Understanding Josephson Oscillations: Solving for the Second Derivative of Theta

[CMJ96]

Homework Statement

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Homework Equations

Sorry about this, I had to put this into wolfram alpha as for some reason it would not work in latex

The Attempt at a Solution

Assuming Ec and Ej are not dependent on time, I have differentiated the first term in equation 10 with respect to time, and replaced the relevant terms with equation 9. The second term I have differentiated the fraction containing N's using the expression in the "relevant equations" section . This has resulted in the following expression

$$ \frac{d^2 \theta}{dt^2} = -E_c E_J \left(1-N^2 \right)^{\frac{1}{2}} sin(\theta) - \frac{E_{J}^2 sin(\theta) cos(\theta)}{1-N^2} $$
Can someone help me figure out why I am missing a factor of $1/(1-N^2)$ in the first term and a factor of $1+N^2$ in the second term? My equation looks similar but I appear to have missed a step

 

[DrClaude]

It appears you forgot the derivative of the $\cos \theta$ term when you took the derivative of eq. (10).

 

[CMJ96]

I see, would I have to apply the product rule to the second term in order to differentiate the $cos \theta$ properly?

$$\frac{d^2 \theta}{dt^2}= -E_c E_J (1-N^2)^{\frac{1}{2} } sin \theta - \frac{E_J ^2 sin \theta cos \theta
}{1-N^2} + \frac{E_J N sin \theta } {(1-N^2)^{\frac{1}{2}} }$$

 

[DrClaude]

I see, would I have to apply the product rule to the second term in order to differentiate the $cos \theta$ properly?

Yes, properly, meaning
$$
\frac{d}{dt} \cos \theta = -\sin \theta \frac{d \theta}{dt}
$$

 

[CMJ96]

Yes, properly, meaning
$$
\frac{d}{dt} \cos \theta = -\sin \theta \frac{d \theta}{dt}
$$

Ahhhh yes, silly mistake... now I'm getting the correct second term (I just have to use the trig identity to tidy it up) , the only issue is that the first term is slightly wrong, I can't see how I can get the $(1-N^2)^{\frac{1}{2}}$ as a denominator

$$\frac{d^2 \theta}{dt^2} = -E_c E_J (1-N^2)^{\frac{1}{2}} sin \theta - \frac{E_J E_c N^2 sin \theta}{(1-N^2)^{1/2}} - \frac{-E_J ^2 (1+N^2) sin \theta cos \theta}{1-N^2} $$

 

[DrClaude]

Ahhhh yes, silly mistake... now I'm getting the correct second term (I just have to use the trig identity to tidy it up) , the only issue is that the first term is slightly wrong, I can't see how I can get the $(1-N^2)^{\frac{1}{2}}$ as a denominator

Then you seem to have still forgotten a term. When calculating
$$
\frac{d}{dt} \cos \theta = -\sin \theta \frac{d \theta}{dt}
$$
the $\frac{d \theta}{dt}$ should bring a term in $E_C$, which eventually will modify the term in $E_C E_J$ in your equation for $\frac{d^2 \theta}{dt^2}$.

 

[CMJ96]

Then you seem to have still forgotten a term. When calculating
$$
\frac{d}{dt} \cos \theta = -\sin \theta \frac{d \theta}{dt}
$$
the $\frac{d \theta}{dt}$ should bring a term in $E_C$, which eventually will modify the term in $E_C E_J$ in your equation for $\frac{d^2 \theta}{dt^2}$.

Ah yes, sorry, I forgot to include the 4th term in my reply, I have added it now, the denominator should be $(1-N^2)^{\frac{1}{2}}$ but for some reason latex won't display it

 

[DrClaude]

Ah yes, sorry, I forgot to include the 4th term in my reply, I have added it now, the denominator should be $(1-N^2)^{\frac{1}{2}}$ but for some reason latex won't display it

I fixed it for you.

You now simply have to combine the two $E_C E_J$ terms into one and you'll be done.

 

[CMJ96]

I fixed it for you.

You now simply have to combine the two $E_C E_J$ terms into one and you'll be done.

I've got it now, thank you very much for the assistance!