How Does Temperature Change When Two Halves of a Heated Cylinder Are Rejoined?

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SUMMARY

The discussion focuses on the thermal dynamics of a homogeneous cylinder that has been cut in half, with one half heated and the other cooled. Upon rejoining the cylinder, the temperature distribution is analyzed using the equation for temperature as a function of radius, angle, and time, specifically $$T(r,\varphi ,t)=\sum _{m,n}J_{m,n}(\xi _{m,n}\frac r R)[B_m cos(m\varphi )+C_msin(m\varphi )]e^{-i\omega t}$$. The key point raised is the determination of coefficients B_m and C_m, particularly why B_m are zero, which relates to the boundary conditions and symmetry of the system. The discussion emphasizes the importance of understanding boundary conditions in thermal analysis.

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Homework Statement


A very long homogeneous cylinder is cut in half along its axis. One half s than equally heated, while the other half is equally cooled. How does the temperature change when the two parts are joined back together, if the cylinder is well isolated?

Homework Equations

The Attempt at a Solution



Hmmm,
I know I can work with $$T(r,\varphi ,t)=\sum _{m,n}J_{m,n}(\xi _{m,n}\frac r R)[B_m cos(m\varphi )+C_msin(m\varphi )]e^{-i\omega t}$$where ##\xi _{m,n}## is m-th zero of n-th Bessel function, but what I do not understand at this point is why all ##B_m## are zero?

Because if I am not mistaken, one boundary condition is $$j=-\lambda \frac{\partial }{\partial r}T(r=R,\varphi t)=0$$ and the other should be something at ##\varphi =0## or ##\varphi =\pi ##. I guess? I assume this second bondary condition will also tell me why the ##cos## is gone in the first equation.
 
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After you rejoin the cylinder, there really is no boundary in the ##\phi## coordinate. If the ##B_m## or ##C_m## are zero depends on how you put your coordinate system. The type of argument to be used to see this is based on symmetry (or actually performing the integrals to solve for the coefficients).
 

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