Prove that Planck function increases with temperature

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SUMMARY

The discussion focuses on proving that the Planck function, defined as Bv(T) = 2hv³c⁻²(e^(hv/kT) - 1)⁻¹, increases monotonically with temperature (T). Participants analyzed the behavior of the function by examining the term e^(hv/kT), which decreases as T increases, leading to the conclusion that the overall Planck function increases with T. A derivative approach was also discussed, with the correct derivative being dBv(T)/dT = 2h²v⁴c⁻²k⁻¹(e^(-hv/kT) + 1)⁻¹, confirming that the gradient remains positive as T increases, thereby solidifying the proof.

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  • Understanding of the Planck function and its formula
  • Familiarity with concepts of thermodynamics and blackbody radiation
  • Basic calculus, including differentiation and the chain rule
  • Knowledge of constants such as Planck's constant (h), Boltzmann's constant (k), and the speed of light (c)
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  • Learn about the implications of blackbody radiation in thermodynamics
  • Explore the application of the chain rule in calculus
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ck99
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Homework Statement



Prove that the Planck function increases monotonically with temperature.

Homework Equations



Bv(T) = 2hv3c-2(ehv/kT - 1)-1

The Attempt at a Solution



I first went through this piece-by-piece, but I am not a mathematician so I don't know if this constitutes "proof"!

1) First consider ehv/kT and note that this function will decrease as T increases.

2) This means (ehv/kT - 1) will also decrease.

3) Therefore (ehv/kT - 1)-1 will increase with T.

4) The other elements are independent of T, so the Planck function will increase with T.

I also thought of taking the derivative with respect to T of the Planck function, to see if the gradient ever reached zero to indicate a stationary point. After doing a couple of substitutions, I got

dBv(T)/dT = 2h2v4c-2k-1(e-hv/kT + 1)-1

I'm not sure that is correct, and I think it tells me that gradient gets smaller as T increases. This doesn't help my argument much!
 
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ck99 said:

Homework Statement



Prove that the Planck function increases monotonically with temperature.

Homework Equations



Bv(T) = 2hv3c-2(ehv/kT - 1)-1

The Attempt at a Solution



I first went through this piece-by-piece, but I am not a mathematician so I don't know if this constitutes "proof"!

1) First consider ehv/kT and note that this function will decrease as T increases.

2) This means (ehv/kT - 1) will also decrease.

3) Therefore (ehv/kT - 1)-1 will increase with T.

4) The other elements are independent of T, so the Planck function will increase with T.

The above argument sounds good to me.
I also thought of taking the derivative with respect to T of the Planck function, to see if the gradient ever reached zero to indicate a stationary point. After doing a couple of substitutions, I got

dBv(T)/dT = 2h2v4c-2k-1(e-hv/kT + 1)-1

That's not the correct result for the derivative. Ignoring the constants out front, you have

B = (x-1)-1 where x = ehv/kT.

Try using the chain rule: dB/dT = (dB/dx)##\cdot##(dx/dT)
 
Thanks TSny, I managed to compute the derivative correctly and that can also be put in terms of positive, increasing functions of T so I am sure that is adequate proof.
 

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