# Thermal Expansion Question

#### 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 = $\alpha$Lo$\Delta$T

and $\Delta$V = $\beta$Vo$\Delta$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/

doublefelix92

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#### Doc Al

Mentor
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 = $\alpha$LodT
dL/dT = $\alpha$Lo

dV/dT = (dV/dL)(dL/dT) = 3L2 * $\alpha$Lo = 3$\alpha$Vo. 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.

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#### 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 = (LO+$\Delta$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?

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#### Doc Al

Mentor
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$\alpha$VO$\Delta$T, as long as the equation for linear expansion holds equally in every dimension.

#### Doc Al

Mentor
Or is it just true that regardless of the shape, it's always approximately true that $\Delta$V = 3$\alpha$VO$\Delta$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.)