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what is energy?


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I think it is the potential of a body to do work.

 

Types include Kinetic energy, potential energy, heat energy (which is sort of kinetic energy at the molecular level)..

 

A wave is not energy but will have energy based on what it is and how fast it is propagating.

 

Other thoughts are that it is always conserved... if you end up with less energy in a system after an event then it has been lost somewhere, even if you can't account for it, it went somewhere in the form of heat, sound, direction change etc..

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I think it is the potential of a body to do work.

 

Types include Kinetic energy, potential energy, heat energy (which is sort of kinetic energy at the molecular level)..

 

A wave is not energy but will have energy based on what it is and how fast it is propagating.

 

Other thoughts are that it is always conserved... if you end up with less energy in a system after an event then it has been lost somewhere, even if you can't account for it, it went somewhere in the form of heat, sound, direction change etc..

Energy must be confined to convert it into a force and for that; The matter is obviously necessary.

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!

Moderator Note

Roger do not answer mainstream questions and threads with your personal misconceptions.

If you don't know the scientific accepted answer do not not respond.



Energy is best left defined as "the ability to perform work" it is a property Edited by Mordred
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Is it merely a wave resonating in a vacuum at it's simplest form?

 

 

 

No, just don't go there.

 

Having got that out of the way your other two questions are entirely reasonable, but very difficult to pin down.

 

 

How do you define exactly what energy is?

 

 

 

As already noted by others, energy is a property, but you might reasonably ask a property of what?

 

 

Well energy is a property possessed by a system by virtue of its configuration.

Generally there is an energy change associated with a change to that configuration.

 

We observe this in many ways

 

A ball rolls off a table and falls to the floor

A kettle of water boils to produce steam

A fast radiation particle precipitates a droplet trail in a cloud chamber

Sodium carbonate and calcium chloride solutions precipitate calcium carbonate solid particles when they are mixed.

The thickness of the brake pads in your car (I hope they are finer than 36 grit) decreases with use.

 

It is also said to be the capacity of a system to perform work.

 

This is the definition where energy is usually first introduced in elementary Physics.

 

But then we go on to talk of available and unavailable energy (Clausius and Maxwell) in more advanced work and then zero point energy (Plank) in even more advanced work.

 

This shows that not all energy can be used to perform work, even in theory.

 

So where are we?

 

Well the answer raises more questions than it answers.

 

And we have yet to define a system, or configuration.

 

 

 

 

How many types of energy is there that we know of?

 

 

I can't say the exact number, partly because there are so many and partly because there is some overlap in usage.

 

Expressions for energy appear in pretty well all branches of physical science, each one tailored for use in its particular area.

 

A great deal of work over several centuries has taken place to make sure that they are all consistent today.

Edited by studiot
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Seems to me that, it depends on which side of the looking glass you're standing on. If energy is a property of mass and mass is made out of energy, Then isn't mass a property of energy?

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Seems to me that, it depends on which side of the looking glass you're standing on. If energy is a property of mass and mass is made out of energy, Then isn't mass a property of energy?

 

 

You are confusing mass and matter; they are not the same thing. Mass is one form of energy. Both are properties of matter.

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Yes but Einstein gave us the "general" formula

of rest mass energy, as

E=m*(c^2).

Is it reversable?

Can we say any energy form has a rest mass?

 

(Matter is a wave, with (repelling) particle properties.)

(Is not matter simply a property also?)

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Can we say any energy form has a rest mass?

 

 

No, we can't.

 

The full equation is E2 = m2c4 + p2c2

 

That reduces to the familiar form for an object at rest. But center-of-mass motion is accounted for separately, i.e. translational KE is not included in rest mass (hence the name). A photon's energy is entirely due to the momentum term. Again, no rest mass energy.

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No, we can't.

 

The full equation is E2 = m2c4 + p2c2

 

That reduces to the familiar form for an object at rest. But center-of-mass motion is accounted for separately, i.e. translational KE is not included in rest mass (hence the name). A photon's energy is entirely due to the momentum term. Again, no rest mass energy.

So !~ the mass start increasing at xo,yo ?

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Can we say any energy form has a rest mass?

No, we can't.

 

The full equation is E2 = m2c4 + p2c2

 

Ok, so does that mean Einstein's (E=m*(c^2)) formula is wrong (as an equality, equation)?

& should be written so

E=~m*(c^2),

as an approximation instead, or

E~m*(c^2)?

 

That reduces to the familiar form for an object at rest.

But center-of-mass motion is accounted for separately,

i.e. translational KE is not included in rest mass (hence the name).

Yes, but if we start with the rest mass speed c,

& remove (=subtract) the translational KE speed v,

do we not decrease the rest mass's energy?

(A mass in motion can surely not be declared completely at rest.)

Let v'=c-v.

A photon's energy is entirely due to the momentum term.

Ok, but isn't momentum

mom=m*v

"mass" m multiplied by speed v?

Isn't there a (rest) mass term? (to assume?).

 

(You did say momentum, instead of impulse, or momentum impulse, or impulse momentum.)

Edited by Capiert
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Can we say any energy form has a rest mass?

Ok, so does that mean Einstein's (E=m*(c^2)) formula is wrong (as an equality, equation)?

& should be written so

E=~m*(c^2),

as an approximation instead, or

E~m*(c^2)?

No, it's not wrong. It was derived for a particle at rest, and that's how it must be applied.

 

 

Yes, but if we start with the rest mass speed c,

& remove (=subtract) the translational KE speed v,

do we not decrease the rest mass?

(A mass in motion can surely not be declared completely at rest.)

No. The translational kinetic energy is accounted for separately, as I said. Rest mass energy is completely separate.

 

Being able to say a mass is at rest in its own frame is one of the founding ideas of relativity.

 

Ok, but isn't momentum

mom=m*v

"mass" m multiplied by speed v?

Isn't there a (rest) mass term? (to assume?).

 

(You did say momentum, instead of impulse, or momentum impulse, or impulse momentum.)

For non-relativistic objects you use mv. For massless objects (which are necessarily relativistic), it's E/c

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No, it's not wrong. It was derived for a particle at rest, and that's how it must be applied.No. The translational kinetic energy is accounted for separately, as I said. Rest mass energy is completely separate.Being able to say a mass is at rest in its own frame is one of the founding ideas of relativity.For non-relativistic objects you use mv.

Why don't we use mv universally (instead), for non_relativistic objects too?

(E.g. You physicists have a dark big hole, in your understanding

about dark energy & dark matter, e.g. a mass defect.

Wouldn't it be easier to drop the (energy as) problem?

E.g. Does momentum have the same dark (=unknown) mass?

For massless objects (which are necessarily relativistic), it's E/c

How do we know light is massless if it has energy E? Edited by Capiert
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Why don't we use mv universally (instead), for non_relativistic objects too?

Because mv is not correct for relativistic objects.

 

How do we know light is massless if it has energy E?

Theory predicts it and measurement confirms it.

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Because mv is not correct for relativistic objects.

Uh? Please explain.

 

#12

No, we can't.

 

The full equation is E2 = m2c4 + p2c2

 

A photon's energy is entirely due to the momentum term.

You just stated before (#12 formula with rho), &

#16

For non-relativistic objects you use mv. For massless objects (which are necessarily relativistic), it's E/c

mv (=momentum)

is used for photons;

which are light speed particles (if I may say).

Isn't moving at the speed of light also

at least relativistic?

 

If we use momentum

to derive energy

for the (relativistic speed) photon;

but momentum is not correct (for relativistic (speed) particles)

then our energy must also be NOT correct.

 

(Why not then throw relativity & energy away?)

 

Theory predicts it and measurement confirms it.

Please explain. Edited by Capiert
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As you've already noticed, mv doesn't work for massless objects Therefore it can't be used universally. Further, for massive objects traveling at speeds close to c, the equation is (gamma)mv, where gamma is the well-known relativistic correction term used for time dilation and length contraction.

If photons had mass they wouldn't travel at c, and that has other implications for how they would behave. Scientists have checked on this, and confirmed that within experimental bounds, the mass is zero.

https://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass



You just stated before (#12 formula with rho)

 

That's not a rho, it's a p (Roman/Latin alphabet, not Greek). Momentum.


mv (=momentum)
is used for photons;
which are light speed particles (if I may say).
Isn't moving at the speed of light also
at least relativistic?

 

mv is NOT used for photons.

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Thanks for the excellent summary.

 

Thank you also for pointing out my type error p,

(I would have continued so without knowing

because I use it so rarely).

Edited by Capiert
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E = hv = hc/lambda (also pc, obviously)

Momentumwise a wave('s amplitude)

goes back & forth (up & down if you will)

like bumping (bouncing) around.

 

How do we know

we are not dealing with

light's (e.g. a photon's) momentum is

mom=k*f

where k is some tiny amount

of a universal momentum?

 

In other words

the frequency f

is a direct relation,

& can be manipulated (=changed)

with (extra) bumps (=bumping, bounces).

Edited by Capiert
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Momentumwise a wave('s amplitude)

goes back & forth (up & down if you will)

like bumping (bouncing) around.

 

How do we know

we are not dealing with

mom=k*f

where k is some tiny amount

of a universal momentum?

 

In other words

the frequency f

is a direct relation,

& can be manipulated (=changed)

with (extra) bumps (=bumping, bounces).

 

 

If k has units of momentum, then kf will not be. It has the wrong units.

 

If you want to relate photon momentum to frequency, E = pc = hv, so one can write p = hv/c

 

There's no need to make up new equations out of whole cloth. Existing physics works quite well.

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