7 votes
Accepted

How can I resolve this apparent paradox about the average age and the average lifetime?

The average lifetime of a particle "at birth" is $1/\lambda$. The average lifetime of the collection of particles that are alive at a given time is longer, because that is a biased sample. ...
  • 1,317
6 votes

Is the concept of entropy a result of limited technology?

I know that entropy is the energy within a system that is unable to do useful work. This doesn’t really bear scrutiny. Entropy and energy have different units, for one. But even an approximate ...
3 votes

Is the concept of entropy a result of limited technology?

Totally wrong. Entropy is NOT "the energy within a system that is unable to do useful work." Instead entropy is the thermal charge that does work as it moves from a higher temperature to a ...
  • 12.2k
2 votes

The average of a continuous value: $\overline{O} = \int O(x) \rho(x) dx$, but coordinate invariant

lpz is correct that you are including an erroneous factor of $|dy/dx|$ in your transformation of $O(x)$. You may be getting misled by the transformation of the density, which uses the same symbol for ...
  • 546
2 votes

Doubt in expansion of integral

I believe they are assuming that $H(\epsilon)$ becomes a constant above $\epsilon > \epsilon_{f}$, so you can split the integral and the integral above $\epsilon_{f}$ when integrating you just get $...
  • 2,006
2 votes

Is the concept of entropy a result of limited technology?

I'll address the "limited technology" part of the question. To begin with the specific "brand" of entropy being dealt with here is that having to do with the operation of heat ...
1 vote

Why is the holographic encoding from Bekenstein-Hawking entropy remarkable?

If I have a cubic lattice of spin 1/2 particles in a volume $V$ with lattice spacing $a$ then the number of spin states is $2^{V/a^3}$, and the entropy is $V/a^3\ln(2)$. I guess I'm saying the ...
1 vote

What is the force exreted by gas particles per unit area on the walls of the container?

It's pressure. That's what pressure is - it's the definition of pressure. Force, energy, and pressure are different things. They even have different units. You can't say "my room is 12 hours long....
1 vote

Will the ideal gas equation change if we consider non equillibrium condition and then use kinetic approach and thermodynamic approach?

Thermodynamics is a phenomenological approach - the ideal gas equation is simply postulated (or derived from more elementary laws of Charles, Gay-Lussac and Boyle-Mariotte). In statistical physics the ...
  • 51.6k
1 vote

Will the ideal gas equation change if we consider non equillibrium condition and then use kinetic approach and thermodynamic approach?

The ideal gas equation doesn't change during non equilibrium conditions, it just doesn't apply during non equilibrium conditions. It only applies under equilibrium conditions. Once equilibrium is re-...
  • 62.9k
1 vote

Why does R. Feynman states that in a low-density gaseous enviornment, the pressure exerted by the molecules is proportional to density?

The first point is summarized by the ideal gas equation, $$ PV=nRT, $$ where the pressure $P$ is proportional to the density $n$. The argument regarding the attraction between the molecules is one of ...
  • 51.6k
1 vote

Why are the microstates of a harmonic oscillator considered equally probable, when the particle spends more time at locations of zero momentum?

You are missing the fact that there are more low-momentum states than high-momentum states. Therefore you are still more likely to observe a low momentum, although all states are equally probable. It’...
  • 5,735
1 vote
Accepted

Summing graphs in the partition function (statistical physics)

To understand this type of computation, you should write down the details explicitly. To lighten the notation, I set $a=z/\lambda^3$ and $b_l=U_l/l!$. Then, writing down explicitly the constraint as ...
  • 8,890
1 vote

What's the relation of entropy with synchronisation?

Pendulum synchronization happens because of energy exchange between two oscillators. In a perfectly frictionless, isolated system this would not increase entropy at all, and the pendulums would keep ...

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