# Why C in E=MC^2

1. Feb 20, 2006

### Darwin

Why "C" in E=MC^2

What special property of the speed of light imposes itself on the amount of energy in a mass? Why should that energy be an exact multiple of the speed of light squared? Why not a multiple of some other constant like the Bohr radius, the Boltsmann constant, or the free electron g factor, or just some other number, such as my shoe size?

2. Feb 20, 2006

### Janus

Staff Emeritus
The special property of Light is that its relative velocity is invarient for inertial reference frames. IOW, no matter who measures the speed of light, or what their relative velocity to each other is, or their relative velocity to the source of the light, they will always get the same value for the speed of light relative to themselves.

With this and the other postulate of Relativity, that the laws of Physics are the same for all inertial reference frames, you can derive the time dilation and length contraction formulas. In turn, with these you can derive the formula for the kinetic energy of a moving object according to Relativity.
(as energy is measured in units of $\frac{md^2}{t^2}$ and time(t) and distance(d) are effected by relative velocity, it stands to reason that the kinetic energy of the object will likewise be effected.)
If you then set v in this formula to 0, (for an object at rest) you are left with
$$e=mc^2$$

3. Feb 20, 2006

### rbj

Darwin, i would suggest learning a little something about basic dimensional analysis (dimensionally, it has to be some velocity squared) and also Planck units. Wikipedia has articles in both.

4. Feb 20, 2006

### robphy

Although we call "c" the "speed of light", it might be better to think of "c" as the "[invariant] maximum signal speed", which is a property of structure of space-time. It just so happens that "light", a phenomena of the electromagnetic field, propagates at this maximum signal speed.

With the issue framed in this way, those other parameters you mentioned [Bohr radius, Boltzmann constant, g-factor, your shoe size,...] are not distinguished in any way.

It might be interesting to note that Galilean/Newtonian physics has no analogous [finite, dimensionful] parameter like "c".

5. Feb 20, 2006

### topsquark

In addition to these sage comments, the c in mc^2 comes from the "gamma" function: (I hope the LaTeX works this time!)
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$
(It worked! Yay!)
Anyway, that's the definition of gamma. When we express the energy function we get an expansion in terms of c^2.

The other way to do it is to look at the 4-vectors. The energy-momentum 4-vector always takes the form: (E/c,px,py,pz) The c is required here to make sure that the "square" of this 4-vector takes the appropriate constant. Squaring the vector we get $$\frac{E^2}{c^2}-p^2=m^2c^2$$, which is the general form of E=mc^2.

-Dan

6. Feb 20, 2006

### bernhard.rothenstein

.........................................
http://arxiv.org/abs/physics/0505025

7. Feb 21, 2006

### Sumo

I've always wondered why its c squared. Apart from the fact that mathematically it works out that way, is there any understanding for why you would want to square a velocity to find energy? Or why it appears again in other relativity equations: sqrt(1 - v^2/c^2) ?

8. Feb 21, 2006

### leandros_p

9. Feb 21, 2006

### leandros_p

The kinetic energy of a moving object is equal to (m*v^2)/2
m= mass
v= velocity

So, it is not *strange* to relate energy with the square of velocity.

Leandros

10. Feb 21, 2006

### Staff: Mentor

In both non-relativistic and relativistic mechanics, kinetic energy is derived via the work-energy theorem: the change in an object's kinetic energy equals the work done by an external force.

$$\Delta K = W = \int {Fdx} = \int {\frac {dp}{dt}} dx$$

In the relativistic case you have to use relativistic momentum $p = m \gamma v$, which brings in the $\gamma$ factor that you mention above.

In turn, the relativistic momentum can be derived by analyzing a collision carefully, taking length contraction and time dilation into account. This is how $\gamma$ gets into the relativistic momentum.

Finally, to see how the $\gamma$ factor comes about in the first place, see (for example) the common derivation of the time-dilation equation using a light-beam clock:

http://www.phys.unsw.edu.au/einsteinlight/jw/module4_time_dilation.htm

I'm sure there are other ways of doing all this, but this is probably the most common chain of logic that you'll see in introductory "modern physics" textbooks.

Last edited: Feb 21, 2006
11. Feb 26, 2006

### Sumo

Ok, but what I mean is, for example, if we square a distance its pretty easy to understand what that means. But I cant understand what squaring a velocity means, or how it has anything to do with energy.

12. Feb 26, 2006

### Staff: Mentor

Physically, it's the same as force times distance and again, getting the units/math to work out is the entire point here. The word "energy" doesn't have some mystical meaning, we use the word to describe a useful mathematical expression.

Also - how does squaring a distance make sense where squaring a velocity doesn't?