March 5th, 2019, 02:10 PM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  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 20190305 at 5.09.23 PM.jpg Last edited by skipjack; March 6th, 2019 at 02:42 AM. 
March 5th, 2019, 04:54 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; March 6th, 2019 at 02:44 AM.  
March 6th, 2019, 01:12 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
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 $\displaystyle \phi$ $\displaystyle \frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right) = \frac{e}{kT_e}$ $\displaystyle T_e = \frac{e}{k}\left\{\frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right)\right\}^{1}$ Last edited by Benit13; March 6th, 2019 at 01:16 AM. 
March 6th, 2019, 07:30 AM  #4  
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  Quote:
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!  
March 6th, 2019, 07:36 AM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Thanks! Dan  
March 6th, 2019, 08:30 AM  #6  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
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). Last edited by Benit13; March 6th, 2019 at 08:33 AM.  
March 6th, 2019, 08:36 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
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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