•

# Photons, mass, and energy

18 replies to this topic

### #1 VeraZwicky

VeraZwicky

Mariner 2

• topic starter
• Posts: 204
• Joined: 26 Jan 2018
• Loc: Nashville, Tennessee

Posted 02 March 2021 - 12:22 PM

E=MC2.  Photons have no mass. They do have plenty of momentum but it's unclear to me how this creates energy. Explanations are appreciated!

### #2 lee14

lee14

Apollo

• Posts: 1,221
• Joined: 19 Dec 2009
• Loc: CNY

Posted 02 March 2021 - 12:47 PM

That's not the appropriate formula to calculate the energy of a photon. A photon is pure energy, and E=mc^2 is an equivalency, and only applicable with mass greater than zero. The energy of a photon is defined by a formula using Planck's constant and the wavelength, or frequency;  E=hv=hc/lambda, (v is frequency, lambda is wavelength, and h is Planck's constant).

Lee

Edited by lee14, 02 March 2021 - 01:28 PM.

• Miguelo and VeraZwicky like this

### #3 Dark Photon

Dark Photon

Sputnik

• Posts: 27
• Joined: 17 Oct 2020
• Loc: CT

Posted 02 March 2021 - 12:49 PM

E=m c^2 works only for particles (bodies) at rest. Anything moving has a momentum, in which case E^2 =(m c^2)^2 + (vector(p) c)^2. Photons have no mass, so E= |p| c.

Note that even for massive particles, when their momentum is much larger than m c, we have to a good approximation the same relation E~ |p| c .

• VeraZwicky and Tangerman like this

### #4 TOMDEY

TOMDEY

Cosmos

• Posts: 9,277
• Joined: 10 Feb 2014
• Loc: Springwater, NY

Posted 02 March 2021 - 12:57 PM

E=MC2.  Photons have no mass. They do have plenty of momentum but it's unclear to me how this creates energy. Explanations are appreciated!

The disconnect is in the contextual meaning of the word ~mass~. Gauge Bosons that identically travel at c (like light) have zero mass else they would necessarily also have infinite energy. So we reasonably compute their energy-equivalent mass to be E/c2, where E is (of course) given by hν. It all ties together quite nicely! Physicists rarely mention contextual meaning conditionals, assuming the audience is sufficiently-immersed to follow the flow. This also explains why ~popular~ books on arcane subjects exhibit a lot of poetic license is bringing their subject matter to the general public.    Tom

• VeraZwicky likes this

### #5 Eye stein

Eye stein

Explorer 1

• Posts: 94
• Joined: 02 Sep 2014
• Loc: South Florida

Posted 02 March 2021 - 01:07 PM

Right on Lee---and when the photon strikes something the wave collapses and that energy is transferred (either in part or totally) .

That's how a solar sail works.

Not quite as simple as that but you get the idea.

Joe

• lee14 likes this

### #6 VeraZwicky

VeraZwicky

Mariner 2

• topic starter
• Posts: 204
• Joined: 26 Jan 2018
• Loc: Nashville, Tennessee

Posted 02 March 2021 - 01:58 PM

Ok, thanks everybody. I think I'm a little closer to understanding. So, photons have no rest mass and always travel at C. Thus, we place our calculations regarding their energy in the context of relativity. In this regime, the photon gets its energy from momentum, not rest mass. And so, regarding light, does frequency always correlate to momentum?

### #7 EJN

EJN

Aurora

• Posts: 4,849
• Joined: 01 Nov 2005
• Loc: Between eigenstates

Posted 02 March 2021 - 02:42 PM

Momentum of a photon is:   p = hv/c   where h is Planck's constant, v is the frequency, c = speed of light.

Since a photon has no rest mass but has energy, p = E/c2 * c = E/c
By substitution of E = hv,
p = hv/c

### #8 TOMDEY

TOMDEY

Cosmos

• Posts: 9,277
• Joined: 10 Feb 2014
• Loc: Springwater, NY

Posted 02 March 2021 - 02:45 PM

Ok, thanks everybody. I think I'm a little closer to understanding. So, photons have no rest mass and always travel at C. Thus, we place our calculations regarding their energy in the context of relativity. In this regime, the photon gets its energy from momentum, not rest mass. And so, regarding light, does frequency always correlate to momentum?

Once you change the units, the phrase ~correlation~ gets somewhat nebulous. But precise (and closely related) would be the converse statements that energy is the conjugate attribute of frequency and that momentum is the conjugate attribute of position. The binding constant is then h. There are an infinite number of such selectable conjugate attribute pairs. When we run experiments, the walking-in decision is which attribute we choose to measure.   Tom

### #9 Otto Piechowski

Otto Piechowski

Vanguard

• Posts: 2,412
• Joined: 20 Sep 2005
• Loc: Lexington, KY

Posted 02 March 2021 - 03:27 PM

May we backtrack to VeraZwicky's opening statement in which she mentions the property called momentum.  It has been said throughout this thread that a photon has momentum and that a photon has no mass.

I just went online to something called the physics classroom which gave the following definition of momentum "Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. ... Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion."

In this definition, momentum is defined as "mass in motion".  Since a photon has no mass but has momentum, it would seem that this definition (mass in motion) is not quite accurate for all phenomena (e.g. a photon traveling from alpha centauri to my eyeball).  How might, how should this definition of momentum (mass in motion) be modified or understood so as to include the phenomenon of the momentum of the massless photon?

Thank you.

Otto

Edited by Otto Piechowski, 02 March 2021 - 03:29 PM.

• Miguelo and VeraZwicky like this

### #10 bob71741

bob71741

Apollo

• Posts: 1,312
• Joined: 16 Feb 2008

Posted 02 March 2021 - 04:00 PM

The definition that I got many years ago is that a photon does not have any REST mass; therefore cannot use e=mc2; e=hv is applicable when in motion.

### #11 EJN

EJN

Aurora

• Posts: 4,849
• Joined: 01 Nov 2005
• Loc: Between eigenstates

Posted 02 March 2021 - 04:41 PM

How might, how should this definition of momentum (mass in motion) be modified or understood so as to include the phenomenon of the momentum of the massless photon?

In quantum mechanics, momentum is defined as a self adjoint operator on the wavefunction. Since that is a

purely mathematical definition you probably won't find it helpful.

May we backtrack to VeraZwicky's opening statement in which she mentions the property called momentum.

I think VeraZwicky is a name-meshing of Vera Rubin & Fritz Zwicky, who separately discovered the "missing mass."

Edited by EJN, 02 March 2021 - 05:02 PM.

• VeraZwicky likes this

### #12 Dark Photon

Dark Photon

Sputnik

• Posts: 27
• Joined: 17 Oct 2020
• Loc: CT

Posted 02 March 2021 - 11:25 PM

In this definition, momentum is defined as "mass in motion".  Since a photon has no mass but has momentum, it would seem that this definition (mass in motion) is not quite accurate for all phenomena (e.g. a photon traveling from alpha centauri to my eyeball).  How might, how should this definition of momentum (mass in motion) be modified or understood so as to include the phenomenon of the momentum of the massless photon?

Thank you.

Otto

There is an abstract way of defining momentum that relies on conservation. It boils down to saying momentum is a vector that does not change as time goes by, for example during collisions. (It's the total momentum of a system that's conserved, but that total momentum is a sum of momenta of individual objects.) From that statement one can find that at low velocities p=m v, or at relativistic velocities p= m v/sqrt(1-v^2/c^2). For photons, one can find that p=E/c.

This is actually a precise statement that's called Noether's theorem. Amazingly, a simple assumption "physics should not change if you move the origin of a coordinate system from point A to point B" leads to momentum conservation via Noether's theorem.

Once momentum is considered in the context of its conservation, momentum of a photon is really no different than momentum of a massive particle. Let's say an atom at rest emits a photon. The atom recoils and gets some momentum p. The photon must have momentum -p, which we could determine just by measuring the velocity of the atom and its mass. This picture is self consistent if photon momentum equals its energy/c.

Edited by Dark Photon, 03 March 2021 - 01:19 AM.

• VeraZwicky and bitnick like this

### #13 Otto Piechowski

Otto Piechowski

Vanguard

• Posts: 2,412
• Joined: 20 Sep 2005
• Loc: Lexington, KY

Posted 03 March 2021 - 12:01 PM

Dark Photon et alii, thank you.  I believe I understand a good bit more now than before.

But just to make sure I can put the nail into the coffin, as it were....

(1) The reason we can say that a photon has momentum is because in the formulas p=E/c and E=mc(squared), E is a form of mass and in that sense, the photon has mass and therefore we have mass in motion.  Is my understanding correct or incorrect or in need of better nuance?

(2) The reason we can say a photon at rest has no mass is because a photon at rest has not yet departed from the atom; that atom having reached an excited state which would cause a photon to be formed/emitted.  Because it has not been formed/emitted it does not yet exist.  Because it does not yet exist, it necessarily has no properties including mass.  Is my understanding correct or incorrect or in need of better nuance?

Thank you.

Otto

Edited by Otto Piechowski, 03 March 2021 - 12:03 PM.

### #14 Dark Photon

Dark Photon

Sputnik

• Posts: 27
• Joined: 17 Oct 2020
• Loc: CT

Posted 03 March 2021 - 11:03 PM

There is no sense in which photons have mass.

Let's go back to the relativistic formula for energy of any object: E=sqrt[ (m c^2)^2 + (p c)^2 ].

Massive objects can be at rest. While at rest they have no momentum and then E=m c^2. Massless particles, like photons, cannot be at rest, but setting m=0 in the above formula gives  E=p c.

The two terms in the formula for energy have different interpretations. The first term, m c^2, is the energy at rest and photons don't have any. The second term is related to the kinetic energy -  that term depends on motion. Massive objects can have any momentum, including zero. Photons cannot have zero momentum because they always move. If they don't move they do not exist.

Having no mass is equivalent to always moving at the speed of light. Photons are born that way: it's their intrinsic property.

What I was trying to mention in an earlier post is that photon momentum is absolutely real and no different than any other form of momentum. Bouncing a ball against an object pushes it back.  Similarly bouncing a photon from a mirror pushes the mirror back.

I am not sure if this is clear enough. Let's hope someone else will try to do better.

Edited by Dark Photon, 04 March 2021 - 10:46 AM.

• Miguelo and VeraZwicky like this

### #15 VeraZwicky

VeraZwicky

Mariner 2

• topic starter
• Posts: 204
• Joined: 26 Jan 2018
• Loc: Nashville, Tennessee

Posted 04 March 2021 - 01:22 PM

There is no sense in which photons have mass.

Let's go back to the relativistic formula for energy of any object: E=sqrt[ (m c^2)^2 + (p c)^2 ].

Massive objects can be at rest. While at rest they have no momentum and then E=m c^2. Massless particles, like photons, cannot be at rest, but setting m=0 in the above formula gives  E=p c.

The two terms in the formula for energy have different interpretations. The first term, m c^2, is the energy at rest and photons don't have any. The second term is related to the kinetic energy -  that term depends on motion. Massive objects can have any momentum, including zero. Photons cannot have zero momentum because they always move. If they don't move they do not exist.

Having no mass is equivalent to always moving at the speed of light. Photons are born that way: it's their intrinsic property.

What I was trying to mention in an earlier post is that photon momentum is absolutely real and no different than any other form of momentum. Bouncing a ball against an object pushes it back.  Similarly bouncing a photon from a mirror pushes the mirror back.

I am not sure if this is clear enough. Let's hope someone else will try to do better.

What is it about the photon that causes change in other subatomic particles? Is it the kinetic energy of the momentum? Is it like adding different colors of paint together? What's a good metaphor for the energy of light?

• lee14 likes this

### #16 lee14

lee14

Apollo

• Posts: 1,221
• Joined: 19 Dec 2009
• Loc: CNY

Posted 04 March 2021 - 02:24 PM

What is it about the photon that causes change in other subatomic particles? Is it the kinetic energy of the momentum? Is it like adding different colors of paint together? What's a good metaphor for the energy of light?

That depends on what it strikes. If it interacts instead of reflecting, it adds energy to the system.  It might be absorbed by an electron of an atom, raising it to a higher level (or shell). If it's otherwise absorbed, it will raise the temperature of the substance it hits.

Lee

• VeraZwicky likes this

### #17 gilbert21

gilbert21

Lift Off

• Posts: 4
• Joined: 23 Jan 2021

Posted 11 April 2021 - 08:10 AM

Momentum via Lenz’s law

A plane wave impinging on a boundary will scatter some energy, down-convert some energy, and exert a force on the boundary.    A boundary in this instance is a plane that divides two isotropic mediums that have different electromagnetic properties.

An example, is a plane wave in a vacuum, interacting with a highly polished silver surface.  A current will be induced into the surface, in such a direction that the resultant magnetic field will be equal to the plane wave’s magnetic component, and repulsion will occur ( Lenz's law).   The remaining energy in the wave will be scattered or reflected.   Energy will be divided between reflection and the momentum imparted to the metal surface.
If we apply carbon black to the metal surface we get half the momentum and significant heating of the carbon. The carbon will now radiate in the infrared.

If we do not constrain the polished metal surface from translational movement then the reflected wave will be at a slightly longer wavelength (loose energy) to account for the energy imparted in the form of momentum to the metal. on the other hand if we constrain the metal surface to a stiff mass the reflected wavelength will be almost equal  to the incident wavelength.

Any and all material’s can be analysed in this way using modern EM theory.

Edited by gilbert21, 11 April 2021 - 08:19 AM.

### #18 astroneil

astroneil

Vanguard

• Posts: 2,307
• Joined: 28 Jul 2009

Posted 11 April 2021 - 06:48 PM

The OP might want to study the work of Arthur Holly Compton:

https://opentextbc.c...compton-effect/

### #19 gilbert21

gilbert21

Lift Off

• Posts: 4
• Joined: 23 Jan 2021

Posted 12 April 2021 - 01:43 AM

The OP might want to study the work of Arthur Holly Compton:

https://opentextbc.c...compton-effect/

I have studied the work of Arthur Holly Compton.  His work is not related to specular reflection but more related to scattering from a grid, crystal or grating.   The dimensions of the scatter's and their spacing are larger than the wavelength of the incident radiation.  Modern EM theory can handle this.   My example demonstrates how specular reflection behaves and solves the problem of transferring  the momentum of a mass-less "photon" to a boundary.  I suggest  that EM theory is superior to QM and is hampered by Einstein's "photon".

Have you any problems with the details of my post relating transfer of momentum to Lenz's law?

Edited by gilbert21, 12 April 2021 - 01:51 AM.

## Recent Topics

 Cloudy Nights LLC Cloudy Nights Sponsor: Astronomics