Massless "particle?"

[Capiert]

It is next to impossible to prove experimentally that anything is zero. We have shown in a lab that the mass is less that 10^-13eV/c^2 and via observations of galatic magnetic fields that the mass should be less than 10-27 eV/c^2. But you cannot prove a zero - because there are always experimental imprecisions and errors; this will lead to error bars and your figure could lurk in the error bars.

 

Theoretically big portions of modern physics would fail - quantum chromo/electro-dynamics would fall over and no longer be re-normalisable and thus no longer give the astonishingly accurate results that it does.

10-27 eV/(c^2) is mighty small,

(I must admit);

but if it (=the photon mass) lies at e.g. 10-32 eV/(c^2)

then we will not know

till we can achieve (better than) that precision.

Edited May 11, 2017 by Capiert

[Strange]

I cannot imagine

momentum

without a (physical) mass m

 

What you can imagine is hardly relevant.

 

 

 

That is its definition,

or should I ignore that?

 

That is one definition. For massive particles.

 

For massless particles it has a different definition.

[Capiert]

Please proceede. Explain a bit (if you want).

What you can imagine is hardly relevant.

I agree, imagination & that we can understand

is the hard(est )core

to connecting

& making sense of things.

 

E.g. We do not write these threads

for machines;

they are written that people will understand,

particularly the 1's asking.

I find that most relevant.

 

(Good that you pointed that out:

Ridgidly relevant,

(I like that).)

Edited May 11, 2017 by Capiert

[Sensei]

"Photon has zero mass" is shortcut from "photon has zero rest-mass".

Which means there is no frame-of-reference in which photon is at rest.

 

Particles with non zero rest-mass, you can accelerate, deaccelerate etc. etc.

Edited May 11, 2017 by Sensei

[frankglennjacobs@gmail.com]

As I understand it, the double-slit exercise proves (over and over) that light is not a particle -- and not a wave, either.

 

But an individual photon causes a visible chemical reaction on photographic film in the double-slit exercise.

 

And light causes the black and white vanes inside the trick light-bub thingies (photometer?) to go around.

 

And just yesterday my mind was farther blown by the news that electrons have no size at all.

 

(Next thing you know we will find out that the earth is NOT flat, and that space is NOT full of Aether, and that fire has nothing to do with Phlogistine.)

 

I am a relic of a past Century, dragged kicking and screaming into these modern times!

 

 

[Strange]

Please proceede. Explain a bit (if you want).

 

 

We know that energy is given by: [latex]e = \sqrt{m_0^2 c^4 + p^2c^2}[/latex] (http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4)

When the rest mass [latex]m_0[/latex] is zero then the energy is just: [latex]e = p c[/latex]

So: [latex]p = \frac e c[/latex]

Or (using the relationship found by Planck: http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3 )

[latex]p = \frac e c = \frac {h \nu} c = \frac h \lambda[/latex]

[Capiert]

"Photon has zero mass" is shortcut from "photon has zero rest-mass".

Which means there is no frame-of-reference in which photon is at rest.

Moment (=stop please).

Isn't (at the speed of light, c,)

light standing still

wrt to the other photons?

(E.g. those photons going in the same direction wrt the earth.)

Isn't speed c

also a reference frame?

What do you mean by shortcut?

Please explain (a bit).

 

 

~Particles with [non zero] rest-mass, can be accelerated & deaccelerated, etc.

What is the etc? (constant speed?)

Edited May 11, 2017 by Capiert

[Strange]

Isn't (at the speed of light, c,)

light standing still

wrt to the other photons?

 

 

No. You can't compare the speed of photons because ...

 

 

 

Isn't speed c

also a reference frame?

 

No. The speed of light is not a valid reference frame. Try it and you will end up dividing by zero.

[Capiert]

As I understand it, the double-slit exercise proves (over and over) that light is not a particle -- and not a wave, either.[

As far as I know, the double slit experiment (results) confirms

(or at least convinces me)

light is a wave.

Maybe a half wavelength

could be interpretted

as a (particlelike) impulse?

 

But an individual photon causes a visible chemical reaction on photographic film in the double-slit exercise.

?

 

And light causes the black and white vanes inside the trick light-bub thingies (photometer?) to go around.

Impulse?

 

And just yesterday my mind was farther blown by the news that electrons have no size at all.

That's wild, & blows mine too.

Some of the professionals

will have to comment there,

I don't know that new background.

 

 

(Next thing you know we will find out that the earth is NOT flat, and that space is NOT full of Aether, ..

I prefer Maxwell's Ether opinion, too.

 

and that fire has nothing to do with Phlogistine.)

 

I am a relic of a past Century, dragged kicking and screaming into these modern times!

 

Yes, variety is the spice of life,

see that you don't suffer (integestion, heartburn)

from too much spice (=new changes).

[Mordred]

Capiert would you please look up the definition of mass under physics.

 

Your basic problem is your not considering what mass is defined as.

 

Mass is resistance to inertia change. If you can't even pay attention to the very definition of mass then your never going to understand the key differences between massive and massless particles.

Edited May 11, 2017 by Mordred

[Capiert]

No. You can't compare the speed of photons because ...

Yes, ?, please continue.

 

No. The speed of light is not a valid reference frame. Try it and you will end up dividing by zero.

Please explain.

You gave us a rooted (energy) equation,

modelled perhaps?

Edited May 11, 2017 by Capiert

[Mordred]

What Strange posted is correct he also posted relevant links to each equation

Edited May 11, 2017 by Mordred

[Capiert]

What Strange posted is correct he also posted relevant links to each equation

I don't doubt he is telling standard physics.

[Strange]

Yes, ?, please continue.

 

Please explain.

You gave us a rooted (energy) equation,

modelled perhaps?

 

 

If you try and use the Lorentz transform to convert between frames of reference when [latex]v = c[/latex] then

[latex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - c^2/c^2}} = \frac{1}{\sqrt{1 - 1}} = \frac 1 0 [/latex] = undefined

 

Therefore v=c is not a valid frame of reference and you cannot compare the speed of photons.

[Capiert]

Capiert would you please look up the definition of mass under physics.

Your basic problem is your not considering what mass is defined as.

Mass is resistance to inertia change. If you can't even pay attention to the very definition of mass then your never going to understand the key differences between massive and massless particles.

In physics, mass is a property of a physical body. It is the measure of an object's resistance to acceleration (a change in its state of motion) when a net force is applied.[1] It also determines the strength of its mutual gravitational attraction to other bodies. The basic SI unit of mass is the kilogram (kg).-Wiki

 

https://en.m.wikipedia.org/wiki/Mass

Edited May 12, 2017 by Capiert

[Strange]

In physics, mass is a property of a physical body. It is the measure of an object's resistance to acceleration (a change in its state of motion) when a net force is applied.[1] It also determines the strength of its mutual gravitational attraction to other bodies. The basic SI unit of mass is the kilogram (kg).-Wiki

 

 

And light does not accelerate and so it does not respond to force and so it has no mass.

[Mordred]

acceleration is inertia change

[Capiert]

If you try and use the Lorentz transform to convert between frames of reference when [latex]v = c[/latex] then

[latex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - c^2/c^2}} = \frac{1}{\sqrt{1 - 1}} = \frac 1 0 [/latex] = undefined

 

Therefore v=c is not a valid frame of reference and you cannot compare the speed of photons.

I don't use the Lorentz transform in momentum

because (it's redundant, when)

I set the limit speed to c=v+v'.

Similar can be done with KE

(but energy gives wrong answers,

m*E should be used instead).

The Galilean transform

was never invented

with a speed limit (e.g. c)

because they thought

light's speed was infinite.

We know better, now

& can change that.

Edited May 12, 2017 by Capiert

[Mordred]

Then your not using the correct methodology and therefore getting the wrong answers.

 

Use the correct formulas as described by relativity and you might have a chance..

 

Start with listing the difference in the transformation equations of Galilean relativity and Lorentz transforms.

 

Secondly understand that the Lorentz transforms are under constant velocity.

I set the limit speed to c=v+v'.

 

This makes absolutely no sense. All observers will always measure the velocity of a photon at c.

 

So why would you ever add two observer velocities if both observers measure the signal at c your essentially adding c to c to equal c...

 

obviously incorrect

Edited May 12, 2017 by Mordred

[Sensei]

 

"Photon has zero mass" is shortcut from "photon has zero rest-mass".

Which means there is no frame-of-reference in which photon is at rest.

 

Moment (=stop please).

Isn't (at the speed of light, c,)

light standing still

wrt to the other photons?

(E.g. those photons going in the same direction wrt the earth.)

Isn't speed c

also a reference frame?

What do you mean by shortcut?

Please explain (a bit).

 

"Mass" can be understood as "rest-mass", or "relativistic-mass". It depends on who is using it, and in which context.

 

You should start from reading articles below:

 

Rest-mass aka invariant-mass

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

 

Relativistic-mass

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

 

~Particles with [non zero] rest-mass, can be accelerated & deaccelerated, etc.

What is the etc? (constant speed?)

etc. = Latin Et cetera

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

Edited May 12, 2017 by Sensei

[swansont]

Well... In reality massless particles shouldn't exist so I still think that photons have some form of mass albeit unmeasurably small, however you might as well say they are massless. Question is is there any way to prove photons have a miniscule amount of mass, Perhaps a stream of condensed photons that impacts a silicon wafer nanometers across in a magnetic bowl (Antimatter trap)? If it moves more than a control object being put through the same experiment without the stream of photons then it would be provable through thousands of more tests that photons have miniscule amounts of mass? I personally think they do have a miniscule amount of mass as something without mass shouldn't really exist. However I could be wrong...

 

 

If the mass of the photon were not identically zero, it would have measurable consequences.

It seems opinions are changing.

What was said in the past

is not true now.

(What a mess.)

Is it possible

we don't have enough accuracy yet?

 

 

They aren't opinions.

 

momentum

mom=m*v.

I cannot imagine

momentum

without a (physical) mass m

(& speed v).

That is its definition,

or should I ignore that?

Would you please explain that exclusion method a bit?

I'm not clear yet.

 

 

 

Momentum is only mv in the classical limit, as I have explained. It is an approximation, not the definition.

[Capiert]

What is your model, and/or what are your testable predictions based on this idea?

 

We know, for example, that the momentum of a photon with a wavelength of 780 nm is 8.5 x 10^28 kg-m/s

Don't you surely mean 8.5*(10^(-28)) kg*m/s, instead? (Negative exponent -28, instead of positive 28.)

 

(1.56 eV=2.55*(10^(-19)) J.

mom~2*E/v.)

 

What is its mass and how fast is it moving?

 

The photon is moving at c light's speed.

The photon's mass is (the momentum

divided by it's speed v=c)

m=mom/v=8.5*(10^(-28)) [kg*m/s]/(2.99*(10^8)) m/s)=2.85*(10^(-36)) kg.

Edited May 12, 2017 by Capiert

[swansont]

Don't you surely mean 8.5*(10^(-28)) kg*m/s, instead? (Negative exponent -28, instead of positive 28.)

 

(1.56 eV=2.55*(10^(-19)) J.)

 

The photon is moving at c light's speed.

The photon's mass is (the momentum

divided by it's speed v=c)

m=mom/v=8.5*(10^(-28)) [kg*m/s]/(2.99*(10^8)) m/s)=2.85*(10^(-36)) kg.

 

 

Yes, the "-" didn't register.

 

2.85*(10^(-36)) kg gets you the 1.59 eV I mentioned. But experiment says that if it's nonzero, it must be 13 orders of magnitude smaller than this. Other observations say it's 27 orders of magnitude. So your model has been shown to be wrong, in spectacular fashion.

[Capiert]

And light does not accelerate and so it does not respond to force and so it has no mass.

I think your problem (there) is

you don't want to admit

gravity's free fall g=-9.8 m/(s^2)

is an acceleration

(when you say that);

light falls (like any other moving object, (ruffly) without air resistance);

sound does NOT.

 

That means light accelerates.

Yes, the "-" didn't register.

2.85*(10^(-36)) kg gets you the 1.59 eV I mentioned. But experiment says that if it's nonzero, it must be 13 orders of magnitude smaller than this. Other observations say it's 27 orders of magnitude. So your model has been shown to be wrong, in spectacular fashion.

Please explain a bit (those experiments, so I can grasp what's happening). Spectacular sounds interesting. Edited May 12, 2017 by Capiert

[Strange]

I think your problem (there) is

you don't want to admit

gravity's free fall g=-9.8 m/(s^2)

is an acceleration

(when you say that);

light falls (like any other moving object, (ruffly) without air resistance);

 

 

Free fall is not acceleration. (Free fall means you are experiencing no force so if you had an accelerometer with you, it would read zero.)