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When the light comes out form the space craft, the light speed will be C at the reference frame.

The light speed is C inside the space craft as well. Learn some physics.

 

E light, space craft = Elight, reference frame + E kinetic energy difference

hv space craft + E kinetic energy = hv reference frame

No. Learn some physics.

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then how come light bends due to gravity? and how light got sucked into black hole?

What Einstein wrote in 1906/8 is not a disproof of whether Einstein accepted the modern interpretation in 1948.

Well this post is a good excuse as any to come back from my sabatical. What you claimed about Einstein (but again never proved) is wrong. In one of his early articles published between 1906 and 1908 E

The light speed is C inside the space craft as well. Learn some physics.

 

 

No. Learn some physics.

 

When the light comes out form the space craft to the reference empty space, it's speed will be C , too.

 

E light, space craft = Elight, reference frame + E kinetic energy difference

 

hv space craft +( V space craft / C) hv space craft = hv reference frame

 

V; frequency

 

At the front of the moving direction wave length of the light will be more shorter than the one in the space craft.

 

The photon in the star is not different from it in the empty space.

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E light, space craft = Elight, reference frame + E kinetic energy difference

 

No, the energy is not the same in a frame where p is zero than in a frame where it is not.

 

The photon in the star is not different from it in the empty space.

 

There is only a kind of photons.

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Emphasis mine:

When the light comes out form the space craft to the reference empty space, it's speed will be C , too.

 

The highlighted phrase is nonsense. It represents a very serious misunderstanding of physics. This misunderstanding led you to say earlier in post 48 When we travel in the space by space craft at the speed 99/100 C. This too is nonsense. The spacecraft is moving at 0.99 c with respect to what? You seem to think there is some universal reference frame against which all speeds can be referenced. Your notion is that of a luminiferous aether. It doesn't exist.

 

Relativity 101:

  • The laws of physics are the same in all inertial reference frames.
    This concept goes way back in time to Galileo. It's implicit in Newtonian mechanics. While Newton did occasionally write about absolute motion ("God's frame"), he also noted that this absolute motion is unknowable. All we mere mortals can see is relative motion because the laws of physics are the same in all inertial frames.
     
  • The speed of light has the same value the same to all inertial observers.
    The invariance of the speed of light is implicit in Maxwell's equations and has been confirmed by a huge number of experiments, starting with the Michelson-Morley experiment. However, the invariance of the speed of light is very much at odds with Newtonian mechanics. Newtonian mechanics would say that the speed of light in your spaceship going at 0.99 c is only 1/100 c. This is not the case. They will see the speed of light as being c.

 

Explaining this conflict between Newtonian mechanics and electrodynamics (Maxwell's equations) was the most important problem facing late 19th century theoretical physicists. The current inability to join general relativity and quantum mechanics pales in comparison to this late 19th century conflict. The problem with general relativity versus quantum mechanics is one of incommensurability. The problem with Newtonian mechanics and electrodynamics is one of complete disagreement on the predicted outcome of easily conducted experiments. Because it was the key problem of the time, it attracted the attention of many physicists of the time. Poincare and others tried to whitewash the problem away. Einstein's insight was to simply take Maxwell's equations and it's invariance of the speed of light at face value and see where that led him.

 

The takeaway point here is that the speed of light is the same to all inertial observers. Taking this one step further, the speed of light as measured by a local experiment is the same to all observers, inertial or not. You do not appear to understand this crucial concept.

 

 

E light, space craft = Elight, reference frame + E kinetic energy difference

This is nonsense even in Newtonian mechanics.

 

 

hv space craft +( V space craft / C) hv space craft = hv reference frame

This is the Newtonian point of view. It is wrong.

 

 

The photon in the star is not different from it in the empty space.

The only difference is that photons in a star travel but a short distance before being absorbed. Otherwise, photons are photons. They are indistinguishable particles. Photons always move at c. Always.

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No, the energy is not the same in a frame where p is zero than in a frame where it is not.

Same photon is moving from the space craft to the empty space.

 

There is only a kind of photons.

 

The energy state is different.

In contains star specific energy term, too. For example, ( V star velocity / C) hv star

 

Emphasis mine:

 

The highlighted phrase is nonsense. It represents a very serious misunderstanding of physics. This misunderstanding led you to say earlier in post 48 When we travel in the space by space craft at the speed 99/100 C. This too is nonsense. The spacecraft is moving at 0.99 c with respect to what? You seem to think there is some universal reference frame against which all speeds can be referenced. Your notion is that of a luminiferous aether. It doesn't exist.

 

 

 

Thank you for good answer.

Is C(light speed) not the reference?

Edited by alpha2cen
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No, the energy is not the same in a frame where p is zero than in a frame where it is not.

Same photon is moving from the space craft to the empty space.

 

Yes, I said this to you in the quote below "There is only a kind..."

 

There is only a kind of photons.

The energy state is different.

 

Yes, I said this to you in the quote above "the energy is not the same..."

Edited by juanrga
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There is only a kind of photons.

This doesn't make sense grammatically. I suspect you meant to say "There is only one kind of photons". (In other words, "photons are photons".) Using the indefinite article a instead of the definite one changes the meaning entirely.

 

Note well: The "this doesn't make sense" problem here is due to your native language not being English. It's not a technical "this doesn't make sense" problem.

 

 


 

 

Is C(light speed) not the reference?

 

This doesn't make sense, period.

 

Here's a simple thought experiment. Spaceships A and B are at rest with respect to one another. Scientists on each spaceship conduct experiments to measure the speed of light. Each sees no variation of the speed of light with direction, and the two agree on the speed they measure. Now spaceship B accelerates to a high speed and then cuts its space drive, ending with a velocity equal to 99/100 the speed of light with respect to spaceship A. After cutting the engines, the scientists on spaceship B once again conduct their experiments to measure the speed of light.

 

What do you think they will they see, alpha2cen? Will the measured speed of light vary with direction?

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This doesn't make sense grammatically. I suspect you meant to say "There is only one kind of photons". (In other words, "photons are photons".) Using the indefinite article a instead of the definite one changes the meaning entirely.

 

Note well: The "this doesn't make sense" problem here is due to your native language not being English. It's not a technical "this doesn't make sense" problem.

 

Thanks! +1 for you.

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Can we make one photon particle?

 

[latex]\bigodot ... \bigodot ...... \bigodot [/latex]

 

What is the minimum unit to make or to detect?

 

the mass of the Sun = total mass of the Sun - photon mass in the Sun

Do we have no problem in this equation?

I don't know what this is supposed to say. It appears to say

 

Total Mass of Sun = Mass of Sun + Photon Mass of the Sun

 

The mass of the sun is a complicated function of the mass-energy of the particles in the plasma that mack up all the matter of the sun, radiation included. This includes the rest mass of all the nuclei in the sun plus the kinetic energy of all those nuclei. This includes the contribution to the mass of the sun by the radiation in the sun, i.e. photons. To be precise, pressure is also a source of gravity. It has more relevance for objects such as neutron stars. The active gravitational mass of a perfect fluid is

 

[math]\rho_g = \rho + 3p[/math]

 

where [math]\rho_g[/math] is active gravitional mass, [math]\rho[/math] = proper energy density and p = pressure. For disordered radiation both [math]\rho[/math] and p are non-zero.

 

A photon has no "at rest" mass ..

Nice! The point you made is the same point made in the physics literature.

 

I prefer the term proper mass ove rest mass anyday. Even though the proper mass of photons is zero we still have

 

1) Inertial mass (aka relativistic mass) of photon is non-zero

2) Passive gravitational mass of light is non-zero

2) Active gravitational mass of light is non-zero

 

... Learn some physics...No. Learn some physics.

Ummmm ........ he's here to learn physics. Hence his questions.

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The mass of the sun is a complicated function of the mass-energy of the particles in the plasma that mack up all the matter of the sun, radiation included. This includes the rest mass of all the nuclei in the sun plus the kinetic energy of all those nuclei. This includes the contribution to the mass of the sun by the radiation in the sun, i.e. photons. To be precise, pressure is also a source of gravity. It has more relevance for objects such as neutron stars. The active gravitational mass of a perfect fluid is

 

[math]\rho_g = \rho + 3p[/math]

 

where [math]\rho_g[/math] is active gravitional mass, [math]\rho[/math] = proper energy density and p = pressure. For disordered radiation both [math]\rho[/math] and p are non-zero.

 

First, it is is difficult to believe that Sun is a perfect fluid.

 

Second, you do not give correct dimensions: mass does not have units of pressure and your [math]\rho_g[/math] cannot be a mass as you believe.

 

Third, the source of gravity in general relativity is the stress-energy-momentum tensor and for a perfect fluid it is

 

[math]T^{\alpha \beta} = \left(\rho + {p \over c^2}\right)u^{\alpha}u^{\beta} + p g^{\alpha \beta}[/math]

 

with [math]\rho[/math] mass density.

 

I prefer the term proper mass ove rest mass anyday. Even though the proper mass of photons is zero we still have

 

1) Inertial mass (aka relativistic mass) of photon is non-zero

2) Passive gravitational mass of light is non-zero

2) Active gravitational mass of light is non-zero

 

Modern physics uses a fundamental concept of mass which is invariant and denoted by m. For a photon m=0; as is well-known photons are "massless particles".

 

"Proper mass" is misleading, it could seem that m is the mass of a object only when measured in the proper frame, which is not true. m is also the mass of the object in the laboratory frame or in any other frame.

Edited by juanrga
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We all agree that light has momentum. (m for notation sake only - Mass will use M)

We all agree that it has a velocity of C.

The mass of an object is the momentum divided by its velocity. (in units weight time interval) (foot pound seconds for projectiles)

This is converted to slugs or newtons by way of reducing weight to mass.

 

Without assuming a mass what is the momentum of a photon?

 

now for the sake of measuring the very small let us use a scale that that multiplies the number (not the actual momentum just the number used to define it) by 1000000. We can call this scale nx's and note that a proton would have many millions of nx units as its mass.

 

so, take the momentum of a photon, multiply by 1000000 and divide by C.

What number do you end up with?

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Without assuming a mass what is the momentum of a photon?

 

The special relativistic expression for the energy is [math]E^2= m^2c^4 + \mathbf{p}^2 c^2[/math]. For massless particles this reduces to [math]E = p c[/math], which gives you the relation of the momentum of the photon to its energy. Further using the relation between the energy of a photon and its frequency [math]\nu[/math] you can obtain [math]p=h\nu/c[/math] where h is Planck constant

 

http://en.wikipedia....ical_properties

Edited by juanrga
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The special relativistic expression for the energy is [math]E^2= m^2c^4 + \mathbf{p}^2 c^2[/math]. For massless particles this reduces to [math]E = p c[/math], which gives you the relation of the momentum of the photon to its energy. Further using the relation between the energy of a photon and its frequency [math]\nu[/math] you can obtain [math]p=h\nu/c[/math] where h is Planck constant

 

http://en.wikipedia....ical_properties

 

 

[math]E^2= m^2c^4+\mathbf{p}^2c^2[/math] reduces to [math]E= mc^2 + \mathbf{p}c[/math] not E=pc

 

you seem to be missing part of the formula?

Edited by PaulS1950
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[math]E^2= m^2c^4+\mathbf{p}^2c^2[/math] reduces to [math]E= mc^2 + \mathbf{p}c[/math] not E=pc

 

you seem to be missing part of the formula?

 

No. If you set [math]m=0[/math] in [math]E^2= m^2c^4+\mathbf{p}^2c^2[/math] you obtain [math]E=pc[/math].

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So you start with the assumption that the mass is zero in an equation to find the mass?

 

How absurd is that?

 

If I assume the mass is [math] 10 ^-1000000000000[/math] then I will "prove" that it does have mass.

 

Start with the values of momentum and / or energy of a photon within its reference frame and calculate mass only from that data.

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[math]E^2= m^2c^4+\mathbf{p}^2c^2[/math] reduces to [math]E= mc^2 + \mathbf{p}c[/math] not E=pc

 

No, [math]\sqrt{m^2c^4+p^2c^2}[/math] does not reduce to [math]mc^2 + pc[/math]

 

If you square the latter expression you get a cross-term of 2mpc^3

 

Without assuming a mass what is the momentum of a photon?

 

The momentum of EM radiation is already established as E/c. The conclusion from relativity is that the photon is massless.

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So you start with the assumption that the mass is zero in an equation to find the mass?

 

How absurd is that?

 

No. The mass of a photon is found from the condition that it travels at c. The result is m=0, as is well-known.

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The strongest theoretical limit on the photon mass is from the galactic magnetic field (Chibisov et al), which implies that [math]m_\gamma < 3 \times 10^{-27}\;{\rm eV}[/math] though this relies on some assumptions.

From a theoretical point of view, the Standard Model predicts a photon with exactly zero mass. To have a non-zero mass one would have to break the U(1) symmetry of QED.

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light have a mass as it behaves as a matter.

 

Photons have zero mass. Light behaves as radiation, which is different from matter.

Edited by juanrga
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  • 2 weeks later...

The momentum of EM radiation is already established as E/c. The conclusion from relativity is that the photon is massless.

 

If the momentum is E/c then the mass is[math] E/c^2.[/math]

Edited by PaulS1950
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