1. Jul 15, 2010

### pc2-brazil

Good morning,

I was reading a derivation of equations of the two-body problem and I found the following statement:
$$\mu \frac{d}{dt}\left (\frac{\vec{r}}{r}\right ) = \frac{\mu}{r} \vec{v} - \frac{\mu \dot{r} }{r^2} \vec{r}$$
Where μ is a constant. (If you're interested on where this came from, see page 19 http://books.google.com.br/books?id...resnum=3&ved=0CCUQ6AEwAg#v=onepage&q&f=false").
Calculating this derivative is easy, using the quotient rule. Anyway, I am trying to verify the inverse, that is, calculating the integral of the right member of the equation.
But my integration knowledge is very limited. I've tried using integration by parts, but I got stuck.
Could anyone give an idea on how I should proceed, or what technique I should use?

Last edited by a moderator: Apr 25, 2017
2. Jul 15, 2010

### mathman

Integration by parts should work.

3. Jul 16, 2010

### pc2-brazil

Note: I forgot to inform that $$\vec{v} = \frac{d\vec{r}}{dt}$$. I'm informing this in a reply because I can't edit the main message.

4. Jul 16, 2010

### mathman

When you tried integration by parts where did you get stuck?

5. Jul 16, 2010

### pc2-brazil

Firstly, I did the following:
$$\int \left (\frac{\mu}{r} \vec{v} - \frac{\mu \dot{r} }{r^2} \vec{r}\right ) dt=\mu\int \left (\frac{\vec{v}}{r} - \frac{\dot{r} \vec{r}}{r^2}\right ) dt=\mu \left (\int\frac{\vec{v}}{r}dt - \int\frac{\dot{r} \vec{r}}{r^2}dt\right )$$
(Note that the vectors r and v and their magnitudes are functions of time.)
Now, I try to solve the following integral by integration by parts:
$$\int\frac{\vec{v}}{r}dt$$
First trial:
Setting two functions x and y:
$$x = \vec{v} \Rightarrow dx = \dot{\vec{v}}dt;\ dy = \frac{1}{r}dt$$
The problem starts right here. How do I solve dy = 1/r dt for y?
Second trial:
$$\int \frac{\vec{v}}{r}dt = \int \frac{1}{r}\frac{d\vec{r}}{dt} dt = \int\frac{1}{r}d\vec{r}$$
Using integration by parts:
$$x = \frac{1}{r} \Rightarrow \frac{dx}{d\vec{r}} = \frac{d}{d\vec{r}}\left (\frac{1}{r}\right );\ dy = d\vec{r} \Rightarrow y = \vec{r}$$
The problem is that I don't know how to calculate $$\frac{d}{d\vec{r}}\left (\frac{1}{r}\right )$$.
As you can see, my knowledge of calculus is very limited. What am I doing wrong here?

Last edited: Jul 16, 2010
6. Jul 17, 2010

### mathman

Note: I am weak in Latex, so I'll use capital letters for vectors - also ' for d/dt.
I would suggest the following: u=1/r and dW=Vdt, then du=-r'/r2 and W=R.
∫(V/r)dt = R/r + ∫(Rr'/r2)dt

7. Jul 17, 2010

### DrRocket

You have to be very careful here because of the potential confusion by symbology and the fact that $$\vec{r}$$ and $$r$$ are very different things. In fact $$r = \sqrt {\vec{r} \centerdot \vec{r}}$$.

So probably the easiest way to do the integral is to simply use what you know about the derivative and go in reverse. We'll for get about the constant $$]mu$$ here since it adds nothing.

$$\int \frac{\frac {d \vec{r}}{dt}}{r} - \frac {r \frac{dr}{dt}}{r^2} dt$$ = $$\int \frac {r \frac{d \vec{r}}{dt} - \vec{r}\frac{dr}{dt}}{r^2}$$ = $$\int \frac{d}{dt}[\frac{\vec{r}}{r}] dt$$= $$\frac{\vec{r}}{r}$$

Last edited by a moderator: Apr 25, 2017
8. Jul 18, 2010

### pc2-brazil

Thank you for the responses.
DrRocket: Yes, r and $$\vec{r}$$ are very different things. r is the magnitude of $$\vec{r}$$. For example, $$\vec{r}$$ could vary only in direction; in this case, r wouldn't vary at all, so $$\frac{d\vec{r}}{dt}$$ would be $$\neq \vec{0}$$ and $$\frac{dr}{dt}$$ would be 0.
If I use mathman's approach I will also get to the solution:
$$\mu \left (\int\frac{\vec{v}}{r}dt - \int\frac{\dot{r} \vec{r}}{r^2}dt\right ) = \mu\left (\frac{\vec{r}}{r} + \int\frac{\dot{r} \vec{r}}{r^2}dt - \int\frac{\dot{r} \vec{r}}{r^2}dt \right ) = \mu\frac{\vec{r}}{r}$$
But I have another question. What if I had the situations below?
$$\int r d\vec{r}$$
$$\int \vec{r} dr$$
Does it make sense? How could I solve them?
For the first one, maybe I could use the relation $$r = \sqrt {\vec{r} \centerdot \vec{r}}$$?

9. Jul 18, 2010

### DrRocket

What is $$\int r d\vec{r}$$ supposed to mean ? Here you have something $$d\vec{r}$$ which has no clear meaning. It appears to be some sort of vector-valued measure, and that at best will take some work to even define.

Similarly, you have $$\int \vec{r} dr$$ and while whatever $$dr$$ is, it is apparently scalar valued. But what it is is also a bit mysterious. An ordinary measure perhaps, but defined on what measure space ?

Just because you can write down a bunch of symbols does not imply that those symbols are meaningful.

10. Jul 18, 2010

### pc2-brazil

For example:
$$\int\frac{\dot{r} \vec{r}}{r^2}dt = \int \frac{dr}{dt}\frac{\vec{r}}{r^2}dt = \int \frac{\vec{r}}{r^2}dr$$
Is that right? If so, I could solve that last integral by integration by parts, then could obtain:
$$u = \frac{1}{r^2};\ d\vec{w} = \vec{r}dr \Rightarrow \vec{w} = \int \vec{r} dr$$
I could also try to write the integral below:
$$\int\frac{\vec{v}}{r}dt = \int\frac{1}{r}\frac{d\vec{r}}{dt}dt = \int\frac{1}{r}d\vec{r}$$
(I think that I didn't find $$\int r d\vec{r}$$ anywhere.)

Last edited: Jul 18, 2010
11. Jul 18, 2010

### mathman

In simple terms the concept of dR, where R is a vector has no definition in mathematics.