# Photon Mass

1. Apr 8, 2014

### mbbolton96

Lately, I have been giving a lot of thought to whether or not photons would have mass or not. I know most on here an on other sites say that they do not have mass, but I think differently.

If light has no mass, then theoretically, it shouldn't have any speed limit (186,282 mi/sec) because then there would be nothing to slow it down from being infinitely fast. Therefore, light must have matter, to slow it down the 186,282 mi/sec number.

I even calculated what the mass of a photon would be in a gamma ray. My calculation came out to be a mass of 2.2102189e-34 kg for a wavelength of 10^-5 nm.

Could someone please converse with me about this. I'm really loving this. I think I even have a way to test my theories. But, I am only a senior in high school and am not sure if anything like this has been done before.

Thanks :)

2. Apr 8, 2014

### Staff: Mentor

It sounds as if you're starting from $F=ma$ and concluding that if the mass is zero then no matter how small the force, the acceleration and hence the velocity can be arbitrarily large?

The problem with this line of thought is that $F=ma$ is a classical expression that applies only to bodies of non-zero mass subject to a constant force. Instead, you need to work with the more general relationship between force and acceleration, which is $F=\frac{dp}{dt}$ where $p$ is the momentum.

Now you need the momentum, and for that you need the most general relationship between energy, mass and momentum: $E^2=(mc^2)^2+(pc)^2$. From this you can see that a particle can have zero mass yet non-zero momentum.
(You might want to amuse yourself by seeing what comes out of that equation for a body at rest, so that $p$ equals zero).

How did you do this calculation? Starting with photon energy and then working back from $E=mc^2$?

Last edited: Apr 8, 2014
3. Apr 8, 2014

### mbbolton96

Yes, I started with E=mc^2 and E=(h*c/lambda). Then got to m=(h/c*lambda) them simply plugged and chugged

4. Apr 8, 2014

### phinds

So do you now understand that light CAN (and in fact does) have a speed limit or do you still think that you have overturned modern physics?

5. Apr 8, 2014

### Matterwave

Not to dampen your spirits or anything, but I can assure you that the scientific community is not so incompetent that we cannot solve a two step algebraic equation. If "photons have mass" was so easy to prove as "well E=mc^2=hf->m=hf/c^2", and still we insisted that "photons don't have mass", that would make us quite incompetent don't you think?

Now, where did the error occur? Well, the error occurred because the popular equation E=mc^2 is only a truncated version of the correct equation given by Nurgatory above: E^2=p^2c^2+m^2c^4. A photon has mass m=0, but momentum p=/=0. In fact, for a photon p=E/c.

But don't be discouraged! It is very easy to take the equations given by popular science as fact when in fact they may be only a portion of the truth. It's very good that you are thinking critically about physical problems.

6. Apr 8, 2014

### Staff: Mentor

Right, but $E=mc^2$ only applies when the momentum is zero and the rest mass is non-zero - that's why I suggested that you see what happens when $p$ is zero in $E^2=(mc^2)^2+(pc)^2$. With light, you're dealing with something that has non-zero momentum and zero rest mass.

The basic point here is that you don't need an infinite speed of light to make it all hang together.

7. Apr 8, 2014

### mbbolton96

But, that's not the point I'm trying to make, it was just another side road off my main point, which is there's a sure-fire way to test for photon mass

8. Apr 8, 2014

### Staff: Mentor

Try it. You'll find that no matter how strong the gravitational field and no matter what the photon's energy, the photon takes the exact same amount of time to make the upwards climb.

The photon will lose energy as it climbs uphill, but that doesn't show up as a loss of velocity, it shows up as the photon being redshifted so that the frequency, energy, and momentum are all reduced.

9. Apr 8, 2014

### ZapperZ

Staff Emeritus
You used the wrong formula, and you used it in ways it wasn't defined to be used. When you starting premise is faulty, you will arrive at nonsensical conclusion.

Zz.

10. Apr 8, 2014

### Enigman

11. Apr 9, 2014

### DrewD

Others have stated this, but I'll try to be more explicit. $E=mc^2$ is derived by first assuming that the speed of light is measured as a constant value, $c$, in all reference frames. Another assumption needed to arrive at this equation is that momentum for the object under consideration is zero. The fact that light has momentum can be derived classically (with no mass involved) or measured, so you cannot just plug and chug with this choice of equations. Use the equation that Nugatory gave you to explore possible values for $m$ and $p$ with given values of $c, \hbar,$ and $\lambda$. You will find that $m=0$ is possible. Experiments have shown this to be true (to varying degrees of precision).

It is an honest mistake to make. If you haven't studied special relativity or electromagnetism in depth, the nature of light is not clear. There were arguments about light for a centuries, but they have been (mostly) put to rest.

12. Apr 9, 2014

### Staff: Mentor

Photons do not have mass:

Last edited by a moderator: May 6, 2017
13. Apr 9, 2014

### pervect

Staff Emeritus
Well, if the FAQ didn't convince you, I don't see why you'd listen to anything that anyone else said.

It's one thing to have specific questions about the FAQ, but to say that the FAQ is wrong and then proceed to trot out your own personal theories doesn't go over well with me.

I don't get the impression you are trying to learn :-(.

14. Apr 9, 2014

### Matterwave

Ok guys, I think we've told this guy enough times photons don't have mass...no need to beat a dead horse...don't scare away the new comers D:

15. Apr 18, 2014

### Andrew Mason

The issue is really whether light travels at different speeds to different observers (ie. observers who are in motion relative to each other). Your theory is that a photon does travel at different speeds to differently moving observers.

If you are right, then the arrival of a photon moving in the direction of the earth's motion and being reflected back to a detector would not coincide with the arrival of an identical photon emitted at the same time and travelling the same distance perpendicular to the direction of the earth's motion.

But this experiment has been done over and over again. It is called the Michelson-Morely experiment. No difference has ever been found. Light appears to travel at the same speed relative to all observers regardless of how fast they are moving.

If you are going to maintain that the photon has mass, you will have to explain why we cannot observe it to travel at different speeds for different observers. Gook luck on that!

AM