# Why is the rest mass of photons 0?

1. May 29, 2008

### regev

In the article "Why does light have invarient speed?" on PF (first URL in attached file), there is a statement that for photons,
$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$
is $$0$$ since $$v = c$$. (I'm not sure if the author is stating this or quoting somebody else.) Is this a valid argument? A different source (second URL in attached file) gives a more complex argument, and mentions experiments to determine upper bounds on the rest mass of photons. If the argument were so simple, why does Baez go to the trouble and mention the possibility that it isn't true?

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2. May 29, 2008

### Yor_on

As far as I know photons are massless. They do have a momentum though.

" taken to mean different things if the light is moving freely or trapped in a container. The definition of the invariant mass of an object is m = sqrt{E2/c4 - p2/c2}. By this definition a beam of light, is massless like the photons it is composed of. However, if light is trapped in a box with perfect mirrors so the photons are continually reflected back and forth in the box, then the total momentum is zero in the boxes frame of reference but the energy is not. Therefore the light adds a small contribution to the mass of the box. This could be measured - in principle at least - either by an increase in inertia when the box is slowly accelerated or by an increase in its gravitational pull. You might say that the light in the box has mass but it would be more correct to say that the light contributes to the total mass of the box of light. You should not use this to justify the statement that light has mass in general. "

So the momentum of a photon is transferable to a change of mass.
And that would refer to the photons inherent energy as I think.

3. Nov 3, 2008

### Yor_on

" However, if light is trapped in a box with perfect mirrors so the photons are continually reflected back and forth in the box, then the total momentum is zero in the boxes frame of reference but the energy is not "

When you read this you might wonder why it is so.
i know I did:)

But the definition of momentum is "The mass of an object multiplied by its velocity."
And velocity is not speed...

Speed= "Speed is the rate of motion."
"Or equivalently the rate of change in position, often expressed as distance d traveled per unit of time t."

Velocity ="In physics, velocity is defined as the ...rate of change of position...."
" It is a vector physical quantity; ...both speed and direction are required.... to define it "

So if you drew a line on the floor and then started to run from side to side of that line.
Always coming back to your original position though, you would indeed have a certain speed.
But your velocity would be zero as there would be no change of position (with even turns that is:)
And that explains how the momentum can be seen as zero inside this box.

But the real question here is, is this true?
Isn't this just a play with words and concepts?

Either there is a 'momentum' to light.
A. it's bound to the objects velocity
B. it's a 'mechanism' describing invariant mass and photons behavior when interacting between different reference frames.
That is, something not bound to any object per se.

This is my new 'hobby horse' and I'm wondering?
So how do you see it?

------------

As for why the photon can travel at 'c' my view:) Is that it has to do with it being massless.
Time dilation is a secondary effect coming as it 'speeds' away.
That is also the explanation to why it has an 'instantaneous acceleration'.
which opens for the question how it can have a momentum again:)

To me that imply that I don't see the right question here.
And so I need to see how you think of momentum?
And inertia too perhaps.

Last edited: Nov 3, 2008
4. Nov 4, 2008

Staff Emeritus
By that argument, nobody should be debunking any nonsense. Right? Why go to the trouble if perpetual motion is impossible?

Physics is an experimental science. It's important that we test our understanding by comparing with data.

5. Nov 4, 2008

### Phrak

I'd be thrilled to see what each of the multiple authors think of a photon at rest. (sarcasm included)

6. Nov 4, 2008

Staff Emeritus
Argument by lack of imagination is not a very good one.

As far as we know, photons are exactly massless, and experimentally, the limits are very small. Very small. So small that a) such a photon at rest would carry so little energy there would be virtually no interaction with the environment, and b) the wavelength would be enormous. At mm/s, we're talking millions of miles. So you have a very low-energy diffuse object - hard to tell you have anything at all there.

7. Nov 4, 2008

### Yor_on

There are some things with your view that I'm wondering over Vanadium.
I remember those 'stopping light' experiments and what happened when they stopped stopping it :)
It as you stated 'disappeared' when stopped, but when allowed to 'recuperate' it just continued if I remember right.

When you are discussing that photons wavelength, how do you see it?
As a result of us tapping the photon of its energy, and so 'unbending' its wave?
Or as a result of our defining its position and therefore 'undefining' its wavelength?
Or do you see it in some other way?

8. Nov 4, 2008

Staff Emeritus
What you're talking about has no relationship to what I am talking about.

9. Nov 4, 2008

### Yor_on

I see.
I was hoping for some interesting thoughts here.
Alas, some other time perhaps:)

10. Nov 6, 2008

### edguy99

maybe

C . the mass equivalent of the energy of the photon (e/c^2) multiplied by the speed of the photon through that medium - comes out to about (energy of photon/c)

11. Nov 6, 2008

### Phrak

The most intesting quote in the Baez article, to me is this one:

"A non-zero rest mass would introduce a small damping factor in the inverse square Coulomb law of electrostatic forces. That means the electrostatic force would be weaker over very large distances."

I'd like to hear from anyone who has an idea of how this conclusion is come about.

12. Nov 6, 2008

### rbj

why don't you just put the URLs in the post here so we can just click on them? this futzing around with attached text files is both silly and inconvenient.

it comes from this, which used to be commonly seen in introductory modern physics textbooks until the use of the term "relativistic mass" became deprecated.

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

i think the two equations are equivalent. this depicts the "inertial mass" of a particle, as observed by some inertial observer moving at a velocity of v relative to that particle, where that inertial mass is simply the momentum of that particle (that we all agree on, whether or not you like the use of "relativistic" vs. "rest" mass) divided by the velocity.

if you don't like the concept of relativistic mass, simply multiply the equation above by v and you have an expression of momentum that is accurate and satisfies everybody's semantics.

if you were to look at the second equation and if photons were assumed to move at speed c (for any inertial observer) and the rest mass was not zero, what would be the relativistic mass or the momentum of the photon? it would be infinite, which we know is not true.

because maybe (but not likely) photons might move at an ever-so-slightly slower speed than c. this is not a contradiction in terms ("speed of photons" <> "speed of light") because c is originally defined to be the wavespeed of light. what you get when you solve Maxwell's equations:

$$c = \sqrt{\frac{1}{\mu_0 \epsilon_0}}$$

it is both elegant (and i think accurate) to say that the particle speed of light is the same as the wave speed. i hope it's true, but if it isn't, it's so close to the truth that if photons do have rest mass, i think the experiments (i think for a frequency of around visible light) were that it would come out to be less than 10-54 kg. at least that was the order of magnitude i seem to remember reading.

13. Nov 6, 2008

### Yor_on

Looking it up I found that the classical definition of momentum is p=mv where momentum (p) is mass (m) times velocity (v) and for an electromagnetic wave (photon) the momentum is h/λ where λ is the wavelength and h is the Planck constant.

Both is accurate descriptions it seems, but do they explain the 'force' itself, and should I see that 'force' as belonging to the objects?

And now you give us another edguy99:)
Do you mean that it could be that 'mass equivalent of the energy of the photon' that gives it a momentum?
It breaks my idea that it's the 'masslessness' of the photon that allows it 'c' as well as its instant acceleration then?
Or is 'equivalent' just a indirect expression and description, without involving any mass at all?

We do have invariant mass.
I just need to see the scale weighing myself to know that:)
And we have photons acting as if they had something equivalent.
And if we look at invariant mass moving it will have a 'momentum' too.

As well as inertia.
What I wonder is if momentum and inertia can be seen as a properties of spacetime?
Acting on ? mass ?

14. Nov 7, 2008

### edguy99

Equivalent would involve no mass.

1/ Consider on or near the surface of the sun. An electron in a hydrogen atom drops from the 3rd to the 2nd energy level and emits a red photon (energy about 2.1 evolts and wavelength of 656 nanometers) and this photon happens to head towards earth.

2/ Lets draw a picture of the sun on the left, earth on the right and this "thing" the photon taking about 8 minutes to reach the earth. Since it is massless, it travels at c (300picometers/attosecond).

3/ To "see" the effects of the photons momentum, zoom into the back of your eye if you glance at the sun. That photon hits an electron in a retinal molecule causing the electron to jump to a higher energy state and break a bond in the retinal molecule setting off a chain reaction in your brain causing you to "see" the photon.

The momentum can be seen as this fast little massless photon hitting the big slow electron and causing it to move a bit.

15. Nov 7, 2008

Staff Emeritus
I'm not sure what to tell you other than you solve the relevant equations. A massive photon replaces the Maxwell Lagrangian with the Proca Lagrangian, and when you solve, you find that you have a force that falls off faster than inverse square.

16. Nov 8, 2008

### Phrak

Thanks, Vanadium. That's exactly what I needed. I'm not concerned that photons should be massless, but how those would imply massive/massless.

17. Nov 8, 2008

### Yor_on

Thanks Edguy.
So here energy and invariant mass is clearly divided as being two different subjects then?
As both photons and invariant mass share momentum but seems very different in their properties.
The only thing I see joining them is the idea of energy, but if so energy isn't mass alone, or?
If you see how I think.

18. Nov 8, 2008

### Naty1

I'm guessing, but do not know, the replacement of the Maxwell Lagragian reflects an ever so tiny mass, hence slightly slower interaction via the EM field, or equivalently, a slightly slower photon. A slower photon would also curve a bit more in a gravitational field.

The discovery that light has a finite speed ended the concept of Newton's universal time and clearly defined cosmic distances...they become observer dependent. A related understanding is that Coulomb's law holds rigorously only if the electric charges are at rest with respect to each other.

19. Nov 8, 2008

### Naty1

Everything has energy, therefore everything has relativistic momentum via mass-energy equivalence. It's an underlying symmetry, just like velocity changes distance and time....there are many subtlies our senses are not tuned to detect. Math sometimes helps pluck out some of the ambiguity

20. Nov 8, 2008

### Yor_on

Thanks for that Naty.
But I still don't get it:)
You say "everything has relativistic momentum via mass-energy equivalence"
But what kind of 'mass-energy' conversions are you referring to?
The photon has no mass?