How to find coordinates of a system (I am taking cylinder now)?

[Istiak]

Homework Statement

How to find coordinates of a system (I am taking cylinder now)?

Relevant Equations

Coordinates

A person wrote that $L=\frac{1}{2} m (\dot{r}^2+r^2 \dot{\phi}^2 +\dot{z}^2)$ But, how to find equation of that coordinate system?

 

[PeroK]

Homework Statement:: How to find coordinates of a system (I am taking cylinder now)?
Relevant Equations:: Coordinates

View attachment 288051
A person wrote that $L=\frac{1}{2} m (\dot{r}^2+r^2 \dot{\phi}^2 +\dot{z}^2)$ But, how to find equation of that coordinate system?

Let's see your ideas. There's an obvious geometric idea and an obvious algebraic idea (hint Cartesian).

 

[Delta2]

What is this about, expressing angular momentum of a particle , in cylindrical coordinates?

 

[PeroK]

What is this about, expressing angular momentum of a particle , in cylindrical coordinates?

I took it to be the kinetic energy and $L$ the simple Lagrangian.

 

[Delta2]

I took it to be the kinetic energy and $L$ the simple Lagrangian.

Call me dumb but I still don't understand what the OP is asking about, the Lagrangian (kinetic energy in this case) seems to be in cylindrical coordinates, but the picture he posted seems to show spherical coordinates.

Is it about to prove that the kinetic energy in cylindrical coordinates is equal to what the OP says?

 

[PeroK]

Call me dumb but I still don't understand what the OP is asking about, the Lagrangian (kinetic energy in this case) seems to be in cylindrical coordinates, but the picture he posted seems to show spherical coordinates.

I didn't notice that!

 

[Steve4Physics]

Hi @Istiakshovon. You ask "how to find equation of that coordinate system?". But a coordinate system does not have an equation. What exactly are your trying to find?

 

[Istiak]

Check it : I have added accurate time. I think he took the equation for kinetic and potential energy.

 

[PeroK]

Check it : I have added accurate time. I think he took the equation for kinetic and potential energy.

Okay, that first equation is simply the KE in cylindrical coordinates. The next step is to introduce the constraint involving $\phi$.

However, given that on previous threads you've failed to demonstrate that you understand vectors and vector components, I'm not surprised you're strugging with Lagrangian mechanics!

 

[PeroK]

PS my reply in post #2 stands:

Let's see your ideas. There's an obvious geometric idea and an obvious algebraic idea (hint Cartesian).

 

[ergospherical]

The "trick" for determining the velocity of a given particle in a mechanical system is as follows: for any generalised coordinates $q$, deduce the velocity of the particle which results when all but one of the generalised coordinates are held constant. The actual velocity is simply the vector sum of all these contributions.

For the simple case of spherical coordinates $q = (r,\theta, \varphi)$,
$\theta, \varphi$ constant $\implies \boldsymbol{v}_1 = \dot{r} \boldsymbol{e}_r$
$\theta, r$ constant $\implies \boldsymbol{v}_2 = r \dot{\varphi} \sin{\theta} \boldsymbol{e}_{\varphi}$
$\varphi, r$ constant $\implies \boldsymbol{v}_3 = r \dot{\theta} \boldsymbol{e}_{\theta}$
therefore
$\boldsymbol{v} = \dot{r} \boldsymbol{e}_r + r \dot{\varphi}\sin{\theta} \boldsymbol{e}_{\varphi} + r \dot{\theta} \boldsymbol{e}_{\theta}$
$v^2 = \dot{r}^2 + r^2 \dot{\varphi}^2 \sin^2{\theta} + r^2 \dot{\theta}^2$

this is usually much faster than doing the coordinate transformation (and for more complicated systems you can sketch a quick "velocity diagram")

 

[Istiak]

However, given that on previous threads you've failed to demonstrate that you understand vectors and vector components, I'm not surprised you're struggling with Lagrangian mechanics!

Which book would you suggest me for vector? :)

 

[PeroK]

Which book would you suggest me for vector? :)

Kleppner and Kolenkow is a good introduction to Classical Mechanics. Note that it doesn't cover Lagrangian Mechanics, as that is considered more advanced.

 

[kuruman]

$\dots$ but the picture he posted seems to show spherical coordinates.

That's not how interpret the picture. I see label $z$ which is appropriate to cylindrical and see no label $r$ which would be appropriate to spherical. This problem looks like a bead that is constrained to move inside a cone of half-angle $\phi$. Furthermore, OP used cylindrical coordinates as generalized coordinates to write the Lagrangian.

 

[PeroK]

That's not how interpret the picture. I see label $z$ which is appropriate to cylindrical and see no label $r$ which would be appropriate to spherical. This problem looks like a bead that is constrained to move inside a cone of half-angle $\phi$. Furthermore, OP used cylindrical coordinates as generalized coordinates to write the Lagrangian.

That's exactly what he's doing.

 

[kuruman]

That's exactly what he's doing.

Yes, and it should be noted that what "a person wrote", $L=\frac{1}{2} m (\dot{r}^2+r^2 \dot{\phi}^2 +\dot{z}^2)$, cannot be interpreted as the Lagrangian because it's lacking the potential energy term.

 

[ergospherical]

Yes, and it should be noted that what "a person wrote", $L=\frac{1}{2} m (\dot{r}^2+r^2 \dot{\phi}^2 +\dot{z}^2)$, cannot be interpreted as the Lagrangian because it's lacking the potential energy term.

looks like a perfectly good lagrangian to me

 

[kuruman]

looks like a perfectly good lagrangian to me

Perfectly good, I agree, but not very interesting without gravity.

 

[PeroK]

Yes, and it should be noted that what "a person wrote", $L=\frac{1}{2} m (\dot{r}^2+r^2 \dot{\phi}^2 +\dot{z}^2)$, cannot be interpreted as the Lagrangian because it's lacking the potential energy term.

That's the KE term. He gets to the PE term next.

 

[Steve4Physics]

Hi. Could anyone kindly explain exactly what question @Istiakshovon's is asking?

 

[kumusta]

It is obvious from the posted YouTube that the intention was to write down the Lagrangian, which consistits of two terms, but in order to show how to get the second term, the potential energy term, only the first term had to be written down first. That means that the phrase "a person wrote" in the original problem posted by
Istiakshovon refers to the first term in the Lagrangian only, the kinetic energy term only, since the complete Lagrangian had not been written down yet. So the way that I understand it, Istiakshovon wanted to know the equations of the coordinate transformation used to go from the usual scalar kinetic energy equation in rectangular coordinate system to one in cylindrical coordinate system. Based on the diagram of the cone that he gave, he wanted to know the transformation equations x = (r cursive)cosθ,
y = (r cursive)sinθ, and tan∅ = (r cursive)/z, or z = (r cursive)cot∅.

 

[kuruman]

It is obvious from the posted YouTube that the intention was to write down the Lagrangian, which consistits of two terms, but in order to show how to get the second term, the potential energy term, only the first term had to be written down first. That means that the phrase "a person wrote" in the original problem posted by
@Istiakshovon refers to the first term in the Lagrangian only, the kinetic energy term only, since the complete Lagrangian had not been written down yet.

I agree with the second half of your reply about the transformations. I disagree with your interpretation about the equation $L=\dots$ There is no reason why one has to write the kinetic energy first and then the potential energy. One does not need the transformations from Cartesian to cylindrical to write down the potential energy anyway. I suspect OP is unsure about writing Lagrangians in general or in particular how to write the potential energy term in this case. I think the time has come to hear from @Istiakshovon whether the original question has been answered.

 

[kumusta]

By definition, the Lagrangian L = T - U, where T = kinetic energy KE and U = mgz is the potential energy in a gravitational field. Transforming KE in rectangular coordinate system CS to cylindrical CS, the rectangular coordinate z in U must also be transformed using z = (r cursive)cot∅ which I mentioned in #21.

 

[ergospherical]

by the way $T-U$ is not the definition of the lagrangian, although this quantity will give the correct equations of motion for the mechanical system in many instances

 

[kumusta]

@ergospherical
I'm sorry but I don't know about the lagrangian that you are talkimg about in #24. The Lagrangian that I'm talking about is the one shown below:

 

[ergospherical]

A Lagrangian is merely a differentiable function $L : TM \longrightarrow \mathbb{R}$ and a motion of $L$ is a function $\boldsymbol{\gamma} : \mathbb{R} \longrightarrow M$ which is an extremal of the functional $S[\boldsymbol{\gamma}] = \displaystyle{\int_a^b} L(\dot{\boldsymbol{\gamma}}) dt$.

If the Lagrangian function is equal to the difference $T-U$, where $T = \dfrac{1}{2}mv^2$ and $U: M \longrightarrow \mathbb{R}$ is some differentiable function, then the Lagrangian is called natural. This isn't a requirement and the Lagrangian sometimes does not take this form; $L$ is whatever it needs to be such that Lagrange's equations produce the correct equations of motion.

A simple example: the Lagrangian of a free relativistic particle is $L = -m\sqrt{1-v^2}$.

 

[kuruman]

$\dots$ using z = (r cursive)cot∅ which I mentioned in #21.

The generalized cylindrical coordinates are $r$, $\theta$ and $z$. The equation $z=r\cot\!\varphi$ is a constraint that puts the particle on a cone of half-angle $\varphi$. It is not part of the transformation equations from Cartesian to cylindrical.

 

[kumusta]

@kuruman
I know that z = (cursive r)cot∅ is the equation of constraint but I did not mention it explicitly. Thanks for bringing it out. But now that you talk about it, I realize only now that I have the wrong idea in mind. The d∅/dt appearing in the kinetic energy (the Lagrangian without the potential energy term) given by the problem poster does not refer to the time derivative of the inclination angle ∅ of the cone, as shown in the original figure, but to the time derivative of the polar angle θ. I remember now that while it is customary in mathematics to give the cylindrical coordinates as (r, θ, z), in analogy to the 2D circular polar coordinates (r, θ), it is usually the case in physics that (ρ, ∅, z) is used for cylindrical coordinates; either ρ or cursive r for the radius of the cylinder and ∅ or φ for the polar angle θ in mathematics. The italic r in physics is somehow reserved for the radial distance of a point from the center of a sphere. With the variable z still in the equation then, it means that the kinetic energy expression given by the problem poster is already that one in purely cylindrical coordinate system and does not involve yet the inclination angle ∅ of the cone shown in the figure in his initial post. It appears that he doesn't know that the (∅ dot) in the kinetic energy actually refers to (θ dot) of the circular polar angle. So maybe, what he was asking for was the change in variable (transformation of variable = change in coordinate) needed to order to include the inclination angle of the cone in the equation, which is simply the change in variable z = (cursive r)cot∅ given by the equation of constraint.

@ergospherical
The expression that you gave in #26 for the Lagrangian of a free relativistic particle is not dimensionally correct. The quantity inside the square root sign must be a pure number without any unit.

 

[Steve4Physics]

@ergospherical
The expression that you gave in #26 for the Lagrangian of a free relativistic particle is not dimensionally correct. The quantity inside the square root sign must be a pure number without any unit.

@ergospherical is using natural units, taking c=1. https://en.wikipedia.org/wiki/Natural_units.

The full expression is $L = -mc^2\sqrt{1- {\big( \frac {v}{c} \big) }^2}$.

 

[kumusta]

Yes, I'm aware of such substitution or convention in some books. I just didn't immediately think about that in this case. I'm the kind of guy who wants to see every factor completely written out in full in any equation. That's the problem when someone uses a particular convention without mentioning about its validity. Then, it's okay now.