# How to visualize E-MC2

1. Mar 2, 2006

### Chaos' lil bro Order

How to visualize E-MC2

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Recently I thought of an analogy that I thought might be useful for visualizing how energy and mass are equivalent. It will sound simplistic at first, but it becomes more and more powerful as you continue reading.

Say you are standing in front of a portable swimming pool that has walls 2 meters in height and to which your eye level is also exactly 2 meters high. The pool is filled with water to a height of 1.8 meters. You get four of your friends to each stand at a corner of this square pool and tell them to simultaneously drop a bowling ball into their respective corner of the pool. Each bowling ball impacts the water and sends a wave moving outwards from the point of impact towards the pools center (forget about reflections against the pool's walls and the waves that are not emitted towards the pool's center). You then notice that at the exact center of the pool all 4 waves constructively interfere with one another and create a large swell in the water level that peaks in amplitude at a height of 0.3 meters above pool sea level. Standing outside of the pool and looking straight at it with your eye level still at 2 meters in height, you observe this swell of water as rising 0.1 meters above the pools walls (pool walls = 1.8 meters high, swell = 0.3 meters high, your eye level = 2.0 meters high; so [1.8 + 0.3 - 2.0 = 0.1 meters]).

This swell, this peak is analgous to matter!! When enough waves (analogous to energy) overlap and constructively interfere to create a large enough peak amplitude ('swell') for you to observe, you see this amplitude as matter.

With this analogy in mind, picture all the material objects that you are familiar with in your everyday life as being merely comprised of several waves of ElectroMagnetic energy that constructively interfere with one another to such a great degree that their amplitude forms what you perceive to be as matter.

Good analogy or not?
What do you think?

2. Mar 3, 2006

### topsquark

So long as you take care to remember that it is just an analogy. After all, matter is not made of EM waves! But the concept is otherwise sound.

-Dan

3. Mar 3, 2006

### pervect

Staff Emeritus
I don't get the point of the analogy at all.

E = mc^2 is just a special case of E^2 - (pc)^2 = (mc^2)^2, when the momentum, p = 0.

The above is the special relativistic formula that relates the energy, E, momentum, p, and invariant mass m of a particle. It is true for any paticle in special relativity.

When p=0, we find that E is not zero, but instead E = mc^2.

This doesn't have anything at all to do with constructive interference of waves.

4. Mar 3, 2006

### Chaos' lil bro Order

I think your math is wrong pervect.

Non-relativistic energy is given by: E=mp*2, where E is energy, p is momentum, m is the mass of the object. In other words, Energy quadruples as momentum doubles.
Or, if you prefer and example on earch E=mgd (energy expended = mass x 9.8m/s x distance travelled).

E=mc*2 simply uses the Gamma function to show how as velocity approaches C, energy approaches infinity, since mass approaches infintity.

Remember, Pair-Production creates particles out of thin air by converting EM waves into particles. Example, 1GeV photon can create an electron/ positron pair (0.511MeV + 0.511MeV). So in fact, you can say that matter is made of EM waves.

5. Mar 3, 2006

### Pengwuino

Chaos,

$$E = mp^2$$
or
$$E = m^3 v^2$$

is meaningless as far as i can see. It is not the total energy of a mass nor is it kinetic nor any weird combination of the two. Also, pair production does not happen out of thin air… unless we literally mean air at low pressure which means that it indeed can happen since it is in matter

6. Mar 4, 2006

### pervect

Staff Emeritus
Whoops! The non-relativistic formula is actually E = p^2/2m, though in spite of this error your remark that the non-relativistic energy quadruples as momentum doubles is correct. Substitute the non-relativistic formula for momentum in the above, and you'll see that p^2 / 2m = m v^2 / 2, while your expression gives the wrong answer.

This has little to do with the correct relativistic formula for energy in terms of momentum, however.

The formula I quoted is a very standard relativistic formula. If you'd like some references, see for instance:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html
or

http://scienceworld.wolfram.com/physics/Energy.html
(equation 9).

The origin of the formula is especiallly clear when the 4-vector formalism is used. Mass (by which I mean invariant mass, aka rest mass) in SR is then defined as the Lorentz invariant quantity associated with the energy-momentum 4-vector.

In a system of units where c=1 (known as geometric units) the formula becomes especially simple:

E^2 - p^2 = m^2

The fact that energy is not zero when momentum is zero is a consequence of this relativistic formula for mass - this is historically how Einstein originally came up with the formula. The formula E=mc^2 also gives the correct energy release for the energy released in matter/antimatter reactions (pair production or pair anhillaiton).

7. Mar 6, 2006

### Chaos' lil bro Order

pervect

I'm having trouble following your posts.

Here is what my point was....

If two Gamma Rays with energies of 0.511 MeV annihilate one another, they create an electron and positron pair each with 0.511 MeV or rest mass, respectively. If the two Gamma Rays had energies of say 0.600 MeV, the extra energy would be conserved by turning into the kinetic energy of the electron and positron, respectively.
In this way, isn't matter created from from energy? The electron from the photon? This is the equivalence that my analogy tried to show. Perhaps the analogy would have been better served to say two waves approach eachother from opposite directions and where they meet a nodal point occurs, both waves are annihilated and matter is pair-produced at this nodal point?

8. Mar 6, 2006

### ZapperZ

Staff Emeritus
Can you show me experimental evidence for this "gamma rays annihilation"? Do you think this is how we currently create an electron-positron pair for a particle accelerator/collider?

Zz.

9. Mar 6, 2006

### pervect

Staff Emeritus
I'm just trying to point out that the origin of E=mc^2. Aside from the references I quoted earlier, you might try the wikipedia article

http://en.wikipedia.org/wiki/E=mc²

It all boils down to the formula that (assuming c=1), E^2-p^2=m^2, which implies that a body with a non-zero mass must have energy even when it's not moving.

10. Mar 10, 2006

### Chaos' lil bro Order

I can't tell if you are being sarcastic or not.
I 'think' electon-positron pairs are produced a number of ways. Gamma annilation being one way it can happen. Colliding a proton into a gold nucleus should also produce them along with various mesons and baryons depending on the collision energies.

Since you have almost 5000 posts I'm assuming that you are just berating me, if so why? If not, share your knowledge and correct me.

I'm not aware of any experimental proof of gamma annihation/ electron-positron pair production. If you know of any, post it, i'd be curious to read it.

11. Mar 10, 2006

### Chaos' lil bro Order

pervect

Suppose the 'raised bump' was a product of such great wave constructive interference that it 'lapped' up out of the water (production) into the air for a brief moment before plopping back down in the water (annihilation). Once it rejoined the water it would in effect send waves outwards (gamma production) akin to the waves that originally produced it (albeit smaller waves).

Let me preempt your response by saying, the positron is the dibbit in the water from where the electron 'lapped' upwards.

12. Mar 10, 2006

### ZapperZ

Staff Emeritus
No, I actually DID want you to point out to me if we have any experimental evidence for a photon-photon anhilation. The physics of gamma-gamma collider is only PREDICTED by QED, meaning we have zero such experimental evidence. A gamma-gamma collider is one of the possiblility that is envisioned at LHC but still up in the air. The cross-section for scattering even for gamma is extremelly small.

Hence, this is still speculative, very much like the Higgs. So your claim that such a thing can produce e-p pairs is not supported by solid physics, but rather still highly speculative. Now if you want to do this, fine, but at least be aware that you ARE doing speculative work (and we'll apply our Guidelines to your post). But don't confuse this as having the same validity as the typical way to created e-p pairs.

So would you want to argue for the validity of your idea using an unverified, speculative example?

Zz.

13. Mar 10, 2006

### Ich

Annihilation is possible, and those (EM) processes are all reversible in time. So pair creation must be also possible. I don´t know about any evidence for it, but I recall from my astrophysics lessons that the intensity (or energy) of jets is supposed to be limited by the onset of pair production.

14. Mar 10, 2006

### ZapperZ

Staff Emeritus
But pair production from EM is NOT restricted to photon-photon anhilation. Pass gamma rays through tungsten. You can get e-p pair EASILY! In fact, this is THE prefered way of generating e-p pair with a high quality of positron. This is the method that the International Linear Collider working group is going to use if the ILC ever gets built. This is a proven technology.

But this is NOT photon-photon anhilation. It is a photon producing e-p pair near a medium (tungsten solid) that is required for conservation purposes. We have no photon-photon collier and what comes out of it are still an aspect of QED that is still being studied.

Zz.

15. Mar 10, 2006

### EL

In theory, Ich, at the moment just in theory...(however I don't doubt its existence at all, but for the final verification we may have to wait a few years)

Last edited: Mar 10, 2006
16. Mar 10, 2006

### Ich

I found the old lecture notes of professor Schönfelder (MPE, Garching). In the section about quasars he writes (my translation):
The gamma spectrum must be cut-off above a threshold energy due to photon-photon interaction. Gamma quanta can interact with other photons and create electron-positron pairs. 100 MeV gamma quanta are mainly absorbed by x-quanta above 2 keV.

That´s not evidence, but it seems to be commonplace for astrophysicists.
http://arxiv.org/PS_cache/astro-ph/pdf/0509/0509095.pdf" [Broken] is an example of the application of this effect. There are not only human-built colliders.

Last edited by a moderator: May 2, 2017
17. Mar 13, 2006

### Chaos' lil bro Order

Thanks Ich. I think we are right in this case.

Aside from experimental evidence, every astrophysics paper I've ever read has considered gamma e-p pair production an axiom. Also, I would site CPT Invariance as indirect evidence that gamma e-p production is a fact, since e-p annihilation produces two gammas.

18. Mar 13, 2006

### Chaos' lil bro Order

zapper

'But pair production from EM is NOT restricted to photon-photon anhilation'

Ummm, so you just admitted that gamma-gamma annihilation does produce e-p pairs, albeit so does gamma-tungsten interactions. Guess we agree then?

Also, for the record, its spelled 'annihilation' not 'anhilation' as you've written more than once. (I'm not trying to be picky, I just thought you should learn how to spell it correctly)

19. Mar 13, 2006

### ZapperZ

Staff Emeritus
Then I suggest someone tell those silly folks at CERN to stop trying to come up with an upgrage to the LHC to allow for a gamma-gamma collider, since such a thing is a done deal already from astrophysics "observation". In fact, ALL high energy physics experiments should be stopped. It's a waste of astronomical amount of money to build something to reproduce all those results already found in those quasars.

Sometime, people, we really need to distinguish the degree of certainty in things like this, and whether such an observation is really that certain.

Zz.

Last edited: Mar 14, 2006
20. Mar 13, 2006

### ZapperZ

Staff Emeritus
No we don't. Unlike you, before I base my "theory" or "ways to visualize" on something, I want to make sure that "something" is well-verified. I never said anything about gamma-gamma interaction that so-and-so cannot happen. I clearly said such a thing are still in the "theorized" or speculative state.

That's why I asked you if you would like to based such a thing on speculation. If you are comfortable with that, then be my guest. But don't think that such a thing has been verified. It hasn't!

Zz.