What is usually the time-derivative of i-hat?

[lightlightsup]

Homework Statement

$\dot{\hat{i}} = 0?$

Relevant Equations

$\dot{\hat{i}} = 0?$

$\dot{\hat{i}} = 0?$

I'm trying to understand how to derive the entire polar vector system from $\hat{i}$ and $\hat{j}$ and since I'm new to all of this, I didn't realize that you could take the time-derivative of a unit-vector like like $\hat{r}$ or $\hat{θ}$.
This led me to asking the question above.
I believe the answer is 0 because the x-axis or $\hat{i}$ usually does not change with time but $\hat{r} = \cos{θ(t)}\hat{i} + \sin{θ(t)}\hat{j}$ does change with time because $θ$ changes with time/$t$.

 

[jedishrfu]

Here's a more thorough discussion of it beyond what you asked:

https://core.ac.uk/download/pdf/73340532.pdf

 

[haruspex]

Here's a more thorough discussion of it beyond what you asked:

https://core.ac.uk/download/pdf/73340532.pdf

No, that is using hats to mean something entirely different (fractional rate of change).

 

[Orodruin]

Homework Statement:: $\dot{\hat{i}} = 0?$
Homework Equations:: $\dot{\hat{i}} = 0?$

$\dot{\hat{i}} = 0?$

I'm trying to understand how to derive the entire polar vector system from $\hat{i}$ and $\hat{j}$ and since I'm new to all of this, I didn't realize that you could take the time-derivative of a unit-vector like like $\hat{r}$ or $\hat{θ}$.
This led me to asking the question above.
I believe the answer is 0 because the x-axis or $\hat{i}$ usually does not change with time but $\hat{r} = \cos{θ(t)}\hat{i} + \sin{θ(t)}\hat{j}$ does change with time because $θ$ changes with time/$t$.

You are correct. As long as you keep your coordinate system fixed, $\hat i$ does not change as it is the same everywhere. For a curvilinear coordinate system, basis vectors change from point to point and if you move around and change point, the basis vector at the point you are at will generally change as you change the point.

Here's a more thorough discussion of it beyond what you asked:

https://core.ac.uk/download/pdf/73340532.pdf

This does not really cover what the OP is looking for, which is the derivatives of basis vectors in a Cartesian coordinate system.

 

[haruspex]

I believe the answer is 0 because the x-axis or $\hat{i}$ usually does not change with time

It is zero because $\hat{i}$ never changes.

I came across this question a while back: https://www.physicsforums.com/threads/whats-the-integral-of-a-unit-vector.980604/. The answer was very interesting - see post #8 there.

 

[Orodruin]

It is zero because $\hat{i}$ never changes.

I came across this question a while back: https://www.physicsforums.com/threads/whats-the-integral-of-a-unit-vector.980604/. The answer was very interesting - see post #8 there.

The result in #10 of that thread should not come as a surprise as $\hat t\, ds = d\vec x$ by definition.

 

[PeroK]

Homework Statement:: $\dot{\hat{i}} = 0?$
Homework Equations:: $\dot{\hat{i}} = 0?$

$\dot{\hat{i}} = 0?$

I'm trying to understand how to derive the entire polar vector system from $\hat{i}$ and $\hat{j}$ and since I'm new to all of this, I didn't realize that you could take the time-derivative of a unit-vector like like $\hat{r}$ or $\hat{θ}$.
This led me to asking the question above.
I believe the answer is 0 because the x-axis or $\hat{i}$ usually does not change with time but $\hat{r} = \cos{θ(t)}\hat{i} + \sin{θ(t)}\hat{j}$ does change with time because $θ$ changes with time/$t$.

You have to distinguish between things as they are defined in the 2D plane and how things change as a particle moves about in a 2D plane.

Neither Cartesian nor polar basis vectors change over time. You have a fixed coordinate system in each case.

But, the polar unit vectors change with position. The vectors $\hat r$ and $\hat \theta$ do not point in the same direction at each point in the plane. The vectors $\hat i$ and $\hat j$ do, of course, point in the same direction at each point.

This means that for a particle moving in the plane the local polar unit vectors change direction as the particle changes its position. And, in fact, the unit vectors change depending on the position and velocity of the particle at each point in its path.

 

[Orodruin]

And, in fact, the unit vectors change depending on the position and velocity of the particle at each point in its path.

Clarification, they only depend on the position. Their rate of change depends on velocity as that is the rate of change in the position.

 

[lightlightsup]

You have to distinguish between things as they are defined in the 2D plane and how things change as a particle moves about in a 2D plane.

Neither Cartesian nor polar basis vectors change over time. You have a fixed coordinate system in each case.

But, the polar unit vectors change with position. The vectors $\hat r$ and $\hat \theta$ do not point in the same direction at each point in the plane. The vectors $\hat i$ and $\hat j$ do, of course, point in the same direction at each point.

This means that for a particle moving in the plane the local polar unit vectors change direction as the particle changes its position. And, in fact, the unit vectors change depending on the position and velocity of the particle at each point in its path.

I think we're almost saying the same thing.
The way I'm mathematically modeling movement at this level is as some function of time:
$\hat{r} = \cos{θ(t)}\hat{i} + \sin{θ(t)}\hat{j}$
So, in my case, if $θ(t) = ωt$, where $ω$ is a constant, then: position is being modeled as a function of time. The direction of $\hat{r}$ is dependent on $θ$ which is dependent on $ωt$.
$\hat{r}$ is a unique unit-vector because, unlike $\hat{i}$ and $\hat{j}$, it is described in terms of other unit-vectors ($\hat{i}$ and $\hat{j}$)?

Please let me know if there is something incorrect about any of this.
Thank You.

Deriving and using the polar vector system blew me away.
It is cool, sophisticated, and seemingly obvious at the same time.

 

[Chestermiller]

In polar coordinates, the unit vectors are $$\hat{r}=\cos{\theta}\hat{i}+\sin{\theta}\hat{j}$$and$$\hat{\theta}-\sin{\theta}\hat{i}+\cos{\theta}\hat{j}$$So, $$d\hat{r}=(-\sin{\theta}\hat{i}+\cos{\theta}\hat{j})d\theta=\hat{\theta}d\theta$$and$$d\hat{\theta}-(\cos{\theta}\hat{i}+\sin{\theta}\hat{j})d\theta=-\hat{r}d\theta$$

 

[rude man]

@OP,
Unit vectors $\hat{r}$ and $\hat{\theta}$ can have finite time derivatives, unlike cartesians i and j.
They are $\dot{\hat{r}} = \dot{\theta} ~ \hat{\theta}$ and $\dot{\hat{\theta}} = - \dot{\theta} ~ \hat{r}$

So your assumptions are correct.

 

[lightlightsup]

@OP,
Unit vectors $\hat{r}$ and $\hat{\theta}$ can have finite time derivatives, unlike cartesians i and j.
They are $\dot{\hat{r}} = \dot{\theta} ~ \hat{\theta}$ and $\dot{\hat{\theta}} = - \dot{\theta} ~ \hat{r}$

So your assumptions are correct.

Am I also correct in assuming that $\hat{r}$ and $\hat{\theta}$ are actually always defined in terms of $\hat{i}$ and $\hat{j}$?
They just greatly help simplify the description of complex circular motion? Which is the reason why they are used?

 

[Chestermiller]

@OP,
Unit vectors $\hat{r}$ and $\hat{\theta}$ can have finite time derivatives, unlike cartesians i and j.
They are $\dot{\hat{r}} = \dot{\theta} ~ \hat{\theta}$ and $\dot{\hat{\theta}} = - \dot{\theta} ~ \hat{r}$

So your assumptions are correct.

Try \boldsymbol{\theta}

 

[Orodruin]

They just greatly help simplify the description of complex circular motion? Which is the reason why they are used?

That is one of many applications where polar coordinates (or cylinder coordinates in 3D) are useful.

 

[rude man]

Try \boldsymbol{\theta}

Thanks Chet.

 

[rude man]

Am I also correct in assuming that $\hat{r}$ and $\hat{\theta}$ are actually always defined in terms of $\hat{i}$ and $\hat{j}$?
They just greatly help simplify the description of complex circular motion? Which is the reason why they are used?

Polar (and cylindrical; same thing except the z axis is added to accommodate 3-D situations) ) coordinates are used when the geometry suggests it. E.g in constant-speed circular motion Newtons equaton reduces to just $F_r = - r{\dot{\theta} }^2$ i.e just one component of force. In x-y the solution is significantly more involved, with two (x and y) components.

You can define polar coordinates in cartesian i and j (see post 10) but they really exist stand-alone.

There are other coordinate systems: spherical (3-D) and tangential/normal coordinates. As if that isn't enough, there are the general coordinates (Lagrangians) where you actually design your own coordinate system to fit the problem! This sounds wild but after learning it I was able to go back in my textbook and solve earlier-chapter problems much more easily that way.

The above refers to kinematics. Cartesian, cyhlindrical and spherical coordinate systems are all of them equally indispensable when you want to solve problems in thermodynamics, electricity & magnetism, and light, such as Laplace's equation. Again, geomtry is the usual determinant for choice.