Unit vectors that change as a function of time

  • #1
tragtf
5
0
Homework Statement
Express the unit vectors which change with time (like those of plane polar coordinates) as a function of time and integrate them.
Relevant Equations
r hat = cos(theta)i hat + sin(theta)j hat
theta hat = - sin(theta)i hat + cos(theta)j hat
we can express theta as a product of the angular velocity and time thus plugging this in space of theta in the above equations we can express r hat and theta hat as a function of time. But i am having problems in integrating them and again converting them back to the terms of r hat and theta hat from i hat and j hat
 
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  • #2
Is this in the context of a specific motion or for an arbitrary motion?
What quantity is being integrated, a velocity?
 
  • #3
It's just for an arbitrary motion. Yes, the quantity being integrated here is velocity in plane polar coordinates
 
  • #4
Velocity in polar coordinates is [itex]\dot r \hat r + r\dot\theta\hat\theta = \frac{d}{dt}(r\hat r).[/itex]
 
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  • #5
tragtf said:
It's just for an arbitrary motion. Yes, the quantity being integrated here is velocity in plane polar coordinates
You integrate vectors. The coordinates do not enter in. No matter what coordinates you use, the integral of a vector-valued function will be unchanged.

You may object because you are entirely accustomed to separating a vector into cartesian components, integrating the components separately and taking the results as the coordinates of the vector integral.

That works. Integration distributes over cartesian coordinates.

Integration does not distribute over polar coordinates.
 
  • #6
So how can we express quantities like average velocity and average acceleration in a changing co-ordinate system like the plane polar coordinates system
 
  • #7
tragtf said:
So how can we express quantities like average velocity and average acceleration in a changing co-ordinate system like the plane polar coordinates system
You either express vector addition and multiplication by a scalar in terms of your chosen coordinates or you convert to cartesian coordinates.

I am not sure I know what you mean by a "changing" coordinate system. Polar coordinates are unchanging. Perhaps you mean that the metric distance between two points in the space is not a function of the differences between the coordinates of the points.

In cartesian coordinates, the metric distance is a function of the differences: ##\Delta s = \sqrt{{\Delta x}^2 + {\Delta y}^2}##
 
  • #8
I meant a coordinate system where the base vectors are changing with time
 
  • #9
tragtf said:
I meant a coordinate system where the base vectors are changing with time
What do you mean by the base vectors for a vector space? Do you mean a set of linearly independent vectors that span the space, i.e. a "basis"?

In what sense do the base vectors for polar coordinates change with time? Are you imagining a rotating coordinate system? A translating coordinate system? A system whose scale changes over time?

I do not see a reasonable way to take ##\hat{r}## and ##\hat{\theta}## as unit vectors such that every vector in the space can be expressed as a linear combination of the form: ##k_1\hat{r} + k_2\hat{\theta}##. Yes, you can parameterize the space with ordered pairs like those. But the vector addition operation that you induce: ##\vec{(x_1, y_1)} + \vec{(x_2,y_2)} = \vec{(x_1+x_2, y_1+y_2)}## would be wrong. So you would not have a "basis" for the vector space.

Note that the basis vectors that you choose for a vector space have nothing to do with the coordinate system (if any) that you use. You can choose polar coordinates and express your basis as $$\lbrace \vec{(1,0)}, \vec{(1.0,\pi/2)} \rbrace$$Or you can choose cartesian coordinates and express the same basis as $$\lbrace \vec{(1,0)}, \vec{(0,1)} \rbrace$$
 
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  • #10
Forget it that was a typo i meant the unit vectors which change with time as the body moves
 
  • #11
tragtf said:
Forget it that was a typo i meant the unit vectors which change with time as the body moves
I still do not know what you mean.

Since you say "unit vector" then we are talking about something that cannot change magnitude with time. So it must merely change direction with time.

Are we, perhaps, talking about parameterizing the motion of a moving body in terms of its tangential velocity (along the path) and the current angle of the path? Both as functions of time perhaps?
 
  • #12
tragtf said:
It's just for an arbitrary motion. Yes, the quantity being integrated here is velocity in plane polar coordinates
Didn't post #4 answer that question?
 
  • #13
Are you asking how to differentiate the unit vectors for velocity and then integrate them to recover the original unit vectors?

If so let me do the first one

Given

##\hat{r} = \cos \left( \omega t\right) \hat{i} + \sin \left( \omega t\right) \hat{j}##
##\hat{\theta} = -\sin \left( \omega t\right) \hat{i} + \cos \left( \omega t\right) \hat{j}##

To find ##\frac{d \hat{r}}{dt}## differentiate component wise wrt ##t##

##\frac{d \hat{r}}{dt} = - \omega \sin \left(\omega t \right) \hat{i} + \omega \sin \left( \omega t \right) \hat{j}##

Notice that this equals ##\omega \hat{\theta}## by inspection

##\frac{d \hat{r}}{dt} = \omega \hat{\theta}##

Now integrate component wise

##\hat{r} = \omega \int \hat{\theta} \,dt = \omega \left[ -\int \sin \left( \omega t \right)\,dt\right] \hat{i} + \omega \left[ \int \cos \left( \omega t\right)\,dt\right] \hat{j}##

##= \cos \left( \omega t \right) \hat{i} + \sin \left( \omega t \right) \hat{j}##

##= \hat{r}##


You can do the same for finding an expression for ##\frac{d \hat{\theta}}{dt}##
 
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  • #14
tragtf said:
So how can we express quantities like average velocity and average acceleration in a changing co-ordinate system like the plane polar coordinates system

The polar basis varies time only because it varies with position. Even simple operations like vector addition or subtraction cannot be done component-wise in this basis; it follows that differentiation and integration cannot be done component-wise either.

Average acceleration is change in velocity divided by time taken.
Average velocity is change in position divided by time taken.

The only question for a non-constant basis is whether you express vector quantities relative to the basis at the start or relative to the basis at the end.
 

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