Vector squared in polar coordinates

[dyn]

Hi
I was always under the impression that i could write
a² = a.a = a² Equation 1

where a⋅ is a vector and a is its modulus but when it comes to the kinetic energy term for a particle in plane polar coordinates I'm confused ( i apologise here as i don't know how to write time derivative with the dot on top).
In this case you get
r(dot)² = r(dot)² + r²θ(dot)²
which doesn't seem to fit with Equation 1. I know how to derive this equation but i.m confused as to why it doesn't fit with Equation 1
Thanks

 

[PeroK]

Hi
I was always under the impression that i could write
a² = a.a = a² Equation 1

where a⋅ is a vector and a is its modulus but when it comes to the kinetic energy term for a particle in plane polar coordinates I'm confused ( i apologise here as i don't know how to write time derivative with the dot on top).
In this case you get
r(dot)² = r(dot)² + r²θ(dot)²
which doesn't seem to fit with Equation 1. I know how to derive this equation but i.m confused as to why it doesn't fit with Equation 1
Thanks

What doesn't fit? In Cartesian cordinates we have: $$\mathbf{\dot r}^2 = \dot x^2 + \dot y^2$$

 

[fresh_42]

If we have a vector $\vec{a}=(a_1,\ldots,a_n)$ in Cartesian coordinates, then the "multiplication with itself", i.e. the scalar product with itself represents the scalar $\vec{a} \cdot \vec{a}^\tau=\sum_{i=1}^n a_i^2.$

In case $\vec{a}=\vec{r} \cdot \cos\varphi$ is given in polar coordinates, we get $\vec{a}\cdot\vec{a}^\tau=\vec{r}\cdot\vec{r}^\tau \cdot \cos^2\varphi .$

 

[dyn]

I was always under the impression that i could write
a² = a.a = a² Equation 1

where a⋅ is a vector and a is its modulus

Is Equation 1 above always valid or only valid in certain circumstances ?

 

[fresh_42]

Is Equation 1 above always valid or only valid in certain circumstances ?

Depends on whether you manage to make it a correct equation. Neither bold and non-bold letters nor the dot has been defined. It is guesswork when we read it as $\vec{a}\cdot\vec{a}^\tau.$

 

[PeroK]

Is Equation 1 above always valid or only valid in certain circumstances ?

Equation 1 is a definition, so yes.

 

[dyn]

If equation 1 is always correct ; then in polar coordinates it should ber(dot)² = r(dot)² Equation A

but actually the correct equation is r(dot)² = r(dot)² + r²θ(dot)²
I understand how to derive this but my question is ; if equation 1 is correct why is Equation A not correct ?

 

[Gaussian97]

Well, simply. Because the derivative of the module is not the module of the derivative.
To help visualize let's change a little the notation, let's say we have a vector $\vec{r}$ and we define it time derivative $\vec{v} = \frac{d\vec{r}}{d t}$.
Then its true that
$$r^2 = \vec{r}\cdot \vec{r} = \vec{r}^2, \qquad v^2= \vec{v}\cdot \vec{v} = \vec{v}^2$$
which is your "equation 1". What it's not true though is
$$v = \frac{dr}{dt}$$
understanding the symbol without an arrow as the module.
What the "equation A" is telling you is that
$$v^2 = \left(\frac{dr}{dt}\right)^2 + r^2 \left(\frac{d\theta}{dt}\right)^2$$

 

[dyn]

Thanks for your replies. I think I'm having a bit of a brain freeze on this at the moment

 

[PeroK]

Thanks for your replies. I think I'm having a bit of a brain freeze on this at the moment

First, if we use the dot product then there is no confusion about what $\mathbf{\dot r} \cdot \mathbf{\dot r}$ means. We also recognise this as the magnitude squared of the vector, which we can write as: $$|\mathbf{\dot r}|^2 \equiv \mathbf{\dot r} \cdot \mathbf{\dot r}$$ It's also common to drop the modulus symbols and write: $$\mathbf{\dot r}^2 \equiv |\mathbf{\dot r}|^2 \equiv \mathbf{\dot r} \cdot \mathbf{\dot r}$$ Finally, it's common to write the modulus of a vector not in boldface, so we write:$$v^2 \equiv \mathbf{\dot r}^2 \equiv |\mathbf{\dot r}|^2 \equiv \mathbf{\dot r} \cdot \mathbf{\dot r}$$ Where we have essentially defined $$v \equiv \sqrt{\mathbf{\dot r} \cdot \mathbf{\dot r}}$$
Now, there is a point over which we must be careful, because we also have: $$r \equiv \sqrt{\mathbf{r} \cdot \mathbf{r}}$$ And, of course, we can take the derivative of that: $$\dot{r} \equiv \frac{dr}{dt} = \frac{d}{dt} \sqrt{\mathbf{r} \cdot \mathbf{r}}$$ And now we must be careful as $$\dot r \ne v$$ For example, in Cartesian coordinates we have: $$v = \sqrt{\dot x^2 + \dot y^2 + \dot z^2}$$ and $$\dot r = \frac{d}{dt}\sqrt{x^2 + y^2 + z^2}$$ Which are clearly not the same thing.

 

[dyn]

I think my confusion comes from the point that usually we write the modulus of vector a as a but in polar coordinates the modulus of vector r(dot) is not r(dot). It contains an extra term with θ(dot) in it
Hence when writing the kinetic energy of a point particle i should write (1/2)m r(dot)² and not write it as (1/2)m r(dot)² as that is not correct in polar coordinates