Mass-Energy Equivalence#

Special Theory of Relativity#

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An event can be defined by place and time, like birthday party at my house at 10:00 pm. This can be encapsulated in what is called a four-vector, with three space coordinates and a time variable {x,y,z,t}. We want to use special relativity to determine these values.

To do this, we need to use the Lorentz Factor:
\(\gamma=1/\sqrt{1-\frac{v^2}{c^2}}\)

This factor, also called the gamma factor, demonstrates how mass and spacetime are effected by high velocity.

Suppose a spacecraft was heading to Proxima Centauri at half the speed of light. The stationary observers on the Earth kept track of the ships distance and the duration of the trip. The moving observer inside the spaceship does the same.

Here's what both observers agree upon:

• The speed of light is: \(c=299,792,458\) meters per second.
• The gamma factor is: \(\gamma=1/\sqrt{1-\frac{v^2}{c^2}}\)
• The velocity of the spaceship is: \(v=0.5c=149,896,229\) meters per second.

Here is where they disagree when the spacecraft has reached its destination:

• The distance as measured by the stationary observer is: \(d=4.018\times 10^{16}\) meters or \(4.247\) light years.
• The distance as measured by the moving observer is: \(D=\frac{d}{\gamma}=\frac{d}{1/\sqrt{1-\frac{v^2}{c^2}}}=d\times \sqrt{1-\frac{v^2}{c^2}}\)
  \(=4.018\times 10^{16}\times \sqrt{1-\frac{149,896,229^2}{299,792,458^2}}\)
  \(=3.480\times 10^{16}\) meters or \(3.68\) light years.
• The duration of the trip as measured by the stationary observer is \(t=\frac{d}{v}=\frac{4.018\times 10^{16}}{149,896,229}=298,052,107\) seconds or \(8.5\) years.
• The duration of the trip as measured by the moving observer is \(T=\frac{t}{\gamma}=\frac{t}{1/\sqrt{1-\frac{v^2}{c^2}}}=t\times \sqrt{1-\frac{v^2}{c^2}}\)
  \(=298,052,107\times \sqrt{1-\frac{149,896,229^2}{299,792,458^2}}\)
  \(=232,139,934\) seconds or \(7.361\) years.

	stationary observer	moving observer	space contraction / time expansion
distance (ly)	4.247	3.680	87%
time (years)	8.5	7.361	115%

Both observers are correct, as are their calendar markings and clocks. High speed warps spacetime, causing them to disagree upon an arrival event. Each observer will have a unique 4-vector {x,y,z,t} for the event.

• The stationary observer will note that the arrival event was at \(\{4.247,y,z,8.500\}\)
• The moving observer will note that the arrival event was at \(\{3.680,y,z,7.361\}\)

Of course, this scenario is simplified, since we are not taking into account acceleration, and have set the coordinate system so the x-axis is in line of site from the Earth to Proxima Centauri.

The effect becomes more pronounced at greater velocities.

vel	dist ly	time yrs
0.5c	3.678	7.36
0.75c	2.819	5.62
0.9c	1.851	3.71
0.999c	0.190	0.380
0.999999c	0.006	0.012

For instance, notice that at 0.999999c, the distance is only 0.006 light years - approximately 3.8 times the distance to Voyager 1 (about 378 AU). And the trip duration is down from 4.2 years to 0.012 years or 4.38 days!

It seems incredible to imagine traveling just over 4 days to reach the nearest star, but it would be possible at that speed. The caviat would be that upon return to the Earth, after a total trip duration of 8 days, 18 hours, you would find that 8.5 years had still past on Earth. If you had a twin back on Earth you would return from a busy week+ in space to find that your twin sibling was now 8 years, 5 months, and 26 days older than you!

This is fascinating, but what does it have to do with Mass-Energy equivalency?

Energy, mass, and the Lorentz factor#

Kinetic energy is moving energy, according to this equation:
\(E=\frac{1}{2}mv^2\)

Note, if the velocity is zero, energy is zero. But something odd happens when we account for how mass changes when a massive object moves at high velocity. This situation is similar to time dilation and space contraction in that it uses the Lorentz factor to calculate a change in mass, as viewed by a stationary observer.

For example, the international space station has a mass of 419,725 kg. If the station were accelerated to 80% the speed of light, Astronauts would notice no mass increase. Their thrusters would work the same way as they did in low Earth orbit. Measured from the Earth (assuming such a measurement was technically possible) the mass would appear to have changed as follows:

\(M=m\gamma=419725\times \sqrt{1-\frac{0.8^2}{1^2}}=699,542\) kg

The ISS would have 1.7 times the mass as it did in orbit, according to the stationary observer. How can an object have two masses, and both be correct? The strange truth is that an object can have an infinite number of masses from its rest mass to its mass at some percentage of the speed of light - which can be arbitrarily close (0.9, 0.99, 0.999, 0.9999...).

This discovery from Special Relativity must be integrated with the Newtonian kinetic energy formula, in order to describe how the mass is changing at high velocities. We need to substitute the moving mass for the stationary mass:

\(E=\frac{1}{2}m\gamma v^2\)

If we expand the gamma factor, we get:
\(E=m\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}v^2\)

Notice that when v is very low relative to c, the mass change is negligible. This makes the change in energy negligible. This allows us to calculate kinetic energy at relativistic and non-relativistic speeds.

This is what happens to a 1 kg mass when its velocity increases.

		kinetic energy
velocity	gamma	with relativity	without relativity	difference
0.000001	1.0000000000005	44937.75893686	44937.75893684	0.00000002
0.00001	1.0000000000500	4493775.8939	4493775.8937	0.0002
0.0001	1.0000000050000	449377591.6	449377589.4	2.2
0.001	1.0000005000004	44937781406	44937758937	22469
0.1	1.0050378152592	4.52E+14	4.49378E+14	2.26E+12
0.25	1.0327955589886	2.90E+15	2.80861E+15	9.21E+13
0.5	1.1547005383793	1.30E+16	1.12344E+16	1.74E+15
0.75	1.5118578920369	3.82E+16	2.52775E+16	1.29E+16
0.9	2.2941573387056	8.35E+16	3.63996E+16	4.71E+16
0.99	7.0888120500834	3.12E+17	4.40435E+16	2.68E+17
0.999	22.3662720421294	1.00E+18	4.48479E+16	9.58E+17

At slow speeds, the mass differences between using relativity or not is negligible; but that quickly (exponentially) changes at higher velocities. Newton's mechanics did not allow for mass to change in this manner, or for the kinetic energy to grow. This is a more complete solution for all velocities.

Using Calculus to account for a changing mass#

One way to formulate the Work-Energy Theorem is as follows:

\(W = \delta K = K_f - K_i = \int_{x_i}^{x_f} F(x)\cdot \ x\)

Work \(W\) is done on an object, and is defined as the dot product of the force and displacement vectors. \(\delta K\) is the change in kinetic energy, implied from the final energy \(K_f\) minus the initial energy \(K_i\) (if the object was already in motion). \(F(x)\) is the force acting on an object as a function of distance. The variables \(x_i\) and \(x_f\) are the starting and ending places for the move.

The changing variable is the distance. This creates a curve for the force as it increases or decreases over that distance interval. The area under the curve is the work done.

\(F=ma\) is a special case of \(F=\frac{d}{dt}(mv)\). The more familiar first formula only works when the mass is constant. If the mass isn't constant, it's necessary to use the more generalized formula.