Doubt in definition of linear thermal expansion It's stated in my textbook that upon heating a rod , the linear change in its length is directly proportional to its original length and the change in temperature , therefore the equations should look like these $$L_2-L_0\propto L_0  (T_2 -T_0)$$$$L_2-L_0=\alpha L_0  (T_2 -T_0)$$ , where $\alpha$ is the coefficient of thermal expansion , what I want to know is , how could this be true?When the rod is expanding then the "original length" is also varying instantaneously , so if we could start our observation at some temperature between $T_0$ and $T_2$ the original length is no longer  $L_0$.Moreover here is a peice of calculation I have done , Lets suppose an intermediate temperature $T_1$,Let us take the length at  that temperature to be $L_1$ $$L_1=L_0(1+\alpha(T_1-T_0))$$ Now if I start with this as my "original length" and calculate $L_2$ $$L_2=L_0(1+\alpha(T_2-T_0))+\alpha^2(T_1-T_0)(T_2-T_1)$$If I had calculated $L_2$ from $T_0$ and $T_2$ $$L_2=L_0(1+\alpha(T_2-T_0))$$The above two are not equal for sure if we accept that we are not considering any approximations, so have I kind of proved that if $\alpha$ exists then its value must change, otherwise the above two can never be equal.
 A: The linear expansion model is an approximation. The more "complete" model assumes a small change $\Delta L$ in the length due to a small change in temperature from $T$ to $T+\Delta T$ is proportional to the product of $\Delta T$ with the length at temperature $T$, $L(T)$. If you choose the proportionality factor to be $\alpha$ and assume it doesn't depend on the temperature, in the limit of vanishingly small changes it's possible to show the length will depend on the temperature with an exponential behavior:
$$
L(T) = L_0 e^{\alpha T},
$$
where $L_0$ is the length at $T=0$ on your temperature scale. If the argument in the exponential is small (and in nature we typically find values  such that $\alpha \ll 1$), we may expand the exponential in it's series expansion and keep only the lowest terms:
$$
L=L_0 e^{\alpha T} = L_0 \left ( 1 + \alpha T + \frac{(\alpha T)^2}{2} + ... \right ) \approx L_0(1+\alpha T),
$$
(if $\alpha \ll 1$, it's not hard to convince yourself that $\alpha^2 \ll \alpha$, and so forth for any higher power of $\alpha$, so it's legitimate to neglect those terms - provided $T$ is not too big). Since we have neglected all terms with power higher than one in $\alpha$, you should also neglect it on your calculations. So the two calculations are equal up to first order in $\alpha$, which is where the linear approximation works anyway.
A: If the relationship between length and temperature was linear then $\dfrac{\Delta L}{\Delta T}$ would be constant and you are looking at two values of $\alpha$ with two starting lengths $L_1$ and $L_2$ at two different temperature $T_1$ and $T_2$ respectively.  
$\alpha_1= \dfrac {1}{L_1}\dfrac{\Delta L}{\Delta T}$ and $\alpha_2= \dfrac {1}{L_2}\dfrac{\Delta L}{\Delta T} \Rightarrow \alpha_1-\alpha_2 =  \left(\dfrac {L_1-L_2}{L_1\,L_2}\right )\dfrac{\Delta L}{\Delta T}$ 
If the two temperature at which the length were measured were not too far apart then $L_1-L_2 \ll L_1L_2$ and the term in brackets is very small so $\alpha_1 \approx \alpha_2$.
If fact a graph of length against temperature is not a straight line but its gradient does not change very much over a small temperature range so that using $L\approx L_0(1+\alpha \,\Delta T)$ around temperature $T_0$ is a reasonably accurate approximation.  
Finding $\alpha$ at a particular temperature $T_0$ can be done by finding the gradient of the chord which connects two data points around that temperature with $\alpha_{\rm mean} = \dfrac {1}{L_1}\dfrac{L_1-L_2}{T_1-T_2}$ which produces a mean value of $\alpha$ over that temperature range, or by finding the gradient of the tangent to the graph of length against temperature at temperature $T_0$ resulting in $\alpha_0=\dfrac {1}{L_0}\left( \dfrac{dL}{dT}\right)_0$.  
The main problem is that the graph of length against temperature is not a straight line and so $\alpha$ is a function of temperature.  
Tables of how $\alpha$ varies with temperature are published eg in Kaye & Laby 
 
and to find $\alpha$ at other temperatures it is suggested that a polynomial fit of the form $\alpha(T) = a + b\,T + c \, T^2 + . . . $ is used.  
