An ion moves along the x-axis of a black box with a speed $V$ and returns in a time $$T=a V^b$$ where $a$ and $b$ are some known constants. Having this, can we reproduce the dependence of a field potential $U(x)$ of this box?

So far I have managed to do this:

Adding to an initial velocity some $dV$ we get the increase in a time $dt$ and so $$T+dt=a (V+dV)^b \, .$$ We can derive $dt$ by subtracting the initial $T$ value $$dt=a(V+dV)^b-aV^b \, .$$ Using the equivalence formula for $dV$ approaching zero we find $$dt=aV^b(1+b\frac{dV}{V})-aV^b=abV^{b-1}dV \, .$$ We can find the acceleration $$\frac{dV}{dt}=\frac{V^{1-b}}{ab} \, .$$

Similarly, as $V=(T/a)^{\frac{1}{b}}$ we have \begin{align} dV &= \left(\frac{T+dt}{a} \right)^\frac{1}{b} - \left(\frac{T}{a} \right)^\frac{1}{b} \\ &= \left(\frac{T}{a} \right)^\frac{1}{b} \left(1+\frac{dt}{bT} \right) - \left( \frac{T}{a} \right)^\frac{1}{b} \\ &=\frac{T^\frac{1-b}{b}dt}{ab} \, . \end{align} This yields again $$\frac{dV}{dt}=\frac{T^\frac{1-b}{b}}{ab} \, .$$ Now I assume it is our acceleration which the field imparts to the particle, thus $$\frac{dV}{dt}=\frac{f}{m}=\frac{dU}{dx m} \, .$$ From there I am not sure whether or not I can integrate the equation $$dU=m\frac{T^{\frac{1-b}{b}}}{ab} \, dx \, .$$ So, are the time $T$ there is a constant in relation to $dx$ or not? The answer I get from the last equation is $$U(x)=U(0)+m\frac{T^{\frac{1-b}{b}}}{ab}x \, .$$ I am confused by two things here:

  1. The difference between the answers derived from the second and the first approach: $$\frac{dV}{dt} = \frac{V^{1-b}}{ab} \quad \text{and} \quad \frac{dV}{dt} = \frac{T^\frac{1-b}{b}}{ab}$$

  2. The possibility of integrating that way.

Are there other ways to get the definite answer for this task?


No, I don't think that you proceed correctly. You need the relationship between force and distance, and this is what you should integrate.

So, please follow my formulas. So, I understand that $V$ is the velocity of the ion (not of the box). From your formula $T=aV^b$, I deduce the acceleration $A$ acquired at the end of the trip, by assuming something that is not written in the exercise, that inside the box the velocity obeys the same rule as we see at the end of the box. As Lionel says, between $v = V$ and $v = 0$ passes only half of the time, $T_1 = T/2$, s.t.

$$V=\left( \frac{2T_1}{a}\right)^{1/b} \Longrightarrow \, \, A= \frac{dV}{dT_1} = 2^{1/b}\frac{1}{ab} \left( \frac{T_1}{a}\right)^{1/b-1}$$

Given the acceleration, the force is $F = mA$, as you say.

Now, you have to calculate the space that the ion travelled inside the box.

$$ X = \int_0 ^{T_1} V \ dT' = 2^{1/b}a\int_0 ^{T_1} \frac {dT'}{a} \ \left( \frac{T'}{a}\right)^{1/b} = 2^{1/b}\frac {ab}{1 + b} \ \ \left( \frac{T_1}{a}\right)^{(1+b)/b}.$$

Now you can express the force as a function of $X$ by eliminating between them the time $T_1$, and then integrate and find the potential. Let's do it:

$$\left( \frac{T_1}{a}\right) = 2^{-1/b}\left( \frac {1+b}{ab} X\right) ^{b/(1+b)}$$

Substituting in the field strength $E = mA/q$, where $q$ is the charge of the ion,

$$ E = 2^{-1/b} \frac {m}{q} \frac {1}{ab} \left( \frac {1+b}{ab} X\right) ^{(1-b)/(1+b)}$$

This can be easily integrated over $X$

$$U(X) = 2^{1-1/b} \frac {1+b}{ab} \frac {m}{q} \left( \frac {1+b}{ab} \right) ^{(1-b)/(1+b)} X^{2/(1+b)} $$

$$ = 2^{1 -1/b} \frac {m}{q} \left( \frac {1+b}{ab} X \right) ^{2/(1+b)} $$


As it follows from the Mechanics by Landau-Lifshitz (which can be found here — page 28), the desired dependence can be found in the form of

$$x(U)=\frac{1}{2\pi \sqrt{2m}}\int_{0}^{U} \frac{T(E)dE}{\sqrt{U-E}}$$

Where $T(E)$ from the law of conservation of energy and the initial statement $T_{½}=\alpha V^\beta$

$$T(E)=2\alpha (\frac{2E}{m})^\frac{\beta}{2}$$

The period $T$ given there is twice as big as a time of return $T_{½}$ due to the fact, that ion has done only a half of its full way by the time it returns to initial point.

From now I am not familiar with methods of integrating such equations, so I have used Mathematica database for this purpose (please, let me know if the answer is wrong, for I cannot make sure that the answer is correct):

$$U(x)=\frac{1}{2}m \left ( \frac{2x\pi}{\alpha}\right ) ^{\frac{2}{\beta+1}}\frac{\Gamma(\frac{\beta+3}{2})}{\Gamma(\frac{\beta+3}{2})}$$

$\Gamma$ here is a gamma function. Note that the answer $U(x)$ is dimensionally correct.

  • $\begingroup$ Are you sure $T$ here is multiplied by 2, or is it just the period? $\endgroup$ – lionelbrits Dec 26 '14 at 0:19
  • $\begingroup$ @lionelbrits Yes, I should have marked that the period is actually twice as big as the time needed for ion to return from $x_0$ through $x_1$ back to $x_0$. It is the same situation as in the case of a simple pendulum. $\endgroup$ – mikeonly Dec 26 '14 at 10:14

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