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"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?

I can't get the actual concept or meaning of various forms of energies . Actually what is energy ? My textbook defines energy as capacity to do work.Most of the websites define energy in the same way. But i can't understand any thing from that definition. I also want to know what is work.Actually, why do we need these quantities?Thanks in advance.[MOD EDIT - PLEASE PHRASE YOUR POST TITLES AS QUESTIONS. THANKS. CHRIS]

It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and we add it all together it gives “28” - always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

It's funny how someone can post a logical irrefutable fact and it means nothing to the discussion underway.

Recoil reduces energy of a photon. Your laser gun has no recoil. Therefore energy of your photons is not reduced.

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.

and matter is composed of radiation, E = mC^2.

Then the equation must be wrong? I agree that photons are massless, the equation only shows the energy required to produce that photon.

Quote from: simplified on 03/10/2010 15:19:26Recoil reduces energy of a photon. Your laser gun has no recoil. Therefore energy of your photons is not reduced.Ok, but what this has to do with what pushes away a photon from light source?

Let's consider analogy to a lightning. A cloud beats ground by negative charge, in the answer the ground beats the cloud by positive charge. When a particle and an antiparticle meet, they beat radiator by gravitational charge. In the answer the radiator beats them by charge of energy.

Quote from: simplified on 09/10/2010 07:24:11Let's consider analogy to a lightning. A cloud beats ground by negative charge, in the answer the ground beats the cloud by positive charge. When a particle and an antiparticle meet, they beat radiator by gravitational charge. In the answer the radiator beats them by charge of energy.hmmm...maybe you should change your nickname from "simplified" to "complicated" []

Quote from: lightarrow on 09/10/2010 20:18:23Quote from: simplified on 09/10/2010 07:24:11Let's consider analogy to a lightning. A cloud beats ground by negative charge, in the answer the ground beats the cloud by positive charge. When a particle and an antiparticle meet, they beat radiator by gravitational charge. In the answer the radiator beats them by charge of energy.hmmm...maybe you should change your nickname from "simplified" to "complicated" []Then maybe you a ruddy critic with superficial judgement []

And so if a mass is a neutralized energy then energy is an activated mass []Photon has 100% of activated mass and 0% of usual mass.Usual object has usual mass and activated mass = (Lorentz's coefficient-1)* usual mass .Usual mass isn't relative thing. Activated mass is just relative thing.I only can not understand that why photons take and mass of Sun and energy of Sun? They must to take only mass,because mass turned into their energy.

Quote from: simplified on 17/10/2010 07:21:13And so if a mass is a neutralized energy then energy is an activated mass []Photon has 100% of activated mass and 0% of usual mass.Usual object has usual mass and activated mass = (Lorentz's coefficient-1)* usual mass .Usual mass isn't relative thing. Activated mass is just relative thing.I only can not understand that why photons take and mass of Sun and energy of Sun? They must to take only mass,because mass turned into their energy.I'm sorry to tell you, but if you talk of "neutralized energy" or "activated mass", you are not talking of physics.

A vertical force can never cause a horizontal displacement; thus, a vertical force does not do work on a horizontally displaced object!!

Quote from: yor_on A vertical force can never cause a horizontal displacement; thus, a vertical force does not do work on a horizontally displaced object!!Imagine a situation in which you are standing under an object which is on rails, such that it will move only horizontally. The underside of this object slopes. You use a pole to apply vertical force against that slope, and the object responds by moving horizontally. Is this not a vertical force causing horizontal displacement?

If your physics don't answer questions then I say.

Thanks Geezer. I tried thinking of a situation in which the poll is constrained to move only vertically, but I guess whatever was constraining it would be converting vertical to horizontal force; right?What about orbital motion, though. E.g. does the Earth's gravitational attraction not act perpendicular to the direction of motion of the moon?

The gravitational force between the Moon and the Earth just maintains the Moon in an orbit around the Earth.

Quote from: Ron Hughes on 08/10/2010 17:29:09Then the equation must be wrong? I agree that photons are massless, the equation only shows the energy required to produce that photon.That equation is wrong for photons. The correct one, valid for all particles and regardless if stationary or moving (in SR) is this:E^{2} = (mc^{2})^{2} + (cp)^{2}p = momentum.For light, in classical EM, E = cp. If you apply it to photons (certainly you will agree on the fact this is allowed) you find:(cp)^{2} = (mc^{2})^{2} + (cp)^{2}=> 0 = (mc^{2})^{2}so m = 0.

E=m'c² p=(m+m')* vE - energy for any objectp - momentum for any objectm - usual mass,this mass has gravitation fieldm'- activated mass(mass turned into energy),this mass has not gravitation fieldc - light speedv - speed of object

Quote from: simplified on 20/10/2010 03:51:59 E=m'c² p=(m+m')* vE - energy for any objectp - momentum for any objectm - usual mass,this mass has gravitation fieldm'- activated mass(mass turned into energy),this mass has not gravitation fieldc - light speedv - speed of objectnonsense.

Your definition is a law for dark people.

Can you disprove the formulae?

Sorry - but is this forum now using "for dark people" as a metaphor for bad science and ignorance. FFS I thought there were rules about racism. I have reported it to the moderators and I hope they do something about it. Matthew

"What we've got here is failure to communicate"...sorry, you cannot view external links. To see them, please REGISTER or LOGINPerhaps, due to a translation problem, Simplified meant "people in the dark"

Quote from: Geezer on 22/10/2010 05:58:04"What we've got here is failure to communicate"...sorry, you cannot view external links. To see them, please REGISTER or LOGINPerhaps, due to a translation problem, Simplified meant "people in the dark"People without light ideas.

Energy is communication of object with the past

Quote from: lightarrow on 08/10/2010 19:20:39Quote from: Ron Hughes on 08/10/2010 17:29:09Then the equation must be wrong? I agree that photons are massless, the equation only shows the energy required to produce that photon.That equation is wrong for photons. The correct one, valid for all particles and regardless if stationary or moving (in SR) is this:E^{2} = (mc^{2})^{2} + (cp)^{2}p = momentum.For light, in classical EM, E = cp. If you apply it to photons (certainly you will agree on the fact this is allowed) you find:(cp)^{2} = (mc^{2})^{2} + (cp)^{2}=> 0 = (mc^{2})^{2}so m = 0.Well. But which units I should use in this formula?

Object travels with speed of 1 m/s.Mass of the object=1kgI want to calculate energy of this object. E ≈ mv²/2 = 0,5 J

Quote from: simplified on 31/10/2010 05:14:04Object travels with speed of 1 m/s.Mass of the object=1kgI want to calculate energy of this object. E ≈ mv²/2 = 0,5 JWrong. What you have written is *not* the energy E. With the term "energy" in relativity and in particle physics, we intend "total energy". What you have written is *just* kinetic energy. E = kinetic energy + mc^{2} (in absence of a potential or other kinds of energy).(Let's say is a language problem, as Geezer said... [])