# 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

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

21. Apr 15, 2008

### pmb_phy

Actually it is true. The rest (aka invariant) mass of a system is the total (as in sum of masses) mass of the system as measured in the zero momentum frame. It can be readily shown that if 3-momentum is conserved in all inertial frames then mass is also conserved in all inertial frames, the zero mometum frame being one such frame. In that frame the mass of the system is called the "rest mass" of the system, even in those cases when none of the particles are at rest!

Were you aware of the fact that rest mass and invariant mass are often used as synonyms?

Pete

22. Apr 15, 2008

### RandallB

That is not quite the whole picture that Tachyonie was thinking of. There are at least three unique cases that do have real rest mass aka invariant mass changing as some mass is disappearing.
(I would not use ‘relativistic mass’ here; I’d consider that an abstract number that does change – an other issue dealing with momentum).

Those three cases include Fission Fusion and the annihilation process Tachyonie mentioned. When you have matter and anti matter particles combine the “annihilation” that results means just what is says, invariant mass in the system has disappeared. In a similar manner mass disappears in both fusion and fission reactions.
Note the atomic mass of He is less than the mass of the H atoms that fuse to make it. Thus fusion in the sun means a loss of weight from the fusion.

One of the points of Relativity is that the rule of conservation of mass is violated and that mass in not always conserved. The conservation law was replaced or better stated as “updated” to say that the net of Mass and Energy must be conserved. Thus any loss of invariant mass in a system is matched by an increase of energy in the form of massless photons.
In the case of the sun, energy that departs the local system of the sun.

23. Apr 15, 2008

### Antenna Guy

That example lends itself well to illustrate the direction of "local time" (albeit on a macroscopic scale).

Regards,

Bill

24. Apr 15, 2008

### MeJennifer

The direction of local time?
What do you mean by that?

25. Apr 15, 2008

### pmb_phy

I'm not a mind reader answer questions that are asked and the question Tachyonie asked was I am a littlebit confused here. What happens in annihilation process then? (matter + antimatter). As far as what he asked then one can only assume that he was referring to the idea that the sum of the proper masses has changed since he was responding to my comment which was
The question regarding whether mass can be converted to energy is question which needs to be stated more clearly. That's why I added the comment regarding the change from proper mass to mass of motion. Actually to be precise I should have referred to the sum of the proper masses. The rest mass of a system or particles is invariant and conserved. To precisely understand this assertion one must first understand what the exact meaning of rest mass of a system or particles. So let me state that now. The rest mass of a system of particles is defined as "the total energy of the system as measured in a frame of reference in which the total momentum is zero"/c2. Since energy is conserved then it follows that the rest mass of the system of particles is conserved. This does not mean that the sum of the rest masses of the particles is conserved, which is probably what Tachyonie was thinking about. Since he didn't respond to my answer I assumed he either hasn't read it yet or if he did he either understood it or simply chose not to respond.
That is impossible for a closed system. It simply can't happen. The invariant mass of a system is the magnitude of the 4-vector obtained by adding the 4-momenta of all the particles in the closed system. Since 4-momentum is conserved it follows that the invaraint mass cannot change.
That's your choice of course. As far as abstract, I see no reason to refer to it as such. The only thing that is measureable are kinematic quantities. Dynamics quantities are defined in terms of the measureable quantities and therefore things like 3-momentum, 4-momentum, 3-force, 3-force, invariant mass, Electric field, magnetic rield, EM field etc. are defined quantities just as relativistic mass is. Therefore there is no reason to think of relativistic mass as abstract and 3-momentum as not abstract. But as I said, its your choice as what you yourself use but I have very good reasons for using the terms as I do. I myself don't like the term but it brings home what quantity I'm speaking about since the term "mass" doesn't really mean one particular thing. When it appears in a paper or a text one can always tell by the context in which it is used, or the author explicity explains what they mean by it.
That is a misconception since the invariant mass in that case is conserved. The energy measured in the zero momentum frame remains conserved and therefore the systems rest mass (aka invariant mass) is also conserved.
Whether that is true or not will depend on how one defines the the term mass. Onky if one uses the term mass to refer to the sum of the proper masses of the particles can the "mass" changes with time. However I never saw anyone use the term in that sense.

Consider how this question has been answer in Spacetime Physics - 2nd Ed., by Taylor and Wheeler. On page 248 (note that mi as used in the text refers to the proper mass of the ith particle)
I don't see anything wrong with how the authors explain this. It is precisely correct .. which shows why this text is so good!
No such update has ever occured. People have misunderstood this for a very long time. The answer has always been the same. The mass of a closed system is conserved. Nothing has changed that.
In case you didn't know, photons have a finite, non-zero, inertial mass. It is the proper mass that is zero for a photon. The inertial mass of a photon is m = hf/c2 (h = Planck's constant and f = frequency ofthe photon). The reason it has inertial mass is because inertial mass is defined as the m in p = mv. Since a photon has momentum it also has inertial mass. Some people, like myself, use the term "mass" to refer to "inertial mass." Using it in anyother way, in my opinion, is an extremely bad idea in general. It should only be used that way when someone gets tired of saying "rest/proper mass" and wants to simply say "mass" instead. So long as its clear what it means then there is no problem. And in all cases I've read to date it has always been clear what has been meant by the term "mass." Its not always easy to see it but it can be determined by the content in which its being used.
When one is speaking of conservation of energy or mass one is usually referring to a closed system and as such the photons+sun is a closed system and therefore the systems mass is conserved.
That's a bad idea for the following reason. The rest mass of a system is not always proportional to the rest energy of the system. This is especially true for non-closed systems, such as a dielectric in an electric field in which case the dielectric becomes polarized by the field and stress is induced into the system. In such case the rest energy of the dielectric is not proportional to the rest mass of the dielectric.

Pete