# Dot product question

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## 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.

## Answers and Replies

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} ##

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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.

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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.
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.

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.