Can pressure be negative?

[Perpetual Student]

try F = -dE/dx. This is the standard definition of force in one dimension. I cited it as tension (a specific example of a force that's easy to picture in 1D) just for pedagogical purposes. Anyway, if you fiddle with F=Ma and E= 1/2 mv^2 you can derive this for yourself.

I keep getting F = 2dE/dx. Does it matter?

 

[69dodge]

try F = -dE/dx. This is the standard definition of force in one dimension. I cited it as tension (a specific example of a force that's easy to picture in 1D) just for pedagogical purposes. Anyway, if you fiddle with F=Ma and E= 1/2 mv^2 you can derive this for yourself.

1/2 mv^2? Aren't we talking about potential energy of a stretched cord, not kinetic energy of a moving object?

Anyway, perhaps F = dE/dx is more familiar to Perpetual Student as dE = F dx? In words, it takes energy to exert a force while moving through a distance. Or something like that.

 

[ben m]

I keep getting F = 2dE/dx. Does it matter?

The factor of 2 indicates that you've made a mistake somewhere.

 

[rwguinn]

Do it with potential energy...
PE=1/2*(k*x^2)
dPE/dx=k*x

 

[Perpetual Student]

OK,


F = m(dv/dt), and
E = 1/2m(dx/dt)²
So, F/E = 2(dv/dt)(dt/dx)² = 2dvdt/(dx)² = 2dx/(dx)² = 2/dx.

F = 2E/dx. No?

 

[Perpetual Student]

OK,


F = m(dv/dt), and
E = 1/2m(dx/dt)²
So, F/E = 2(dv/dt)(dt/dx)² = 2dvdt/(dx)² = 2dx/(dx)² = 2/dx.

F = 2E/dx. No?

The above obviously makes no sense. 2/dx is meaningless and what does it mean to divide 2E by an infinitesimal? Where did I go wrong?

 

[69dodge]

OK,


F = m(dv/dt), and
E = 1/2m(dx/dt)²
So, F/E = 2(dv/dt)(dt/dx)² = 2dvdt/(dx)² = 2dx/(dx)² = 2/dx.

F = 2E/dx. No?

It's true that v dt = dx. But you have dv dt = dx, which is wrong.
Also, you're trying to find dE/dx, not E/dx.

I think the following works:

E = (1/2) mv².
So, dE/dx = mv dv/dx.
F = m dv/dt.
F / (dE/dx) = m (dv/dt) / (mv dv/dx) = (1/v)(dv/dv)(dx/dt) = (1/v)(1)(v) = 1.
So, F = dE/dx.

ETA:
That could be made simpler.
dE/dx = mv dv/dx = m (dx/dt) (dv/dx) = m dv/dt = F.

 

[rwguinn]

A brief search for tension = dE/dx has yielded no results. It does not seem to be part of Hooke's law or the definition of tension shown in the wikipedia article. Can anyone link a source showing the derivation of tension as dE/dx?

Do it with potential energy...
PE=1/2*(k*x^2)
dPE/dx=k*x

So I'm chopped liver? Or do you fizicissts always want to do it the hard way? Just because I'm a macro guy does not mean I am stupid or uninformed in my field, which is Mechanical Engineering, specifically, loads, stress and dynamics.
Using KE to derive Hooke's law is rather silly. It's a quasi-static process, not a dynamic one.
Yes, ideally (disregarding losses) KE(max)=PE(max) for a dynam8ic system, and force is dE/dx, both ways.

 

[Perpetual Student]

It's true that v dt = dx. But you have dv dt = dx, which is wrong.
Also, you're trying to find dE/dx, not E/dx.

I think the following works:

E = (1/2) mv².
So, dE/dx = mv dv/dx.
F = m dv/dt.
F / (dE/dx) = m (dv/dt) / (mv dv/dx) = (1/v)(dv/dv)(dx/dt) = (1/v)(1)(v) = 1.
So, F = dE/dx.

ETA:
That could be made simpler.
dE/dx = mv dv/dx = m (dx/dt) (dv/dx) = m dv/dt = F.

Thank you. :

 

[Perpetual Student]

OK, given F = dE/dx, how do I use (interpret) that expression if I am pulling on a rigid rod (where there is no Δx)?

 

[rwguinn]

OK, given F = dE/dx, how do I use (interpret) that expression if I am pulling on a rigid rod (where there is no Δx)?

dE/dx=kx=F
if dx=0, k=1/0 (I don't know how to lay an "8" on its side....)
The interpretation is that:
Any force you apply, when k=1/0 will not result in any delta-x

 

[Perpetual Student]

dE/dx=kx=F
if dx=0, k=1/0 (I don't know how to lay an "8" on its side....)
The interpretation is that:
Any force you apply, when k=1/0 will not result in any delta-x

Sorry, I'm not a physicist or an engineer, but I do know some mathematics.
What is kx? Is x a point? a length? is k some constant, if so what constant?

 

[Ziggurat]

Sorry, I'm not a physicist or an engineer, but I do know some mathematics.
What is kx? Is x a point? a length? is k some constant, if so what constant?

For a harmonic oscillator, x is the displacement from equilibrium and k is the spring constant.

And 69dodge rwguinn messed up somewhere (I think by conflating potential energy and kinetic energy): F = -dE/dx, not dE/dx, where E refers to potential energy, so the force from a spring is F=-kx (it's a restoring force).

 

[rwguinn]

Sorry, I'm not a physicist or an engineer, but I do know some mathematics.
What is kx? Is x a point? a length? is k some constant, if so what constant?

You were the one looking for a derivation conforming to Hooke's Law...

x is the same thing you have been going on about getting the derivative of E wrt.
K is a constant, in Hooke's law it is also9 know as STIFFNESS, or 1/flexibility.

V=dx/dt
A=d^2x/dt^2
where t is time.

 

[rwguinn]

For a harmonic oscillator, x is the displacement from equilibrium and k is the spring constant.

And 69dodge rwguinn messed up somewhere (I think by conflating potential energy and kinetic energy): F = -dE/dx, not dE/dx, where E refers to potential energy, so the force from a spring is F=-kx (it's a restoring force).

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.

 

[Perpetual Student]

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?

 

[rwguinn]

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?

it can't. You determine it the same way a scale can indicate the force that is called your weight.
Don't get cause and effect bass-ackward.
Force is applied. dE/dx is a RESULT of that force.

 

[Perpetual Student]

it can't. You determine it the same way a scale can indicate the force that is called your weight.
Don't get cause and effect bass-ackward.
Force is applied. dE/dx is a RESULT of that force.

OK, that makes sense. Thanks.
I am referring back to the comment made by ben m that:

Try some analogies on other quantities for which the derivatives make sense. Tension in an elastic rope, for example, is F = dE/dx. Does that mean a rope that's not *actually in the process of stretching* doesn't have a tension ("there's no dx")? The slope of a mountainside is dy/dx. Does that mean the mountain-slope is only really there when you're climbing it, and ceases to be a slope when you stop? No.

So, F = dE/dx applies when we have "an elastic" object. I used his response inappropriately.

 

[Perpetual Student]

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?
From ben m:

A better way to think about it is from Zig's explanation. A real, macroscopic system, internally in thermal equilibrium, will have some stockpile of energy. That energy is almost-but-not-quite uniformly spread---the distribution jitters around a bit due to statistical fluctuations---and those statistical fluctuations correspond to entropy changes. So both of the quantities in the derivative---the dE and the dS---are being "explored" by various parts of the system, even if it's not losing net energy.

The jittering around of statistical fluctuations in E and S is a tough concept to get around to define T.

I come from a world of unambiguous mathematical concepts. This stuff is fuzzy, to say the least.

 

[rwguinn]

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?
From ben m:


The jittering around of statistical fluctuations in E and S is a tough concept to get around to define T.

I come from a world of unambiguous mathematical concepts. This stuff is fuzzy, to say the least.

You and me, both.
I'm still trying to grasp an increase in entropy when there's no where for it to increase in...