# 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.

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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:

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

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BvU
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$$m( r \ddot \theta + 2 \dot r \dot \theta) = 0$$ ...
(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
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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:

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

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

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

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TSny
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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.

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PeroK
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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: $\frac{L}{T^2}$

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

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
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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
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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
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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
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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
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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 $r$.

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

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

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

$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big)$$
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
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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,
$$\frac{m}{2} \dot{\vec{x}}^2 + V(r)=E=\text{const},$$
$$\frac{m}{2} (\dot{r}^2 + r^2 \dot{\theta}^2)+V(r)=E. \qquad (**)$$
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
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Sure, but this you can address easier as soon has you have found $F(r)$ or $V(r)$ ;-).

PeroK
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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
Yes, and this you get after you've found $V(r)$ by solving the equation of motion for $r$.