My Math Forum How can I deduce this?

 Physics Physics Forum

 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 2019-03-05 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: 2,305
Thanks: 962

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by SenatorArmstrong 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

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: 2,164 Thanks: 736 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 $\displaystyle \phi$ $\displaystyle \frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right) = \frac{e}{kT_e}$ $\displaystyle T_e = \frac{e}{k}\left\{\frac{d}{d\phi}\left(\ln \left(J_p + J_i\right)\right)\right\}^{-1}$ Thanks from topsquark and SenatorArmstrong 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:
 Originally Posted by Benit13 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!

March 6th, 2019, 07:36 AM   #5
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,305
Thanks: 962

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by Benit13 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

March 6th, 2019, 08:30 AM   #6
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,164
Thanks: 736

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by SenatorArmstrong 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).

Last edited by Benit13; March 6th, 2019 at 08:33 AM.

 March 6th, 2019, 08:36 AM #7 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 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. Thanks from topsquark and SenatorArmstrong

 Tags deduce

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post JOHNNZ Algebra 2 November 10th, 2016 06:37 AM jiasyuen Trigonometry 1 January 30th, 2015 02:10 PM precalciskillingme Algebra 2 December 12th, 2012 06:09 PM modulus1 Algebra 8 November 13th, 2011 06:51 PM rollerJ Number Theory 6 May 4th, 2011 10:02 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top