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Can pressure be negative?

Which direction? Pull or push? he queried....
Once again. we get to sign conventions. I did NOT equate PE to KE, except as it pertains to the (theoretical) total energy in an oscillating system. So please read what I say, rather than your pre-conceived notion of what I know.

Take it easy. I just thought you dropped a minus sign, and that's the sort of minor mistake we all make from time to time. I may have misinterpreted what you meant, but I wasn't trying to comment on what you know.

As for sign conventions, F = -kx is the standard convention for Hooke's Law, and so is F = -dE/dx, where E is the potential energy. If you want to know how force relates to work done ON a system (for example, the change in kinetic energy), then one generally writes F = dW/dx (or dW = F dx), where W is the work. You may be clear about which one you mean when you write E, but others (like me, apparently) can get easily confused about what you mean if you don't use this more standard notation.
 
I am not trying to be difficult. I am trying to get this. I have a rigid rod suspended (tied to and hanging from a horizontal beam) with a weight pulling it down. Is there tension on that rod? If there is no dx, how can F = dE/dx define that tension? :confused:

You seem to be suggesting that if a rod is rigid, it will not deform in response to an applied force.

This is not the case. Nothing is perfectly rigid. In fact, special relativity forbids perfect rigidity, since perfect rigidity would actually violate causality. Deformations may be very small for a very rigid object (corresponding to very steep potential wells and hence large derivatives at small displacements), but they aren't zero.
 
So, back to T = dE/dS. It's a very difficult concept to grasp when you want T to mean something other than a measure of "one object is hotter than another." If T is to have some meaning for an object without reference to another, how does dE/dS do that? There is such a thing as absolute temperature, right?

S, E, and T are all state variables of your system. They can be defined independently of any other system. The significance of the quantity T comes largely from interactions between systems, but the quantity itself is perfectly well defined for a single system. And that's what we want, really: a quantity that I can calculate for one system on its own, which will then tell me how it interacts with another system whose temperature I can also calculate independently. The beauty of it comes precisely from the fact that I don't need to consider the two systems as a whole explicitly. It would be far less useful if I had to.
 
S, E, and T are all state variables of your system. They can be defined independently of any other system. The significance of the quantity T comes largely from interactions between systems, but the quantity itself is perfectly well defined for a single system. And that's what we want, really: a quantity that I can calculate for one system on its own, which will then tell me how it interacts with another system whose temperature I can also calculate independently. The beauty of it comes precisely from the fact that I don't need to consider the two systems as a whole explicitly. It would be far less useful if I had to.
The ability to really understand why dE/dS defines temperature seems to require a lot more training and background in physics than I have. So be it; I'm too old to be anything more than a physics dilettante.
Nevertheless, here is my amateur's take on this. (I would appreciate it if you would comment on this.) Looking at dE/dS as a tangent to a curve, (E is the vertical axis and S is horizontal), temperature is greater if the slope is steeper. So, when the rate of energy increase is greater than the rate of entropy increase the temperature is higher. On the other hand, if temperature is lower, entropy is growing at a rate greater than that of energy. If this is a reasonable interpretation, is there some intuitive way to look at this relationship to understand why temperature is the result of the relative rate of change of these quantities?
 
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is there some intuitive way to look at this relationship to understand why temperature is the result of the relative rate of change of these quantities?

If you're comfortable with calculus, this might be really easy, otherwise it may be a little technical, but I'll offer it in case it sticks, because if it does, it may satisfy.

We always want to maximize total entropy. For two systems in thermal contact, the total entropy is simply S = S1 + S2. To find the maximum of this function, we find where the derivative with respect to the relevant parameter, and set that equal to 0. So if we pick, say, the energy E1 in system 1, we find the value of E1 where
dS(E1)/dE1 = 0
This will tell us how much energy to put in system 1 versus system 2 in order to maximize total entropy (Note that this is NOT the total energy, so this quantity is NOT related to the total temperature either). Now let's do some manipulations:
dS/dE1 = dS1/dE1 + dS2/E1 = 0
dS1/dE1 = -dS2/E1
But conservation of energy requires that dE2 = -dE1, so
dS1/dE1 = dS2/E2
And that defines thermal equilibrium. But we know that temperatures are equal in thermal equilibrium, so this quantity must have something to do with temperature.

If you're NOT at thermal equilibrium, but off of it, then energy should move from 1 to 2 if dS1/dE1 < dS2/dE2, because that means you lose less entropy in 1 than you gain in 2, thus increasing total entropy. But we want energy to flow from 1 to 2 if T1 > T2, so
T1 = 1/(dS1/dE1)
T2 = 1/(dS2/dE2)
Now the > vs < thing works the way we want, plus we still maintain temperatures being equal at equilibrium.
 
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Force & Temperature Basics

I have a rigid rod suspended (tied to and hanging from a horizontal beam) with a weight pulling it down. Is there tension on that rod? If there is no dx, how can F = dE/dx define that tension? :confused:

The correct way to think of the force is that it is the gradient of a potential. So when you see the form F = dE/dx the "E" must always be a potential energy, not a kinetic energy. When you say "If there is no dx ..." the implication is that you want dx to represent a motion, something moving over a distance dx, but that is a conceptual mistake. Nothing is moving (just because a force is applied does not require anything to move, just push your car with the brake set and see what happens). In this case dx is just an incremental length of string and dE is the increment of the total energy stored in the string, that is stored in that particular dx of string, or rod. Pull on the rod with a stronger force, and any dx increment of the rod will carry a larger dE increment of internal, potential energy. Pull hard enough to break the rod (or string) and the potential energy will become kinetic.

"Work" is defined as a force applied over a distance (force times distance in common parlance). Now something has to move. Push on your car with the brake set until you collapse in exhaustion, but you have still done no work because the car has moved zero distance.

In the case of Newtonian mechanics, just note that momentum ("P") is mass times velocity ("mv") and force is mass times acceleration ("ma"). But ma = m dv/dt so the force is the derivative of the momentum, or F = dP/dt. This is the common way to describe force in Newtonian mechanics, but you have to remember that the concept of a field, and therefore the gradient of a potential, did not yet exist when Newton came up with his laws (we owe the concept of fields primarily to James Clerk Maxwell). So in old style Newtonian mechanics the force is defined in terms of its effect on moving bodies, as the first derivative of the momentum. But in a more modern formalism that includes particles and fields, we can define the force independently of its effect on things, and define it instead in terms of what it is intrinsically or where it comes from, and that is the gradient of a potential.

As for the temperature, the correct full form of the equation is dE = TdS - pdV where E is internal energy, T is temperature, S is entropy, p is pressure and V is volume. So the form T = dE/dS is valid only if the volume is held constant, so that dV = 0 (there are a few similar looking definitions using enthalpy, Helmholtz functions, Gibbs functions, and maybe more I am unaware of). Basically, the equation tells you that for the given increment in energy, a small increment in entropy represents a larger temperature, while a larger increment in entropy represents a smaller temperature (since the volume is constant the system can do no work expanding itself, so all of the energy has to become internal energy).

It's easier to understand by going back to the original definition of thermodynamic beta as the inverse temperature (beta = 1/kT = d/dE (log[N]) where N is the number of microstates, or the number of ways that one can arrange the quanta of energy in the system, E is the energy, T is the temperature and k is Boltzmann's constant (a number you can look up)). Now you can see the same thing, the rate of change as a function of energy, of the ways one can arrange the quanta of energy internal to the system. N is just a number, the count of available microstates, but it is a function of energy. Now suppose the system "saturates", so that even as more energy is pumped in, there are no more changes possible in microstates. The rate of change then becomes zero, beta becomes zero, and the temperature necessarily becomes infinite.

Now we all know that if you stick a mercury thermometer into what ever it is, the mercury column will not streak to infinity, and if we stick a digital thermometer into what ever it is, we won't see a sideways 8. That's what some people are thinking when they insist that the temperature cannot be infinite, and to that extent they are correct. But of course, those instruments don't measure temperature anyway, they measure something else and that is in turn interpreted as a temperature, over some limited and defined set of conditions. Temperature, like anything else in real physics, is whatever the equations define it to be. As you can see from the example above, the temperature becomes infinite, but there is nothing even close to "physically infinite" in the system.

Maybe that helps clarify things a bit? It's 1:45 AM here, so I might have slipped somewhere.
 
You seem to be suggesting that if a rod is rigid, it will not deform in response to an applied force.

This is not the case. Nothing is perfectly rigid. In fact, special relativity forbids perfect rigidity, since perfect rigidity would actually violate causality. Deformations may be very small for a very rigid object (corresponding to very steep potential wells and hence large derivatives at small displacements), but they aren't zero.
We use rigid elements all the time in stress and dynamic analysis. The equation is X1=RBT*x2, where RBT is the rigid body transformation.
However, we do it for a very specific reason, that being the difference in elemental stiffness being extremely large.
Again, assumptions that we know are not absolutely correct (there's that A-word, again!) but are "good enough". Anybody who want's to argue about a couple of parts in 100,000 can do it without me...:D
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Take it easy. I just thought you dropped a minus sign, and that's the sort of minor mistake we all make from time to time. I may have misinterpreted what you meant, but I wasn't trying to comment on what you know.

As for sign conventions, F = -kx is the standard convention for Hooke's Law, and so is F = -dE/dx, where E is the potential energy. If you want to know how force relates to work done ON a system (for example, the change in kinetic energy), then one generally writes F = dW/dx (or dW = F dx), where W is the work. You may be clear about which one you mean when you write E, but others (like me, apparently) can get easily confused about what you mean if you don't use this more standard notation.
Nope- I simply assumed that Force is a vector, and assigned a direction consistent with my assumptions.
 
Nope- I simply assumed that Force is a vector, and assigned a direction consistent with my assumptions.

The sign convention I used is still the normal sign convention when dealing with vectors. The only difference is one uses F = -grad E. But the minus sign is still there. If you're taking the direction change to be implicit, don't be surprised when someone doesn't know you're doing so. You might be able to keep it straight in your own head, but it's a terrible way to teach it to someone else, because they likly won't.
 
The sign convention I used is still the normal sign convention when dealing with vectors. The only difference is one uses F = -grad E. But the minus sign is still there. If you're taking the direction change to be implicit, don't be surprised when someone doesn't know you're doing so. You might be able to keep it straight in your own head, but it's a terrible way to teach it to someone else, because they likly won't.
Your knowledge and expertise are exceeded only by your condescending attitude and superiority complex.
Carry on.
 
Your knowledge and expertise are exceeded only by your condescending attitude and superiority complex.
Carry on.

Yes, every so often someone tries to teach you something you already know. Or at least you think you know, or think that you're the sole intended audience.

I don't think "putting down rwguinn" was ever the goal or the result. There certainly does seem to be a sign convention even if ignoring it is not technically wrong, and you're not the sole audience.

Semi-lurker thanks for all the discussion - I've learned something about temperature!
 
If you're comfortable with calculus, this might be really easy, otherwise it may be a little technical, but I'll offer it in case it sticks, because if it does, it may satisfy.
Calculus should not be the problem.

We always want to maximize total entropy. For two systems in thermal contact, the total entropy is simply S = S1 + S2. To find the maximum of this function, we find where the derivative with respect to the relevant parameter, and set that equal to 0. So if we pick, say, the energy E1 in system 1, we find the value of E1 where
dS(E1)/dE1 = 0
So far, so good.

This will tell us how much energy to put in system 1 versus system 2 in order to maximize total entropy (Note that this is NOT the total energy, so this quantity is NOT related to the total temperature either).
Whoops, you lost me. How can the above say anything about system 1 vs. system 2 since system 2 is not in the expression?
I can't continue until this is clarified.
By the way, many thanks for doing this!
 
The correct way to think of the force is that it is the gradient of a potential. So when you see the form F = dE/dx the "E" must always be a potential energy, not a kinetic energy. When you say "If there is no dx ..." the implication is that you want dx to represent a motion, something moving over a distance dx, but that is a conceptual mistake. Nothing is moving (just because a force is applied does not require anything to move, just push your car with the brake set and see what happens). In this case dx is just an incremental length of string and dE is the increment of the total energy stored in the string, that is stored in that particular dx of string, or rod. Pull on the rod with a stronger force, and any dx increment of the rod will carry a larger dE increment of internal, potential energy. Pull hard enough to break the rod (or string) and the potential energy will become kinetic.

"Work" is defined as a force applied over a distance (force times distance in common parlance). Now something has to move. Push on your car with the brake set until you collapse in exhaustion, but you have still done no work because the car has moved zero distance.

In the case of Newtonian mechanics, just note that momentum ("P") is mass times velocity ("mv") and force is mass times acceleration ("ma"). But ma = m dv/dt so the force is the derivative of the momentum, or F = dP/dt. This is the common way to describe force in Newtonian mechanics, but you have to remember that the concept of a field, and therefore the gradient of a potential, did not yet exist when Newton came up with his laws (we owe the concept of fields primarily to James Clerk Maxwell). So in old style Newtonian mechanics the force is defined in terms of its effect on moving bodies, as the first derivative of the momentum. But in a more modern formalism that includes particles and fields, we can define the force independently of its effect on things, and define it instead in terms of what it is intrinsically or where it comes from, and that is the gradient of a potential.

That is extremely helpful, thanks. My confusion was in thinking about these infinitesimals as "change in" instead of "increment of."
Now onto the more difficult part: dE/dS as temperature. Still working...
 
Whoops, you lost me. How can the above say anything about system 1 vs. system 2 since system 2 is not in the expression?
I can't continue until this is clarified.
By the way, many thanks for doing this!

We have to systems in thermal contact with each other but thermally isolated from the rest of the universe. System 1 has energy E1 and entropy S1, and system 2 has energy E2 and entropy S2. The total entropy of the combined system is S = S1 + S2, and the total energy is E = E1 + E2.

The total energy is a conserved quantity. We can move energy from system 1 to system 2, or the reverse, but it can't go anywhere else. To describe the allocation of energy between system 1 and system 2, it's enough to specify just the energy in system 1, since system 2 just holds the remainder. So a description of the energy of system 2 is implicit in the description of the energy of system 1, because total energy is conserved.

The total entropy, on the other hand, is NOT a conserved quantity. It will vary as we move energy back and forth between system 1 and system 2. So we can describe how the total entropy depends on the energy allocation between the two subsystems by looking at the function S(E1). To maximize a function with respect to a variable, we take the derivative of the function and set it to zero. And that's why we do what I described above. Does that help?
 
We have to systems in thermal contact with each other but thermally isolated from the rest of the universe. System 1 has energy E1 and entropy S1, and system 2 has energy E2 and entropy S2. The total entropy of the combined system is S = S1 + S2, and the total energy is E = E1 + E2.

The total energy is a conserved quantity. We can move energy from system 1 to system 2, or the reverse, but it can't go anywhere else. To describe the allocation of energy between system 1 and system 2, it's enough to specify just the energy in system 1, since system 2 just holds the remainder. So a description of the energy of system 2 is implicit in the description of the energy of system 1, because total energy is conserved.

The total entropy, on the other hand, is NOT a conserved quantity. It will vary as we move energy back and forth between system 1 and system 2. So we can describe how the total entropy depends on the energy allocation between the two subsystems by looking at the function S(E1). To maximize a function with respect to a variable, we take the derivative of the function and set it to zero. And that's why we do what I described above. Does that help?

Got it! That helps a lot.
 
If you're comfortable with calculus, this might be really easy, otherwise it may be a little technical, but I'll offer it in case it sticks, because if it does, it may satisfy.

We always want to maximize total entropy. For two systems in thermal contact, the total entropy is simply S = S1 + S2. To find the maximum of this function, we find where the derivative with respect to the relevant parameter, and set that equal to 0. So if we pick, say, the energy E1 in system 1, we find the value of E1 where
dS(E1)/dE1 = 0
This will tell us how much energy to put in system 1 versus system 2 in order to maximize total entropy (Note that this is NOT the total energy, so this quantity is NOT related to the total temperature either). Now let's do some manipulations:
dS/dE1 = dS1/dE1 + dS2/E1 = 0

Just to be certain, I assume that last expression should be dS2/dE1, the omitted d being an oversight.
 
...
We always want to maximize total entropy.
...

Thanks to your help and Tim Thompson's, I think I'm getting somewhere. Question: How should I interpret the above? Why do we "want" to maximize entropy? Is it because over time it will "seek a maximum?
 
Thanks to your help and Tim Thompson's, I think I'm getting somewhere. Question: How should I interpret the above? Why do we "want" to maximize entropy? Is it because over time it will "seek a maximum?

Is it because the combined system can only be in equilibrium when entropy is maximized?
 
Entropy

Is it because the combined system can only be in equilibrium when entropy is maximized?

I think that is correct. The fabled 2nd law of thermodynamics tells us that in the case of any isolated system, the change in entropy will always be positive, except for idealized reversible processes, where it can be zero, but it will never be negative. If the two systems combined are treated as a third, isolated system, then the total entropy of that third system must be greater than the sum of the entropies of the two systems we started with, before they came into contact.

I wrote a few webpages about entropy some years back which may or may not shed light on the subject: Adventures in Entropy. Any system in thermodynamic equilibrium should have a maximum entropy, if I am not mistaken.
 

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