# Re: Why?

1. Feb 12, 2005

### godzilla7

Re: mmmmmm

can anyone explain why we consider light to be massless is this a philosophy question then?

just trying to understand how something can be a particle and a wave yet have no mass. Just starting physics degree and need to get to grips with some of the fundamentals is this a difficult question, or do we just not know?

Last edited by a moderator: Feb 12, 2005
2. Feb 12, 2005

### Staff: Mentor

3. Feb 12, 2005

### DB

"UV photons are more energetic than visible photons, and so are more "massive" in this sense, a statement which obscures more than it elucidates"

Am I right here?
UV photons are more energetic because a stronger energy or interaction sent them off. Their energy is dependent on only frequency and momentum. If we were to say they were more massive then visible photons, this would have a huge effect on their speed and frequency. And you couldn't consider them light. So saying that UV photons are more massive, is strictly relativistic and isn't worth saying right?

4. Feb 12, 2005

### dextercioby

No,the adjective "massive" strictly refers to "rest mass".Period...

Daniel.

5. Feb 12, 2005

### DB

So you're saying UV photons have a higher rest mass? Is that possible?

6. Feb 12, 2005

### dextercioby

If zero UV is greater than zero visible,then it must be so...

It means that we've invented the mathematics which sets two zero-s unequal...

Daniel.

7. Feb 12, 2005

### DB

Wow thats new for me and hard to understand. Is there a name for such a concept or any keyword that I could google and read about?

8. Feb 12, 2005

### Staff: Mentor

Right about what? You are quoting a line from the second article I linked. The line before this one sets the context: "On the other hand, the "relativistic mass" of photons is frequency dependent." The author's point is that the concept of "relativistic mass" is old-fashioned and confusing; don't use it unless you know what you are doing.

The energy and momentum of a photon depend on its frequency.
Huh? They have a greater "relativistic mass", but so what? The invariant mass of both is zero (that's the one that counts). Their speed is still c.
Reread those articles. I'm not sure you got the message.

9. Feb 12, 2005

### dextercioby

Ouch,if you didn't get my joke,it must have been really bad,huh...? :yuck: No,there's no math.And two zero's will be always be the same.
Photons are massless,because the equations which tell it are confirmed by experimental results...
We have no reason to think otherwise.

Daniel.

10. Feb 12, 2005

### Staff: Mentor

I thought it was funny... but maybe that says something about me.

11. Feb 12, 2005

### DB

I ment am i right about the following paragraph i wrote, not the quote.

12. Feb 12, 2005

### Staff: Mentor

OK. (Sorry if I mistook your statement.) My comments on that paragraph are above. No, I'd say you were not right (mostly). While it is perfectly possible to describe things using "relativistic mass" (if you know how that's defined), I would advise against it: stick to the usual meaning of mass (the invariant mass or "rest mass"). All particles, including photons, have a fixed mass that is independent of inertial reference frame.

13. Feb 12, 2005

### dextercioby

BTW,the concept of "relativistic mass" is rather fuzzy,meaning that you 2 options to think about when hearing it:the mass from SR (namely with the 'unaltered' gamma factor),or the one from GR (with the 'altered' gamma factor involving the gravitational potential)...
So "mass" means "rest mass" in all contexts..

Daniel.

14. Feb 13, 2005

### godzilla7

Re: Thanks

Thanks for your replies, don't encourage dexter you'll give him a big head; besides I am the smartest person on this forum

15. Feb 13, 2005

### godzilla7

Re: interesting

I see, we are not saying absolutely that light has no mass, but we believe it to be zero or if not very very small.

What if light Had mass, any implications other than those in the 2 articles.

Also the bit about why light is bent by objects, that is because an object such as the moon bends space time and so the light is not attracted by the gravity merely deviated by the warping of the fabric of space, this is true is it not?

Anyway thanks again.

16. Feb 13, 2005

### masudr

No we are saying that light has no rest mass. The trick is to realise that mass is nothing special. It's mass-energy that we are really interested in, since one can change into another. It is all contained in the equation $$E^2=(m_0c^2)^2+(pc)^2$$.

17. Feb 14, 2005

### godzilla7

Re: mass

Yes I can see that now, rest mass, relative mass; what would happen if we could stop light, I.e. make it travel at a velocity of 0? what would happen to the wave, would it cease to exist? I've heard that containing light or at least controlling it would be a rather usefull skill to master. I know they have stopped light or at least slowed it to very low speeds, can we use this to get any insight into the way light propogates?

18. Feb 14, 2005

### masudr

The speed of light will never change in any inertial frame (i.e. any co-ordinate system which is not accelerating). When they say they have stopped light, what they really mean is that they have got a bunch of photons that are being absorbed and re-emitted within the atoms of some material such that the group of photons don't propagate through the material at light speed. The photons themselves actually do travel at light speeds; it's the added delay in absorption and re-emission that "slows" the light down.

In fact, light has energy given by $$E=hf$$ and according to the equation I have put above, knowing that $$m_0=0$$ for photons, we have

$$hf=(pc)^2\Rightarrow p^2=\frac{hf}{c^2}$$

showing that light can never be stopped.

19. Feb 15, 2005

### godzilla7

Light can't be stopped!

Hey when I turn off the light switch it stops it thanks for posting, interesting stuff.

20. Feb 15, 2005

### masudr

Sorry, I made a rather large error in my previous post.

I said $$E=hf$$ so what we actually get is

$$(hf)^2=(pc)^2 \Rightarrow p=\frac{hf}{c}$$

As an interesting aside, we remembering that $$v=f\lambda$$ for a wave, and that for light v=c we get $$fc^{-1}=\lambda^{-1}$$

and finally putting that together with what we had at the start gives us

$$p=\frac{h}{\lambda}$$

which is the de Broglie hypothesis. No problem for the post, by the way.