# E=mc^2 can C^2 be described as a physical process?

1. Apr 10, 2010

### rasp

I have always had trouble appreciating E=mc^2 because I can't relate C^2 to a physical process, like velocity or acceleration. How should one imagine the dimensional characteristics of C^2?
Along the same vein, is E=mc^2 a mathematical conversion only, or does C^2 describe an actual process by which mass can be converted to energy? e.g. spin 1 piece of matter around a cyclotron at a speed approaching C and another piece of matter in the opposite direction at a speed approaching C so that when they collide they will hit at C^2?

2. Apr 10, 2010

### Theheretic

I always figured the c2 was just an arbitrary part of the equation put in there to reach a needed number depending on what values are used but I really don't know.

3. Apr 10, 2010

### Pengwuino

No no, it has nothing to do with the velocity of the atoms. It's simply, in a sense, a proportionality constant. For example, the electric field of a charged particle is, in SI units, $$\frac{q}{4\pi \epsilon_0 r^2}$$. What is $$\frac{1}{4 \pi \epsilon_0}$$? Just a proportionality constant, it doesn't represent a physical process, it has nothing to do with anything other then the fact that $$E = \frac{q}{r^2}$$, by itself, makes no sense. It also has nothing to do with the speed of the atoms. Infact, it's the energy mass has at rest!

4. Apr 10, 2010

### Theheretic

The question is, if it does not represent any physical tangible process but rather is an abstract and arbitrary equation, then how does one arbitrarily come up with such an equation as Einstein did?

5. Apr 10, 2010

### Staff: Mentor

It was anything but arbitrary: he derived it.

Also, note the similarity between that and the Newtonian kinetic energy equation: e=.5mv^2... In order to get the dimensions of energy (kg-(m/s)^2), the velocity has to be squared.

6. Apr 10, 2010

### bcrowell

Staff Emeritus
FAQ: Where does E=mc2 come from?

Einstein found this result in a 1905 paper, titled "Does the inertia of a body depend upon its energy content?" This paper is very short and readable, and is available online. A summary of the argument is as follows. Define a frame of reference A, and let an object O, initially at rest in this frame, emit two flashes of light in opposite directions. Now define another frame of reference B, in motion relative to A along the same axis as the one along which the light was emitted. Then in order to preserve conservation of energy in both frames, we are forced to attribute a different inertial mass to O before and after it emits the light. The interpretation is that mass and energy are equivalent. By giving up a quantity of energy E, the object has reduced its mass by an amount E/c2, where c is the speed of light.

Why does c occur in the equation? Although Einstein's original derivation happens to involve the speed of light, E=mc2 can be derived without talking about light at all. One can derive the Lorentz transformations using a set of postulates that don't say anything about light (see, e.g., Rindler 1979). The constant c is then interpreted simply as the maximum speed of causality, not necessarily the speed of light. We construct the momentum four-vector of a particle in the obvious way, by multiplying its mass by its four-velocity. (This construction is unique in the sense that there is no other rank-1 tensor with units of momentum that can be formed from m and v. The only way to form any other candidate is to bring in other quantities, such a constant with units of mass, or the acceleration vector. Such possibilities have physically unacceptable properties, such as violating additivity or causality.) We find that this four-vector's norm equals E2-p2c2, and can be interpreted as m2c4, where m is the particle's rest mass. In the case where the particle is at rest, p=0, and we recover E=mc2.

A. Einstein, Annalen der Physik. 18 (1905) 639, available online at http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

Rindler, Essential Relativity: Special, General, and Cosmological, 1979, p. 51

7. Apr 10, 2010

### Count Iblis

You can have different views about the status of c as a genuine physical constant, see here:

http://arxiv.org/abs/physics/0110060

The position taken by Michael Duff is basically that the constants c, hbar and G are simply irrelevant rescaling constants that appear in equations because we choose to express certain physical quantities in different units. In his opinion, there is nothing fundamental about the dimensions that we have assigned to physical quantities.

8. Apr 11, 2010

### yuiop

The full equation is:

$$E = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$

The often quoted E=mc^2 is just the special case for a stationary particle. There is a physical process to relate to in the full equation (v^2). The c^2 term is just a physical constant or scaling factor, a bit like the G in the Newtonian equation F = GMm/r^2. The G in this equation is not a physical process but a physical constant like c^2.

This is way off the mark.

9. Apr 11, 2010

### rasp

Ok, what I get is C^2 is a special quantity derived from theory to solve a set of equations and different than any other V^2, and it is not a process. Yet does E vary as a quantity in proportion as mV^2 approaches mc^2?

Any thoughts on how e=mc^2 helped the engineers start on a path to actually convert m to E?
How would a physicist categorize a law such as E=mC^2 versus say F=ma, which are all physical processes?

10. Apr 11, 2010

### Count Iblis

The meaning of E = m c^2 is simply that the mass of an object is its rest energy. Or put differently, if you have a closed box at rest and the total energy content of the box is E, then it has a mass of E/c^2, regardless of composition or any other physical properties. I.e. mass is determined by energy and nothing else but energy.

Then what is the c^2 doing in E = m c^2? Well, had we known that mass and rest energy are the ame thing (and also that space and time are different components of the same quantity), we would not have invented different units for mass and energy and for distances and time intervals.

If you have different units for the same physical quantities (because these quantities manifest themselves in seemingly unrelated ways) and only much later you discover phenomena that link the two quantities, then the equation that will link these will have to contain conversion factors (unless society were flexible enough to abandon the old unit system)

It is just like saying that the total distance from a point in France to a point in Britain is:

L1 + L2

But if you insist that L1 be expressd in miles and L2 in kilometers, because you are not supposed to denote distances in Britain by kilometers, then a conversion factor of 1.609 km/mile will pop up in the equation itself.

11. Apr 11, 2010

### Pengwuino

You're trying to equate two different "kinds" of energy. Your kinetic energy, .5mv^2, is the energy an object has because it is in motion (this is also a low-speed approximation to kinetic energy). The rest energy, mc^2, is the intrinsic energy something has simply because it has mass. The total energy of a particle in motion is the sum of this rest energy and the kinetic energy. At low speeds, the kinetic energy is far less than the rest energy but at relativistic speeds, it can easily exceed the rest energy. There's also nothing special about when an object has a kinetic energy equal to its rest energy.

12. Apr 11, 2010

### Staff: Mentor

I tend to think of c² as simply a conversion factor that is necessary only because we have conventionally chosen to measure energy and mass in different units. Similarly with any other dimensionful universal constant.

13. Apr 11, 2010

### Count Iblis

I agree. Unfortunately, this point of view is completely ignored in many textbooks. Even in some advanced books were natural units are used, this is pretended to be just a trick.

I.m.o., there are some non trivial points when deriving classical mechanics from special relativity that are completely ignored in textbooks precisely because the factors c are already present. So, all you see are trivial arguments based on an expansion in powers of v/c.

Instead, one should start with relativity where c = 1 in all equations. With those equations one has to derive classical mechanics by studying the low velocity limit. This requires one to introduce a rescaling parameter and study the limit in which this tends to infinity. But doing this correctly is not a trivial exercise. Using c as it appears in equations in SI units and then letting c go to infinity amounts to cheating.

14. Apr 12, 2010

### rasp

Ok, so do i understand correctly from what has been said that e=mc^2 is simply a conversion statement similar to miles= 1.609 km and not like a physical process such as F=ma?
Did the e=mc^2 statement lead scientists to develop a process to actually convert mass to energy. I thought according to SP that as a particle approaches the speed of light its rest mass is converted to energy?

15. Apr 12, 2010

### Frame Dragger

No, the EQUATION shows mass-energy equivalence; why this fixation on 'c'?

The part in bold is genuinely nonsesical, or you want an answer other than "rest+kinetic".

As for it leading to conversion of mass to energy, I don't know that it LED them, but have you heard of thermonuclear weapons, the NIF, ITER...?

16. Apr 12, 2010

### utesfan100

Kinetic energy is defined to be the integral of VdP from 0 to V. Using SR this comes out to be:

K = gamma*mc^2 - mc^2

By adding the constant mc^2 to the kinetic energy one gets E = gamma*mc^2, which eliminates the pesky constant, but then the object has an energy of mc^2 at rest.

Then we said "I wonder if mass really contains that much energy" and blew up chunks of it to see that it was real energy.

17. Apr 12, 2010

### yuiop

I don't think this directly gave them recipe of how to convert mass to energy. I am not sure (because this is basically a historical question and I not sure of their thought processes), but I think one clue was that the masses of elements in the periodic table did not have an exact integer sum of the masses of their constituent components (electrons, protons and neutrons). The natural decay of some elements indicated that the sum of the rest masses of the decay products was not exactly the same as the mass of the original particle. The difference in mass was tiny, but the E=mc^2 relationship indicated that the "lost rest mass" could represent a lot of energy.
As FD said this is just wrong. I guess you could say that a rocket powered by a nuclear reactor, converts some of its nuclear fuel rest mass to energy, but this is not true for a single particle being accelerated by an external device.