# Dot product question

• I

## Summary:

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

## Main Question or Discussion Point

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
Mentor
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.

They're not including the cos of theta. ???

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##."

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
Mentor
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.

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
Mentor
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.

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.

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} ##

Last edited:
• Delta2
jedishrfu
Mentor
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##.

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

Mark44
Mentor
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.
dimitri151 said:
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.

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

Mark44
Mentor
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>##
dimitri151 said:
##\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|##.

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

Mark44
Mentor
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 - ??
dimitri151 said:
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}##.

Last edited:
I'll recapitulate.
"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.

• hutchphd
Mark44
Mentor
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 and berkeman
jedishrfu
Mentor
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
Mentor
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.