Originally Posted by Physicsguru
The condition for an object to have a temperature which is independent of time (i.e. constant temperature), is for its temperature to be absolute zero degrees kelvin. This is impossible, therefore temperature must be a function of time.
ZapperZ said:
This is bogus. Again, you have made a statement without justification. Show me exactly where it has been shown that what you said is true.
Well suppose that (as Crosson said) the definition of temperature for the case of constant mass M (therefore dM/dt=0), and constant volume V (therefore dV/dt=0) is given by:
\frac{1}{T} = \frac{dS}{dU}
With density \rho defined as
\rho = \frac{M}{V}
Using symbols consistent with the paper cited by Healy.
Now focus on the formula 1/T=dS/dU
Clearly the LHS is undefined for the case of T=0. Most clearly stated, division by zero error must lead to a contradiction, so that the definition above, on its own forbids a temperature of absolute zero from being realized. Not even the vacuum has a temperature of absolute zero, certainly material objects immersed in it won't have a temperature of absolute zero either, in fact they will be hotter than the vacuum. And so, a material object in the vacuum will radiate energy, so that its temperature will drop, and the vacuum local to it will gain energy, assuming the basic laws of thermodynamics are correct. This having been said, consider the RHS, which is dS/dU... the derivative of entropy with respect to energy.
Suppose that U and S, can be related through the variable t, therefore we have two unknown functions U(t), and S(t).
We can take the derivative of S with respect to t, to obtain dS/dt, and we can take the derivative of U with respect to t, to obtain dU/dt. We can then solve for temperature as follows:
\frac{1}{T} = \frac{dS}{dU} = \frac{dS}{dt} \frac{dt}{dU}
Suppose that the temperature T can be constant. In this case, the derivative with respect to time of the LHS above, will necessarily be zero.
The derivative with respect to time of the RHS is given by:
\frac{d}{dt} [\frac{dS}{dt} \frac{dt}{dU} ]
Now use the product rule of the differential calculus to obtain:
\frac{dt}{dU} \frac{d^2S}{dt^2} + \frac{dS}{dt} \frac{d}{dt} \frac{dt}{dU}
So if the temperature can be constant in time, then the formula above must be equal to zero that is:
\frac{dt}{dU} \frac{d^2S}{dt^2} + \frac{dS}{dt} \frac{d}{dt} \frac{dt}{dU} = 0
From which it follows that
\frac{dt}{dU} \frac{d^2S}{dt^2} = - \frac{dS}{dt} \frac{d}{dt} (\frac{dt}{dU})
Now, suppose instead that we already do have a formula for entropy in terms of energy U, and that we can just differentiate S with respect to U. In this case, we have again:
1/T = dS/dU
So that if temperature is to be constant then we must have:
0 = \frac{d}{dt} \frac{dS}{dU}
Realizing that the dt is superfluous, we really just need to look at the differential of the formula for dS/dU, that is the condition for constant temperature is also given by:
0 = d ( \frac{dS}{dU})
One way for the statement above to be true, is for a system to have a constant entropy. In that case, dS=0, independently of U. Suppose that the entropy of an object must necessarily increase over any two consecutive moments in time. Thus, even though dS is necessarily nonzero, the formula above could still be true if dS must be constant.
PHYSICS,please...
