# If light has momentum then wouldn't it have force?

1. Aug 18, 2010

### zeromodz

We know that light has momentum, but doesn't have mass. So if light reflects of say a mirror. The change in momentum is zero. If it is zero then we have a problem.

ΔP = ΔFT
ΔT = ΔP / F
ΔT = 0 / F
ΔT = 0

This shows that light is instantaneous because the time it takes for light to bounce of something is zero. Is there something I am doing wrong here? And if I am wrong and there is a change in momentum then wouldn't light create a force when it reflects?

Last edited: Aug 18, 2010
2. Aug 18, 2010

### AJ Bentley

You are, there is and it does. Look up 'solar sails' in Google.

3. Aug 18, 2010

### LostConjugate

I see your error. Light has relativistic mass, otherwise known as kinetic mass, or kinetic energy, or just mass. You may have heard that light is like an object that has 0 rest mass, or no mass in it's rest frame, that does not mean that light doesn't have mass. Light in many ways is the most fundamental way to understand what it means to have mass.

This is all well explained in a Special Relativity course.

4. Aug 18, 2010

### diazona

Energy, not mass. Most modern physicists would say that light has energy, not mass, and that the term "relativistic mass" is potentially misleading and discouraged.

But other than the choice of terminology, you're right, of course

5. Aug 18, 2010

### Andy Resnick

You are on the right track. It's more appropriate to consider the *pressure* (force/area) that light can exert on ponderable matter; pressure is a more appropriate concept because the field is distributed over space, not concentrated at a point.

6. Aug 18, 2010

### zeromodz

So light itself exerts a pressure on you. That is really interesting.

7. Aug 19, 2010

My understanding is light can exert pressure to mass through coupling. Kinetics does not apply because Photons have zero mass. I think that the transfer is mediated by the W and Z bosons. The kinetics can however be applied when the momentum is transferred from the W and Z bosons to mass because they both have mass.
TM

8. Aug 19, 2010

### Dr Lots-o'watts

A metal having been hit by a high energy laser pulse can sound as though it's been hit by some sort of small hammer.

9. Aug 19, 2010

### LostConjugate

Since when is there a difference between energy and mass? Perhaps before 1905.

10. Aug 19, 2010

### Dr Lots-o'watts

Not a difference, but a factor of c^2, when momentum is 0.

Saying mass and energy are the same because of e=mc^2 is like saying energy and height are the same because e=mgh. Don't get things mixed up. Interchangeable does not mean the same.

11. Aug 19, 2010

### diazona

Um... since probably around the 1950s or '60s? I think :shy: Actually Dr. Lots-o-watts is completely correct, there is a factor of c², but I'm talking about more than that...

It's most clear to talk about it in mathematical terms. The formula that relates the two is
$$E^2 = m^2c^4 + p^2c^2$$
(direct from SR) In modern terminology, E is the energy and m is the mass. In older terminology, m was "rest mass", and there was this thing called "relativistic mass" (which is what you are calling "mass") that was defined as E/c².

12. Aug 19, 2010

### LostConjugate

Ok, first of all the factor c^2 should not be there, unless you are working in different units of time and space, but thats like working with meters and inches. In SR you should always work with c = 1 unless your working on a special project that involves meters or feet or whatever.

So we have $$E^2 = m^2 + (mv)^2$$

But $$m$$ and $$E$$ are equivalent, so $$(mv)$$ is just $$E_2v$$ (where $$E_2$$ is some energy in the object)

and $$v$$ is just d/dt which is another form of $$E$$, lets say $$E_3$$. So we have $$E - (E_2E_3)^2 = m$$

So $$E_t = m$$ where $$E_t$$ is the total energy of the object.

Mass is the result of field permutations, objects that was see and feel as tangible solid objects are just persistent illusions.

Last edited: Aug 19, 2010
13. Aug 19, 2010

### ZapperZ

Staff Emeritus
You should not confuse "short-formed notations" with actual physics. In physics, we use MANY shortcuts. We often set h=c=k=1. This means nothing physically other than an easy form of notation. Simply looking at the dimensions alone is sufficient to indicated that they are not identical.

There IS a difference between mass and energy. An electron isn't JUST made up of mass that can be converted into energy. It's spin must also be taken into consideration, and here, a one-to-one conversion of energy into mass cannot occur without other external factors coming into play. The same with a photon being converted into mass. You can't just take a photon and completely turn it into just one particle.

Zz.

14. Aug 19, 2010

### LostConjugate

Well I wasn't trying to say that we as humans have the technology to make these conversions, just that we first defined mass, then we defined energy, then we found that they are one and the same. Mass is just a word we use, but it means no more than internal energy of the object.

I don't think we set c=1 just for convenience, that is much more natural unit to work with, meters is just a random length we made up to measure things on earth. c was found empirically.

15. Aug 19, 2010

### ZapperZ

Staff Emeritus
Yes, it is found empirically, and it's number isn't 1, and it has a dimension.

I work with stuff a lot by setting h or h-bar=1. Look in any photoemission papers and you'll see that "energy" has a symbol of omega, which is angular frequency. This isn't physics. It is short-hand notation.

Mass also is not a word. In the Standard Model, mass can interact with the Higgs field, while photons do not! So already here, the physics is different!

Zz.

16. Aug 19, 2010

### LostConjugate

I didn't know about that. I thought the Higgs field was still just theory. I will look into that, a photon should be effected by it just as it is by the gravitational field. I don't know anything about the Higgs field though other than LHC has not been able to prove its existence.

-- I was trying to say because it was found empirically it should be normalized.

17. Aug 19, 2010

First, you're missing $$\gamma$$ factor in momentum. Second, you're mixing rest mass (first equation) and relativistic mass (last equation). Third, that mumbo jumbo in the middle is just wrong. (How did you come to conclusion that velocity is energy?) The dimensions are completely off, which is why you should keep the c factors in the equations, unless you know what you're doing (hint: E - E^4).

18. Aug 19, 2010

### LostConjugate

All I did was remove c^2 from their equation.

velocity is just another form of energy, the exact amount may be 1/2v^2 per unit mass but that is irrelevant.

Either way, I am not sure if your trying to say that mass is something more than just energy or not.

19. Aug 19, 2010

I understand very well what you did.
Saying that velocity is just another form of energy is one thing (philosophy), and making equivalence between E and v is another (bad maths).
That's hardly irrelevant. The square there is very important as well as that 'per unit mass'.
I consider questions like this to be too philosophical for my tase. I am just pointing what I think are errors in your derivation.

I understand what your point is, but why are you trying to make it more rigorous by showing horrible mathematical skills?

20. Aug 19, 2010

### LostConjugate

I was trying to show that the equation provided instead of E = Mc^2 is really just the same thing, it still shows that energy and mass are equivalent.