Kinetic energy and E=MC^2

  • #1
For a photon the energy is given by
$$E = h\nu$$
Does this includes both kinetic energy and energy due to mass?
If so then de Broglie gave the equation λ = h/p by equating hν with mc2 but since E = mc2 just gives the energy due to relativistic mass at that speed it would not include kinetic energy of photon and thus deBroglie equation may be wrong.
 

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  • #2
Drakkith
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Does this includes both kinetic energy and energy due to mass?
No, a photon has neither. Since photons are massless, that pretty much precludes them from having kinetic energy. Especially since photons with different wavelengths have different energies but travel at the same speed.
 
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  • #3
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For a photon the energy is given by
$$E = h\nu$$
Does this includes both kinetic energy and energy due to mass?
If so then de Broglie gave the equation λ = h/p by equating hν with mc2 but since E = mc2 just gives the energy due to relativistic mass at that speed it would not include kinetic energy of photon and thus deBroglie equation may be wrong.

Fact 1: Planck's constant times frequency is the total energy of a photon.

Fact 2: Zero mass times speed of light squared is the total rest energy of a photon.

From fact 2 we can see that a photon only has energy that is not rest energy.

Now we replace "not rest energy" with another term "kinetic energy", so we get: Photon only has energy that is kinetic energy.
 
  • #4
Drakkith
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Now we replace "not rest energy" with another term "kinetic energy", so we get: Photon only has energy that is kinetic energy.
That's not correct as far as I know. A photon does not have kinetic energy. At least, not kinetic energy like a ball or a bullet has. It certainly doesn't depend on velocity. A photon moving through a medium, such as glass or water, has the same energy as a photon of the same frequency moving through space, despite the former being slower than the latter.
 
  • #5
sophiecentaur
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Fact 1: Planck's constant times frequency is the total energy of a photon.

Fact 2: Zero mass times speed of light squared is the total rest energy of a photon.

From fact 2 we can see that a photon only has energy that is not rest energy.

Now we replace "not rest energy" with another term "kinetic energy", so we get: Photon only has energy that is kinetic energy.
I class that as a GIGO argument. It presupposes that the only alternative to rest energy is Kinetic Energy and why would that necessarily be so. It's Electromagnetic Energy - if you want to classify it- and nothing to do with mass or mechanics.
 
  • #6
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I class that as a GIGO argument. It presupposes that the only alternative to rest energy is Kinetic Energy and why would that necessarily be so. It's Electromagnetic Energy - if you want to classify it- and nothing to do with mass or mechanics.
Does a sack filled with light have kinetic energy or electromagnetic energy, when the sack is moving very fast?
 
  • #7
sophiecentaur
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Neglecting the mass of the sack and any reflection loss, it has to be EM energy because there's no KE in the photons.
Why are you questioning this fundamental idea?
 
  • #8
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Neglecting the mass of the sack and any reflection loss, it has to be EM energy because there's no KE in the photons.
Why are you questioning this fundamental idea?
Because, maybe massless things never have kinetic energy, and all the other things have kinetic energy, when they are moving?

Light in a sack is a thing with a rest mass. m=E/c2, where E is measured in the sack's rest frame.
So maybe we can give some kinetic energy to this thing with a rest mass?
 
  • #9
Khashishi
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I shall have to disagree with Drakkith. Light has no rest mass in vacuum, but it does have kinetic energy. It is conventional to assign all of the energy of light to kinetic energy. It's neater this way, since you can change the kinetic energy by means of Lorentz boosts.
Dexter, you still made a mistake since ##E = mc^2## only applies to an object at rest. (Light is never at rest.)
The full equation is ##E^2 = m^2 c^4 + p^2 c^2##
Since ##m=0##, we have ##E= pc##
 
  • #10
Drakkith
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I shall have to disagree with Drakkith. Light has no rest mass in vacuum, but it does have kinetic energy. It is conventional to assign all of the energy of light to kinetic energy.
Perhaps. I'm not an expert in this area. Got a link or reference? I'd like to read more on this.
 
  • #11
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I am not saying that light is not electromagnetic energy, because that sounds like something that might be wrong. I am saying that light has kinetic energy, because that sounds true.

Let's say a satellite sends out a 10 W beam of microwaves. A still standing satellite disk will receive 10 W of radio waves.

But a satellite disk moving at speed 0.9 c away from the satellite will receive 1 W of radio waves and 9 W of pushing work.
 
  • #12
Khashishi
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But a satellite disk moving at speed 0.9 c away from the satellite will receive 1 W of radio waves and 9 W of pushing work.
No, that's not right. Let's use the transformation laws. Let x be in the direction of motion.
The 4-force in the satellite frame is: F = (10W, 10W, 0, 0)
##F_1' = \gamma (F_1 - \beta F_2) = \frac{1}{\sqrt{1-0.9^2}} (10W - 0.9*10W) = 2.29W##
##F_2' = \gamma (F_2 - \beta F_1) = \frac{1}{\sqrt{1-0.9^2}} (10W - 0.9*10W) = 2.29W##
The satellite dish will receives 2.29W in its frame of reference. I don't know what you mean by pushing work, but the force is 10W/c in the satellite frame of reference.
 
  • #13
Khashishi
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Perhaps. I'm not an expert in this area. Got a link or reference? I'd like to read more on this.
I don't have a reference, but I'm sure I've come across it several times. For massless particles, the rest energy is zero and the total energy is all kinetic energy.
 
  • #14
Drakkith
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I don't have a reference, but I'm sure I've come across it several times. For massless particles, the rest energy is zero and the total energy is all kinetic energy.
Alright. I'll try to find some information on my own then. Thanks.
 
  • #15
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No, that's not right. Let's use the transformation laws. Let x be in the direction of motion.
The 4-force in the satellite frame is: F = (10W, 10W, 0, 0)
##F_1' = \gamma (F_1 - \beta F_2) = \frac{1}{\sqrt{1-0.9^2}} (10W - 0.9*10W) = 2.29W##
##F_2' = \gamma (F_2 - \beta F_1) = \frac{1}{\sqrt{1-0.9^2}} (10W - 0.9*10W) = 2.29W##
The satellite dish will receives 2.29W in its frame of reference. I don't know what you mean by pushing work, but the force is 10W/c in the satellite frame of reference.
Yeah, for some reason I started talking about power, when energy is the thing we are discussing.

So I should have said this: Let's say a satellite sends out a 10 J pulse of microwaves. A still standing satellite disk will receive 10 J of radio waves.

But a satellite disk moving at speed 0.9 c away from the satellite will receive 1 J of radio waves and 9 J of other energy, this other energy is kinetic energy, the disk is being pushed to higher speed.
 
  • #16
Khashishi
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That's still wrong. The radio waves have energy and momentum. The acceleration of the "disk" depends on whether it absorbs the microwaves or reflects them, but it is simple to calculate using conservation of momentum.
 
  • #17
I shall have to disagree with Drakkith. Light has no rest mass in vacuum, but it does have kinetic energy. It is conventional to assign all of the energy of light to kinetic energy. It's neater this way, since you can change the kinetic energy by means of Lorentz boosts.
Dexter, you still made a mistake since ##E = mc^2## only applies to an object at rest. (Light is never at rest.)
The full equation is ##E^2 = m^2 c^4 + p^2 c^2##
Since ##m=0##, we have ##E= pc##
Since total energy of a photon is given by ##E=pc## which is only the kinetic energy and kinetic energy is ##E=\frac{1}{2} (\gamma m)c^2## which would give us ##E = \frac{pc}{2}## where does the other half of the energy goes if pc/2 is the kinetic energy?
 
  • #18
Vanadium 50
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and kinetic energy is E=12(γm)c2E=\frac{1}{2} (\gamma m)c^2
Where did you get that equation?
 
  • #19
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That's still wrong. The radio waves have energy and momentum. The acceleration of the "disk" depends on whether it absorbs the microwaves or reflects them, but it is simple to calculate using conservation of momentum.
The dish absorbed the waves, that's why it gained 10 J of energy from a 10 J pulse.

(kinetic energy increased by 9 J)
 
  • #20
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