How much does the photon weighs?

1. Jun 7, 2004

tumor

How much does the photon weighs?

2. Jun 7, 2004

Staff: Mentor

A photon has no mass. "weight" doesn't apply.

3. Jun 7, 2004

tumor

Can you explain litlle bit more please?

4. Jun 8, 2004

chroot

Staff Emeritus
Photons do not have mass, but they do have energy. Energy couples gravitationally in the general theory of relativity, and thus light does have weight. A box full of light actually weighs (very slightly) more than an identical empty box.

- Warren

5. Jun 9, 2004

turin

tumor,
Or you may be thinking of radiation pressure. This is caused by an impulse that a photon imparts to an object upon reflection or absorption.

6. Jun 9, 2004

Blackforest

Is the frequency of a unique photon "traveling" in vacuum far from any gravitational source a constant? Thanks for any explaination

7. Jun 9, 2004

Staff: Mentor

Oops - I knew that. :grumpy:

8. Jun 10, 2004

JohnDubYa

Weight is a layman's term, not a scientific term. You need to be more precise as to what you are trying to measure. Its mass? The magnitude of the graviational force acting on it? The magnitude of any normal force acting on it?

9. Jun 11, 2004

Clausius2

Hi men,

Talking about this, when I saw for the first time the Einstein momentum formula for relativistic dynamic:

p=mv/[1-(v/c)^2]^(1/2)

I wondered what happens with photons:

v=c and m=0, so It is not determined.

But it has to agree with the De Broglie Principle: p=h/L
where p=photon momentum,
h=Planck constant
L=wave lenght of the photon's associated wave.

What happens?. How can I calculate the limit when v-->c and m--->0?

10. Jun 11, 2004

theFuture

Consider m^2c^4 = E^2 - (pc)^2 where m is the rest mass. A photon has no rest mass, so E=pc. This allows you to talk about the energy and momentum of photon.

11. Jun 11, 2004

baffledMatt

That's an interesting question. Does anybody know the answer?

I would guess that to make it meaningful you have to specify the order of the limits, ie something like

m/(1-v/c) -> something.

Can someone make sense of this? Or at least tell me whether this is just nonsense

Matt

12. Jun 11, 2004

Clausius2

To be honest, my physics tutor did it some time ago, but I can't remember how. Somebody asked him this in class, and he answered something by means of L'Hopital rule.

But I have been trying to employ L'Hopital rule for this limit without success.

13. Jun 11, 2004

Clausius2

What you are saying is agree with De Broglie principle:

E=pc=h/L

and quantum Planck principle too.

But for relativistic momentum ¡t is not obvious.

14. Jun 11, 2004

theFuture

Sure it is. Asume you know the speed (energy) of the photon you now know the momentum.

15. Jun 12, 2004

Clausius2

How can I do that? Do I have to introduce c=E/p in the relativistic momentum formulae?.

The problem here is how to match Einstein relativistic momentum with De Broglie's principle, and I cannot see how.

16. Jun 12, 2004

theFuture

What I posted was a lorentz invariant (m^2c^4 is invariant under lorentz transformations). So, assuming you know the wavelength you know the energy. You know the energy, you know the momentum.

17. Jun 27, 2004

relativelyslow

if energy is the same thing as mass, couldn't you convert lights energy to mass? what would it be?

18. Jun 27, 2004

chroot

Staff Emeritus
Yes, as I said, you can certainly use radiation to produce pairs of particles -- one matter and one anti-matter. If you smash gamma rays into a block of lead, for example, you'll get a shower of matter particles out of it.

- Warren

19. Jun 27, 2004

pmb_phy

I don't believe the weight of a photon is a well defined quantity in that different people mean different things by the term weight of a photon. Normally the term "weight" refers to the force required to support an object in a gravitational field. I suppose you could say that the gravitational force on a photon is the "weight". But I personally stay away from that kind of definition myself.
By "a photon has no mass" it is meant that the proper mass of a photon is zero. However a photon has inertial mass (aka "relativistic mass" which is has due to the fact that a photon has momentum), active gravitational mass (due to the fact that light generates a gravitational field) and passive gravitational mass (due to the fact that light is deflected in a gravitational field). Keep in mind the famous experiment by Pound and Rebka on gravitational redshift. The paper they wrote on this topic was entitled "On the Weight of Photons". Also recall Lev Okun's paper "The Concept of Mass" that appeared in Physics Today in June 1989. In that paper Okun calculates the gravitational force on two photons moving in a Schwarschild field, both in the radial direction and the horizontal direction. Okun concludes ..., a horizontally moving photon ... is twice as heavy."
For a calcultion which demonstrates that a box of photons weighs more than the same box with no photons in it see the paper "On the concept of mass in relativity" which is listed at http://www.geocities.com/physics_world/
The weight of an object is a well defined quantity in physics as well as in general relativity. I like the following definition. Its what people use the most and its quite reasoable. From The equivalence principle and the question of weight, Kenneth Nordtvedt, Am. J. Phys. 43(3), March 1975
The most rigorous definition of momentum is found by the implicit rule for defining (relativistic) mass. I.e. mass, m, is defined such that the quantity mv is conserved. The quantity mv is then defined to be the momentum. The relation p=mv/[1-(v/c)^2]^(1/2) is derived by assuming the particle can travel at speeds less than c (i.e. its derived only for tardyons). If you want to express the momentum in terms of energy then its E = pc, which is derived using Maxwell's equations. It then follows that m = p/c. If you want to express it in terms of mass then defined m = E/c2.

Pete

20. Jun 28, 2004

Moe

Why? Photons have no mass. They also have no "size" in any real sense. So most of the relativity stuff just doesn't apply, like the lorentzian contractions for instance.