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Capiert
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On 5/26/2022 at 10:04 PM, Capiert said:

Isn't
 (height h=) distance d
 divided
 by time t
 an average_speed
 va=d/t?

 

What

has this

to

do with

power?

?

 

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Posted (edited)

(James Watt's, mechanical)
 Power

 P=F*d/t, va=d/t

 is the Force (e.g. Weight Wt=m*g)
 F=m*g
 multiplied
 by (e.g. the height h=) distance d
 per time t.

That works out
 to, Power
 P=F*va
 is the (e.g. weight Wt=m*g)
 Force (F=m*a)
 multiplied
 by the average_speed va=d/t=h/t;
 instead of

 F*v=(p^2)/(m*t)
 which
 is the momentum_squared (p^2)=mom^2=(m*v)^2


 (for the mass m;
 multiplied
 by the speed(_difference) v=vf-vi
 (of final_speed vf
 minus
 the initial_speed vi));


 divided by both:
 mass m
 & time t

(I.e.
 F*v is definitely
 NOT Power;
 although
 it might seem similar.)

Force
 F=mom/t (=p/t)
 is the
 (change
 in) momentum mom=m*v
 per time t. 

Thank you
 for asking.
(I thought NOBODY
 would dare,
 (for at least 3 days).)

James Watt's
 (definition
 of) Power
 is (the rate
 of doing work),
 where he defined Work(_Energy)
 WE=Wt*d
 as lifting
 weight Wt (=m*g, Force F=m*a, let (linear_)acceleration a=g)
 to a specific
 height (h=d distance).

(We call that
 (kind of work(_energy))
 Potential_Energy
 PE=m*g*h.)

That (work_(energy))
 done
 (with)in
 a specific
 amount
 of time t
 is (his) Power
 P=WE/t.

Converting
 work_energy
 WE=KE
 to (moving)
 kinetic_energy,
 we get

 KE=m*v*va,

 where with the speed(_difference) v=vf-vi
 & the average_speed va=d/t=h/t
 that gives
 KE=m*(vf-vi)*(vf+vi)/2, or

 KE=m*(vf2-vi2)/2.

 

Disclaimer:

(To me)
 it looks like
 somebody goofed
 on (producing, e.g. creating)
 James Watt's (formula) syntax
 which needs "average" ((for the) speed).

I suspect most people
 missed that (detail).

(Easily found
 with (simple) algebra.)

I hope that answers
 your question.

Btw.
Even, the
 (Ewert's 1996, universal)
 conservation of mass*Energy
 m*E=mom*moma
 m*E=(m*v)*(m*va)
 m*E=m*m*v*va
 which correctly proportions
 the mass_squared m2
 with (respect to)
 the (single, non_squared)
 height h
 in m*PE,


 (indirectly denying (
 https://en.wikipedia.org/wiki/Julius_von_Mayer

 Julius Robert von Mayer (25 November 1814 – 20 March 1878)
 (remarkable) 1841 conservation
 of Energy,
 as a(n absurd) bunder=Fake,
 due to lack of mass);


 (but)
 produces similar (confirming) results
 m*PE=m*KE
 m*m*g*h=m*m*v*va.

(Why we are (supposedly)
 allowed to cancel mass m
 & (then, attempt to) maintain
 the (mass's, (fallen)) height h
 is a riddle to me.

 (..because..)

It'( i)s physically wrong.

Meaning it (=cancelling mass)
 can fail
 (the proportioning relation
 to height)!

But does NOT always.

Of course (naturally) everybody has heard, of, conservation of mass. ?)

Noether's Theory
 is NOT going
 to help you,
 if you CAN'T even get the basics right=correct.


It's (=Noether's theorem:
 which uses distance (e.g. (Work_)Energy WE=F*d, e.g. PE=m*g*h);
 instead of (per) time (e.g. (average_)momentum moma=m*d/t=m*va)

 is) based (most probably) on NOTHING
 (but (maybe (crumbling, unreliable)) NONSENSE). ?
But more interesting
 might be where that (Noether theorem) went wrong,
 (when the simplest
 of algebra
 can prove an error
 that, that theorem did NOT)
 e.g. why you guys & gals
 put all your eggs (e.g. hope(ful assumptions))
 in that 1 basket.

(Why do things simply;
 when you can do them (more) complicated?
 (..so NOBODY can see (thru) the ERRORS)?
NOBODY is PERFECT.
NOT even me.
Why NOT strive for error reducing methods, instead?)


E.g. (Correct is, that:)
 Fallen height distance
 is (=must be (made))
 wrt mass_squared m2;
 NOT mass m(1) ONLY.

 

https://en.wikipedia.org/wiki/Conservation_of_energy

Main article: Noether's theorem

Emmy Noether (1882-1935) was an influential mathematician known for her groundbreaking contributions to abstract algebra and theoretical physics.

The conservation of energy is a common feature in many physical theories. From a mathematical point of view it is understood as a consequence of Noether's theorem, developed by Emmy Noether in 1915 and first published in 1918. The theorem states that every continuous symmetry of a physical theory has an associated conserved quantity; if the theory's symmetry is time invariance then the conserved quantity is called "energy". The energy conservation law is a consequence of the shift symmetry of time; energy conservation is implied by the empirical fact that the laws of physics do not change with time itself. Philosophically this can be stated as "nothing depends on time per se". In other words, if the physical system is invariant under the continuous symmetry of time translation then its energy (which is the canonical conjugate quantity to time) is conserved. Conversely, systems that are not invariant under shifts in time (e.g. systems with time-dependent potential energy) do not exhibit conservation of energy – unless we consider them to exchange energy with another, an external system so that the theory of the enlarged system becomes time-invariant again. Conservation of energy for finite systems is valid in physical theories such as special relativity and quantum theory (including QED) in the flat space-time.

If we take again
 P=F*va, swap sides
 F*va=P, /F

 va=P/F, va=d/t
 P/F=d/t, *t*F
 both sides produce energy
 P*t=F*d, P*t=E & F=mom/t
 E=(mom/t)*d, rearrange
 E=mom*d/t, va=d/t
 E=mom*va, mom=m*v

 E=m*v*va, is the Kinetic Energy
 KE=m*v*va.

 We do NOT need
 Noether's theorem
 to show
 the "connection"
 between Force (F=P/va) & Power (P=F*va)
 or distance (d=t*P/F) versus time (t=d*F/P)
 as average_speed va=d/t (=P/F).

 

https://en.wikipedia.org/wiki/Julius_von_Mayer

It (German knighthood) for caloric,
Mayer was the first person to state the law of the conservation of energy, one of the most fundamental tenets of modern day physics. The law of the conservation of energy states that the total mechanical energy of a system remains constant in any isolated system of objects that interact with each other only by way of forces that are conservative. Mayer's first attempt at stating the conservation of energy was a paper he sent to Johann Christian Poggendorff's Annalen der Physik, in which he postulated a conservation of force (Erhaltungssatz der Kraft). However, owing to Mayer's lack of advanced training in physics, it contained some fundamental mistakes and was not published.

 

..(Mayer) examined experimentally; for example, if kinetic energy transforms into heat energy, water should be warmed by vibration

 

Since he (=Mayer) was not taken seriously at the time, his achievements were overlooked and credit was given to James Joule. Mayer almost committed suicide after he discovered this fact. He spent some time in mental institutions to recover from this and the loss of some of his children. Several of his papers were published due to the advanced nature of the physics and chemistry. He was awarded an honorary doctorate in 1859 by the philosophical faculty at the University of Tübingen. His overlooked work was revived in 1862 by fellow physicist John Tyndall in a lecture at the London Royal Institution. In July 1867 Mayer published "Die Mechanik der Wärme." This publication dealt with the mechanics of heat and its motion. On 5 November 1867 Mayer was awarded personal nobility by the Kingdom of Württemberg (von Mayer) which is the German equivalent of a British knighthood.

Edited by Capiert
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Posted (edited)
31 minutes ago, Capiert said:

KE=m*(vf2-vi2)/2.

It looks like you are saying an object moving at a constant velocity has zero KE, since vf = vi   

That's a problem. 

Edited by Bufofrog
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Posted (edited)

I think power is the time derivative of work. In order to make the simplification like:

svg.image?P=Fv

you are presuming a constant force is being applied over a distance (I'd say change in position).

When you want to take an instantaneous power output, I think it makes sense there is no conservation of energy. You'd need to extend the time frame, and perhaps incorporate the other present bodies and conditions to know how the conservation may or may not weigh out.

 

As an exercise, can you tease out the difference between change of position vs. distance (e.g.: "Why can Work be said to be 0 despite obvious evidence to the contrary, in a self-referential frame?") and velocity vs. speed. You mention speed at one point and then use v in your equation: velocity implies a direction against a map, and a change in direction gives us an acceleration, so when you have your instantaneous speed this option for change of direction as acceleration is removed from the equation (i.e.: svg.image?P=Fv can be said to exist presumptively also in the case of a "point"..).

Edited by NTuft
clarity; spelling.
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Posted (edited)
1 hour ago, Bufofrog said:

It looks like you are saying an object moving at a constant velocity has zero KE, since vf = vi   

That's a problem. 

Please explain that problem.

Kinetic_Energy
 is (already) relative
(with respect)
 to the initial_speed vi.

(-c<)vi<c
 can be (almost) anything
 less than light's_speed (+/-)c.

That means the initial_speed vi
 is excluded
 in that (amount
 of) "kinetic_energy('s)"
change
 of speed.

E.g.
2 masses moving
 at the same speed
 (wrt each other),
 have NO speed_difference v=vf-vi
 (wrt each other),
 thus a constant distance
 is maintained (=kept)
 between them.

Only if a speed_difference v
 exists between them
 can they affect each other
 in e.g. a collision,
 to change the other's speed
 e.g. via Newton's 3rd law,
 (equal & opposite) reaction,
 Repulsion.

E.g. Assuming an (EM_)Field
 wrt (a decreasing) distance,
 to ((elastically) repulsively) bounce.

The catch there
 is electric_repulsion
 is inversely proportional
 to the radial_distance "squared"!

That is NO longer a linear relation;
 but instead exponential
(wrt distance)!
 

Edited by Capiert
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19 minutes ago, Capiert said:

E.g.
2 masses moving
 at the same speed
 (wrt each other),
 have NO speed_difference v=vf-vi
 (wrt each other),
 thus a constant distance
 is maintained (=kept)
 between them.

Directionless points, or are we to assume they're "moving on parallel lines"?

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We are assuming
 they are moving
 on parallel lines,
 in the same direction.

Can (=May)
I pacify
 that argument(?)
 by saying:
 (I acknowledge)
 an "initial" Kinetic_Energy KEi=m*vdi*vai
 where
 the intial_speed vi=vi-0
 is (simply)
 extrapolated
 from a predacesser
 speed_difference
 with its maximum_speed
 being (only) that excluded
 initial_speed vdi=vi-0
(whatever zero=0 speed
 should be
 e.g. relative to something else('s motion_speed)).
 vai=(0+vi)/2
 (analogy (similar to))
  va=(vf+vi)/2).

 e.g. minus zero,
 where zero
 is that (next, smaller)
 reference.

That is all done so
 because KE
 is (already) relative
 to its (own) initial_speed vi.

Thus the kinetic_energies
 can be added (sequentially).
 
 

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Posted (edited)
3 hours ago, NTuft said:

I think power is the time derivative of work. In order to make the simplification like:

svg.image?P=Fv

you are presuming a constant force is being applied over a distance (I'd say change in position).

Yes.

3 hours ago, NTuft said:

When you want to take an instantaneous power output, I think it makes sense there is no conservation of energy.

I agree that there is NO such thing
 as conservation of Energy (COE);
 although conservation of mass*Energy m*E
 might exist.

COE is a FAKE,
 because it often FAILS,
 although NOBODY
 has the guts
 to kick it OUT;
 they all still (perhaps naively) defend it
 (as traditional BRAINWASHING,
 e.g. NOT to upset things=tradition).

Conservation of ENERGY
 does NOT deserve
 to be mentioned
 except for its (crude) history
 e.g. development
 of Physics.

3 hours ago, NTuft said:

You'd need to extend the time frame,
 and perhaps incorporate the other present bodies
 and conditions
 to know how
 the conservation may or may not weigh out.

(That is:)
 Wasted time,
 considering
 my (just=immediately) previous comment.

But what do you mean
 by extend
 the time frame?

3 hours ago, NTuft said:

As an exercise,
 can you tease out the difference
 between change of position
 vs. distance

Distance
 d (x,y,z),
 d=P2(x2,y2,z2)-P1(x1,y1,z1),
 is the difference
 between 2 position(s
 points):
 e.g. P2(x2,y2,z2)
 minus P1(x1,y1,z1) ;
 (where P1(x1,y1,z1)
 is the origin(al starting point)
 which determines
 the positive direction). 

 e.g.
 x=x2-x1
 y=y2-y1
 z=z2-z1.

3 hours ago, NTuft said:

 (e.g.:
 "Why can Work be said to be 0
 despite obvious evidence
 to the contrary,
 in a self-referential frame?")

Please explain.

(e.g. "contrary"
 to "self-referential" frame).

(What is that?)
(E.g. A Point source?)
(Is ego supreme?)

Everything exists
 "in" the universe;
 NOT OUTSIDE
 of it (=the universe).

Everything is "a" part
 of the universe (=connected);
 NOT "apart" !.

(E.g. I am in the universe;
 NOT independent
 from the universe).

(Please give me a clue
 (as)
 to help understand you better.)

But if I understand you correctly
 then the "initial" kinetic_energy KEi
 might help fix=remedy things;
 for the total=final kinetic_energy KEf=KEi+KE.

E.g.
Kinetic_Energy
 KE=KEf-KEi
 is the difference
 between
 final (kinetic_energy KEf)
 & initial (kinetic_energy KEi)

 similar
 to speed(_differrence vd
 vd=v)
 v=vf-vi
 where the subscripts are:
 final f
 & initial i.

3 hours ago, NTuft said:

and

 ("you can tease out the difference between")

3 hours ago, NTuft said:

 velocity vs. speed. 

I'( a)m sorry
 (but) you have lost me (there).

I can NOT imagine
 any speed
 without direction.

"I DON'T know where
 we'( a)re going Captain;
 but

(at warp_speed)
 we are getting there
 in a he(ck) of a hurry!
"-Scotty.

Direction (e.g. angle)
 is also relative.
E.g. to a line.
E.g. 2 points (P1 & P2, each an x,y,z);

 but the 1st (point)
 must be established
 from the 2nd,
 to complete
 that relation
 (for positve,
 (versus negative)
 direction)
 i.e. relativity.

3 hours ago, NTuft said:

You mention speed "at" one point
 and then use v(elocity)
 in your equation:
 velocity implies a direction against a map,

I will assume
 (with (the word) map)
 you are implying x,y,z
 (e.g. wrt to some other reference
 (of similar structure, e.g. Ref, e.g Frame)).

I (can still)
 consider
 a speed(_difference) v (=vd),
 to be
 (composed
 of) components
 vx, vy, & vz.

3 hours ago, NTuft said:

 and a change in direction gives us an acceleration,
 so when you have your instantaneous speed

(But)
 I do NOT use instantaneous_speed;
 I use average_speed va=d/t,
 instead.

3 hours ago, NTuft said:

 this option
 for change
 of direction
 as acceleration

 is removed
 from the equation
 (i.e.: svg.image?P=Fv can be said
 to exist presumptively also
 in the case of a "point"..).

I(' am sorry, I) DON'T follow (you).

My (speed_difference) v
 is a mixture
 of x,y,z speeds.

Where is the problem?

(I suspect you are adding
 unneeded complexity;
 where NONE is needed.

Am I right or wrong?
If wrong please explain.)

 

Edited by Capiert
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3 hours ago, Capiert said:

Please explain that problem.

The KE equation you wrote would mean that an object moving at a constant speed would have zero KE.  That is incorrect, hence the equation is incorrect.

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Posted (edited)
20 hours ago, Capiert said:

But what do you mean
 by extend
 the time frame?

Integrate over a section of the function that is describing the Work output.

 

21 hours ago, Capiert said:

We are assuming
 they are moving
 on parallel lines,
 in the same direction.

And so you are modelling a system that has no physical importance or corollary, it seems.

 

20 hours ago, Capiert said:
23 hours ago, NTuft said:

to the contrary,
 in a self-referential frame?")

Please explain.

From a singular reference we would measure displacement, change of position, distance, as zero if it returned to its initial position, thus no work is seemingly done in the system; this is a common teaching point on a quirk about Work.

 

20 hours ago, Capiert said:

Kinetic_Energy
 KE=KEf-KEi
 is the difference
 between
 final (kinetic_energy KEf)
 & initial (kinetic_energy KEi)

No, kinetic energy is... let's say a function of momentum, a vector quantity.  A seemingly 'conserved' physical quantity, too.

A vector implies a magnitude (e.g. speed is a magnitude) and a direction. 

I think these are necessary complexifications. You keep bringing up "d/t", but even the classical mechanics you're looking to do here need for you to be familiar with d/dt.

Edited by NTuft
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26 minutes ago, NTuft said:

No, kinetic energy is... let's say a function of momentum, a vector quantity.

Just to be clear to Capiert, that does not mean KE is a vector, it isn't.

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Quote

 

There are two main descriptions of motion: dynamics and kinematics. Dynamics is general, since the momenta, forces and energy of the particles are taken into account. In this instance, sometimes the term dynamics refers to the differential equations that the system satisfies (e.g., Newton's second law or Euler–Lagrange equations), and sometimes to the solutions to those equations.

However, kinematics is simpler. It concerns only variables derived from the positions of objects and time. In circumstances of constant acceleration, these simpler equations of motion are usually referred to as the SUVAT equations, arising from the definitions of kinematic quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).

Physics:Equations of motion : https://handwiki.org/wiki/Physics:Equations of motion

 

Perhaps you are looking to deal in kinematics, and I am overcomplicating things.

However, to impart a speed onto a mass (and you supposing this is a mass that has been put in motion, yes?) requires an application of force for it to acquire kinetic energy. Once it has been brought up to speed, there is not a constant force being applied; I conceptualize the force acting may now be zero.

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9 hours ago, Bufofrog said:

Just to be clear to Capiert, that does not mean KE is a vector, it isn't.

For a non-rotating rigid body or a point object, the directionality is built into the unit of the joule:

Quote

Joule

The joule (/dʒaʊl, dʒuːl/ jowl, jool;[1][2][3] symbol: J) is a derived unit of energy in the International System of Units.[4] It is equal to the amount of work done when a force of 1 newton displaces a body through a distance of 1 metre in the direction of the force applied. It is also the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second.

But, yes, in this case it is a dot product, a scalar: 

Quote

[...] The distinction may be seen also in the fact that energy is a scalar quantity – the dot product of a force vector and a displacement vector. By contrast, torque is a vector – the cross product of a force vector and a distance vector. Torque and energy are related to one another [...]

Joule: https://handwiki.org/wiki/Joule

so, I think a KE system accounting for objects experiencing rotational force does have a vector component, but do correct me if I'm wrong.

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19 minutes ago, NTuft said:

For a non-rotating rigid body or a point object, the directionality is built into the unit of the joule:

But, yes, in this case it is a dot product, a scalar: 

so, I think a KE system accounting for objects experiencing rotational force does have a vector component, but do correct me if I'm wrong.

Energy does not have a direction. That should be obvious to you, if you think about it clearly. 

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6 hours ago, NTuft said:

so, I think a KE system accounting for objects experiencing rotational force does have a vector component, but do correct me if I'm wrong.

Torque is a vector and energy is not. The units for energy and torque are equivalent, but the convention is that torque uses the units N-m rather than joules, because torque is not energy.

15 hours ago, NTuft said:

Perhaps you are looking to deal in kinematics, and I am overcomplicating things.

However, to impart a speed onto a mass (and you supposing this is a mass that has been put in motion, yes?) requires an application of force for it to acquire kinetic energy. Once it has been brought up to speed, there is not a constant force being applied; I conceptualize the force acting may now be zero.

This is correct, and is described by the work-energy theorem.

W = integral of F.dx (dot product, i.e. due to the component of the force in the direction of displacement)

F = ma = m dv/dt so we are integrating m dv/dt dx

Let's drop the vector notation and look at the components (i.e. we've done a dot product) We rearrange this to be m dx/dt dv

which is mv dv

Integrate and you get W = 1/2 mv^2, evaluated at some initial and final v. Work is the change in kinetic energy

 

(note that capiert has a tendency to use equations while ignoring any constraints and initial conditions, and try to apply those equations in general, which leads to problems. Also the use of non-standard terminology, labeling, and symbols)

On 5/31/2022 at 5:32 PM, Capiert said:

I agree that there is NO such thing
 as conservation of Energy (COE);
 although conservation of mass*Energy m*E
 might exist.

COE is a FAKE,
 because it often FAILS,
 although NOBODY
 has the guts
 to kick it OUT;
 they all still (perhaps naively) defend it
 (as traditional BRAINWASHING,
 e.g. NOT to upset things=tradition).

It's neither "tradition" nor "brainwashing"

It has a very sound basis (time translation symmetry) and protocols on how to apply it. If you ignore those protocols, of course, you get the wrong answer. We don't kick things out because a few people fail to understand them.

And mass-energy is famously not a conserved quantity.

On 5/31/2022 at 5:32 PM, Capiert said:

E.g.
Kinetic_Energy
 KE=KEf-KEi
 is the difference
 between
 final (kinetic_energy KEf)
 & initial (kinetic_energy KEi)

∆KE = KEf-KEi

i.e. KEf-KEi is the change in kinetic energy, not the kinetic energy. This is a very important distinction.

On 5/31/2022 at 1:07 PM, Capiert said:

Power

 P=F*d/t, va=d/t

 is the Force (e.g. Weight Wt=m*g)
 F=m*g
 multiplied
 by (e.g. the height h=) distance d
 per time t.

Here is an example of using a specific equation and trying to apply it in general.

Power applies to more than the situation of an object falling under gravity, so one cannot make a general claim that force is weight, since there are other forces. If you use weight, then the equation will only apply to a falling object, and also only if you can assume that this is a constant force.

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Posted (edited)
15 hours ago, exchemist said:

Energy does not have a direction. That should be obvious to you, if you think about it clearly. 

I am thinking that everything is torque-ing, and multiplying (or defining directionality in the same direction as force applied) by a "dimensionless angle" re-vivifies an aspect of directionality in 3-D of all KE; spin? 

 

@swansont,

Wow! Thank you! I think this is the best lesson on learning calculus; I ought to review it again and again. What better way than with real world-scaled models to illustrate what calculus is doing on a function that describes motion???

I will refer to the degree of freedom that torque implies: multiplying the units by a dimensionless quantity (perhaps rapidity needs units?) comes through with implied directionality again, as from vectors that's a bugaboo... I think with quaternions we can translate around self-refential centers in 3-D, what with the extra "degree of freedom"? I need to read your post more thoroughly,

 

And try to figure out what @exchemist is saying, exactly.

 

No I do not understand time translation symmetry, but I doubt it.

 

Also, we are assuming a constant, non-interactive mass? Bah. I think "the most zeroest of relative momentum frames" would be needed to assess when if ever interconversion conservation of mass and energy is a "law"...

Edited by NTuft
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12 hours ago, NTuft said:

I will refer to the degree of freedom that torque implies: multiplying the units by a dimensionless quantity (perhaps rapidity needs units?) comes through with implied directionality again, as from vectors that's a bugaboo... I think with quaternions we can translate around self-refential centers in 3-D, what with the extra "degree of freedom"? I need to read your post more thoroughly,

For a torque to become energy (i.e. do work) it needs to happen through some rotation (analogous to a force exerted through a displacement). The angle is unitless, but again it's a dot product (the torque is a vector but typically the angle can only be with the same handedness or opposite, which tells you if you add or subtract energy)

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On 5/26/2022 at 11:04 PM, Capiert said:

Isn't
 (height h=) distance d
 divided
 by time t
 an average_speed
 va=d/t?

 

What's the speculation here?

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  • 2 weeks later...
On 6/1/2022 at 11:13 AM, NTuft said:
On 5/31/2022 at 1:18 PM, Capiert said:

We are assuming
 they are moving
 on parallel lines,
 in the same direction.

And so you are modelling a system that has no physical importance or corollary, it seems.

 

This is an erroneous statement on my part. In particular, I found mention of something like "runaway pair", or "breakaway pair", but I cannot find the correct term now -- it was specifically a pair moving on parallel lines with force interactions offset to maintain distance -- so perhaps someone can fill in that blank; @joigus. It's likely there are other paired systems I'm not aware of that make my statement false.

This also seems relevent, regarding a free particle:

uncertainty.pdf 1

So I say that it may be imprecise to make the assumption that you can determine that the particles are both on parallel lines (determining position) and moving with constant speed (or, the kinetic energy or momentum they gained in getting up to speed; determining momentum).

 

On 6/2/2022 at 3:28 AM, swansont said:

(note that capiert has a tendency to use equations while ignoring any constraints and initial conditions, and try to apply those equations in general, which leads to problems. Also the use of non-standard terminology, labeling, and symbols)

On 5/31/2022 at 2:32 PM, Capiert said:

I agree that there is NO such thing
 as conservation of Energy (COE);
 although conservation of mass*Energy m*E
 might exist.

COE is a FAKE,
 because it often FAILS,
 although NOBODY
 has the guts
 to kick it OUT;
 they all still (perhaps naively) defend it
 (as traditional BRAINWASHING,
 e.g. NOT to upset things=tradition).

It's neither "tradition" nor "brainwashing"

It has a very sound basis (time translation symmetry) and protocols on how to apply it. If you ignore those protocols, of course, you get the wrong answer. We don't kick things out because a few people fail to understand them.

@Capiert can you elaborate on time translation symmetry, as swansont introduced to give reference to your formulations, and if that is somehow requisite in your formulation which I would now phrase as: Conservation Of Energy (COE) is fake by some kind of classical derivations relating to Power. Is that a correct phrasing for your argument? Perhaps @exchemist knows about this time translation symmetry business as his question is formulated with odd spacing as you employ. I hope my egregious use of parentheses (as is my custom) is similarly well met.

Here also (the second half of the page is) on Work as a dot product:

1.pdf 1

@swansont:

On 6/2/2022 at 3:04 PM, swansont said:

For a torque to become energy (i.e. do work) it needs to happen through some rotation (analogous to a force exerted through a displacement). The angle is unitless, but again it's a dot product (the torque is a vector but typically the angle can only be with the same handedness or opposite, which tells you if you add or subtract energy)

 2.pdf1

Since, "Torque can be defined as the rate of change of angular momentum, analogous to force."2, I think we have to settle on it being a pseudovector (which is news to me). I may not understand fully but the reference .pdf specifically qualifies it as a cross product, not dot product. 

 

@exchemist 

On 6/1/2022 at 10:13 PM, exchemist said:

Energy does not have a direction. That should be obvious to you, if you think about it clearly.

Here is another page on this section about vectors (forgive me, but it brings in HCl which you mentioned specifically in the London Dispersion/Van der Waals forces thread, so I include it here for our point of contention):

3.pdf

So, to the point: if we have a separated pair of opposite point charges there should be potential energy between (yes?), and the directionality again (like with the joule) is built into the notion of the dipole here with energy going from negative to positive by convention. You could be right, maybe energy has no directionality, but in a system where it dissipates I imagine it goes outward radially, and in a system where energy is transferred otherwise I most always seem to imagine it has directionality (but it is often built into the units or conventions under consideration).

1McQuarrie, Donald A. (Donald Allan). Quantum Chemisty. 2nd ed., University Science Books, 2008.

2Handwiki.org : Physics:Angular momentum

 

Edited by NTuft
corrected attachment #2; added #3 + exchemist
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5 hours ago, NTuft said:

This is an erroneous statement on my part. In particular, I found mention of something like "runaway pair", or "breakaway pair", but I cannot find the correct term now -- it was specifically a pair moving on parallel lines with force interactions offset to maintain distance -- so perhaps someone can fill in that blank; @joigus. It's likely there are other paired systems I'm not aware of that make my statement false.

This also seems relevent, regarding a free particle:

uncertainty.pdf 391.01 kB · 1 download  1

So I say that it may be imprecise to make the assumption that you can determine that the particles are both on parallel lines (determining position) and moving with constant speed (or, the kinetic energy or momentum they gained in getting up to speed; determining momentum).

 

@Capiert can you elaborate on time translation symmetry, as swansont introduced to give reference to your formulations, and if that is somehow requisite in your formulation which I would now phrase as: Conservation Of Energy (COE) is fake by some kind of classical derivations relating to Power. Is that a correct phrasing for your argument? Perhaps @exchemist knows about this time translation symmetry business as his question is formulated with odd spacing as you employ. I hope my egregious use of parentheses (as is my custom) is similarly well met.

Here also (the second half of the page is) on Work as a dot product:

1.pdf 277.95 kB · 0 downloads  1

@swansont:

  2.pdf 645.23 kB · 1 download 1

Since, "Torque can be defined as the rate of change of angular momentum, analogous to force."2, I think we have to settle on it being a pseudovector (which is news to me). I may not understand fully but the reference .pdf specifically qualifies it as a cross product, not dot product. 

 

@exchemist 

Here is another page on this section about vectors (forgive me, but it brings in HCl which you mentioned specifically in the London Dispersion/Van der Waals forces thread, so I include it here for our point of contention):

3.pdf 322.27 kB · 0 downloads

So, to the point: if we have a separated pair of opposite point charges there should be potential energy between (yes?), and the directionality again (like with the joule) is built into the notion of the dipole here with energy going from negative to positive by convention. You could be right, maybe energy has no directionality, but in a system where it dissipates I imagine it goes outward radially, and in a system where energy is transferred otherwise I most always seem to imagine it has directionality (but it is often built into the units or conventions under consideration).

1McQuarrie, Donald A. (Donald Allan). Quantum Chemisty. 2nd ed., University Science Books, 2008.

2Handwiki.org : Physics:Angular momentum

 

You are not thinking clearly. As a result, you are confusing force, which certainly has a direction, with energy, which does not. You have a force between 2 charged particles, which diminishes with increasing separation. An increase in energy is the work done by moving them a certain distance against that force. That energy has no direction.

Edited by exchemist
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9 hours ago, NTuft said:

Since, "Torque can be defined as the rate of change of angular momentum, analogous to force."2, I think we have to settle on it being a pseudovector (which is news to me). I may not understand fully but the reference .pdf specifically qualifies it as a cross product, not dot product. 

Who is saying it's a pseudovector? I don't see that in your book excerpt. It's a vector, following the protocol of a cross product.

Quote

So, to the point: if we have a separated pair of opposite point charges there should be potential energy between (yes?), and the directionality again (like with the joule) is built into the notion of the dipole here with energy going from negative to positive by convention. You could be right, maybe energy has no directionality, but in a system where it dissipates I imagine it goes outward radially, and in a system where energy is transferred otherwise I most always seem to imagine it has directionality (but it is often built into the units or conventions under consideration).

The dipole can have any orientation and the result will be the same. There's no directionality relative to a coordinate system. It's a scalar.

The dissipation of energy follows certain rules, and it depends on the system. The gradient of potential energy gives you the force; the gradient is where directionality comes in, not the energy itself. Knowing that e.g. the electrostatic potential energy depends on the separation tells you the gradient is radial, and the force is indeed radial.

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16 hours ago, NTuft said:

Since, "Torque can be defined as the rate of change of angular momentum, analogous to force."2, I think we have to settle on it being a pseudovector (which is news to me).

Whether torque is a vector or a pseudovector is a question that has nothing to do with energy being a scalar that comes from time-translation invariance. I'm not sure how relevant this is to the discussion, as I'm not sure of what's being discussed, and I cannot formulate what the claimed speculation is.

This distinction vector/pseudovector has nothing to do with energy stemming from time-translation invariance. Noether's theorem is concerned with continuous symmetries, while the "pseudo" in "pseudovector" has to do with rotation-related vectors (angular momentum, torques, angular velocities) being perfect vectors with respect to rotations (also called SO(3), but behaving differently to normal vectors (position, velocity, etc.) with respect to so-called improper rotations; that are the product of rotations by inversions of coordinates.

Normal vectors (SO(3) vectors; polar vectors being another name) change the sign of their components when we invert the sign of the coordinates. Pseudo-vectors do not, because they are really "external products of vectors".

I hope that helps, but some parts of this thread sound like dadaist poetry to me. And I don't mean either Swansont of exchemist's contributions.

I don't mean to be facetious. I really would like to understand what point is being made to be of any help at all.

 

Edited by joigus
minor correction
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Quote

Power?

Power is energy divided by time.

P=E/t

 

Energy is:

E=Q*U

Q is integer multiply of e (elementary charge)

i.e., the kinetic energy of a single electron:

E=e*U

Therefor unit of energy in quantum physics is eV (electronvolt)

https://en.wikipedia.org/wiki/Electronvolt

 

Power is current multiplied by voltage.

P=I*U

(Q=I*t)

The kinetic energy of the particles can be summed up (there are billions of them). How do we know what their energy is? Because they dissipate their kinetic energy as they flow through the medium, heating it, colliding with particles in the medium. The temperature rise can be read with a thermometer. The XIX century unit of energy was the calorie,  the amount of energy needed to heat 1 gram of water by 1 C.

https://en.wikipedia.org/wiki/Voltage_drop

(This is why copper wires are so desirable, as they have very low resistance, resulting in a small voltage drop when current flows)

(This is why quantum physicists want to create superconductors and dream of superconductors at ambient temperatures)

 

Edited by Sensei
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I think it’s also important to mention here that we don’t actually live in a Newtonian universe, except as an approximation in the low-energy, low-velocity regime. What this means is that Noether’s theorem actually applies to an action within at least a 4D Minkowski spacetime (ignoring gravity for now) - in which case the conserved quantity associated with time-translation invariance is not just ‘energy’, but the full stress-energy-momentum tensor. This should be obvious, since ‘energy’ on its own depends on the observer, and is thus conserved only within a given frame.

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