Fundamentals of Astrodynamics: Dot Product Question

[dimitri151]

TL;DR

Question about dot product of vector and vectors time derivative from reading

I'm reading Fundamentals of Astordynamics by Bate, Mueller, White and having trouble with this passage (pg15):
"2. Since in general a⋅a' = a a'..."
I don't think that this is the case. For instance in uniform circular motion r⋅r' = 0.

Would appreciate if anyone has some insight into this.

 

[jedishrfu]

Aren’t they just writing the dot product definition?

$$a \cdot a’ = |a| |a’| cos \theta $$

where $\theta$ is the angle between them.

in the case of circular motion r and r’ are perpendicular and hence $\theta$ is $\pi / 2$ radians and so the dot product is zero.

 

[dimitri151]

They're not including the cos of theta. ?

 

[dimitri151]

A more complete quote of the passage:
"1. Dot multiplying equation (1.3-4) by $\dot{r}$
$\dot{r}\cdot\ddot{r}+\dot{r}\cdot \frac{\mu}{r^3} r=0$
2. Since in general $\boldsymbol{a\cdot\dot{a}}=a\dot{a}$, $\boldsymbol{v=\dot{r}}$, and $\boldsymbol{\dot{v}=\ddot{r}}$, then
$\boldsymbol{v\cdot\dot{v}}+\frac{\mu}{r^3}\boldsymbol{r\cdot\dot{r}}=0$, so
$v\cdot\dot{v}+\frac{\mu}{r^3}r\dot{r}=0$."

 

[dimitri151]

Ok, solved. If you let $\vec{r}=r \cos\theta \cdot \hat{i}+r sin\theta \cdot\hat{j}$, then calculate $\vec{r}\cdot \frac{d \vec{r}}{dt}$ you do indeed get $r\cdot \frac{dr}{dt}$ (scalar).

 

[Mark44]

It would have been helpful for us to see equation 1.3-4. Presumably the left side of equation 1.3-4 is $\ddot r + \frac \mu {r^3} r$, but I can't say what the right side of this equation is.

A more complete quote of the passage:
"1. Dot multiplying equation (1.3-4) by $\dot{r}$
$\dot{r}\cdot\ddot{r}+\dot{r}\cdot \frac{\mu}{r^3} r=0$
2. Since in general $\boldsymbol{a\cdot\dot{a}}=a\dot{a}$, $\boldsymbol{v=\dot{r}}$, and $\boldsymbol{\dot{v}=\ddot{r}}$, then
$\boldsymbol{v\cdot\dot{v}}+\frac{\mu}{r^3}\boldsymbol{r\cdot\dot{r}}=0$, so
$v\cdot\dot{v}+\frac{\mu}{r^3}r\dot{r}=0$."

For item 2, the only way that $\boldsymbol u \cdot \boldsymbol v = uv$ is if the angle between the two vectors is zero.

 

[dimitri151]

For item 2, the only way that is if the angle between the two vectors is zero.

I thought this as well, but look at this.
Let $\vec{r}=r \cos\theta \cdot \hat{i}+r \sin\theta \cdot\hat{j}$.
Then $\frac{d\vec{r}}{dt}=(\frac{dr}{dt} \cos \theta -r \sin\theta \frac{d\theta}{dt})\hat{i}+(\frac{dr}{dt}\sin\theta +r \cos\theta \frac{d\theta}{dt})\hat{j}$.
And $\vec{r}\cdot\frac{d\vec{r}}{dt}=r \cos\theta(\frac{dr}{dt} \cos \theta -r \sin\theta \frac{d\theta}{dt})+r \sin\theta (\frac{dr}{dt}\sin\theta +r \cos\theta \frac{d\theta}{dt})=r\cdot\frac{dr}{dt}$

 

[Mark44]

I thought this as well, but look at this.
Let $\vec{r}=r \cos\theta \cdot \hat{i}+r \sin\theta \cdot\hat{j}$.
Then $\frac{d\vec{r}}{dt}=(\frac{dr}{dt} \cos \theta -r \sin\theta \frac{d\theta}{dt})\hat{i}+(\frac{dr}{dt}\sin\theta +r \cos\theta \frac{d\theta}{dt})\hat{j}$.
And $\vec{r}\cdot\frac{d\vec{r}}{dt}=r \cos\theta(\frac{dr}{dt} \cos \theta -r \sin\theta \frac{d\theta}{dt})+r \sin\theta (\frac{dr}{dt}\sin\theta +r \cos\theta \frac{d\theta}{dt})=r\cdot\frac{dr}{dt}$

There are two forms of the dot product: a coordinate form, and a coordinate-free form.
Here's a simple example of the coordinate form. If $\vec u = <u_1,u_2>$ and $\vec v = <v_1, v_2>$, then $\vec u \cdot \vec v = u_1\cdot v_1 + u_2 \cdot v_2$
For the coordinate-free form, $\vec u \cdot \vec v = uv \cos(\theta)$. Here $u = |\vec u|$ and similar for the other vector.
So I say again, if $\vec u \cdot \vec v = uv$, then necessarily $\cos(\theta) = 1$, which means that the angle between the two vectors is zero.

 

[dimitri151]

To me its clear $\vec u \cdot \vec v =\vert u\vert\vert v\vert$ implies $\theta=0$. None the less, I can't find the hole in the proof in #7 which says $\vec{r}\cdot\frac{d\vec{r}}{dt}=r \frac{dr}{dt}$ for a general vector $\vec{r}=r \cos\theta \cdot \hat{i}+r \sin\theta \cdot \hat{j}$. I don't believe that that should be generally true, but that's what is shown.

 

[dimitri151]

Ok, got it.
$\boldsymbol{a\cdot\dot{a}}=a\dot{a}$ means $\boldsymbol{a\cdot\dot{a}}=\vert a\vert \dot{\vert a \vert}$ not $\boldsymbol{a\cdot\dot{a}}=\vert a\vert \vert \dot{a} \vert$
Then if:
$\boldsymbol{a}=x \cdot \hat{i} +y \cdot \hat{j}$,
$\boldsymbol{\dot{a}}=\frac{dx}{dt} \cdot \hat{i}+\frac{dy}{dt} \cdot \hat {j}$
$\boldsymbol{a} \cdot \boldsymbol{\dot{a}}=x \frac{dx}{dt} +y \frac{dy}{dt}$
and
$\vert \boldsymbol{a} \vert \dot{\vert \boldsymbol{a} \vert }=(x^2+y^2)^{\frac{1}{2}} \frac{1}{2}(x^2+y^2)^{-\frac{1}{2}} (2 x \frac{dx}{dt}+ 2 y \frac{dy}{dt})=x \frac{dx}{dt} +y \frac{dy}{dt}$

 

[jedishrfu]

Are you missing some parentheses around the 2x and 2y terms?

It looks like you’ve taken the time derivative of the length of $\vec a$ to be the same as the length of the time derivative of $\vec a$.

 

[dimitri151]

Yes, was missing parentheses. I'm taking the length of $\vec a$ and multiplying by the derivative of the length of $\vec a$.

 

[Mark44]

Let $\vec{r}=r \cos\theta \cdot \hat{i}+r \sin\theta \cdot\hat{j}$.

You're assuming that $\theta$ is a function of t, but there's no indication here of any such relationship. Without such a relationship $\theta$ is a constant, which makes $\vec r$ a constant.
In your vector equation above, $\vec r$ is the vector that extends from the origin out to a point (x, y).

In post #1 the work refers to equation 1.3-4. We need more information.

I can't find the hole in the proof in #7

I don't either, but I don't think I'm seeing all of the information.

 

[dimitri151]

Yes, r, $\theta$ functions of t. See first line of equations in #10. I believe that solves it.

 

[Mark44]

See first line of equations in #10. I believe that solves it.

Equation #10 ??
Proof in #7 ??
Equation 1.3-4 - ??
I don't have that book!

Ok, got it.
$\boldsymbol{a\cdot\dot{a}}=a\dot{a}$ means $\boldsymbol{a\cdot\dot{a}}=\vert a\vert \dot{\vert a \vert}$ not $\boldsymbol{a\cdot\dot{a}}=\vert a\vert \vert \dot{a} \vert$
Then if:
$\boldsymbol{a}=x \cdot \hat{i} +y \cdot \hat{j}$,

I find it much more convenient to write using vector notation rather than dragging along the unit vectors $\hat i$ and $\hat j$. IOW, like this:
$\boldsymbol{a}= <x, y>$

$\boldsymbol{\dot{a}}=\frac{dx}{dt} \cdot \hat{i}+\frac{dy}{dt} \cdot \hat {j}$
$\boldsymbol{a} \cdot \boldsymbol{\dot{a}}=x \frac{dx}{dt} +y \frac{dy}{dt}$
and
$\vert \boldsymbol{a} \vert \dot{\vert \boldsymbol{a} \vert }=(x^2+y^2)^{\frac{1}{2}} \frac{1}{2}(x^2+y^2)^{-\frac{1}{2}} (2 x \frac{dx}{dt}+ 2 y \frac{dy}{dt})=x \frac{dx}{dt} +y \frac{dy}{dt}$

I don't know what you're doing here.
You have $\boldsymbol{a}= <x, y>$, so $|\boldsymbol a| = \sqrt{x^2 + y^2}$,
and $\boldsymbol{\dot{a}} = <\frac{dx}{dt}, \frac{dy}{dt}>$, so $|\boldsymbol {\dot a}| = \sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2}$.
Then $\boldsymbol{a} \cdot \boldsymbol{\dot{a}} \ne |a||\dot a|$.

 

[dimitri151]

I think it's not the length of the time derivative of $a$, but the time derivative of the length of $a$.

 

[Mark44]

Please show me the equations and proofs you have cited.
If you don't provide this information, I'm going to give up on this thread.

Equation #10 ??
Proof in #7 ??
Equation 1.3-4 - ??

I think it's not the length of the time derivative of a, but the time derivative of the length of a.

The notation you used, $|\boldsymbol {\dot a}|$, means the length of the derivative of a, which is $\sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2}$.

 

[dimitri151]

I'll recapitulate.
My original question was about this in Fundamentals of Astrodynamics, by Bate (pg15):
"in general $\boldsymbol{a\cdot\dot{a}}=a \; \dot{a}$."
Here's a picture of the page:
************************************************************************************

*************************************************************************************
It didn't make sense because I interpreted $\dot{a}$ to mean the length of the time derivative of vector $a$ in which case we should have $\boldsymbol{a} \cdot \boldsymbol{\dot{a}} = a \; \dot a \; cos \theta$.
But since posting this question, I realized that the author intends $\dot{a}$ to mean the time derivative of the length of $\boldsymbol{a}$ not the length of the time derivative of $\boldsymbol{a}$.
I demonstrate that this indeed the case in #10 of the thread above (obviously I'm referring to the thread, what else would I be referring to, there is no #10 in the quote from the book).
Just check #10 above, where $x$ and $y$ are obviously implied to be functions of $t$ since when I took their time derivative they didn't disappear. In the last line of #10, I multiply the length of $\boldsymbol{a}$ by the time derivative of the length of $\boldsymbol{a}$ and this equals $\boldsymbol{a\cdot\dot{a}}$. Since $\boldsymbol{a}$ is a general vector this does prove the author's claim.
You don't really need 1.3-4 to solve the problem.

This must be a common question because I found it in two other places, once on Stack Exchange where I found the solution, and once on this site where it wasn't really clearly resolved.

Cheers.

 

[Mark44]

But since posting this question, I realized that the author $\dot{a}$ to mean the time derivative of the length of $\boldsymbol{a}$ not the length of the time derivative of $\boldsymbol{a}$.

That makes sense, now, but his notation is far from clear. This would have been much clearer:
$\boldsymbol{a} \cdot \boldsymbol{\dot{a}} = |\boldsymbol a|\frac d {dt} \left|\boldsymbol a \right|$.

That would have made it clear that he's taking the time derivative of the magnitude of a.
The notation $\dot a$ suggests to me that this is what is happening: $|\boldsymbol {\dot a}|$. IOW, taking the derivative first, and then getting the magnitude.

 

[jedishrfu]

Sometimes math notation is ambiguous like this especially in the vector world where boldface letters can represent a vector without the pesky arrow hat And non-boldface to represent its length.

 

[Mark44]

Sometimes math notation is ambiguous like this especially in the vector world where boldface letters can represent a vector without the pesky arrow hat And non-boldface to represent its length.

You missed what the problem was here. I'm very comfortable with either kind of notation for a vector: $\boldsymbol v$ or $\vec v$.
My complaint about the notation was using $\dot a$ to mean $\frac d {dt}|\boldsymbol a|$; i.e., taking the derivative of the magnitude of a vector, as opposed to the magnitude of the derivative of the vector.