Physics, asked by randycunningham3315, 11 months ago

Entropy change independent of diffusion coefficient?

Answers

Answered by aman3495
0
hey your ans is
Assume we have a system with particle density nn, such that nn fulfils the diffusion equation with DD as the diffusion coefficient. Initially n(r⃗ ,t=0)=n0δ3(r⃗ )n(r→,t=0)=n0δ3(r→), which implies
n(r⃗ ,t)=n0(4πDt)3/2exp(−r24Dt).
n(r→,t)=n0(4πDt)3/2exp⁡(−r24Dt).
If we define the entropy as
S=−∫d3rn(r⃗ ,t)ln(n(r⃗ ,t)Λ3),
S=−∫d3rn(r→,t)ln⁡(n(r→,t)Λ3),
we get that the time derivative of the entropy is given by
dSdt=3n02t.
dSdt=3n02t.
However, this expression does not depend on DD which seems strange. If we would rescale D→D′=2DD→D′=2D and also rescale t→t′=t/2t→t′=t/2 we get the same equation for n(r⃗ ,t)n(r→,t). Hence we expect the rescaled system to behave exactly as the system before rescaling, but the time evolution is twice as fast. At a specific time t0t0 we expect the two systems with DD and D′D′ to have evolved differently much. Does this imply that the entropy time derivatives for the different systems are different.

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Answered by Anonymous
1

Wave-function is a complex number because of two properties it should meet. On the one hand it's modulus square is observable and thus should be real (it gives probability density). ... This is possible only in case such a boost is simply a phase of the complex number, which does not affect its modulus.

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