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March 5th, 2019, 02:10 PM   #1
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How can I deduce this?

I have included an image from my textbook. I have tried a lot of different things, but keep getting stuck. I have no clue how a derivative suddenly pops up in here as well. Any expertise would be greatly appreciated since obviously I am missing something here.

Thank you!

Edit: Link to image here since the attached image is poor quality.

https://i.imgur.com/SFslOqx.png

Screen Shot 2019-03-05 at 5.09.23 PM.jpg

Last edited by skipjack; March 6th, 2019 at 02:42 AM.
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March 5th, 2019, 04:54 PM   #2
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Originally Posted by SenatorArmstrong View Post
I have included an image from my textbook. I have tried a lot of different things, but keep getting stuck. I have no clue how a derivative suddenly pops up in here as well. Any expertise would be greatly appreciated since obviously I am missing something here.

Thank you!

Edit: Link to image here since the attached image is poor quality.

https://i.imgur.com/SFslOqx.png
It's probably a typo. Where the d/d(phi) is there should be a (phi) and there should be a 1/J0e inside the logarithm.

-Dan
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Last edited by skipjack; March 6th, 2019 at 02:44 AM.
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March 6th, 2019, 01:12 AM   #3
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I don't think there are typos...
$\displaystyle J_p = J_{e0} \exp\left(\frac{e\phi}{kT_e}\right) - J_i$
$\displaystyle J_p + J_i = J_{e0} \exp\left(\frac{e\phi}{kT_e}\right)$
$\displaystyle \ln \left(J_p + J_i\right) = \ln \left[J_{e0} \exp\left(\frac{e\phi}{kT_e}\right)\right]$
$\displaystyle \ln \left(J_p + J_i\right) = \ln J_{e0} + \ln \left[\exp\left(\frac{e\phi}{kT_e}\right)\right]$
$\displaystyle \ln \left(J_p + J_i\right) = \ln J_{e0} + \frac{e\phi}{kT_e}$

Differentiate both sides with respect to $\displaystyle \phi$

$\displaystyle \frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right) = \frac{e}{kT_e}$
$\displaystyle T_e = \frac{e}{k}\left\{\frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right)\right\}^{-1}$
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Last edited by Benit13; March 6th, 2019 at 01:16 AM.
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March 6th, 2019, 07:30 AM   #4
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Originally Posted by Benit13 View Post
I don't think there are typos...
$\displaystyle J_p = J_{e0} \exp\left(\frac{e\phi}{kT_e}\right) - J_i$
$\displaystyle J_p + J_i = J_{e0} \exp\left(\frac{e\phi}{kT_e}\right)$
$\displaystyle \ln \left(J_p + J_i\right) = \ln \left[J_{e0} \exp\left(\frac{e\phi}{kT_e}\right)\right]$
$\displaystyle \ln \left(J_p + J_i\right) = \ln J_{e0} + \ln \left[\exp\left(\frac{e\phi}{kT_e}\right)\right]$
$\displaystyle \ln \left(J_p + J_i\right) = \ln J_{e0} + \frac{e\phi}{kT_e}$

Differentiate both sides with respect to $\displaystyle \phi$

$\displaystyle \frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right) = \frac{e}{kT_e}$
$\displaystyle T_e = \frac{e}{k}\left\{\frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right)\right\}^{-1}$
Huge thanks! This was the very last step in a derivation I spent a good bit of the day yesterday following so seeing the dots connect is a huge relief.

I am not sure why this result is useful as opposed to simply solving the current density formula for temperature. For one thing, I suppose you no longer have to worry about your initial current density, but at the cost of the derivative w.r.t potential showing up in the result. Perhaps this result is more useful from the experimental side of things. Any thoughts on this? I am going to bring it up during the lab meeting today.

Again, thanks a lot!
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March 6th, 2019, 07:36 AM   #5
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Quote:
Originally Posted by Benit13 View Post
I don't think there are typos...
$\displaystyle J_p = J_{e0} \exp\left(\frac{e\phi}{kT_e}\right) - J_i$
$\displaystyle J_p + J_i = J_{e0} \exp\left(\frac{e\phi}{kT_e}\right)$
$\displaystyle \ln \left(J_p + J_i\right) = \ln \left[J_{e0} \exp\left(\frac{e\phi}{kT_e}\right)\right]$
$\displaystyle \ln \left(J_p + J_i\right) = \ln J_{e0} + \ln \left[\exp\left(\frac{e\phi}{kT_e}\right)\right]$
$\displaystyle \ln \left(J_p + J_i\right) = \ln J_{e0} + \frac{e\phi}{kT_e}$

Differentiate both sides with respect to $\displaystyle \phi$

$\displaystyle \frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right) = \frac{e}{kT_e}$
$\displaystyle T_e = \frac{e}{k}\left\{\frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right)\right\}^{-1}$
A very interesting tactic. I'll have to remember this method.

Thanks!

-Dan
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March 6th, 2019, 08:30 AM   #6
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Originally Posted by SenatorArmstrong View Post
I am not sure why this result is useful as opposed to simply solving the current density formula for temperature. For one thing, I suppose you no longer have to worry about your initial current density, but at the cost of the derivative w.r.t potential showing up in the result. Perhaps this result is more useful from the experimental side of things. Any thoughts on this? I am going to bring it up during the lab meeting today.
I don't really know what's going on, but it seems like there is a plasma and some form of probe placed within it in which is measuring its properties by measuring the current induced in the probe.

Perhaps the idea here is that the temperature of the plasma is being related to measured current densities in regions with different $\displaystyle \phi$.

Then the plan could be to change the $\displaystyle \phi$ by placing the probe in different positions, measure $\displaystyle J_p$ and $\displaystyle J_i$ and the use the results to obtain a measured plasma temperature without having to know anything about the number density of the plasma, its constituents or its specific distribution (because it's exponential and that's all that matters).
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Last edited by Benit13; March 6th, 2019 at 08:33 AM.
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March 6th, 2019, 08:36 AM   #7
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It seems to be useful only when $\displaystyle \frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right)$ is a constant in the area of interest. The textbook implies that this should be noticeable in a plot of experimental data, but doesn't publish such a plot.
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