# Light has mass?

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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

If Higgs field really exists, light mass will be zero. Because light is not affected by Higgs field.

This is an assumption made about untested data. No one has determined the properties of the Higgs Boson or the higgs field. No one knows what effects the field may or may not have - it is all conjecture.

Paul

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This is an assumption made about untested data. No one has determined the properties of the Higgs Boson or the higgs field. No one knows what effects the field may or may not have - it is all conjecture.

Paul

Then, what is mass? This explanation is based on the recent result.

Edited by alpha2cen
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If the momentum is E/c then the mass is$E/c^2.$

How does that jibe with $E^2 = p^2c^2 + m^2c^4$? I get zero.

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If the momentum is E/c then the mass is$E/c^2.$

No, as swansont has said to you if the momentum is E/c then the mass is zero: m=0.

One can also make the inverse computation and show that if the mass is $E/c^2$ then the momentum is zero: p=0.

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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. ...

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

It depends on what terms you wish to express the momentum in. As you said above, m = p/v therefore p = mv where m is the relativistic mass of the body, not its rest mass.

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?

Sorry but I don't understand all that. Can you be clearer?

This is an assumption made about untested data. No one has determined the properties of the Higgs Boson or the higgs field. No one knows what effects the field may or may not have - it is all conjecture.

Paul

The problem here is that you're confusing rest mass with relativistic mass.

How does that jibe with $E^2 = p^2c^2 + m^2c^4$? I get zero.

The problem is obvious, isn't it? He's talking about relativistic mass and you're talking about rest mass. The rest mass of a photon is zero and its relativistic mass is E/c2

Edited by Autumn_Man
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Because gravity bends space/time and light will always follow the geodesic.

Gravity can't BEND AN EMPTY SPACE.

Gravity can't BEND TIME BECAUSE, TIME IS A CONCEPT.

lol.

Pawel Kolasa

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Gravity can't BEND AN EMPTY SPACE.

Gravity can't BEND TIME BECAUSE, TIME IS A CONCEPT.

lol.

Pawel Kolasa

Space ≠ Spacetime. Time ≠ Spacetime.

LOL, Pawel Kolasa!

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Does light have a case dependent mass different on the situation?

When it collide with photoelectric device, it has a mass.

But, it does not have a mass when it moves in the space.

Edited by alpha2cen
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It depends on what terms you wish to express the momentum in. As you said above, m = p/v therefore p = mv where m is the relativistic mass of the body, not its rest mass.

The problem is obvious, isn't it? He's talking about relativistic mass and you're talking about rest mass. The rest mass of a photon is zero and its relativistic mass is E/c2

Both used the term "mass" and denoted it by m. And as it is usual in physics "mass" means what you call "rest mass". You seem to believe that "relativistic mass" and "rest mass" have the same status in physics but this is not true. This is why "mass" means the latter for any physicist.

Moreover I would like to know how you obtain the relativistic mass M = E/c2 from the 'definition' M= p/v. Where, in your own words, p is momentum [*].

[*] I may confess this is a tricky question Does light have a case dependent mass different on the situation?

When it collide with photoelectric device, it has a mass.

But, it does not have a mass when it moves in the space.

What part of photons mass is always zero m=0 you do not still understand? When it collides or when it moves in space the photon has zero mass.

TIME IS A CONCEPT.

Time is a physical quantity.

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What part of photons mass is always zero m=0 you do not still understand? When it collides or when it moves in space the photon has zero mass.

From the point of view of particle, i.c., particle basis, light mass is zero.

But, from the point of view of light, the mass is only very short period physical property.

So, when light collide with photoelectric device surface, or similar situation, light mass is not zero and has particle property.

And, the other situation, light mass is zero and wave property.

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So, when light collide with photoelectric device surface, or similar situation, light mass is not zero and has particle property.

Light is made of photons and photons are massless. When photons participate in the photoelectric effect ("collide" in your language) they continue having zero mass.

Edited by juanrga
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From the point of view of particle, i.c., particle basis, light mass is zero.

But, from the point of view of light, the mass is only very short period physical property.

So, when light collide with photoelectric device surface, or similar situation, light mass is not zero and has particle property.

And, the other situation, light mass is zero and wave property.

On the particle point of view the relativistic mass of a photon is finite and has the value m = p/c = E/c2at all times. The proper mass or a photon is zero at all times.

When it comes to the invariant mass of disordered radiation there is a zero-momentum frame of the radiation and when there is energy and a zero momentum frame the invariant mass of that radiation is not zero. Tjhe invariant mass of a system of particles is defined the same way as proper mass is only with E replaced by the sum of all energies of the particles and p replaced by the sum of all 3-momentun of the particles.

That means that for a system of two photons moving parallel to each other in the same direction the invariant mass is zero. If the photons are moving in the oppsite direction of each other then the invariant mass of the two photons is non-zero

Until you realize that the proper mass of a photon is always zero and the relativistic mass of a photo is never zero people are going to run you around in circles.

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On the particle point of view the relativistic mass of a photon is finite and has the value m = p/c = E/c2at all times.

In #85 I asked you to provide an explicit derivation of relativistic mass from your definition as m = p/v. I notice three things: (i) you ignored my invitation to provide us a derivation, (ii) you have changed the definition from your previous m = p/v to your new m = p/c --i.e., you changed a velocity by a speed-- and (iii) your new definition, m = p/c, continues being incorrect: i.e.; p/c ≠ E/c2 , because p, which you defined as momentum, is a vector. Indeed the energy of a photon is E = |p|c.

I find curious that you are repeating the same kind of trivial mistakes than a recently banned user who presented himself as an 'expert' in the concept of mass. He also liked to say us that the relativistic mass of a photon was "finite" as if he believed that zero is an infinite quantity! I suppose that both of you studied the same literature.

Edited by juanrga

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