with the time-dependent rotation matrix, i.e.,
$$V_j' \vec{e}_j'=V_j ' D_{ij} \vec{e}_i = V_i \vec{e}_i\; \rightarrow V_i=D_{ij} V_j'$$
or in matrix-vector notation for the components
$$\vec{V}=\hat{D} \vec{V}'.$$
The other way around is
$$\vec{V}'=\hat{D}^{-1} \vec{V} = \hat{D}^{\text{T}} \vec{V},$$
where in the final step we have used that $\hat{D} \in \mathrm{SO}(3)$, i.e., $\hat{D}^{-1}=\hat{D}^{\text{T}}$.

 

[paradisePhysicist]

If you have an inertial frame of reference you can define an origin and a righthanded Cartesian basis and then describe position and momentum (and with them all other relevant quantities) in terms of the corresponding components. This I write as $\vec{x}$ and $\vec{p}$ (considering one particle, but that's enough to get the concept). Since the basis vectors are time-independent the components of the time derivative of these vectors are just the time derivatives of the components, and the Newtonian laws of motion are written always in reference to an inertial frame, particularly the equation of motion for the components of $\vec{x}$ simply reads
$$m \ddot{\vec{x}}=\dot{\vec{p}}=\vec{F}(\vec{x}).$$
Now we refer to an arbitrary reference frame which is accelerated and rotating relative to the inertial frame. The non-inertial frame is uniquely determined by an origin $\vec{R}(t)$ (components with respect to the inertial frame) and a right-handed basis $\vec{e}_j'$ fixed in the non-rotating frame. Thus there is a time-dependent rotation matrix such that
$$\vec{e}_j' = D_{ij}(t) \vec{e}_i.$$
To describe the motion in terms of components wrt. the non-inertial frame, we still have to use Newton's Laws in the inertial frame. So we write (Einstein summation for Cartesian components applies)
$$\vec{x}(t)= x_j'(t) \vec{e}_j'(t) = D_{ij}(t) x_j'(t) \vec{e}_i.$$
Now consider an arbitrary vector with components $V_i$ wrt. the inertial basis and $V_j'$ wrt. the non-inertial basis. Then
$$\dot{V}_i=(D_{ij} \dot{V}_j'+\dot{D}_{ij} V_j') \vec{e}_i.$$
Now since $\hat{D} \in \mathrm{SO}(3)$ we have $D_{ij} D_{ik}=\delta_{jk}$ and thus $\vec{e}_i=D_{ik} \vec{e}_k'$. So we get
$$\dot{V}_i= (D_{ij} \dot{V}_j' + \dot{D}_{ij} V_j') D_{ik}) \vec{e}_k'.$$
So the components of the time derivative wrt. the non-inertial basis is
$$\mathrm{D}_t V_k'=\dot{V}_k' + D_{ik} \dot{D}_{ij} V_J'.$$
So we get in addition to the normal time derivative of the component $V_k'$ an extra term from the rotation of the non-inertial basis relative to the inertial basis. To analyse this extra term we note that
$$D_{ik} D_{ij}=\delta_{kj} \; \Rightarrow \; \dot{D}_{ik} D_{ij}+D_{ik} \dot{D}_{ij}=0$$
and thus
$$\Omega_{kj}'=D_{ik} \dot{D}_{ij}=-\dot{D}_{ik} D_{ij}=-\Omega_{jk}'.$$
So $\Omega_{kj}'$ is antisymmetric and thus can be written in terms of three vector components $\omega_i'$ as
$$\Omega_{kj}'=\epsilon_{k i j} \omega_i',$$
from which our "covariant time derivative" gets
$$\mathrm{D}_t V_k' = \dot{V}_k' + \epsilon_{kij} \omega_i' V_j'.$$
This can be finally written in vector notation
$$\mathrm{D}_t \vec{V}'=\dot{\vec{V}}' + \vec{\omega}' \times \vec{V}'.$$
So our Newtonian equation of motion now reads
$$m \mathrm{D}_t^2 \vec{x}'=\vec{F}'(\vec{x}').$$
Now
$$\vec{v}'=\mathrm{D}_t \vec{x}' = \dot{\vec{x}}' + \vec{\omega}' \times \vec{x}'$$
and then
$$\mathrm{D}_t^2 \vec{x}' = \mathrm{D}_t \vec{v}' = \ddot{\vec{x}}' + 2 \vec{\omega}' \times \dot{\vec{x}'} + \vec{\omega} ' \times (\vec{\omega}' \times \vec{x}') + \dot{\vec{\omega}}' \times \vec{x}'.$$
So the acceleration written in terms of the non-inertial components gets pretty involved. In the equation of motion one often brings all the terms from the acceleration on the left-hand side execept the first one to the right-hand side, i.e., one writes an equation which looks like Newton's equation in non-inertial frames with additional "force terms" (the socalled "inertial forces") on the right-hand side:
$$m \ddot{\vec{x}}'=\vec{F}'(\vec{x}') - 2m \vec{\omega}' \times \dot{\vec{x}}' -m \vec{\omega}' \times (\vec{\omega}' \times \vec{x}') - m \dot{\vec{\omega}}' \times \vec{x}'.$$
The inertial forces are named as Coriolis force, centrifugal force, and angular acceleration force.

Sometimes in addition you have to separate the position vector in terms of the position vector of the origin of the non-inertial frame, $\vec{R}$ and the position vector wrt. this origin, i.e.,
$\vec{x}=\vec{R}+\vec{r}$. Then the eom. for the non-inertial position components becomes even more complicated.

I am a bit confused, does the uppercase omega denote something different than the lowercase?

 

[vanhees71]

You have $\Omega_{ij}'=\epsilon_{ikj} \omega_k'=-\epsilon_{ijk} \omega_k'$. Thus $\vec{\omega}'$ is the negative of the "Hodge dual" of $\hat{\Omega}'$. The inverse of this mapping between antisymmetric tensors and (axial) vectors is
$$\omega_k'=-\frac{1}{2} \epsilon_{ijk} \Omega_{jk}'.$$
Of course both do the same to the vector components, i.e.,
$$\hat{\Omega}' \vec{V}'=\vec{\omega}' \times \vec{V}'.$$

 

[paradisePhysicist]

The Euler equation is nothing else than the equation of motion for the angular momentum. It's time derivative is the torque, and you have to take the "covariant" time derivative for components wrt. a rotating reference frame,
$$\mathrm{D}_t \vec{J}=\dot{\vec{J}}+\vec{\omega} \times \vec{J}=\vec{\tau}.$$

I guess I'm confused at this. J I assume means total angular momentum, but in this chart it has a little caret on top of it so I am not sure. http://www.cmat.uni-halle.de/~hsl/PoM-files/Symbols.pdf

Also, I am a beginner of calculus and not very expert at it. I am a bit confused as to what you mean by a time derivative. In calculus a derivative returns the point slope of any given function, I do not understand how you can get a time derivative from a vector.

 

[vanhees71]

The symbols in different textbooks are different. There are so few symbols in the Latin and Greek alphabet that one has to make decisions for how to name what, and in different scientific communities you have different nomenclature and conventions. That's why you have to carefully check the meaning of each symbol in each textbook or paper to make sure you understand the meaning the author had in mind when writing it down.

For physics you need from day 1 on multi-variable calculus and (Euclidean) vectors. Of course, I don't get a time derivative from a vector but use time derivatives of vectors.

In classical mechanics of a point particle you can describe the motion of the particle by first defining a frame of reference, i.e., a fixed point in space ("the origin", $O$) and a Cartesian right-handed basis, i.e., three vectors $\vec{e}_j$ with $j \in \{1,2,3\}$. Then you can describe the location of any point $P$ in space uniquely by the vector $\vec{x}=\overrightarrow{OP}$. Then the motion of this point particle is uniquely described by $\vec{x}(t)$, i.e., a function that gives for any time $t$ the position of the particle in terms of the particle's position vector.

Now you can decompose any vector uniquely using the Cartesian basis, i.e., $\vec{x}=\sum_j x_j \vec{e}_j$. By definition the $\vec{e}_j$ are fixed in space, i.e., time-independent. Then the time derivative of a vector is defined in close analogy to the derivative of a usual function $f(t)$ by the limit of the difference quotient,
$$\dot{\vec{x}}(t)=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}(t) = \lim_{\Delta t \rightarrow 0} \frac{\vec{x}(t+\Delta t)-\vec{x}(t)}{\Delta t}.$$
Since the basis vectors are by definition time-independent you have
$$\dot{\vec{x}}(t)=\sum_{j=1}^3 \dot{x}_j(t) \vec{e}_j,$$
and here $x_j(t)$ are three real functions, and of course you can use the rules you learned in calculus of a single function to the components of vectors.

So for classical mechanics, if you know calculus of real functions, you just need to learn to handle vectors and their components with respect to a basis. Then you can use your calculus of real functions to take derivatives and integrals of components.

The only thing you'll also need is calculus with many independent variables, because soon you deal with functions of a vector or equivalently functions of the three vector components with respect to a Cartesian basis.

 

[paradisePhysicist]

Hmm, I understand most of that, except its not really "clicking" for me so to speak, like I'm not feeling like I have a firm grasp of the bigger picture. Maybe its the summer heat lol. I sniffed some essential oils to help me concentrate but it smells off, I wonder if the high temperatures disrupted the mixture. Also, having more letters in the alphabet would indeed be nice, a lot of the letters are used multiple times and have multiple different definitions.

 

[wrobel]

there are too many omegas in the last formula of #2

 

[vanhees71]

Thanks for pointing this out. The Euler equations of the spinning top is already cumbersome enough in their correct form .

 

[paradisePhysicist]

I don't know what the page is talking about, because the guy didn't tell us to which reference frame he refers to. Obviously he's discussing the spinning top. Then usually the Euler equations are given, and they refer to the body-fixed (non-inertial, rotating) frame of reference. The Euler equation is nothing else than the equation of motion for the angular momentum. It's time derivative is the torque, and you have to take the "covariant" time derivative for components wrt. a rotating reference frame,
$$\mathrm{D}_t \vec{J}=\dot{\vec{J}}+\vec{\omega} \times \vec{J}=\vec{\tau}.$$
On the other hand you have the relation between $\vec{J}$ and $\vec{\omega}$, given by the tensor of inertia (whose components wrt. the body-fixed frame is constant in time, and you can always choose it to be diagonal by choosing the principle vectors of the tensor as the basis of your body-fixed frame):
$$\vec{J}=\hat{\Theta} \vec{\omega}.$$
So your equation reads [EDIT: corrected in view of #13]
$$\hat{\Theta} \dot{\vec{\omega}} + \vec{\omega} \times \hat{\Theta} \vec{\omega}=\hat{\tau}.$$
This coincides with what the guy on the website writes, if he really refers to the body-fixed frame, and it's in any textbook on classical mechanics that's worth the paper it is printed on ;-).

Hmm, I tried searching online but I haven't found that particular symbol of theta with a caret. Does it represent a 3x3 matrix of inertia tensor like in this pdf? https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec26.pdf May I see the website or book you are using the symbols from?

 

[vanhees71]

$\hat{\Theta}$ is my symbol for the $\mathbb{R}^{3 \times 3}$ matrix representing the tensor of inertia with respect to the body-fixed basis.

It's not from a book but from my own lecture notes on mechanics (in German)

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf

 

[wrobel]

ok but from my own lecture notes on mechanics (in German)

page 109
I have got a habit to think that the Noether theorem in the Hamiltonian statement is a very trivial thing:
let $H=H(z),\quad z=(p,q)$ be a Hamiltonian and this Hamiltonian is invariant under the flow $g^s_F(z)$ of some other Hamiltonian system with a Hamiltonian $F(z)$. That is
$$H(g^s_F(z))=H(z),\quad \forall s.$$ Then $$\{F,H\}=0.$$