# Find Slope and Intercept

• I
Gold Member
Hi all, my friend asked me how to find the slope and intercept of ##ln_(tnd)## vs ##ln(\Gamma)^{-2} ## in equation 7 of p 1031 in the file attached. I believe the slope is just ##\frac {B \gamma}{T} ## , the coefficient of ##ln(\Gamma)^{-2} ## and, since the relation goes through the origin, the intercept is ##(0,0)##. This seems sraightforward Mathematics, but since I don't know any Physics beyond a basic level, I may be missing something. Could someone verify/correct? Thanks.

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tnich
Homework Helper
Hi all, my friend asked me how to find the slope and intercept of ##ln_(tnd)## vs ##ln(\Gamma)^{-2} ## in equation 7 of p 1031 in the file attached. I believe the slope is just ##\frac {B \gamma}{T} ## , the coefficient of ##ln(\Gamma)^{-2} ## and, since the relation goes through the origin, the intercept is ##(0,0)##. This seems sraightforward Mathematics, but since I don't know any Physics beyond a basic level, I may be missing something. Could someone verify/correct? Thanks.
I can't read the text in the image. Could you please blow it up a bit?

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I can't read the text in the image. Could you please blow it up a bit?
Sure, sorry

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tnich
Homework Helper
Sure, sorry
This looks exactly the same as the first image.

Gold Member
Ok, please see (7) in page 1031: to find slope of ##ln_(t_{ind})## vs ##(ln(\Gamma))^{-2} ##. I think it is just the coefficient ## \frac {B (\gamma)^3}{T^3} ##

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tnich
Homework Helper
I can't read the text in the image. Could you please blow it up a bit?
What you need to do is put this in the form ##y = mx + b##, where y is your independent variable ##ln_(t_{ind})##, x is your independent variable ##(lnΩ)^{-2}##, m is the slope and b is the intercept. You have almost done that, but you have left out the power of 3 in the slope and it is not clear to me why you assume that ##ln_(t_{ind})=0## when ##(lnΩ)^{-2}= 0##.

WWGD
Gold Member
What you need to do is put this in the form ##y = mx + b##, where y is your independent variable ##ln_(t_{ind})##, x is your independent variable ##(lnΩ)^{-2}##, m is the slope and b is the intercept. You have almost done that, but you have left out the power of 3 in the slope and it is not clear to me why you assume that ##ln_(t_{ind})=0## when ##(lnΩ)^{-2}= 0##.
Thanks. Isn't the coefficient m ##\frac {B (\gamma)^3}{T^3} ##? Also, I guess here is where I need some knowledge of physics to find out intercepts. I have no idea how to solve for the RH side, setting it to 0. Nor for the RH side, while setting ##(lnΩ)^{-2}= 0##

Mark44
Mentor
Hi all, my friend asked me how to find the slope and intercept of ##ln_(tnd)## vs ##ln(\Gamma)^{-2} ## in equation 7 of p 1031 in the file attached. I believe the slope is just ##\frac {B \gamma}{T} ## , the coefficient of ##ln(\Gamma)^{-2} ## and, since the relation goes through the origin, the intercept is ##(0,0)##. This seems sraightforward Mathematics, but since I don't know any Physics beyond a basic level, I may be missing something. Could someone verify/correct? Thanks.
The two variables are ##\ln(t_{ind})## and ##\left(\ln(\Omega)\right)^{-2}##. The coefficient of the latter expression is ##\frac{B\gamma^3}{T^3}##, not ##\frac {B \gamma}{T} ## as you wrote.

You could write the equation in the form of y = mx + b, with y being ##\ln(t_{ind})## and x being ##\ln(\Gamma)^{-2} ##. The slope m would be ##\frac{B\gamma^3}{T^3}##, and presumably the intercept would be the 2nd and 3rd terms on the right side of eqn. 7 of the PDF.

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Mark44
Mentor

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The two variables are ##\ln(t_{ind})## and ##\left(\ln(\Omega)\right)^{-2}##. The coefficient of the latter expression is ##\frac{B\gamma^3}{T^3}##, not ##\frac {B \gamma}{T} ## as you wrote.

You could write the equation in the form of y = mx + b, with y being ##ln_(tnd)## and x being ##ln(\Gamma)^{-2} ##. The slope m would be ##\frac{B\gamma^3}{T^3}##, and presumably the intercept would be the 2nd and 3rd terms on the right side of eqn. 7 of the PDF.
Ah, yes, I wrote it the slope without the cubes in one and with the cube in the other. Yes, I thought of it as a straightforward math pre-calc problem but I was not sure if there would be some Physics-related issues I was not aware of.

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Yes, but you see, there were some additional questions I wanted to ask that are outside of the scope of the Math alone. We take data for values of (x,y) , then we re-scale to (lnx, lny ) and we get a statistically-significant line. BUT there must be some Physics-specific tweaking to find the slope of the line to be the expression ## \frac {B \gamma^3}{T^3} ##. Here ##\gamma## is surface energy and B is some sort of Boltzmann constant/expression, T is temperature. These values do not, AFAIK, pop up from doing a regression; there must be some Physics tweaking, I believe, shaping the choice of these coefficients.

Mark44
Mentor
Yes, but you see, there were some additional questions I wanted to ask that are outside of the scope of the Math alone. We take data for values of (x,y) , then we re-scale to (lnx, lny ) and we get a statistically-significant line. BUT there must be some Physics-specific tweaking...
Seems to me that any "physics-specific tweaking" would have to be mathematically valid.

Gold Member
Seems to me that any "physics-specific tweaking" would have to be mathematically valid.
True, but it requires Physics-specific knowledge too. How do I know that, e.g., I should use the/(a?) Boltzmann constant in this case?

Mark44
Mentor
True, but it requires Physics-specific knowledge too. How do I know that, e.g., I should use the/(a?) Boltzmann constant in this case?
I don't see how a named constant will affect what you're doing to find the slope and intercept of a line.

Gold Member
I don't see how a named constant will affect what you're doing to find the slope and intercept of a line.
The thing is that ( as I understand it) , we are doing a least-squares regression on a collection of data points ##(x_i,y_i)##. This returns a numerical estimate for the slope together with a confidence interval for it. But somehow this numerical value is ultimately substituted by a formula ## \frac {B\gamma^3}{T^3} ## where T is the temperature and B, gamma are constants. These constants are not derived/derivable by least-squares alone. Some knowledge of Physics is necessary to use and adjust these constants. EDIT : With ## x_i :=ln(t_{ind}); y_i= =ln(\Gamma)^{-2} ##

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tnich
Homework Helper
The thing is that ( as I understand it) , we are doing a least-squares regression on a collection of data points ##(x_i,y_i)##. This returns a numerical estimate for the slope together with a confidence interval for it. But somehow this numerical value is ultimately substituted by a formula ## \frac {B\gamma^3}{T^3} ## where T is the temperature and B, gamma are constants. These constants are not derived/derivable by least-squares alone. Some knowledge of Physics is necessary to use and adjust these constants. EDIT : With ## x_i :=ln(t_{ind}); y_i= =ln(\Gamma)^{-2} ##
It looks to me like you have been given a lab exercise dealing with the formation of crystals in which you have varied saturation state ##Ω## and then measured the values of induction time ##t_{ind}##. The journal article you provided explains the physics and shows how to estimate surface energy ##γ## by fitting a line to suitably transformed data point pairs. The idea is not to estimate values of ##B## and ##T##. You need to control temperature ##T## in your experiment, and calculate ##B## from Boltman's constant and a nucleus shape factor. Then you can use that information and the slope of the line to calculate an estimate of ##γ##. So yes, you need to calculate a value of B using some knowledge of the crystal you are dealing with.

By the way, the independent variable in your equation for the line is ##ln^{-2}(Ω)##. That should be your value of x in the equation for the line ##y=mx+b##. You have x and y reversed in your recent edit.

WWGD