Mathematical definition of a rigid body

[wrobel]

I am teaching classical mechanics for students from mathematical dept. Students demand mathematically rigor definition of a rigid body. It is strange but I have not found the one in textbooks.

My attempt :

Df. (kinematical version): Take a non-void open set $D\subset\mathbb{R}^3$ and a pair of functions:
$$ \boldsymbol r_A\in C^2\big([0,T],\mathbb{R}^3\big),
\quad U\in C^2\big([0,T],\mathrm{SO}(3)\big),\quad U(0)=E,\quad \boldsymbol r_A(0)\in \overline D.$$
We shall say that a point $B$ belongs to the rigid body if its motion is described by the following position vector
$$\boldsymbol r_B(t)=\boldsymbol r_A(t)+U(t)\big(\boldsymbol r_B(0)-\boldsymbol r_A(0)\big),\quad t\in[0,T], \qquad(*)$$
where the initial value $\boldsymbol r_B(0)\in\overline D.$
By definition the rigid body consists of such points $B$.

 

[Hill]

I am teaching classical mechanics for students from mathematical dept. Students demand mathematically rigor definition of a rigid body. It is strange but I have not found the one in textbooks.

My attempt :

Df. (kinematical version): Take a non-void open set $D\subset\mathbb{R}^3$ and a pair of functions:
$$ \boldsymbol r_A\in C^2\big([0,T],\mathbb{R}^3\big),
\quad U\in C^2\big([0,T],\mathrm{SO}(3)\big),\quad U(0)=E,\quad \boldsymbol r_A(0)\in \overline D.$$
We shall say that a point $B$ belongs to the rigid body if its motion is described by the following position vector
$$\boldsymbol r_B(t)=\boldsymbol r_A(t)+U(t)\big(\boldsymbol r_B(0)-\boldsymbol r_A(0)\big),\quad t\in[0,T], \qquad(*)$$
where the initial value $\boldsymbol r_B(0)\in\overline D.$
By definition the rigid body consists of such points $B$.

Doesn't the definition above assume a mathematically rigorous definition of "motion"?

 

[wrobel]

I am afraid I do not understand the question

 

[PeroK]

I would have said that a rigid body is a set of particles whose pairwise distances from each other are constant.

 

[Hill]

I would have said that a rigid body is a set of particles whose pairwise distances from each other are constant.

This definition allows for reflection, while rigid body does not.

 

[PeroK]

This definition allows for reflection, while rigid body does not.

Reflection is impossible if each particle obeys Newton's laws of motion.

 

[wrobel]

I would have said that a rigid body is a set of particles whose pairwise distances from each other are constant.

Sure we must just add that this set contains at least three points that do not belong to the same line. Else the angular velocity vector is not defined uniquely.
I usually use the definition you quoted but I do not know a completely formal way to pass from this definition to derivation of the Euler formula:
$$\boldsymbol v_B=\boldsymbol v_A+\boldsymbol\omega\times \boldsymbol{AB}$$
upd: that is why I propose such a version of the definition in the initial post.

 

[wrobel]

You could define strain and set it to zero. This will take care of translation and rotation.

Yes, but the main problem is a way from a definition to the Euler formula.

 

[wrobel]

1) Choose a point on the body (not space) as the origin.
2) After motion, this point on the body is still the origin. In space, this point has undergone a translation only.
3) Since the body is rigid, every other point on the body can only undergo a rotation about the origin point.

isn't it a verbal description of the formula (*) from the initial post?

 

[wrobel]

I just noted that if we want to have a mathematically consistent theory without gaps filled with physical intuition, we have to formalize properly the definition of a rigid body and there is a formalization that gives apparently the shortest way to the main formula of the rigid body's kinematic.

 

[Baluncore]

A rigid body is a system of particles with fixed locations relative to each other.

 

[wrobel]

A rigid body is a system of particles with fixed locations relative to each other.

see #7

 

[Baluncore]

A rigid body is a system of particles with fixed locations relative to each other.
see #7

Two particles could define a rigid body.
There will always be special cases that are inconvenient.
I see no reason to restrict the definition based on what you may, or may not, want to define or compute later.
You may later specify that the body have a centre of mass, and a non-zero volume.

 

[wrobel]

Two particles could define a rigid body.

ok
but this is not the main question of the thread.
How do you derive formula of #7 from this:

A rigid body is a system of particles with fixed locations relative to each other.

by pure mathematical means?
Just bring this derivation here.

Google found this paper

and this paper is irrelevant

Sorry I could not be of more help.

do not be sorry I did not ask for help

 

[Gavran]

I am teaching classical mechanics for students from mathematical dept. Students demand mathematically rigor definition of a rigid body. It is strange but I have not found the one in textbooks.

There is an interesting definition that can help.
A rigid set is a set with the identity function as its only automorphism.

 

[wrobel]

A rigid set is a set with the identity function as its only automorphism.

May I have an example?

 

[Gavran]

May I have an example?

See the images below.

Both functions are automorphisms.

 

[wrobel]

Both functions are automorphisms.

Now I see what you mean. Then the same question as stated in #14

 

[Hill]

I am afraid I do not understand the question

The definition in #1 includes the phrase,

if its motion is described by

which is not mathematically rigorous, I think.

 

[wrobel]

The definition in #1 includes the phrase,

which is not mathematically rigorous, I think.

Doesn't the definition above assume a mathematically rigorous definition of "motion"?

A motion of the point $B$ relative to a frame $Oxyz$ is exactly the mapping $$t\mapsto\boldsymbol r_B(t)=x_B(t)\boldsymbol e_x+y_B(t)\boldsymbol e_y+z_B(t)\boldsymbol e_z.$$

 

[Hill]

Does the definition need $C^2$ functions? Isn't $C$ sufficient?

 

[wrobel]

Does the definition need C2 functions? Isn't C sufficient?

I impose $C^2$ just because it is mechanics and we must consider velocities and accelerations as well.

 

[Hill]

I impose $C^2$ just because it is mechanics and we must consider velocities and accelerations as well.

I see.
What happens when two rigid bodies collide?

 

[Hill]

I impose $C^2$ just because it is mechanics and we must consider velocities and accelerations as well.

I'm truly sorry that you've ignored my question above and I hope you reconsider and reply. Let me elaborate.
I'm asking about two rigid bodies colliding because as I understand their velocities are discontinuous in this case which contradicts their motion being described by the $C^2$ functions.

 

[pinball1970]

I impose $C^2$ just because it is mechanics and we must consider velocities and accelerations as well.

Is there a reason why your mathematics threads are in General rather than one of the maths SF? They are all very mathsy!

 

[weirdoguy]

I guess most people wouldn't see them in Math forums.