Linear coefficient of thermal expansion

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  • #1
Nathanael
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Main Question or Discussion Point

The "coefficient of linear expansion" ([itex]≡k[/itex]) was defined in my book by the following relationship:
##\Delta L=Lk\Delta T##
Where L is length and T is temperature

I'm wondering, is this just an approximation? Because, if you were to increase the temperature by [itex]\Delta T[/itex] and then calculate the new length, and then decrease the temperature by the same [itex]\Delta T[/itex] and calculate the new length again, you would not get back to your original length.

Wouldn't the "symmetrical" definition of [itex]k[/itex] be
##L_{f}=L_{0}e^{k\Delta T}##

This leads me to the question:
Is the reason they don't define it like this because the idea of 'linear thermal expansion' is not true to that degree of accuracy?
(In other words, finding the new length (to that accuracy) is not as simple as using a single constant?)
 

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  • #2
Bystander
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not as simple as using a single constant?
It's temperature dependent, and generally so small that the dependence isn't measureable before things melt/decompose.
 
  • #3
Stephen Tashi
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The "coefficient of linear expansion" ([itex]≡k[/itex]) was defined in my book by the following relationship:
##\Delta L=Lk\Delta T##
That's only an approximation if we treat the [itex] \Delta[/itex]'s as finite, which is the traditional interpretation of [itex] \Delta[/itex]'s.

I notice the Wikipedia article defines the coefficient of linear expansion using the notation for derivatives. http://en.wikipedia.org/wiki/Thermal_expansion
 
  • #4
Nathanael
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That's only an approximation if we treat the [itex] \Delta[/itex]'s as finite, which is the traditional interpretation of [itex] \Delta[/itex]'s.

I notice the Wikipedia article defines the coefficient of linear expansion using the notation for derivatives. http://en.wikipedia.org/wiki/Thermal_expansion
Thanks, I didn't see that. Everywhere I looked kept using [itex] \Delta L[/itex] and [itex] \Delta T[/itex]. Next time I'll be sure to check Wikipedia.
 
  • #5
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I think it was very perceptive of you to intuitively realize that it should, more precisely, be expressed differentially.

Chet
 

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