I How Do I Find the Slope and Intercept of this Crystal Formation Experiment?

[WWGD]

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.

 

[tnich]

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?

 

[WWGD]

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

Sure, sorry

 

[tnich]

Sure, sorry

This looks exactly the same as the first image.

 

[WWGD]

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 $\frac {B (\gamma)^3}{T^3}$

 

[tnich]

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]

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]

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.

 

[Mark44]

Thread moved to General Math section, as what we're talking about here is less about physics than it is about mathematics.

 

[WWGD]

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.

 

[WWGD]

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.

 

[Mark44]

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.

 

[WWGD]

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]

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.

 

[WWGD]

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 $x_i :=ln(t_{ind}); y_i= =ln(\Gamma)^{-2}$

 

[tnich]

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]

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.