Log expansion for infinite solenoid

[Shinobii]

Hello, I found an approximation for this log function:

[tex] log \Bigg(\frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \Bigg), [/tex]

where [itex] \Lambda \rightarrow \infty [/itex]. The above is approximated to the following,

[tex] -log \bigg(\frac{\rho}{\rho_o} \bigg) + log \bigg(\frac{2 \Lambda}{\rho_o} \bigg). [/tex]

How is this done? I tried expanding the [itex] \sqrt{1 + x^2} [/itex] term, but I still don't get how they arrive to the above approximation.

Any help would be greatly appreciated!

Cheers!

I have no idea why this was sent to linear algebra section . . . And I do not know how to move it to classical physics. . .

 

[jbunniii]

What is ##\rho_0##? It appears in the second expression but not the first.

 

[Shinobii]

I do not know what [itex] \rho_o [/itex] is, I assume some constant.

I found this approximation here:

https://www.physicsforums.com/showthread.php?t=119419

 

[Shinobii]

Wow, never mind. Clearly I am being silly here, for [itex] \Lambda \rightarrow \infty [/itex].

[tex] log\bigg( \frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \bigg) \rightarrow log \bigg( \frac{ 2 \Lambda}{\rho} \bigg) \rightarrow log(2 \Lambda) - log(\rho). [/tex]

As for the [itex] \rho_o [/itex] I have no idea why that enters the equation.

 

[blue_raver22]

Hello, thank you for sharing your findings. It seems that you have stumbled upon a simplified approximation for the logarithmic expansion of an infinite solenoid. This type of expansion is often used in physics to simplify complex equations and make calculations easier.

To understand how this approximation is derived, we need to look at the properties of logarithms and the behavior of the function in question. First, let's recall that the logarithm of a product is equal to the sum of the logarithms of its factors. This means that we can rewrite the original log function as:

log \Bigg(\frac{\Lambda}{\rho}\Bigg) + log \Bigg(\sqrt{1 + \frac{\Lambda^2}{\rho^2}} \Bigg)

Next, let's focus on the second term. We can use the binomial theorem to expand the square root:

\sqrt{1 + \frac{\Lambda^2}{\rho^2}} = 1 + \frac{\Lambda^2}{2\rho^2} + O\Bigg(\frac{1}{\rho^4}\Bigg)

Since we are interested in the case where \Lambda \rightarrow \infty, we can ignore the higher order terms and simplify the expression to:

\sqrt{1 + \frac{\Lambda^2}{\rho^2}} \approx 1 + \frac{\Lambda^2}{2\rho^2}

Substituting this back into our original equation, we get:

log \Bigg(\frac{\Lambda}{\rho}\Bigg) + log \Bigg(1 + \frac{\Lambda^2}{2\rho^2} \Bigg)

Now, we can use another property of logarithms, which states that the logarithm of a sum is equal to the logarithm of the individual terms multiplied together. This gives us:

log \Bigg(\frac{\Lambda}{\rho}\Bigg) + log \Bigg(1 + \frac{\Lambda^2}{2\rho^2} \Bigg) = log \Bigg(\frac{\Lambda}{\rho} \cdot \bigg(1 + \frac{\Lambda^2}{2\rho^2} \bigg) \Bigg)

Finally, we can simplify this expression by using the fact that \Lambda \rightarrow \infty. This means that \frac{\Lambda^2}{\rho^