# Central force on a particle following a logarithmic spiral

## Homework Statement:

Given a central force field that allows to keep particles in the following logarithmic spiral $r( \theta ) = k e^{\alpha \theta}$ (where $k$ and $\alpha$ are constant).

a) Determine $r(t)$ and $\theta (t)$.

b) Give the equation of the force.

c) Discuss the nature of the force fields associated to the obtained force.

## Relevant Equations:

$$m( \ddot r - r \dot \theta^2 ) = F(r)$$
$$m( r \ddot \theta + 2 \dot r \dot \theta) = 0$$
I want to focus this question on understanding the force $F(r)$ I get (thus, I want to focus on c) ). However, below the dashed line, I included steps on how I derived $F(r)$.

We are going to work in polar coordinates.

Knowing that the acceleration is:

$$a = \Big( \ddot r - r \dot \theta^2 \Big) \hat r + \Big( r \ddot \theta + 2 \dot r \dot \theta \Big) \hat \theta$$

Newton's second law ends up with the form:

$$F(r) = m \Big( \ddot r - r \dot \theta^2 \Big) \ \ \ \ (1)$$

$$m \Big( r \ddot \theta + 2 \dot r \dot \theta \Big) = 0 \ \ \ \ \ \ \ (2)$$

a)

$$\theta (t) = \frac{(At + B)}{mr^2} \ \ \ \ \ (3)$$

$$r (t) = k \ exp \Big(\alpha \frac{(At + B)}{mr^2} \Big) \ \ \ \ \ (4)$$

b)

$$F(r) = m \Big[ k\Big( \frac{A \alpha}{m r^2} \Big)^2 e^{\frac{B \alpha}{m r^2}} e^{\frac{A \alpha t}{m r^2}} - \frac{1}{r^3} \Big(\frac{A}{m}\Big)^2 \Big] \ \ \ \ \ (5)$$

c)

Let me define the following quantities to simplify the equation for $F(r)$

$$C = \frac{A}{m}$$

$$D = \frac{B}{m}$$

So we get a central force of the form:

$$F(r) = m \Big[ k\Big( \frac{C \alpha}{ r^2} \Big)^2 e^{\frac{D \alpha}{r^2}} e^{\frac{C \alpha t}{ r^2}} - \frac{C^2}{r^3}\Big] \ \ \ \ \ (6)$$

Alright so one notices that the repulsive inverse square field and the attractive inverse cube field are involved.

However, what can we say about the exponents?

Thanks.

---------------------------------------------------------------------------------------------------------------------------------------------

a)

- Getting $\theta (t)$

OK so $(2)$ can be rewritten as follows:

$$\frac{d}{r dt} \Big( m r^2 \dot \theta \Big) = 0$$

Integrating twice with respect to $t$ we get:

$$\theta (t) = \frac{(At + B)}{mr^2} \ \ \ \ \ (3)$$

Where $A$ and $B$ are constants.

- Getting $r (t)$

By plugging in equation $(3)$ into the given equation $r( \theta ) = k e^{\alpha \theta}$ we get $r (t)$

$$r (t) = k \ exp \Big(\alpha \frac{(At + B)}{mr^2} \Big) \ \ \ \ \ (4)$$

b)

So we just have to use equation $(1)$ to get:

$$F(r) = m \Big[ k\Big( \frac{A \alpha}{m r^2} \Big)^2 e^{\frac{B \alpha}{m r^2}} e^{\frac{A \alpha t}{m r^2}} - \frac{1}{r^3} \Big(\frac{A}{m}\Big)^2 \Big] \ \ \ \ \ (5)$$

## Answers and Replies

Related Advanced Physics Homework Help News on Phys.org
BvU
Science Advisor
Homework Helper
2019 Award
$$m( r \ddot \theta + 2 \dot r \dot \theta) = 0$$ ...
(2) can be rewritten as follows ...
Integrating ${\displaystyle {d\over dt}}\Bigl ( mr^2\dot \theta \Bigr) = 0\$ once with respect to $t$ we get
$\Bigl ( mr^2\dot \theta \Bigr) = A\$, i.e. angular momentum is constant.
How do you get (3) from that ?

Can you elaborate more on your question? Do you want a graph of function?

TSny
Homework Helper
Gold Member
BvU is pointing out that you made a mistake in going from

$$\frac{d}{ dt} \Big( m r^2 \dot \theta \Big) = 0$$
to
$$\theta (t) = \frac{(At + B)}{mr^2} \ \ \ \ \ (3)$$

You said that you integrated the first equation twice with respect to time. But it appears that you treated $r$ as a constant in the second integration.

BvU is pointing out that you made a mistake ... you treated $r$ as a constant in the second integration.
You both are right, let's fix it.

Let me use:

$$C = \frac{A}{m}$$

The idea is to integrate the following equation with respect to time (where both $r$ and $\theta$ are functions of time):

$$\int d \theta= \int \frac{C}{r^2} dt \ \ \ \ (7)$$

We do not know what is the explicit formula for $r(t)$ but we know what's it for $r(\theta )$

$$r( \theta ) = k e^{\alpha \theta}$$

My idea is to use the chain rule to get $r(t)$:

$$\frac{d}{dt} r(t) = \frac{dr}{d \theta}\frac{d \theta}{dt} = k \alpha e^{\alpha \theta} \dot \theta \ \ \ \ (8)$$

The next step is plugging $(8)$ into $(7)$ and integrate.

Multiplying top and bottom of the RHS of $(7)$ by $(\frac{d}{dt})^2$, I get:

$$\frac{(\frac{d}{dt})^2 C dt}{(\frac{d}{dt} r)^2} = \frac{(\frac{d}{dt})^2 C dt}{ (k \alpha e^{\alpha \theta} \dot \theta)^2} = \frac{ C dt}{ (k \alpha e^{\alpha \theta} d\theta)^2}$$

Which leads to:

$$\int k \alpha e^{\alpha \theta} (d \theta)^3 = \int C dt$$

I've done something wrong because $d \theta$ shouldn't be to the power of three.

Do you agree on the idea of using the chain rule though?

PeroK
Science Advisor
Homework Helper
Gold Member
Do you agree on the idea of using the chain rule though?
Another idea is to use or derive the general central force equation:
$$\frac{d^2}{d\theta^2}(\frac 1 r ) + \frac 1 r = -\frac{mr^2}{l^2}F(r)$$

Last edited:
TSny
Homework Helper
Gold Member
My idea is to use the chain rule to get $r(t)$:

$$\frac{d}{dt} r(t) = \frac{dr}{d \theta}\frac{d \theta}{dt} = k \alpha e^{\alpha \theta} \dot \theta \ \ \ \ (8)$$
OK. Note that this may be written as $$\dot r = \alpha r \dot \theta \ \ \ \ (8')$$

The next step is plugging $(8)$ into $(7)$ and integrate.

Multiplying top and bottom of the RHS of $(7)$ by $(\frac{d}{dt})^2$, I get:

$$\frac{(\frac{d}{dt})^2 C dt}{(\frac{d}{dt} r)^2} = \frac{(\frac{d}{dt})^2 C dt}{ (k \alpha e^{\alpha \theta} \dot \theta)^2} = \frac{ C dt}{ (k \alpha e^{\alpha \theta} d\theta)^2}$$
What you did here is a wild ride!
But it's not correct. You cannot treat $(\frac{d}{dt})^2$ this way.

In your first post, you arrived at $\frac{d}{rdt} (mr^2\dot\theta) = 0$ which may be simplified to $\frac{d}{dt}(r^2 \dot \theta)=0$.
So $r^2 \dot \theta = C$, where $C$ is a constant. ($C$ is the angular momentum per unit mass.) Solve this for $\dot \theta$ and substitute into equation (8') above.

Last edited:
PeroK
Science Advisor
Homework Helper
Gold Member
OK. Note that this may be written as $$\dot r = \alpha r \dot \theta \ \ \ \ (8')$$

So $r^2 \dot \theta = C$, where $C$ is a constant.
If you've got this far, perhaps solving for $r$ first is simplest!

What you did here is a wild ride!
But it's not correct. You cannot treat $(\frac{d}{dt})^2$ this way.
I have the impression that mathematicians treat $\frac{d}{dt}$ differently than physicists, don't they?

Try solving this for $\dot \theta$ and then substitute into equation (8') above.
We have:

$$\dot \theta = \frac{C}{r^2} \ \ \ \ (9)$$

Plugging $(9)$ into $(8')$ we get:

$$\dot r - \frac{\alpha C}{r} = 0$$

This is a linear first order homogeneous ODE whose solution is (if you do not agree I'll include the steps):

$$r(t) = \pm \sqrt{2C(\alpha t + E)} \ \ \ \ (10)$$

Where $E$ is the constant of integration.

Plugging $(10)$ into $(9)$ we get:

$$d \theta = \frac{d t}{2C(\alpha t + E)}$$

This is a first order homogeneous ODE whose solution is (if you do not agree I'll include the steps):

$$\theta (t) = \frac{ln(\alpha t + E)}{2 \alpha} + F$$

Where $F$ is the constant of integration.

The double derivative of $(10)$ with respect to time is:

$$\ddot r = \frac{-2 \alpha^2 C^2}{4(2C(\alpha t + E))^2} = \frac{-\alpha^2 C^2}{2r^2}$$

The force $F(r)$ I get is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{2r^2} - \frac{C}{r}\Big)$$

There may be something wrong. Checking dimensional analysis ...

EDIT: As PeroK pointed out, this result is not OK. Dimensional analysis of $\frac{-\alpha^2 C^2}{2r^2}$ yields $L^2$, while it should yield the dimension for acceleration: $\frac{L}{T^2}$

Last edited:
PeroK
Science Advisor
Homework Helper
Gold Member
$$r(t) = \pm \sqrt{2C(\alpha t + E)} \ \ \ \ (10)$$

Where $E$ is the constant of integration.
First, you should be able to simplify this immediately to:

$$r(t) = \sqrt{r_0^2 + 2C\alpha t}$$

Second, you already have an equation relating $r$ and $\theta$, so use that instead of integrating $\theta$.

To get $f(r)$ you can again shortcut things by using the constant angular momentum to avoid differentiating $\theta$.

PS I'd replace $C = l/m$ where $l$ is the angular momentum.

PeroK
Science Advisor
Homework Helper
Gold Member
The force $F(r)$ I get is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{2r^2} - \frac{C}{r}\Big)$$

There may be something wrong. Checking dimensional analysis ...
This can't be correct.

PeroK
Science Advisor
Homework Helper
Gold Member
To get $f(r)$ you can again shortcut things by using the constant angular momentum to avoid differentiating $\theta$.
By this I mean that, using $l = mr^2 \dot \theta$, we have:
$$f(r) = m(\ddot r - r \dot \theta^2) = m(\ddot r - \frac{l^2}{m^2 r^3})$$

Second, you already have an equation relating $r$ and $\theta$, so use that instead of integrating $\theta$.
Which one?

The only equation I see is $\dot r = \alpha r \dot \theta$, and we need to integrate it.

PeroK
Science Advisor
Homework Helper
Gold Member
Which one?

The only equation I see is $\dot r = \alpha r \dot \theta$, and we need to integrate it.
You forgot this one!

Homework Statement:: Given a central force field that allows to keep particles in the following logarithmic spiral $r( \theta ) = k e^{\alpha \theta}$ (where $k$ and $\alpha$ are constant).

OK I made a mistake here:

$$\ddot r = \frac{-2 \alpha^2 C^2}{4(2C(\alpha t + E))^2} = \frac{-\alpha^2 C^2}{2r^2}$$
It is

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$

Thus the force is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{r^{3/2}} - \frac{C}{r}\Big)$$

But this is again wrong (I checked dimensions and do not match).

I do not see why, as I checked several times my computations and they are OK.

EDIT: I am going to try the procedure PeroK suggested

PeroK
Science Advisor
Homework Helper
Gold Member
OK I made a mistake here:

It is

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$

Thus the force is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{r^{3/2}} - \frac{C}{r}\Big)$$

But this is again wrong (I checked dimensions and do not match).

I do not see why, as I checked several times my computations and they are OK.
Sadly, I think both terms are wrong. If $r = (a + bt)^{1/2}$, then time derivatives of $r$ involve (odd) integer powers of $r$.

Alternatively, try implicitly differentiating $r^2$.

I don't see how you got the second term. See post #12.

PS If you get $F(r)$ using post #6, you can see what you're aiming for!

Alright, so from post #12 we see that $F(r)$ is:

$$F(r) = m \Big( \ddot r - \frac{C^2}{r^3}\Big)$$

By differentiating equation (10) we get:

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$

Thus:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{r^{3/2}} - \frac{C^2}{r^3}\Big)$$

Which can be rewritten as:

$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r^{-1/2}} - \frac{1}{r}\Big)$$

I am closer to the equation you gave at #6;

I am sure there has to be a way to go from $\frac{-\alpha^2 }{r^{-1/2}}$ to $\frac{d^2}{d\theta^2}(-\frac 1 r )$... but how? I have been trying to use the chain rule. We also know $r(\theta)$ from the exercise statement. I suspect we could also use it to get rid of $\alpha$

TSny
Homework Helper
Gold Member
Alright, so from post #12 we see that $F(r)$ is:

$$F(r) = m \Big( \ddot r - \frac{C^2}{r^3}\Big)$$
OK

By differentiating equation (10) we get:

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$
The power of $r$ in the denominator is not correct. Once you fix that, you will see that $F(r)$ has a nice dependence on $r$.

The power of $r$ in the denominator is not correct. Once you fix that, you will see that $F(r)$ has a nice dependence on $r$.
Oops my bad. Now fixed:

$$\ddot r = \frac{-\alpha^2 C^2}{r^3}$$

Thus $F(r)$ is:

$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big)$$

This expression is OK. I've noticed that the expression $\frac{-\alpha^2 }{r}$ is OK, as $\alpha$ is dimensionless.

To put it in the same form PeroK suggested in #6, we see that:

$$\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) = \frac{\alpha^2 }{ke^{\alpha \theta}}$$

Thus:

$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big)$$

Thank you all, it was fun!

Nature of the obtained force

Only the attractive inverse cube field is involved.

PeroK
Science Advisor
Homework Helper
Gold Member
$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big)$$

$$\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) = \frac{-\alpha^2 }{ke^{\alpha \theta}}$$

$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big)$$
Just an observation. It looks odd to me that you leave expressions in the form that you do. Maybe it's just style, but I would have written:
$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big) = -\frac{A^2}{mr^3}(\alpha^2 +1)$$
$$\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) = \frac{-\alpha^2 }{ke^{\alpha \theta}} = -\frac{\alpha^2 }{r}$$
$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big) = -\frac{A^2}{mr^3}(\alpha^2 +1)$$

It's almost like you switch off one step before the end. It's a minor point, perhaps.

vanhees71
Science Advisor
Gold Member
2019 Award
I think the issue is a bit overcomplicated in this thread. The point is that you have given the trajectory in terms of $r=r(\theta)$. You don't care about how the trajectory is run through as a function of time. The two "homework equations" are all you really need.

The 2nd equation is nothing else than angular-momentum conservation, i.e., you can integrate it to
$$r^2 \dot{\theta}=\ell=\text{const}. \quad (*)$$
Then you introduce the central potential
$$V(r)=-\int \mathrm{d} r F(r)$$
and use the energy-conservation law,
$$\frac{m}{2} \dot{\vec{x}}^2 + V(r)=E=\text{const},$$
which reads in polar coordinates
$$\frac{m}{2} (\dot{r}^2 + r^2 \dot{\theta}^2)+V(r)=E. \qquad (**)$$
Now you can eliminate $\dot{\theta}$, using (*) and get a differential equation for $\mathrm{d}r/\mathrm{d} \theta=\dot{r}/\dot{\theta}$ from (**), which lets you derive $V(r)$ from the given solution $r=r(\theta)$.

Thank you for your post vanhees71! :)

You don't care about how the trajectory is run through as a function of time.
But note that $a)$ is asking for $r(t)$ and $\theta (t)$.

vanhees71
Science Advisor
Gold Member
2019 Award
Sure, but this you can address easier as soon has you have found $F(r)$ or $V(r)$ ;-).

PeroK
Science Advisor
Homework Helper
Gold Member
I think the issue is a bit overcomplicated in this thread. The point is that you have given the trajectory in terms of $r=r(\theta)$. You don't care about how the trajectory is run through as a function of time.
Part a) in the OP asked for the trajectory as a function of time!

vanhees71
Science Advisor
Gold Member
2019 Award
Yes, and this you get after you've found $V(r)$ by solving the equation of motion for $r$.