What are "energy" and "work" ?

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Offline JP

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What are "energy" and "work" ?
« Reply #50 on: 07/10/2010 10:27:24 »
"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?

You're perfectly right.  A photon's momentum is never zero, so its momentum can always be drawn as an arrow of non-zero length.  What's key here is that the photon didn't exist initially, so the total momentum was zero initially.  When the photon comes out of the source, in order for the total momentum to stay the same, the source has to recoil in the opposite direction of the photon. 

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Offline Pmb

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What are "energy" and "work" ?
« Reply #51 on: 07/10/2010 11:12:19 »
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]
We've discussed this subject in the recent past so you might want to look up that debate. I created a web page on my website at

http://home.comcast.net/~peter.m.brown/mech/what_is_energy.htm


to support the answer that nobody knows what energy is. There is an e-book online which discusses it too. See "What is Energy?, by Dave Broyles. Basically it' what Richard Feynman said (See my web page for reference for this quote and Feynman's reasoning for it)
Quote
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.
In the former discussion I presented some arguments which I believe render the "Energy is the ability to do work" meaningless.

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Offline Ron Hughes

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« Reply #52 on: 07/10/2010 15:39:36 »
It's funny how someone can post a logical irrefutable fact and it means nothing to the discussion underway.
From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other. Sherlock Holmes.

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Offline Pmb

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What are "energy" and "work" ?
« Reply #53 on: 07/10/2010 22:02:19 »
It's funny how someone can post a logical irrefutable fact and it means nothing to the discussion underway.
I posted a response to the comment what is energy?What are you talking about?

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Offline Ron Hughes

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« Reply #54 on: 07/10/2010 23:43:41 »
Pmb, that wasn't pointed at you. Energies lowest common denominator is radiation.Any other explanation requires the use of matter's momentum which is used every day to do work and matter is composed of radiation, E = mC^2.
« Last Edit: 07/10/2010 23:45:17 by Ron Hughes »
From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other. Sherlock Holmes.

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Offline lightarrow

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« Reply #55 on: 08/10/2010 12:19:14 »
Recoil 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?

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Offline lightarrow

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« Reply #56 on: 08/10/2010 12:20:49 »
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.
Photons are massless, and that's all, no possibility of opinions here.

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Offline lightarrow

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« Reply #57 on: 08/10/2010 12:26:23 »
and matter is composed of radiation, E = mC^2.
Wrong. That's only your idea.

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Offline Ron Hughes

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« Reply #58 on: 08/10/2010 17:29:09 »
Then the equation must be wrong? I agree that photons are massless, the equation only shows the energy required to produce that photon.
« Last Edit: 08/10/2010 17:32:23 by Ron Hughes »
From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other. Sherlock Holmes.

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Offline lightarrow

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« Reply #59 on: 08/10/2010 19:20:39 »
Then 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:

E2 = (mc2)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 = (mc2)2 + (cp)2

=> 0 = (mc2)2

so m = 0.
« Last Edit: 08/10/2010 19:25:53 by lightarrow »

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Offline simplified

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« Reply #60 on: 09/10/2010 06:19:02 »
Recoil 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?
Radiator loses thermal energy because  photon steals it.

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Offline simplified

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« Reply #61 on: 09/10/2010 07:24:11 »
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.

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Offline lightarrow

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« Reply #62 on: 09/10/2010 20:18:23 »
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.
hmmm...maybe you should change your nickname from "simplified" to "complicated"  [:)]

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Offline simplified

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« Reply #63 on: 10/10/2010 09:30:21 »
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.
hmmm...maybe you should change your nickname from "simplified" to "complicated"  [:)]
Then maybe you  a ruddy critic with superficial judgement [::)]

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Offline lightarrow

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« Reply #64 on: 10/10/2010 12:31:31 »
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.
hmmm...maybe you should change your nickname from "simplified" to "complicated"  [:)]
Then maybe you  a ruddy critic with superficial judgement [::)]
Ok, I accept your critic. But then can you please explain me what is that you intended? I sincerely have not understood it ( it can be me, of course).

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Offline simplified

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« Reply #65 on: 17/10/2010 07:21:13 »
And so if a mass is a neutralized energy then energy is an activated mass  [:P]
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.

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Offline lightarrow

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« Reply #66 on: 17/10/2010 13:42:32 »
And so if a mass is a neutralized energy then energy is an activated mass  [:P]
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.

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Offline simplified

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« Reply #67 on: 17/10/2010 15:10:38 »
And so if a mass is a neutralized energy then energy is an activated mass  [:P]
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.
And so if a mass is a neutralized energy then energy is an activated mass  [:P]
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.
If your physics don't answer questions then I say.

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Offline Bill S

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« Reply #68 on: 17/10/2010 18:59:50 »
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?

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Offline Geezer

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« Reply #69 on: 17/10/2010 21:11:17 »
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?

Only if you prevent from rotating where you are holding it, but then you are producing a horizontal force. If you allow the pole to pivot freely where you are holding it, it will just tilt over as it goes up the incline and no force will be imparted to the vehicle at all.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline lightarrow

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« Reply #70 on: 18/10/2010 18:20:24 »
If your physics don't answer questions then I say.
Ok, how many angels can stay on a pin's head?
If you can't answer then I say 12.
Is this science?
If you use *scientific*, *precisely defined* terms, you will have answers. If you don't use, you can't have, but not because I don't have answers or because I don't want to answer, but because your question is completely  *undefined*; I don't have the least idea of what you want to ask or if it has a real meaning. Is it more clear now?
« Last Edit: 18/10/2010 18:25:45 by lightarrow »

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Offline Bill S

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« Reply #71 on: 18/10/2010 23:48:19 »
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?   

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Offline Geezer

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« Reply #72 on: 19/10/2010 01:56:22 »
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?  

Well, sort of. You are transferring the horizontal force at the top of the pole to a horizontal force at your wrist.

The Earth does not make the Moon move relative to the Earth. The gravitational force between the Moon and the Earth just maintains the Moon in an orbit around the Earth. If that force suddenly ceased, the Moon would keep on moving in a straight(ish) line.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Bill S

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« Reply #73 on: 19/10/2010 20:49:17 »
Quote from: Geezer
The gravitational force between the Moon and the Earth just maintains the Moon in an orbit around the Earth.

Here we seem to be treating gravity as a force, in spite of G R. This raises the question: If a force is holding the moon in orbit, is it doing any work?

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Offline Geezer

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« Reply #74 on: 19/10/2010 21:02:32 »
Yes, that's the Newtonian view. It still works quite well though!

In Newtonian mechanics there is a force, but as the force acts perpendicular to the motion, it's not doing any work.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline simplified

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« Reply #75 on: 20/10/2010 03:51:59 »
Then 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:

E2 = (mc2)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 = (mc2)2 + (cp)2

=> 0 = (mc2)2

so m = 0.
This is  pompous and complicated formula because you do not use correct scientific terms. [;D]
 
                 E=m'c²

                            p=(m+m')* v

E - energy for any object
p - momentum for any object
m - usual mass,this mass has gravitation field
m'- activated mass(mass turned into energy),this mass has not gravitation field
c - light speed
v - speed of object
                          [:P]

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Offline Geezer

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« Reply #76 on: 20/10/2010 04:15:55 »
WOAHH!!

Simplified,

Please try to be be careful with your translation (I know it's not easy) but "pompous" is probably not what you really meant to say. "Pompous" is a big insult in English.

It means "you are just saying something, but it means nothing."

There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline lightarrow

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« Reply #77 on: 20/10/2010 15:44:55 »
                 E=m'c²

                            p=(m+m')* v

E - energy for any object
p - momentum for any object
m - usual mass,this mass has gravitation field
m'- activated mass(mass turned into energy),this mass has not gravitation field
c - light speed
v - speed of object
nonsense.

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Offline simplified

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« Reply #78 on: 20/10/2010 17:30:32 »
                 E=m'c²

                            p=(m+m')* v

E - energy for any object
p - momentum for any object
m - usual mass,this mass has gravitation field
m'- activated mass(mass turned into energy),this mass has not gravitation field
c - light speed
v - speed of object
nonsense.
Your definition is a law for dark people.

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Offline peppercorn

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« Reply #79 on: 20/10/2010 17:52:59 »
Your definition is a law for dark people.
[???] [???] [???]

What? ...Are you saying this a law for people from the African sub-continent??? [:D]


[To clarify for anyone who might have misunderstood - my intention above is to gently ridicule the poster's English NOT make a racist comment.  Thanks!]
« Last Edit: 21/10/2010 15:57:58 by peppercorn »

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Offline simplified

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« Reply #80 on: 20/10/2010 18:06:36 »


What? ...Are you saying this a law for people from the African sub-continent??? [:D]
[/quote]Can you disprove the formulae?

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Offline peppercorn

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« Reply #81 on: 20/10/2010 18:19:41 »
Can you disprove the formulae?

I wasn't aware that the responsibility was on me to disprove it!

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Offline imatfaal

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« Reply #82 on: 21/10/2010 13:48:49 »
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
There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n

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Offline BenV

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« Reply #83 on: 21/10/2010 13:57:13 »
Simplified - I have sent you a private message.  Please respond to me as soon as you receive it.

In the meantime, this thread will be locked.

This thread is now unlocked to give Simplified the opportunity to clarify his meaning.
« Last Edit: 21/10/2010 16:45:53 by BenV »

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Offline simplified

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« Reply #84 on: 21/10/2010 18:57:19 »
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
I never thought of a literal origin of this expression.I am so sorry.

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Offline imatfaal

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« Reply #85 on: 21/10/2010 19:36:27 »
We all say things in heat of an argument that we might not have fully thought out - the fact that large portions of the world suffer because of prejudice means that I tend to be over-sensitive when I come across the language of bigotry in an arena such as this great forum.  Your last post reads true and with that apology I personally hope that the forum can go back to the cut-and-thrust of academic debate and avoid disharmony .  Regards  Matthew Newell - Imatfaal

ps come to think of it the forum probably has a rule against profanity - and if the derivation of FFS isnt profane then nothing is!
There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

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Offline Geezer

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« Reply #86 on: 22/10/2010 05:58:04 »
"What we've got here is failure to communicate"

http://www.youtube.com/watch?v=1fuDDqU6n4o



Perhaps, due to a translation problem, Simplified meant "people in the dark"
« Last Edit: 22/10/2010 06:00:40 by Geezer »
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline simplified

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« Reply #87 on: 22/10/2010 14:42:50 »
"What we've got here is failure to communicate"

http://www.youtube.com/watch?v=1fuDDqU6n4o



Perhaps, due to a translation problem, Simplified meant "people in the dark"
People without light ideas.

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Offline lightarrow

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« Reply #88 on: 22/10/2010 17:41:05 »
"What we've got here is failure to communicate"

http://www.youtube.com/watch?v=1fuDDqU6n4o



Perhaps, due to a translation problem, Simplified meant "people in the dark"
People without light ideas.
Then is appropriate for those people who write "smoky", "unclear", non-scientific things, with the aim to make other believe they are scientific.
« Last Edit: 23/10/2010 13:10:42 by lightarrow »

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Offline BenV

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« Reply #89 on: 22/10/2010 17:59:37 »
The off topic/misunderstanding posts are shrunk to allow easier reading of the actual science discussion - hopefully folks can pick it up from here...

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Offline simplified

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« Reply #90 on: 23/10/2010 04:28:14 »
If  activated mass does not exist, then communication of object with the past exists. It slows down time of object. [8D]

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Offline simplified

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« Reply #91 on: 23/10/2010 04:34:34 »
Energy is communication of object with the past .

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Offline Bill S

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« Reply #92 on: 27/10/2010 18:15:40 »
Quote from: simplified
Energy is communication of object with the past

Please explain, or change your name from "simplified", it doesn't suit you. [:P

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Offline simplified

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« Reply #93 on: 30/10/2010 17:26:55 »
Then 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:

E2 = (mc2)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 = (mc2)2 + (cp)2

=> 0 = (mc2)2

so m = 0.
Well. But which units I should use in this formula?

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Offline simplified

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« Reply #94 on: 30/10/2010 17:38:45 »
Units of measurement

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Offline lightarrow

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« Reply #95 on: 30/10/2010 22:56:31 »
Then 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:

E2 = (mc2)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 = (mc2)2 + (cp)2

=> 0 = (mc2)2

so m = 0.
Well. But which units I should use in this formula?
It's a strange question. The measurement's units are:
kg for mass,
Joule for energy,
m*s-1 for c 
kg*m*s-1 for p.
Joule = N*m = kg*m2*s-2

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Offline simplified

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What are "energy" and "work" ?
« Reply #96 on: 31/10/2010 05:14:04 »
Object travels with speed of 1 m/s.Mass of the object=1kg
I want to calculate energy of this object.
   E ≈ mv²/2 = 0,5 J
Let's use your formula:
                        E² = (mc²)² + (cp)²

E²=(1*299792458²)²+(299792458*1*1)²=8,077608713*10³³

   E=√(8,077608713*10³³)=89875517873681764,5 J  [:o]

I don't understand where is my mistake? [:(]
 

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Offline Geezer

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What are "energy" and "work" ?
« Reply #97 on: 31/10/2010 05:52:15 »
I'm not sure how to explain this properly, but please try to remember that English may not be first language of some of our posters. Translation can be a tricky business, so please try to make sure there are no misunderstandings because of a translation problem.

Thanks!
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline lightarrow

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What are "energy" and "work" ?
« Reply #98 on: 31/10/2010 10:23:19 »
Object travels with speed of 1 m/s.Mass of the object=1kg
I want to calculate energy of this object.
   E ≈ mv²/2 = 0,5 J
Wrong. 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 + mc2 (in absence of a potential or other kinds of energy).
(Let's say is a language problem, as Geezer said... [:)])
« Last Edit: 31/10/2010 10:27:47 by lightarrow »

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Offline simplified

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What are "energy" and "work" ?
« Reply #99 on: 31/10/2010 15:25:10 »
Object travels with speed of 1 m/s.Mass of the object=1kg
I want to calculate energy of this object.
   E ≈ mv²/2 = 0,5 J
Wrong. 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 + mc2 (in absence of a potential or other kinds of energy).
(Let's say is a language problem, as Geezer said... [:)])
Thank you very much.