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I am just an independent student and was learning thermal expansion with differential equations and i saw someone on the internet solving the differential equation for the law like below:

$$\frac{1}{L}\frac{dL}{dT}=\alpha$$ $$\int_{L_{0}}^{L}\frac{1}{L}dL=\int_{T_{0}}^{T}\alpha \,dT$$ $$[ln(L)]_{L_{0}}^{L}=[\alpha T]_{T_{0}}^{T}$$ $$ln(\frac{L}{L_{0}})=\alpha(T-T_{0})$$ Then, it continues. The part I dont understand is why on the second line you can integrate both sides but with different bounds on each side.

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  • $\begingroup$ That's just the standard method of solving a differential equation by separation of variables, written out explicitly. Here is another discussion. Also here, where one of the answers makes the different limits explicit. $\endgroup$
    – march
    Commented May 13 at 1:32
  • $\begingroup$ One way to think about it is that you have differentials of different variables that you are integrating, but only a single process. The bounds of the integral correspond to the specific process for each variable. $\endgroup$ Commented May 13 at 2:03
  • $\begingroup$ The bounds of the two variables occur at the same states. $\endgroup$ Commented May 13 at 10:50

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It is recommended to use clearer notation such as the one Riley Scott Jacob has introduced, where the limits of integration are $T_A$ and $T_B$, with $L_A=L(T_A)$ and $L_B = L(T_B)$. That way, we are avoiding the abuse of notation where $T$ is both a limit of integration and a dummy variable inside the integral.

Starting with your first equation, with each side viewed as a function of $T$, we can write $$\int_{T_A}^{T_B} \frac 1 {L} L'(T)~dT = \int_{T_A}^{T_B} \alpha~dT$$ where for clarity I've defined $L'(T)=\frac{d}{dT}L(T)$.

Now formally apply the change of variables $T\to L$ to the first integral. The lower and upper bounds become $L(T_A) = L_A$ and $L(T_B) = L_B$, respectively. We also have $dL = L'(T)~dT$, so we are left with the equation you are after: $$\int_{L_A}^{L_B} \frac 1 L ~dL = \int_{T_A}^{T_B} \alpha~dT$$

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  • $\begingroup$ This just "hides" the essence under minor additional manipulations. $\endgroup$ Commented May 13 at 14:12
  • $\begingroup$ @AgniusVasiliauskas The "minor manipulations" are how you formally arrive at the desired result. $\endgroup$
    – Puk
    Commented May 13 at 14:14
  • $\begingroup$ Somewhat agree, but I think change of variables is not always possible, only for trivial integrals. $\endgroup$ Commented May 13 at 14:21
  • $\begingroup$ @AgniusVasiliauskas please, reveal what you think is the hidden essence. What is missing in both this and the argument below is what happens when the function $L(T)$ is not invertible as the case for liquid water would be around $4C$. $\endgroup$
    – hyportnex
    Commented May 13 at 14:21
  • $\begingroup$ @hyportnex Yes, here I'm implicitly assuming $L(T)$ is invertible, as it often is (and seems to be in this case since $\alpha$ is assumed constant). If that is not the case, one would need to break the integral down to a sum of integrals on individual domains in each of which $L(T)$ is invertible. $\endgroup$
    – Puk
    Commented May 13 at 14:27
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Take a slightly relabeled version of your equation (2): $$\int_{L_A}^{L_B}\mathrm{d}L\,\frac{1}{L}=\int_{T_A}^{T_B}\mathrm{d}T\,\alpha$$ Consider that the thermal expansion process exists in some configuration space with parameters $L$ and $T$. The process is just a path between some starting point $A$ and ending point $B$ within this space.

Depiction of the thermal expansion process; own work

What we are really doing, then, is integrating from $A\to B$. It is simply that $T(A)=T(L_A)=T_A$, and $L(A)=L(T_A)=L_A$ (and likewise for $B$).

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  • $\begingroup$ In your case when temperature increases, rod length decreases, which is not the case in usual thermal expansion. You should flip curve vertically. Also in integrals, $T_A, L_A$ represents lower bounds of variables while in your chart, $T_A \gt T_B$ (variables swapped for temperature integral.) $\endgroup$ Commented May 13 at 14:06
  • $\begingroup$ $T_A$ and $L_A$ represent the initial configuration. It may be the case that $T_A$ < $T_B$, but there is certainly no reason it must be. $\endgroup$ Commented May 13 at 15:09
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Integration bounds are strictly mapped to the integration variable. In the left side we are integrating over elementary rod length $dL$, while in the right side we are integrating over elementary temperature $dT$. So bounds in the left side must have length dimensions and bounds in the right side - temperature dimensions.

Now how exact bound values are chosen depends on Physical process at hand, that is you must be aware about the process. In this case we are analyzing thermal expansion, which can be imagined as a process such as when rod temperature changes from $T_0 \to T$, then rod length accordingly protrudes from $L_0 \to L$. That's why such integral bounds.

In my opinion, it is such that good Physicists will tend to choose good integration bounds, and this comes with experience. I'm not totally against other more mathematically exposed explanations, but this time I think we should be more concentrated on Physics, not simple math variable mangling which in principle does not say anything new about the process at hand.

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  • $\begingroup$ this is a Kentucky Bourbon and this is an essence, I would always take the former over the latter. $\endgroup$
    – hyportnex
    Commented May 13 at 14:51
  • $\begingroup$ Well... if that's what physics means to you, I feel sorry for you. $\endgroup$ Commented May 13 at 20:52

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