I Vector squared in polar coordinates

AI Thread Summary
The discussion revolves around the confusion regarding the representation of vectors and their magnitudes in polar coordinates, particularly in the context of kinetic energy. The equation a² = a·a is questioned for its validity in polar coordinates, where the kinetic energy expression includes both radial and angular components. It is clarified that while the equation is a definition, the relationship between the derivative of the modulus and the modulus of the derivative does not hold, leading to the correct kinetic energy equation ṙ² = ṙ² + r²θ̇². The key takeaway is that in polar coordinates, the kinetic energy must account for both radial and angular velocities, which differs from Cartesian coordinates. Understanding these distinctions is crucial for accurate calculations in polar systems.
dyn
Messages
774
Reaction score
63
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
 
Mathematics news on Phys.org
dyn said:
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: $$\mathbf{\dot r}^2 = \dot x^2 + \dot y^2$$
 
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 said:
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 ?
 
dyn said:
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.##
 
dyn said:
Is Equation 1 above always valid or only valid in certain circumstances ?
Equation 1 is a definition, so yes.
 
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 ?
 
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
$$v^2 = \left(\frac{dr}{dt}\right)^2 + r^2 \left(\frac{d\theta}{dt}\right)^2$$
 
  • Like
Likes dyn and PeroK
Thanks for your replies. I think I'm having a bit of a brain freeze on this at the moment
 
  • #10
dyn said:
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.
 
  • Like
Likes dyn
  • #11
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
 
Back
Top