I was studying \(\text{Thermal Expansion}\) from H.C. Verma - Concepts of Physics 2 (if you have the book, page 6 - topic 23.10). If you don't have the book, this is the concerned matter.

Note: If you know this stuff, you can jump to the end.

Consider a rod at temperature \(T\) and suppose its length at this temperature is \(L\). As the temperature is changed to \(T+\Delta T\), the length is changed to \(L+\Delta L\). We define average coefficient of linear expansion in the temperature range \(\Delta T\) as \[\overline{\alpha}=\frac{1}{L}\frac{\Delta L}{\Delta T}\] The coefficient of linear expansion at temperature \(T\) is limit of average coefficient as \(\Delta T\rightarrow0\), i.e., \[\alpha=\displaystyle\lim_{\Delta T\rightarrow0}\frac{1}{L}\frac{\Delta L}{\Delta T}=\frac{1}{L}\frac{dL}{dT}\] Suppose the length of a rod is \(L_{0}\) at \(0^{\circ}C\) and \(L_{\theta}\) at temperature \(\theta\) measured in Celsius. If \(\alpha\) is small and constant over the given temperature interval, \[\alpha=\frac{L_{\theta}-L_{0}}{L_{0}\theta}\] or \[L_{\theta}=L_{0}(1+\alpha\theta)\] The coefficient of volume expansion \(\gamma\) is defined in a similar way. If \(V\) is the volume of a body at temperature \(T\), the coefficient of volume expansion at temperature \(T\) is \[\gamma=\frac{1}{V}\frac{dV}{dT}\] It is also known as coefficient of cubical expansion.

If \(V_{0}\) and \(V_{\theta}\) denote the volumes at \(0^{\circ}C\) and \(\theta\) (measured in Celsius) respectively and \(\gamma\) is small and constant over the given temperature range, we have \[V_{\theta}=V_{0}(1+\gamma\theta)\]

It is easy to show that \(\gamma=3\alpha\).

My question is: How do you prove that \(\gamma=3\alpha\)?

Note for those who jumped: \(\gamma\) is the coefficient of cubical expansion, and \(\alpha\) is the coefficient of linear expansion.

## Comments

Sort by:

TopNewestWe consider a cube of metal with each side equal to L.

If we heat it up, the length of each side will change from L to (L+ΔL), where ΔL = αLΔT.

In other words, the new length is:

L′ = L + αLΔT = L(1 + αΔT)

So the new volume is:

Vo′ = (L′)^3 = [L(1 + αΔT)]^3 = (L^3)(1 + 3αΔT + 3(αΔT)² + (αΔT)^3) = Vo(1 + 3αΔT + 3(αΔT)² + (αΔT)^3)

Now we make an approximation. Since αΔT is quite small compared to 1, we can safely say that the "3(αΔT)²" term and the "(αΔT)^3" term are negligible compared to the "3αΔT" term. Therefore, to a good approximation,

Vo′ = Vo(1 + 3αΔT)

But we already know, from the definition of γ, that:

Vo′ = Vo(1 + γΔT)

So this means that γ = 3α.

I could have written the solution in Latex but I am short on time. (NTSE upcoming) – Ayush Gupta · 2 years, 3 months ago

Log in to reply

– Omkar Kulkarni · 2 years, 3 months ago

Wait did you mean \(\overline{\alpha}\) instead of \(\alpha\)?Log in to reply

– Omkar Kulkarni · 2 years, 3 months ago

Ohh okay! Makes sense now :) Thanks! All the best for NTSE.Log in to reply

Y=3a – Onyx Georgia · 2 years, 1 month ago

Log in to reply

– Omkar Kulkarni · 2 years, 1 month ago

What?Log in to reply