Find Slope and Intercept

  • I
  • Thread starter WWGD
  • Start date
  • #1
WWGD
Science Advisor
Gold Member
5,419
3,680
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.
 

Attachments

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336
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?
 
  • #3
WWGD
Science Advisor
Gold Member
5,419
3,680
I can't read the text in the image. Could you please blow it up a bit?
Sure, sorry
 

Attachments

  • #4
tnich
Homework Helper
1,048
336
Sure, sorry
This looks exactly the same as the first image.
 
  • #5
WWGD
Science Advisor
Gold Member
5,419
3,680
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} ##
 

Attachments

  • #6
tnich
Homework Helper
1,048
336
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##.
 
  • Like
Likes WWGD
  • #7
WWGD
Science Advisor
Gold Member
5,419
3,680
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##
 
  • #8
34,545
6,248
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.
 
Last edited:
  • #9
34,545
6,248
Thread moved to General Math section, as what we're talking about here is less about physics than it is about mathematics.
 
  • #10
WWGD
Science Advisor
Gold Member
5,419
3,680
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.
 
  • #11
WWGD
Science Advisor
Gold Member
5,419
3,680
Thread moved to General Math section, as what we're talking about here is less about physics than it is about mathematics.
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.
 
  • #12
34,545
6,248
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.
 
  • #13
WWGD
Science Advisor
Gold Member
5,419
3,680
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?
 
  • #14
34,545
6,248
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.
 
  • #15
WWGD
Science Advisor
Gold Member
5,419
3,680
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} ##
 
Last edited:
  • #16
tnich
Homework Helper
1,048
336
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.
 
  • Like
Likes WWGD
  • #17
WWGD
Science Advisor
Gold Member
5,419
3,680
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.
Thanks, tnich, this is from a friend who asked me for help. I can help her with Math aspects, but not with the Physics part.. And, yes, I have mixed the two too many times; I need to pay attention more carefully.
 

Related Threads on Find Slope and Intercept

Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
15
Views
11K
Replies
35
Views
6K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
4
Views
4K
Top