# Energy-mass equivalence - E=mc^2

1. Oct 4, 2006

### heliocentricprose

We all know that if we break up an atom, we get a lot of energy in return. I'm having trouble visualizing what this equation actually means. Does it mean that a large amount of energy is required to bring matter together? Does it mean a large amount of energy can be converted mass? Isn't energy a property of mass to begin with (the motion of atoms, the motion of electrons)? What is it?

2. Oct 4, 2006

### chroot

Staff Emeritus
Relativity treats energy and mass as the same "thing," and the equation you posted indicates that one can be converted into the other.

The c^2 is nothing more than a proportionality constant which allows us to use the human units of kilograms and joules. Many physicists prefer to use so-called "natural units," in which c=1 (and thus if unit of time is the second then the unit of distance is the light-second), and the equation becomes the very simple E = m.

- Warren

3. Oct 4, 2006

### Aether

Yes. Sometimes when two particles are smashed together at very high velocity, a new particle(s) can be formed with a total mass that is greater than the sum of the masses of the original particles by an amount that is precisely accounted for by the difference in kinetic energy of the original particles and the final particles.

4. Oct 4, 2006

### Zeit

The equation E = mc² comes from a larger : E² = (mc²)² + (pc)²

When the studied object is motionless, p = 0, so E² = (mc²)². So, even when an object is motionless, it has an energy. So, we can say that mass is a kind of energy.

5. Oct 6, 2006

### pmb_phy

Not always. It can take more energy to split an He atom than to fuse two H atoms to make an He atom. There is a special curved called the "Binding energy per nucleon" curve. It has a peak. Nucleons corresponding to points to the right of the curve will release energy when they split. The opposite is true for nucleons to the left of the curve. Its the difference of getting energy from fission and getting energy from fussion.

For fussion it means that negative work is done bringing the nucleons together for a release of energy and it means that positive work is done fussing two nucleons to release energy. (I'm only 95% correct on this. The positives and negatives of work can confuse me sometimes. Someone please correct me if I'm wrong!)

Energy cannot be converted into mass and mass cannot be converted into energy. The forms of energy may converted from one to the other. One for being mass-energy which is the energy that can be extracted from mass. But the total mass and the total energy is a constant in nuclear reactions such as these.

Pete

6. Oct 6, 2006

### pmb_phy

This is not true. What we can say is what the physics tells us, and the physics tells us here is that if we add energy to a body (e.gl in the form of radiation/heat/etc) then its proper mass will increase. Thus if you take a system of two (+) charges and do work on the system of the amount dE then the change in proper mass dm0 will change in accordance with the relationship

dE = dm0c2

Pete

Last edited: Oct 7, 2006
7. Oct 6, 2006

### pmb_phy

This kind of thinking can lead to trouble, ... and it has. The famous equation only applies to closed systems or isolated particles. Not under the most general conditions. One well-known relativity author though in the above manner and a question in his homework section arrived at a paradox (i.e. a solution which seems wrong) in his homework section. The question pertained to mass density of a magnetic field. The paradox is outlined here

http://www.geocities.com/physics_world/sr/mass_mag_field.htm

An example of the mass-energy relation not holding for a non-isolated body is described here

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

The reference to the correct physics is given below and is from Rindler's SR text. Rindler explains why E does not always equal mc2. I recommend reading it.

Best wishes

Pete

8. Oct 6, 2006

### pervect

Staff Emeritus
Pete is correct when he notes that (p/v) is not equal to E/c^2 for a non-isolated system. However, his defintions of mass (m=p/v) do not appear to me to be standard for relativistic systems. Rather, it is an unforunate carry-over of Newtonian defintions. Not too long ago I went to the trouble of tracing down two textbook quotes (plus the sci.physics.faq) which disagree with Pete's defintion of mass.

So we have three diffrent defintions of mass to confuse the poor innocent newbie. We have invariant mass, relativistic mass, and Pete's mass.

Pete insists that his defintion of mass is the "correct" defintion of relativistic mass, last I heard. I don't use relativistic mass much, greatly preferring invariant mass, but the textbooks I consulted did not agree with Pete on the defintion of relativistic mass.

For the more advanced student, there are some interesting lessons to be leaned regarding the mass of composite systems. My favaorite article that makes similar point (related to the point that Rindler makes) is

http://arxiv.org/abs/physics/0505004

however, while it is interesting to the advanced student, I don't think it will be that useful to someone who is struggling to understand E=mc^2.

9. Oct 6, 2006

### heliocentricprose

e=mc^2 is easy enough to understand... however, how do you define energy? It's a motion of an atom, correct? What about dark energy? Does it have some sort of mass association as well?

10. Oct 6, 2006

### Zeit

Hello,

I don't understand why I should be wrong. Rest mass, as its name suggests me, cannot increase : it is an invariant mass. The "relativistic mass" is, as I understand it, a wrong concept ; we should talk of "relativistic momentum". So, may be is it better to use E² = (mc²)²+(pc)² than E = gamma*mc², where m is the invariant mass.

But, what I see is even when an object as a kinetic energy = 0, these equations tell us there's energy. So, if I'm wrong, tell me please

11. Oct 6, 2006

### DaveC426913

Can you elaborate? It seems to me this explanation is at the crux of the common question "what does the c^2 represent in E=mc^2?"

How did you get from E=mc^2 to E=m?

Oh. I think I see. The unit c is expressed in a unit of our choosing: km/s or mi/s. you've just chosen the units of light-seconds per second. thus: c^2=1.

So, could we literally say that the amount of energy E would literally move a mass m a distance of 1 light second in a time of 1 second?

12. Oct 6, 2006

### pervect

Staff Emeritus
Here is the fine point behind what Pete was saying, in terms of invariant mass (which is both modern and has a totally unambiguous defintion).

The invariant mass of a point mass is Lorentz invariant - it does not change when you change the frame of reference.

However, if you have a system with a non-zero volume, the invariant mass of that system is only invariant under a Lorentz boost if the system is an isolated system.

For one reference, see the paper I quoted earlier.

A vector like the energy-momentum 4-vector is covariant if and only if the norm of that vector, i.e. the mass for the energy-momentum 4-vector, is invariant under a Lorentz boost.

A system with a non-zero volume has an energy-momentum 4-vector, but that vector is not covariant when the system is not isolated. So the name "invariant mass" could be slightly misleading for distributed systems with non-zero volume. The system has a mass, calculated in the usual manner, but that number is only independent of the frame of reference used when the system is isolated, or when the system has zero volume.

If you work at it, you can eventually see why this is so due to the relativity of simultaneity. A non-isolated system is constantly exchanging energy with its surroudnings. The total amount of energy "in the system" depends on the defintion of simultaneity used in the case where the system is interacting with its surroundings. For an isolated system, this isn't a problem, the energy is constant and the energy-momentum 4-vector is covariant.

What is ALWAYS covariant is the stress-energy tensor.

Relativistic mass is wildly coordinate dependent. Invariant mass is better, but it still is coordinate/frame dependent when you have a non-isolated system. The stress-energy tensor is ALWAYS covariant and thus independent of the coordinates used. This , in my opinion, is an excellent reason to stay away from the concept of mass, and (as Einstein did) to stick with the stress-energy tensor as the fundamental description of distributed systems, i.e. systems with a non-zero volume.

13. Oct 6, 2006

### scott1

$$E=mc^2$$ tells us that mass is form of a energy which therefore can be converted into energy and energy has mass.

BTW PF has a cool feture called Latex so you should type it how I did in the above equation https://www.physicsforums.com/showthread.php?t=8997"for more info

Last edited by a moderator: Apr 22, 2017
14. Oct 7, 2006

### pmb_phy

The term invariant means does not change value when the system of coordinates is changed. It does not mean "does not change in time."
Whoever told you that is totally wrong. There is absolutely nothing wrong with the concept. People who would have you believe so choose to define "mass" in another way which makes the other definition incorrect - "By definition." One way can be called the "geometric" view of mass whereas the other way can be called the "analytic" view of mass. Even in the geometric view there is a misuse of the concept of mass. They really should use the term "proper mass."

Regarding definitions - I could call my house a "bottle of milk" and yet every aspect of my "bottle of milk" will be exactly the same as those who call my house a "house."

I have described everything you'd want to know (and then some) in a paper I wrote on the subject. It is located here

http://www.geocities.com/physics_world/mass_paper.pdf

Mass is what defines momentum. Not the other way around.
That is absolutely correct. All bodies with finite proper mass contains a finite amount of energy. The derivation of the E= m0c2 starts out with the assumption that the body posseses energy. This is something which is not proved in the derivation. Einstein assumed that the body had energy. He then showed that when the energy of the body decreases then the proper mass of the body decreases as well in direct proportion. This only demonstrates that a body posseses energy (which Einstein assumed before he started the derivation) and when that energy changes so too does the mass.

A derivation of E0 = m0c2 is on a web site I created at

http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm

It is similar in content as Einstein's derivation. I like this derivation because it is much easier to follow.

For more on the concept of invariant mass (aka "proper mass") please see

http://www.geocities.com/physics_world/sr/invariant_mass.htm

See Fig. 4 especially.

Pete

Last edited: Oct 7, 2006
15. Oct 7, 2006

### pmb_phy

Let me start by quoting my favorite physicist, Richard Feynman. In his lectures he states
That is in The Feynman Lectures on Physics, Vol I - III, Feynman, Leighton, and Sands, Addison Wesley, (1963)(1989).

For more on the concept of mass please see the web page I created for this purpose.

http://www.geocities.com/physics_world/mech/what_is_energy.htm

Best wishes

Pete

16. Oct 7, 2006

### Zeit

Hello,

Thanks for the answers. In my first post, I wanted to help, but I realize the good answer is much more difficult than I thought :shy: Thanks again.

Zeit