Twice the speed of light.

[dart_ship]

Write4U what is the mass (M) of a photon?

 

[Write4U]

Write4U what is the mass (M) of a photon?

@ c , it's mass is the equivalent of its energy divided by c^2

What does E = Mc^2 mean?

"Energy equals mass times the speed of light squared."
On the most basic level, the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing.

https://www.pbs.org/wgbh/nova/einstein/lrk-hand-emc2expl.html

 

[DaveC426913]

A photon at rest has zero mass.

You really don't understand the first thing about this do you. Get a textbook.

Sss's point is that there is no such thing as a photon at rest.

The postulates of SR mean that all photons in a vacuum move at c, as observed in all frames of reference.

"A photon at rest" is an oxymoron.

 

[Ssssssss]

Write4U what is the mass (M) of a photon?

0

@ c , it's mass is the equivalent of its energy divided by c^2

No its mass is 0 always. Dividing energy by \(c^2\) gives you relativistic mass which is a different and pointless thing.

 

[Write4U]

No its mass is 0 always. Dividing energy by \(c^2\) gives you relativistic mass which is a different and pointless thing.

It is mass! You can talk around it all you want.

On the most basic level, the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing.
https://www.pbs.org/wgbh/nova/einstein/lrk-hand-emc2expl.html

 

[Ssssssss]

It is mass! You can talk around it all you want.

No it isn't because mass is \(\sqrt{|p_\mu p^\mu|}\) and relativistic mass (aka total energy divided by \(c^2\)) is \(|p_\mu u^\mu|\) so even if you aren't up on tensor notation you can see that mass depends only on the particle's four momentum \(p\) whereas total energy depends on both the particle's four momentum and the observer's four velocity \(u\) so they are obviously different. The PBS source you keep quoting is wrong and what it should say is something like that separate units for mass and energy was a mistake and that mass is definitely a contribution to the energy of a particle but not that mass is energy. Saying mass is energy is clearly wrong because photons have 0 mass but not 0 energy and if you just say mass is energy there's no consistent way to describe the photon's mass as 0 because you can never bring it to rest and discuss its energy when it's not moving because that's oxymoronic.

 

[Janus58]

It is mass! You can talk around it all you want.

No, it isn't. Not by the modern convention of "mass". When we say "mass" today we mean what used to be qualified as "Rest mass". The term "relativistic mass" has fallen out of use, and just considered "energy". It's just that energy in of itself can imbue momentum without any "mass" being involved.

 

[Write4U]

Saying mass is energy is clearly wrong because photons have 0 mass but not 0 energy and if you just say mass is energy there's no consistent way to describe the photon's mass as 0 because you can never bring it to rest and discuss its energy when it's not moving because that's oxymoronic.

Well, if you cannot have a photon at rest, perhaps it is oxymoronic to say that it has zero rest-mass. It will never have effective zero mass and always effective energetic momentum. When it hits your skin it is causal to heat.

These restrictions are consistent with the constraint that mass and energy that act as causal influences cannot travel faster than the speed of light and/or backwards in time.

https://en.wikipedia.org/wiki/Causality_(physics)

Interestingly:

Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.

Wikipedia

IOW, energy can be converted into mass and mass can be converted into energy.

Are you telling me that mass can be converted into energy, but energy cannot be converted into mass?
E = Mc^2 is a one-way equation?

Please note that I am playing devil's advocate here. I have no objection to the notion of separate properties. It is the logic in the distinctions that interests me.

Bohmian Mechanics holds that photons are particles which acquire mass from momentum.
That effectively does away with that pesky particle/wave duality and the strange phenomenon that the wave collapse creates a "massive" impact.

 

[DaveC426913]

Please note that I am playing devil's advocate here.

Notable.

 

[Ssssssss]

Well, if you cannot have a photon at rest, perhaps it is oxymoronic to say that it has zero rest-mass.

I never use the term rest mass for exactly that reason and I always say either mass or invariant mass.

It will never have effective zero mass and always effective energetic momentum.

Define "effective mass" in mathematical terms.

Are you telling me that mass can be converted into energy, but energy cannot be converted into mass?
E = Mc^2 is a one-way equation?

No because I'm not stupid and I have no idea how you got the idea that I might think that from what I wrote.

Please note that I am playing devil's advocate here.

No you are not because a devil's advocate advances a coherent argument against a position whereas you are simply stating things that are wrong like that photons don't have 0 mass.

Bohmian Mechanics holds that photons are particles which acquire mass from momentum.

Reference please.

 

[Write4U]

Then there is this;

Rest mass of photon on the surface of matter

Highlights

• Mathematically the rest mass of photons is complex number when comes in contact with the surface of matter.
• Mass depends upon scalar curvature of the surface of matter and wavelength of the photon.
• Photon itself reveals illusion posing with a mass on the surface of matter because of wave-particle duality.

Abstract

The behavior of a photon is strange. It possesses both wave nature and particle nature. Some experiments show both behaviors of photons can exist simultaneously, while some other experiment state that both properties do not co-exists simultaneously. According to electromagnetic theory, the rest mass of photon in free space is zero and also photon has non-zero rest mass, as well as wavelength-dependent.

The very recent experiment revealed its non-zero value as 10-54kg.
Even experimental results concluded that within matter (dispersive) the photon shows its imaginary rest mass.

https://www.sciencedirect.com/science/article/pii/S2211379719330943#

I am not talking about photons flying around. I am talking about what occurs when a photon hits a physical surface and with what kind of non-zero value.

We assign zero value to the massive properties of a photon, and then turn around and purpose a non-zero property to a wave function.
If not, where does it get its physical mass from on impact (wave-function collapse)? Is this the moment when the photon becomes a physical particle?
When it becomes a particle does it acquire non-zero force (mass)?

Is this the dilemma?

When the same thing has a dual character, where one state has zero force, and the other has non-zero force, who're you gonna call, the particle or the wave? I see a logical conflict, a differential equation, if you will.

I just happen to like Bohm, who assigned an inherent potential that a zero value particle @ c
acquires mass, whereas the potential of a collapsing wave function @ c, converts energy into mass, however small. We can see the resulting "impact" patterns on the photographic plate in the double slit experiment. The mathematics must equate, therefore pM = eM at the moment the particle strikes the surface. It is a causal event and has an effective result.

 

[Ssssssss]

That paper seems to be misrepresenting what its references are saying and you only have to look at the title of its reference 24 to see that those nonzero values are upper bounds on any possible mass of a photon not a value. We do not simply assume that photons are massless but scientists do experiments do to check and those figures are the limit of what possible nonzero mass the photon could have and be consistent with experiment and they are not the mass which we still believe to be 0. And also they are talking about photons in matter where they are coupled to matter fields and don't travel at c so would be expected to have mass.

 

[Write4U]

those nonzero values are upper bounds on any possible mass of a photon not a value.

a value is not a value? You need to get philosophical here, it's all mathematical and operates in accordance with certain mathematical guiding equations. This is the self-ordering aspect of the Universe.
Everything has a value and when values interact they do so in a mathematical manner 1 + 1 = 2 (even in the Abstract)

An equation means the description of the "same" pattern from two different aspects, but listing the common applicable values to both.

I like Roger Antonsen's Ted Talk video.

 

[Ssssssss]

a value is not a value?

I will rephrase: those numbers are upper bounds on the possible range of mass values and not best estimates of the actual value which is 0 a fact which would have been obvious to you if you'd looked at references 24-26 in the paper you cited instead of trying to score points with cod philosophy.

 

[Ssssssss]

I just happen to like Bohm, who assigned an inherent potential that a zero value particle @ c
acquires mass, whereas the potential of a collapsing wave function @ c, converts energy into mass, however small.

Reference please.

 

[Write4U]

Reference please.

Bohmian Mechanics.

Pilot Wave model of the Universe.

In theoretical physics, the pilot wave theory, also known as Bohmian mechanics, was the first known example of a hidden-variable theory, presented by Louis de Broglie in 1927. Its more modern version, the de Broglie–Bohm theory, interprets quantum mechanics as a deterministic theory, avoiding troublesome notions such as wave–particle duality, instantaneous wave function collapse, and the paradox of Schrödinger's cat. To solve these problems, the theory is inherently nonlocal.

The de Broglie–Bohm pilot wave theory is one of several interpretations of (non-relativistic) quantum mechanics. An extension to the relativistic case has been developed since the 1990s.[3][4][5][6]

According to pilot wave theory, the point particle and the matter wave are both real and distinct physical entities (unlike standard quantum mechanics, where particles and waves are considered to be the same entities, connected by wave–particle duality). The pilot wave guides the motion of the point particles as described by the guidance equation.

Ordinary quantum mechanics and pilot wave theory are based on the same partial differential equation.

The main difference is that in ordinary quantum mechanics, the Schrödinger equation is connected to reality by the Born postulate, which states that the probability density of the particle's position is given by {\displaystyle \;\rho =|\psi |^{2}~.}

Pilot wave theory considers the guidance equation to be the fundamental law, and sees the Born rule as a derived concept.

https://en.wikipedia.org/wiki/Pilot_wave_theory

 

[Ssssssss]

I know that. I was looking for a reference about Bohm's work that said that it "assigned an inherent potential that a zero value particle @ c
acquires mass, whereas the potential of a collapsing wave function @ c, converts energy into mass, however small" which is what you said and none of the excerpts you provided say and I was looking for it because Bohm's work was as far as I know non relativistic and therefore says nothing about photons and was only much later extended by others to cover QFT concepts like photons and I don't think that later work is likely to say anything of the sort anyway.

 

[dart_ship]

Ssssssss light must have mass for black holes can pull it in (and stop it from escaping) and light reflects off objects (even in space)?

 

[Janus58]

Ssssssss light must have mass for black holes can pull it in (and stop it from escaping) and light reflects off objects (even in space)?

No, The idea that you need Mass to have, or be effected by, gravity is an aspect of Newtonian physics. But we have moved past that to Relativistic Physics where Gravity is related to the Energy-stress tensor, of which mass may be a component, but doesn't have to be. In other words, just like in my previous post, where energy alone has momentum, energy alone participates in gravity. Light has energy, so it is effected by gravity.

 

[Write4U]

Light has energy, so it is affected by gravity

Yes and that gravitational effect is identified as energy having "virtual mass".
Else "no mass, no mas"