# Very basic questions about energy

1. Jun 10, 2005

### Ratzinger

Have three easy questions.
1. When converting mass to energy, is this energy equivalent to the term field. To we get photons (lots of them and high energetic) for matter?
2. Creating matter/ particles requires energy. Again, what is meant here by energy? Colliding two particles very fast yields new matter, right? Can high energy fields (gamma rays) create real matter? If so, how and where (in matter and vacuum?)?
3. Virtual matter can be created in vacuum. Does vacuum mean complete nothingness devoid also of photons? Or is it that photons turn into mater for a moment depending on how much energy they carry?

Utterly confused I ask you clever people for help.

2. Jun 10, 2005

### mathman

High energy (above 1.022 Mev) gamma rays can form electron-positron pairs in the presence of matter, such as lead - used for shielding.
Right after the big bang, particle-antiparticle pairs were formed when pairs of gamma rays collided together.

3. Jun 11, 2005

### Ratzinger

hello world?

Thank you mathman for replying, but I also would like to know why and how that’s happening.
Why is no one else answering here?
1. questions too silly
2. too unclear formulated
3.not interesting enough
5. too basic, insult your intelligence

Last edited: Jun 11, 2005
4. Jun 11, 2005

### jackle

There are a few different processes that do it and I'm not sure which you are referring to. Nuclear fission, fusion and decay or annihilation of anti matter can all produce gamma rays (high energy photons). Each of these are a seperate topic.

"Colliding particles" sounds like you are thinking of particle accellerators. If I recall correctly, you can get Muons just from an ordinary electron with a lot of kinetic energy (without any collisions). Sometimes collisions create new particles, but not in everyday situations. Gamma rays with enough energy like to create electrons and positrons. The process is called pair production if you want to look it up.

The definition of a vacuum can be argued over. Vacuum is really complete nothingness but we know that what we normally call "vacuum" can have stuff in it. More stuff, if you include things that aren't there for very long. Also there is a lot of empty space in ordinary stuff. But you won't get any energy or any matter out of a good vacuum though. Not for keeps anyway.

Most ordinary people get confused by this stuff. I'm a novice myself

5. Jun 14, 2005

### Ratzinger

E=mc^2

Thank you jackle for answering. Things getting clearer. So just for the record: the E in E=mc^2 stands for energy in form of photons. So E=mc^2 is the conversion formula of matter into photons? Could just one of you give me confirmation on that, please?

6. Jun 14, 2005

### jackle

E doesn't just apply to energy in the form of photons.

7. Jun 14, 2005

### dlgoff

The correct equation, or should I say complete equation, is:
$E^2=m^2c^4+m^2 p^2$

8. Jun 15, 2005

### Ratzinger

E doesn't just apply to energy in the form of photons.
Yesterday 05:14 PM

Come on, 211 views and no one able to explain E=mc^2???

9. Jun 15, 2005

### mathman

If ALL the mass was converted to energy, then it would be (essentially by definition) photons. However most reactions involve some particles in and other particles out.

10. Jun 15, 2005

Staff Emeritus
Energy is not equivalent to field, or to matter. It is its own "thing" with its own dimensional units. To get an amount of energy that corresponds to an amount of something else you have to multiply by some constant. For example to get the amount of energy corresponding to a certain mass at rest you mutiply by the speed of light squared, a constant which changes the mass units to mass times length squared over time squared units of energy.

Because of conservation of momentum constraints, a single photon, however high its energy, can't create a massive particle pair in the vacuum. However if another massive particle can be part of the interaction then all of the particle's energy, if it is high enough, can be converted into massive particles, together with their kinetic energy.

The photon's energy is proportional to its frequency and the particles' energy is partly in their mass and partly in the kinetic energy of their motion, and the total energy in has to equal the total energy out.

You are correct that colliding a particle and its antiparticle will produce a burst of high energy which will go into particles according to the various conservation laws and calcuable probabilities.

In quantum field theory, the vacuum is empty of OBSERVABLE particles. However there also in QFT are unobservable particles whose product of lifetime by energy is too small to amount to a full quantum, so they can't be observed, and the so-called vacuum is alive with the births and deaths of these particles, which are called virtual.

Hope this helps.

11. Jun 15, 2005

### Kane O'Donnell

Also, treating E = mc^2 as a conversion equation is a little bit more subtle. As dlgoff has already pointed out, it's a reduced form of the more general equation in the case when the object with mass m is at rest. What E = mc^2 says is, if we ignore potential energy, an object with no kinetic energy still posseses some energy by virtue of having a mass, and in fact there is a *lot* of it, because c^2 is a very big number.

The reason people use it as a conversion equation is that in *certain* situations the mass-energy can be converted into other forms, for example into the kinetic energy of a decay product (like an alpha particle or a gamma ray). However, in many other interactions it isn't relevant because the mass-energy is effectively locked away.

Kane

12. Jun 16, 2005

### jackle

What about neutrinos? Aren't they often considered energy?

13. Jun 16, 2005

### mathman

Neutrinos are leptons (like electrons) and are treated as particles. Whenever they are produced, as in many reactions, they carry kinetic energy.

14. Jun 17, 2005

### Ratzinger

Yes, it did. Thank you