What is the Relationship Between Coefficients for Thermal Expansion?

[doublefelix92]

Hi Physicsforums,

I was trying to derive the formula that the coefficient for volume expansion, β, is 3 times the coefficient for length expansion, $\alpha$.

As a reminder, the formulas are:

$\Delta$L = $\alpha$L_o$\Delta$T

and $\Delta$V = $\beta$V_o$\Delta$T

where, supposedly, $\beta$ = 3$\alpha$.

My proof actually gave me a completely different result. I double-checked it a few times, and I can't find the problem. Does anyone know what's wrong?

here it is: http://imageshack.us/photo/my-images/267/img20111212204020.jpg/

thanks in advance,

doublefelix92

 

[Doc Al]

My proof actually gave me a completely different result. I double-checked it a few times, and I can't find the problem. Does anyone know what's wrong?

Get rid of all the terms with higher powers of α. α is usually quite small.

 

[doublefelix92]

Well, then how come in this proof it works out EXACTLY?

V = L³
dV/dL = 3L²

dL = $\alpha$L_odT
dL/dT = $\alpha$L_o

dV/dT = (dV/dL)(dL/dT) = 3L² * $\alpha$L_o = 3$\alpha$V_o. Now just turn dV/dT into $\Delta$V/$\Delta$T (since dV/dT is constant, you can do that), and you have EXACTLY the equation.

It should come out to the same result, regardless of which proof you use, if you do no approximations.

Edit: I'm guessing the error in this derivation is that L is a function of T, so (L^2)*Lo is not Vo. Checking now if that results in the same equation as the original.

 

[Acut]

@doublefelix92:

β≈3α. It's only an approximation. That's why your proof went wrong.
If you neglect any higher order terms of α, then you will find β=3α.

 

[doublefelix92]

Okay. I'm glad that I know it's an approximation. It bothers me, because my textbook (Y&F University Physics) didn't mention that detail WHATSOEVER, and in fact their proof is just plain incorrect; they made an illegal substitution that results in the coefficients being exact multiples.

Anyway - I still have a question. The two ways I'm deriving the equation don't agree. The first method is using V + $\Delta$V = (L_O+$\Delta$L)³, and plugging in. The second way is to derive the equation starting from V = L^3 and the length expansion equation. I have both of my proofs in the following image.

Why do they not agree?? I re-did them multiple times. What step was illegal?

http://imageshack.us/g/849/img20111213012858.jpg/

Also - ALL of these calculations were just for cubes. Nothing is a perfect cube in real life. How do I know that other shapes follow similar rules?

 

[Doc Al]

No reason they should agree. The first involves no approximations at all. In the second method, as soon as you introduce a derivative you've introduced an approximation.

dV = 3L²dL is only true for small changes, yet you treat it as equivalent to ΔV = 3L²ΔL. But that's not quite accurate, as the first method demonstrates. (ΔV really equals (L + ΔL)³ - L³)

As long as you get rid of all higher powers, the two derivations give the same approximate answer.

 

[doublefelix92]

Nice, that makes perfect sense. Thank you so much. Any idea on the other shapes question? Or is it just true that regardless of the shape, it's always approximately true that $\Delta$V = 3$\alpha$V_O$\Delta$T, as long as the equation for linear expansion holds equally in every dimension.

 

[Doc Al]

Or is it just true that regardless of the shape, it's always approximately true that $\Delta$V = 3$\alpha$V_O$\Delta$T, as long as the equation for linear expansion holds equally in every dimension.

Exactly. (You can think of other shapes as composed of small cubes, if you like.)