Science wiz! Please help

I was studying Thermal Expansion\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 TT and suppose its length at this temperature is LL. As the temperature is changed to T+ΔTT+\Delta T, the length is changed to L+ΔLL+\Delta L. We define average coefficient of linear expansion in the temperature range ΔT\Delta T as α=1LΔLΔT\overline{\alpha}=\frac{1}{L}\frac{\Delta L}{\Delta T} The coefficient of linear expansion at temperature TT is limit of average coefficient as ΔT0\Delta T\rightarrow0, i.e., α=limΔT01LΔLΔT=1LdLdT\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 L0L_{0} at 0C0^{\circ}C and LθL_{\theta} at temperature θ\theta measured in Celsius. If α\alpha is small and constant over the given temperature interval, α=LθL0L0θ\alpha=\frac{L_{\theta}-L_{0}}{L_{0}\theta} or Lθ=L0(1+αθ)L_{\theta}=L_{0}(1+\alpha\theta) The coefficient of volume expansion γ\gamma is defined in a similar way. If VV is the volume of a body at temperature TT, the coefficient of volume expansion at temperature TT is γ=1VdVdT\gamma=\frac{1}{V}\frac{dV}{dT} It is also known as coefficient of cubical expansion.

If V0V_{0} and VθV_{\theta} denote the volumes at 0C0^{\circ}C and θ\theta (measured in Celsius) respectively and γ\gamma is small and constant over the given temperature range, we have Vθ=V0(1+γθ)V_{\theta}=V_{0}(1+\gamma\theta)

It is easy to show that γ=3α\gamma=3\alpha.


My question is: How do you prove that γ=3α\gamma=3\alpha?

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

Note by Omkar Kulkarni
4 years, 5 months ago

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We 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 - 4 years, 5 months ago

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Ohh okay! Makes sense now :) Thanks! All the best for NTSE.

Omkar Kulkarni - 4 years, 5 months ago

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Wait did you mean α\overline{\alpha} instead of α\alpha?

Omkar Kulkarni - 4 years, 5 months ago

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Y=3a

onyx georgia - 4 years, 4 months ago

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What?

Omkar Kulkarni - 4 years, 4 months ago

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