# Transform Mass in energy

1. Apr 9, 2008

### eoghan

Hi there!
I've a question about special relativity: how can I transform mass in energy? I mean.. if I take an hammer and I beat a table, I transfer energy to the table, but why this energy isn't transformed in mass?
And if I have a mass, how can I transform it in energy? If I break an atom I free energy, but why can't I have energy breaking a glass or something else?

2. Apr 9, 2008

### mathman

For all day to day activity involving mass-energy conversion, such as heating things up on a stove, the mass change is so small as to be essentially unmeasurable. It is only with nuclear reactions (fission, fusion, radioactive decay) or extreme accelerations (particle physics devices) that mass changes are significant.

3. Apr 10, 2008

### eoghan

Why?

4. Apr 10, 2008

### pmb_phy

Mass is never converted to Energy since both mass and energy are conserved quantities. Only the form of the mass can change, e.g. from proper mass to mass of motion.
The mass of a particle is a function of the particle's speed. The mass is practicly unchanged for velocities which are much less than the speed of light. But even if v = c/1000 the speed is enormous yet the mass will be about the same as the proper mass.

Pete

5. Apr 10, 2008

### Tachyonie

I am a littlebit confused here. What happens in annihilation process then? (matter + antimatter)

6. Apr 10, 2008

### lbrits

Mass is perfectly capable of being transformed into energy and vice versa. Some people confusingly talk about relativistic and rest-masses, but this is not standard practice anymore. The only "mass" a thing has is the number you get when you put it on a scale, i.e., it's rest mass (that is an relativistically invariant quantity and is therefor meaningful).

The thing that is conserved is proper four momentum $$p^\mu$$. So you may take an electron and positron and allow them to annihilate into two photons. In the rest frame of the e-p pair, you'll have

$$p_1^\mu = (m, 0, 0, 0)$$
$$p_2^\mu = (m, 0, 0, 0)$$

and afterwards, the two photons will have proper momenta, say,
$$p_1^{\prime \mu} = (m, m, 0, 0)$$
$$p_2^{\prime \mu} = (m, -m, 0, 0)$$

You'll notice that four-momentum is conserved, but the individual photons don't have mass (you can't go into a photon's rest frame).

7. Apr 10, 2008

### lbrits

I realized that we hadn't actually answered the question. The reason you can't, under normal circumstances, create mass using energy is because you need a lot of energy in concentrated form. The lightest particles (ignoring neutrinos for now) are electrons, with a mass of 0.511 MeV/c^2. On the other hand, visible light comes in energy lumps of around 10eV. So you really have to have very concentrated amounts of energy. Said differently, you'd have to hit your hammer pretty hard before you created electrons.

On the other hand, there is no threshold for creating photons, since they are massless. I guess you could say the mass of a particle is the term that goes in the energy relation $$E = \sqrt{p^2 c^2 + m^2 c^4}$$. So we can create photons of arbitrarily little energy (fortunately, else all would be dark).

Coming back to my photon example, note that although the photons don't have mass, if you had the electron/positron pair in a box, and weighed the box, allowed the pair to annihilate and could somehow weigh the box again before the photons escaped, the box would weigh the same. Roughly speaking, the "net" four momentum is still $$\textstyle\sum p^\mu = (2m, 0, 0, 0)$$ which is of a massive particle at rest. There is a saying that says a hot potato weighs more than a cold potato, and I hope you can see why this is the case =)

8. Apr 10, 2008

### dst

Are you sure about that? Since there is an upper bound on a photon's energy (when wavelength hits planck length, IIRC?) then isn't there a lower bound?

9. Apr 10, 2008

### dst

Not at all, any concentrated energy has a certain mass. Chemical bonds have a characteristic mass even. I don't remember if it's experimentally verified but I imagine so.

10. Apr 10, 2008

### lbrits

We don't know if there's an upper bound or not, since we don't know what happens at the Planck length. Something might take over that keeps everything smooth. On the other hand, I suspect that QED behaves very non-linearly far below the Planck length anyway, so "known" physics alread takes over.

There's no reason to believe that there's a lower bound to the energy a photon can posess. Now, if the photon were placed in a box (or, the universe is of finite size), then you could only create photons with certain wavelengths, so "arbitrarily small" wouldn't be correct. But, as a theory, QED doesn't forbid arbitrarily small wavelengths and actually requires it for gauge invariance, afaik.

11. Apr 10, 2008

### yuiop

Please clarify your definition of rest mass using this example:

Say we have 3 identical flywheels, A, B and C.

Flywheel A is cold and not spinning.
Flywheel B is cold and spinning.
Flywheel C is hot but not spinning.

Flywheels B and C are heavier than A when weighed on scales.
Does this mean flywheels B and C have more rest mass than flywheel A?

Is weight a good definition of mass? All 3 flywheels have zero weight far from gravity or when in freefall, but we still consider tham to have mass. Maybe a better definition of mass is the inertial definition, where the mass of a system is a measure of its resistance to being accelerated?

It could be argued that since flywheels B and C weigh more than flywheel A and require more energy to accelerate to a given linear velocity than flywheel A that they also have more inertial mass (resistance to being accelerated) than flywheel A.

When we refer to the rest mass of a system, we seem to actually mean the "rest energy" of the system as measured by an observer that measures the total momentum of the system to be zero. The "rest mass" of a system is usually quoted in terms of $mc^2$ which is in units of energy rather than units of mass.

Would it not be better to speak of the rest energy of a system when the system has zero linear and angular momentum relative to the observer?

12. Apr 10, 2008

### lbrits

I'm not too concerned with the internal details of your flywheels. When I mean "at rest" I mean we are in the rest frame, i.e., that $$P^i = 0$$. I don't mean that it isn't spinning. If you wanted to go into a frame in which it isn't spinning, then you are in an accelerating coordinate system, so you'd have other problems.

"Would it not be better to speak of the rest energy of a system when the system has zero linear and angular momentum relative to the observer?"

Usually when relativists/field theorists/whomever talk about mass they mean rest mass and in the rest frame, this is simply the rest energy. So yes, when the system has zero linear momentum. I'm going to have to say nay on the angular momentum. That isn't a good rest frame at all (different parts of your coordinate system are traveling at different speeds) =)

For particles this is all a bit overkill, but of course we need to be able to talk about bound states and stuff like that. I invite you to look at the definition of the energy momentum tensor and think of enclosing your system (potato, flywheel) in a volume. Going into the zero linear momentum frame, I believe that if you integrate $$T^{00}$$ over the volume, you will get the number you read on the scale.

But when I'm in a hurry, I simply say that the mass(squared) is whatever you get on the RHS of $$p^\mu p_\mu$$.

13. Apr 11, 2008

### Lojzek

If you hit a table with a hammer, then it's internal energy will increase, so mass will increase acording to E=mc^2 formula: it will be more difficult to accelerate the table after the hit, because it has more energy/mass.
However the increase of mass will be very small, since c is so large: if you hit the table with energy of 100 Joules, the mass will increase by about 10^-15 kg.
Converting mass to energy also happens in classical physics, but the change is again so small that it can be neglected. Breaking glass is not a good example: why would broken glass have less energy than undivided glass. A good example would be cooling of an object, relaxing a spring or slowing down rotation: in all these examples the mass would decrease proportionaly to decrease of energy.

14. Apr 11, 2008

### Lojzek

One more thing: I think that the rest mass-energy should not be considered a new type of energy, but rather one of the known energy types (kinectic, electromagnetic, gravitationat, strong/weak nuclear), which make the system more difficult to accelerate, so the easiest way to calculate them is weighing the whole system and using E=mc^2. The only exception are indivisible particles: we can't break them apart to see which energy contributes to their mass (although it might be possible that those particles are in fact composed of smaller undiscovered particles).

15. Apr 11, 2008

### yuiop

I accept your argument that we should define the rest mass of the system as one where the system has zero linear momentum but NOT zero angular momentum relative to the observer.

You seem to be agreeing that flywheels b and C have greater rest mass than flywheel A, so it appears that if we take a stationary flywheel and spin it (or heat it) we increase the the rest mass of the flywheel system. This is a little confusing because we are constantly told that rest mass can not change. I think we have to make the distinction that the rest mass always remains constant under a Lorentz transformation, but physical processes such as spinning a system or heating it can change the rest mass of the system?

16. Apr 12, 2008

### Staff: Mentor

I typically prefer the term "rest energy" to "rest mass" when describing systems of more than one particle. If you have an ideal gas in a stationary container, and you heat it up it will gain energy. The hot gas will have greater inertia and gravitation, but, there are the same number of particles in the gas as before (i.e. no more matter has been created). "Mass" is commonly understood to be a property of matter, while "energy" is easier to understand as a property of a system and I believe this is a source of confusion here.

A simple example is the photon pair created after anihlation of an electron and a positron. The two photons together have the same energy (converted from matter to radiation) as the original electron and positron, but does it make sense to say that together they have mass? I believe that while it is technically correct it is confusing.

17. Apr 14, 2008

### eoghan

Thanks to everybody

18. Apr 14, 2008

### DrGreg

That is correct. "The rest mass cannot change" is not true. "The rest mass is invariant" is true. "Invariant" has a specific meaning: it means that different inertial observers calculate the same value at a particular event (location+time). It does not mean the value cannot change over time. Such a value would be described as "conserved" rather than "invariant". So rest mass is always invariant but not always conserved.

(When talking about large bodies [or a system of particles] rather than a single point particle, it's probably better to refer to "invariant mass" rather than "rest mass", to avoid any confusion over spinning or internal motion. Or you could take DaleSpam's approach and call it "rest energy".)

19. Apr 14, 2008

### Disgod

You can sort of transform some mass into energy. It is possible to make a fusion reactor, but it is extremely inefficient and the amount of energy you get out of compared to how much you need to put in, is tiny.

Well I was going to give a link to a website that shows a homemade fusion device, but I can't post links yet. If you want to see the website just do a google search for "Homemade Fusion Reactor" and it should be the first link "Fusion is Easy". The website states you can't get any usable energy from the reactor, but you're still transforming some mass into energy, not much, but some.

And then as other people have said, the reason you can't transform mass into energy is that it requires huge amounts of energy to get the process started at a level you can measure, as in nuclear bomb territory. Fusion bombs are actually fueled by an fission explosion first. Fusion works because it uses highly unstable elements and then compresses them under thousands of tons of pressure to start the fusion reaction.

20. Apr 15, 2008

### pmb_phy

The mass remains the same. The only thing that has changed is the sum of the proper masses. The total inertial mass (aka relativistic mass) remains constant.

Pete

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