What Is the Physical Meaning of Helmholtz Free Energy?

asdf1
Messages
734
Reaction score
0
what's the physical meaning of that equation? all i see is a jumble of numbers and equations...
 
Physics news on Phys.org
Are you talking about the equation for a specific system, because if so, you'll need to tell us what it is if you want help. In general, the helmholtz free energy is A=U-TS, where U is the internal energy, T is the temperature, and S is the entropy. This is hardly a jumble of numbers, hence my first comment. It is the function that is minimized at equilibrium for a system held at constant volume and temperature. It is also easily calculated from statistical mechanics by A=-kT ln(Z), where Z is the partition function.
 
what do you usually use that equation for?
 
Like I said, it is the thermodynamic potential (ie, the function that is minimized at equilibrium) for any system held at constant temperature and volume. So you would calculate the free energy and then minimize it with respect to each free variable to find the equilibrium position of the constrained system. It is also used to calculate various other thermodynamic quantities starting from a statistical model of the system, since it is the thermodynamic quantity most easily obtained from statistical methods (more specifically, the canonical ensemble).
 
thank you very much! :)
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top