In volume expansion $\beta\approx3\alpha$, let's suppose we have a rectangular solid with height $h_o$, width $w_o$, and length $l_o$, then $V_o=h_ow_ol_o$

Now, after heating, each side increases by a factor of $\alpha\Delta T$,

$V=h_o(1+\alpha\Delta T)w_o(1+\alpha\Delta T)l_o(1+\alpha\Delta T)$

$\Delta V= V-V_o=h_ow_ol_o(1+\alpha\Delta T)^3-V_o$

We will substitute first equation, since $V_o=h_ow_ol_o$

$\Delta V=V_o(1+\alpha\Delta T)^3-V_o$

$\Delta V= V_o((1+\alpha\Delta T)^3-1)$

$\Delta V= V_o(1+\alpha\Delta T)(1+\alpha\Delta T)(1+\alpha\Delta T)-1)$

$\Delta V=V_o(3\alpha\Delta T+3(\alpha \Delta T)^2+(\alpha \Delta T)^3)$

Now, the last part is the part that I don't get, the last two parts cancel out because if $\alpha$ was small, then we will cancel $\alpha^2 $ because $\alpha^2 $ will get even smaller, and $\alpha^3 $ will be even much smaller, thus we get rid of them, and we are left with $\Delta V=V_o3\alpha \Delta T$.

I don't understand why are we making this assumption here? What if $\alpha$ was big, the last two terms won't really tend to zero... it just doesn't make sense to me in which to why we cancel them out.

Please excuse my ignorance, I would really appreciate it if someone would kindly explain it in a way that it would make sense to me.

  • $\begingroup$ It is an experimental fact that $\alpha \ll 1$. $\endgroup$ – user137289 Feb 12 '18 at 20:41
  • $\begingroup$ @Pieter Thank you, I don't know why I have completely forgotten about this fact, and I rushed into asking without paying attention to it. $\endgroup$ – Dewton Feb 12 '18 at 20:50
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    $\begingroup$ The definition of the volume coefficient of expansion is given here: en.wikipedia.org/wiki/Thermal_expansion . Note that it involves a derivative of the volume with respect to the temperature. In other words, it's considering the rate of change of volume to infinitesimal changes in temperature. So the higher order terms in $\Delta{T}$ drop out, and so only the term linear in $\Delta{T}$ remains and you're basically left with the last equation that you wrote down. $\endgroup$ – user93237 Feb 12 '18 at 21:03
  • $\begingroup$ @SamuelWeir, that's (also) an answer. :) $\endgroup$ – stafusa Feb 12 '18 at 21:21
  • $\begingroup$ @stafusa - OK, I've just been accustomed to writing short comments recently. I'll consider whether something is more appropriate as an answer in the future. $\endgroup$ – user93237 Feb 12 '18 at 21:24

The more advanced way (and more precise way) of treating this is to express the linear expansion as $$\frac{dL}{L}=\alpha dT$$ where dL and dT are differentials. Then by the product rule, $$dV=d(hwl)=wldh+hldw+hwdl=hwl\left(\frac{dh}{h}+\frac{dw}{w}+\frac{dl}{l}\right)=hwl(3\alpha dT)$$So,$$\frac{dV}{V}=3\alpha dT$$

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    $\begingroup$ Chester Miller, the algebra associated with this problem is quite insightful. Thanks for posting such a direct and elegant solution. $\endgroup$ – David White Feb 13 '18 at 1:16

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