# The Naked Scientists Forum

### Author Topic: What are "energy" and "work" ?  (Read 28296 times)

#### lightarrow

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##### What are "energy" and "work" ?
« Reply #25 on: 01/10/2010 16:19:07 »
Energy is ability to overcome resistance. Work is a victory of quantity of resistance.
Define "ability", define "overcome" and define "resistance".

#### lightarrow

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##### What are "energy" and "work" ?
« Reply #26 on: 01/10/2010 16:26:27 »
Energy is the capacity of a system to do work.
This kind of definition is often found in books, but it doesn't say anything; with that "definition" everything has the capacity of doing work and the in-capacity to do it.

Take two mass point: one stationary and the other moving at velocity v. Which has the capacity of doing work? The first? The other has not this capacity? Ok, the two mass point are electrically charged with charge +Q, and the stationary one is fixed spatially in his position. Now which of the two "has the capacity of doing work" on the other?

Furthermore, what can do work is not energy, and not even a system which has energy: what can do work are *forces* (or fields).
« Last Edit: 01/10/2010 16:28:16 by lightarrow »

#### simplified

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##### What are "energy" and "work" ?
« Reply #27 on: 02/10/2010 12:11:00 »
I don't understand energy.What pushes away a photon from light source? Repellent fields or blast wave?

#### yor_on

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##### What are "energy" and "work" ?
« Reply #28 on: 02/10/2010 16:17:09 »
I think that depends on how you choose to define 'forces' myself. If you expect there to be forces as f.ex that 'blast wave' then, under the assumption that a photon is a propagating entity, there will have to be some sort of explanation to why it 'moves'. If you on the other side consider 'forces' to be something that we use, lacking a better/simpler description for the phenomena, and perhaps also wonder about how those 'photons' can 'move' without acceleration, as well as how we ever are going to prove them to be a 'source' more than in a indirect way.. Then you've got me? I don't know either.. The idea of them not 'moving' makes it even weirder :)

#### simplified

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##### What are "energy" and "work" ?
« Reply #29 on: 02/10/2010 16:58:30 »
energy of blast = energy of recoil + energy of photon ?

#### lightarrow

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##### What are "energy" and "work" ?
« Reply #30 on: 02/10/2010 18:38:25 »
I don't understand energy.What pushes away a photon from light source?
Its speed  .  Why do you think it needs something to push it? Photons are massless...

#### yor_on

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##### What are "energy" and "work" ?
« Reply #31 on: 02/10/2010 20:47:02 »
Yeah, it seems to fall back to what mass is?
And why there is that difference, mass needs an acceleration, a mass-less particle is just assumed to 'move' by itself, as if it the idea of an action-reaction have no relevance to it?

And it fits with all experiments we have made up today too?
It's phreakinly weird that one :)

And we use the 'photon frames' invariance (as seen from all frames thought up) as a proof as the universe actually 'compress' itself to our own 'frame of reference'. And that's really impressive if it is correct as we then seem to 'contract' all energy there is, adapting it to our motion/relative mass/momentum.

That is, if you agree with me on it being a real occurrence, distances contracting, as well as time dilation existing for real.
==

And btw: For this one it doesn't matter if you look at it as a 'time dilation' only, or a 'length contraction. You are free to exchange those two I think? To look at it only as a time dilation only f.ex won't change that fact.

When observing the moving twin you can follow him traveling in your super telescope, at no time losing sight of him, but he will still be younger than his twin on Earth. And as you think of it, putting away your telescope, there is only one explanation available. Somehow his motion changed his time, making it slower.

And if you translate that into distance you will find that he must had a shorter journey than the one you thought yourself to measure, relative you. If you exchange that 'slower time', relative you, into his 'distance made', then that distance had to have been shorter for him. And if you ask him he will agree, from his frame the distance actually was shorter although his 'intrinsic frame of times arrow' ::)) never changed for him.

Very weird.
« Last Edit: 02/10/2010 21:10:24 by yor_on »

#### yor_on

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##### What are "energy" and "work" ?
« Reply #32 on: 02/10/2010 21:26:39 »
But if you look at as a length contraction only?
Why would he need to have aged slower?

That falls back to how we measure a distance, we do it using 'times arrow'.
If you want to assume that time is invariant, never changing, and at the same time introduce only a length contraction you are contradicting yourself. The only way we can define a 'distance' is using a clock and, at least, two 'frames of reference'. Your own relative what you measure. Without using that clock distance won't exist, much in the same way as it doesn't seem to exist for our 'photon'.

Or, can you see any other way to define 'distance'?

#### yor_on

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##### What are "energy" and "work" ?
« Reply #33 on: 03/10/2010 00:41:55 »
Although you might be able to argue that time is invariant, if seen from your own frame of reference. Meaning that the twin traveling never experienced any 'slowing' of his time, making his clock the 'universal one'. Against that we have the twin staying at home who then, with as much right, could argue that it was his time that was the 'invariant and universal.'. But they don't fit together, do they?

That's why we use 'frames of reference'.

But it still makes a very strange truth in that your clock, for you, will be the universal one, no matter where you are. Just as mine will make mine 'universal invariant time'. And, that only when being together, as defined by being 'at rest' relative each other, will we ever share that 'universal time', and then only if we also happen to have the exact same mass, it seems to me? And as I said before, as you can change your point of view, observing your own atoms instead, and then define them different 'frames of reference' according to their mass and motion relative each other, how the he* do I define a 'frame of reference'? That one gives me a headache.
===

Maybe I'm looking at it from the wrong point of view?

Maybe there are no 'frames of reference'? We think there are but we use clocks to define them right? And those clocks all ticks differently depending on our 'frame of reference'. And that goes for distances too. So? Is there something wrong in our conception of time? And is there something wrong in our conception of 'distance'. Use the 'forbidden' frame of the photon, and ask yourself what that frame sees? We do use it as we refer to its timelesness as the explanation to why it can keep its 'intrinsic energy' no matter how far it travels. Or do you have a better explanation for that? Length contraction perhaps :) And what would that do SpaceTimes 'geometry'?

Without a clock you can't have a distance.
And that one I'm fairly sure on.
« Last Edit: 03/10/2010 01:22:02 by yor_on »

#### Ron Hughes

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##### What are "energy" and "work" ?
« Reply #34 on: 03/10/2010 04:39:34 »
yor, when was the last time that someone shot a beam of light from here then jumped a billion light years away from Earth and waited to check the validity of the statement that light does not lose energy? I don't know if it does or not. It seems to me that if light lost energy at a rate of say 10^-20Hz/light yr we would never know it. If there is proof that would sway me I would be interested.

#### simplified

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##### What are "energy" and "work" ?
« Reply #35 on: 03/10/2010 06:17:02 »
I don't understand energy.What pushes away a photon from light source?
Its speed  .  Why do you think it needs something to push it? Photons are massless...
Effect of recoil reduces energy of a photon. Therefore.

#### yor_on

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##### What are "energy" and "work" ?
« Reply #36 on: 03/10/2010 07:13:22 »
Yeah Ron, I've started to wonder anew :)

Considering that if a photon would be proofed to have even the slightest of masses it would have to be inside 'fermion time' so to speak.. I've reopened Ethos thread for those questions, and I would be pleased if we started with why mainstream physics consider it to be intrinsically time less, and then take it from there..

I know why I consider it to be so, but I'm not sure of how much of that is just my own conjecture and how much is actual proofs, mathematical or not. And it sure have a relevance.

#### lightarrow

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##### What are "energy" and "work" ?
« Reply #37 on: 03/10/2010 12:00:18 »
I don't understand energy.What pushes away a photon from light source?
Its speed  .  Why do you think it needs something to push it? Photons are massless...
Effect of recoil reduces energy of a photon. Therefore.
And if my laser gun doesn't recoil at all, since its mass >> photon's momentum/c ?  The photon shouldn't be shoot away? I don't understand your reasoning.

#### simplified

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##### What are "energy" and "work" ?
« Reply #38 on: 03/10/2010 15:19:26 »
I don't understand energy.What pushes away a photon from light source?
Its speed  .  Why do you think it needs something to push it? Photons are massless...
Effect of recoil reduces energy of a photon. Therefore.
And if my laser gun doesn't recoil at all, since its mass >> photon's momentum/c ?  The photon shouldn't be shoot away? I don't understand your reasoning.
Recoil reduces energy of a photon. Your laser gun has no recoil. Therefore energy of your photons is not reduced.

#### Ron Hughes

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##### What are "energy" and "work" ?
« Reply #39 on: 03/10/2010 17:58:36 »
yor, take a photon with E = fh and calculate it's mass from m = fh/C^2. The equation is derived from  fh = E = mC^2. As you can see we could claim mathematically that all photons have mass.

#### yor_on

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##### What are "energy" and "work" ?
« Reply #40 on: 03/10/2010 19:34:03 »

I'm already 'arguing' as good as I know there, which doesn't say much ::))
And I'm sure you will get better responses to your equation there as I hope the 'heavy artillery' will roll out to define what that elusive light does and does not :)

Heh.
And ahem :)

#### yor_on

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##### What are "energy" and "work" ?
« Reply #41 on: 03/10/2010 19:39:44 »
Sim have you seen this one?
Measuring the Recoil of Photons. That one seems to say that there is a measurable recoil?

But if there is no acceleration? You could use it as a proof of that a photon have a 'source' that it 'propagates' from, if correct too, it seems? But how does something without a intrinsic 'time frame' leave a recoil? That one I'm not sure how to see at all.
==

And yeah, Ron, I agree, so much for discussing 'photons' at the 'correct' thread.
Ah well :)
« Last Edit: 03/10/2010 19:42:50 by yor_on »

#### syhprum

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##### What are "energy" and "work" ?
« Reply #42 on: 03/10/2010 20:12:32 »
I always understood that a photon gun radiating 300MW had a recoil of 1 Newton, hence the photon drive spaceships beloved by sci fi authours.

#### yor_on

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##### What are "energy" and "work" ?
« Reply #43 on: 03/10/2010 20:19:27 »
That is when 'bouncing' isn't it?
I never thought of it in any other way when reading about Enterprises warpdrive?
Hmm shouldn't the correct sentence be?

"Beam me up Scotty, but hey, watch out for that dam*d recoil please.."

That way it's acceptable for me, but if we consider one photon, just 'materializing'?

#### JP

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##### What are "energy" and "work" ?
« Reply #44 on: 04/10/2010 02:44:15 »
But if there is no acceleration? You could use it as a proof of that a photon have a 'source' that it 'propagates' from, if correct too, it seems? But how does something without a intrinsic 'time frame' leave a recoil? That one I'm not sure how to see at all.

I think the problem is that you're trying to apply classical mental pictures to non-classical processes.  With photons you're dealing with particles (quantum mechanics) that are moving fast (relativity), so classical intuition might not hold.  In this case, it doesn't because the photon has no rest frame (relativity), plus the photon at some point is created by some process (quantum mechanics).  This all together means you don't have to accelerate the photon because it moves at the speed of light always, including when it gets created.  There is no acceleration.

Even quantum mechanics has to obey some familiar rules, though.  You can use conservation of momentum to determine recoil if you know the momentum of the photon.

#### yor_on

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##### What are "energy" and "work" ?
« Reply #45 on: 04/10/2010 15:03:40 »
So the momentum will be 'ring/sphere' formed at its 'materialization' then?
Or how it is thought to work, I always thought of momentum as a 'force' well, at least pointing in the direction of the photons velocity, but it seems hard to state that the recoil lies before the photon?

Maybe that's more correct way to look at it thinking of it, as something evenly distributed? But to talk about a recoil from something just 'coming into existence' should then also mean that what we call that recoil take place everywhere at that photon it seems. The easiest way i can imagine a recoil in this way is if there would be some sort of 'sticky elasticity' involved in its materialization, like boundaries letting go. Does that make sense?
« Last Edit: 04/10/2010 22:00:53 by yor_on »

#### Ron Hughes

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##### What are "energy" and "work" ?
« Reply #46 on: 04/10/2010 20:39:56 »
If the equation above is correct ( and we know it is ) then what ever emits the photon must have a recoil.

#### JP

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##### What are "energy" and "work" ?
« Reply #47 on: 05/10/2010 03:32:10 »
So the momentum will be 'ring/sphere' formed at its 'materialization' then?
Or how it is thought to work, I always thought of momentum as a 'force' well, at least pointing in the direction of the photons velocity, but it seems hard to state that the recoil lies before the photon?

Momentum is a vector, so its like an arrow pointing in the direction of the photon's motion.  Momentum of the photon source is a vector pointing in the direction of its motion.  If the source was initially stationary, its initial arrow was zero length.   Since momentum is conserved, after emitting a photon the arrow of the photon plus the arrow of the source have to add up to zero length (meaning that if you align them tip-to-tail and follow them, you end up where you started after traversing both arrows).

In Newtonian mechanics, change in momentum means that something changed its velocity.  This means acceleration, which by Newton's second law means there was a force.  However, the creation of photons isn't Newtonian mechanics, so the photon doesn't have to experience a force or accelerate.

#### yor_on

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##### What are "energy" and "work" ?
« Reply #48 on: 06/10/2010 18:46:07 »
"If the source was initially stationary, its initial arrow was zero length. Since momentum is conserved, after emitting a photon the arrow of the photon plus the arrow of the source have to add up to zero length (meaning that if you align them tip-to-tail and follow them, you end up where you started after traversing both arrows)."

Makes perfect sense, except for one thing. A photon can't be seen as 'stationary', can it?
Or do we allow it a 'instant' inside our arrow, where it is 'stationary' before it starts to move, without accelerating?

#### Ron Hughes

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##### What are "energy" and "work" ?
« Reply #49 on: 06/10/2010 22:38:20 »
A photon or wave ( http://www.thenakedscientists.com/forum/index.php?topic=34333.0 ) can never be stationary, anytime an electron or proton is moved the change in it's field position will propagate away at C. A photon/wave is only created by the movement of charged particles. If there is another way please tell me.

#### The Naked Scientists Forum

##### What are "energy" and "work" ?
« Reply #49 on: 06/10/2010 22:38:20 »