Central force on a particle following a logarithmic spiral

In summary, the conversation discusses the force ##F(r)## and how it relates to polar coordinates and Newton's second law. The equation for ##F(r)## is derived and simplified using constants ##A## and ##B##. The conversation also explores the use of the chain rule to find the function ##r(t)## and the general central force equation. There is a discussion about a mistake in the integration process and alternative methods to simplify the equations.
  • #1
JD_PM
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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:

$$F(r) = m \Big[ k\Big( \frac{A \alpha}{m r^2} \Big)^2 e^{\frac{B \alpha}{m r^2}} e^{\frac{A \alpha t}{m r^2}} - \frac{1}{r^3} \Big(\frac{A}{m}\Big)^2 \Big] \ \ \ \ \ (5)$$
 
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  • #2
JD_PM said:
$$m( r \ddot \theta + 2 \dot r \dot \theta) = 0$$ ...
(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 ?
 
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  • #3
Can you elaborate more on your question? Do you want a graph of function?
 
  • #4
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.
 
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  • #5
TSny said:
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?
 
  • #6
JD_PM said:
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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  • #7
JD_PM said:
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! :oldsmile:
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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  • #8
TSny said:
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!
 
  • #9
TSny said:
What you did here is a wild ride! :oldsmile:
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?

TSny said:
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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  • #10
JD_PM said:
$$r(t) = \pm \sqrt{2C(\alpha t + E)} \ \ \ \ (10)$$

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.
 
  • #11
JD_PM said:
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.
 
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  • #12
PeroK said:
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:
$$f(r) = m(\ddot r - r \dot \theta^2) = m(\ddot r - \frac{l^2}{m^2 r^3})$$
 
  • #13
PeroK said:
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.
 
  • #14
JD_PM said:
Which one?

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

You forgot this one!

JD_PM said:
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).
 
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  • #15
OK I made a mistake here:

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

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
 
  • #16
JD_PM said:
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!
 
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  • #17
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##
 
  • #18
JD_PM said:
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##.
 
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  • #19
TSny said:
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.
 
  • #20
JD_PM said:
$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big)$$

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

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

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.
 
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  • #21
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 let's you derive ##V(r)## from the given solution ##r=r(\theta)##.
 
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  • #22
Thank you for your post vanhees71! :)

vanhees71 said:
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)##.
 
  • #23
Sure, but this you can address easier as soon has you have found ##F(r)## or ##V(r)## ;-).
 
  • #24
vanhees71 said:
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!
 
  • #25
Yes, and this you get after you've found ##V(r)## by solving the equation of motion for ##r##.
 
  • #26
vanhees71 said:
Yes, and this you get after you've found ##V(r)## by solving the equation of motion for ##r##.
The direct approach could have been done more efficiently:

$$r = ke^{\alpha \theta}$$
$$\dot r = \alpha r \dot \theta = \frac{\alpha l}{mr}$$
$$\frac{d(r^2)}{dt} = 2r \dot r = \frac{2\alpha l}{m}$$
$$r = (r_0^2 + \frac{2\alpha l}{m}t)^{1/2}$$
 
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1. What is a central force on a particle?

A central force on a particle is a type of force that acts on a particle directed towards or away from a central point or axis. Examples of central forces include gravity, electrostatic force, and magnetic force.

2. What is a logarithmic spiral?

A logarithmic spiral is a type of spiral curve that grows or shrinks at a constant rate as it rotates around a central point. It is also known as an equiangular spiral because the angle between any consecutive turns remains constant.

3. How does a particle follow a logarithmic spiral?

A particle following a logarithmic spiral is constantly changing direction and moving towards the center of the spiral at a constant speed. This is due to the balance between the central force acting on the particle and the particle's inertia.

4. What are some real-life examples of central forces on particles following a logarithmic spiral?

Some examples include the spiral arms of galaxies, the formation of seashells, and the growth of plants such as ferns and succulents.

5. How is the central force on a particle following a logarithmic spiral related to the golden ratio?

The golden ratio, also known as the divine proportion, is often found in the growth patterns of logarithmic spirals. This is because the ratio of the distance between any two points on a logarithmic spiral to the distance from the center to the first point is equal to the golden ratio.

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