A member of this forum once said...

Do you really believe that asshole einstein? He just tells you E=mc² and you believe it w/out fully understanding it?

So, you know what? Challenge accepted.

Let's start with the basics. We know that the speed of light, denoted c, is always constant. Whether you're walking at 5 mph, riding in a plane at 500 mph, or on a rocket going 5,000 mph, you're always going to see light travel at the same speed. Why is this? Let's take a closer look.

Speed is defined by a distance covered over time. In the case of the speed of light, it's a distance of 299,792,458 meters covered over a time of one second. If this is always going to be constant, then either the distance has to change, or the time has to change. You guessed it, it's the time that changes.

But let's look at this more analytically. Let's suppose the we're going to shoot a radio transmission to astronauts on the moon, and instantaneously, as soon as they get it, they'll send the signal back to us. Let's also say that the distance to the moon is exactly a distance of d in km. Since we know how fast the signal goes, c, we can expect the transmission to hit the moon at a time t_one-way = d / c. Then, we can expect the time to get our signal back to be t_round-trip = 2d / c, since it has to make a return trip, too. To make things simpler, let's just call t_round-trip to be t₀.

But what if another group of astronauts are on a shuttle going some speed v? Let's look at that.

In this case, the people on the shuttle are going to cover a distance a = dt * v, where dt is the time that passes for them. We also know that the distance the light travels is going to be distance b + distance c, which, is going to be 2 * sqrt(d² + (a/2)²). Using what we know for a, this becomes 2 * sqrt(d² + (dt * v / 2)²).

Since the speed of light is always going to be a constant c, this ultimately means that the light takes dt = 2 * sqrt(d² + (dt * v / 2)²) / c seconds to reach the shuttle again. But for stationary observers, we can simply say that d = c * t₀ / 2 as previously mentioned. Combining the two equations gives: dt = 2 * sqrt((c * t₀ / 2)² + (dt * v / 2)²) / c.

This is a complex equation to solve, so I hope you remember your algebra. With a little bit of jiggery pokery we can move the terms around to find out what dt is. First, square both sides:

dt² = 4 * ((c * t₀ / 2)² + (dt * v / 2)²) / c²

Then expand a little:

dt² = 4 * (c² * t₀² / 4 + dt² * v² / 4) / c²

Then expand a little more to kill off some unnecessary terms:

dt² = t₀² + dt² * v² / c²

Then move both dt terms to the same side:

dt² v² * c^-2 * dt² = t₀²

Now, we can factor out dt:

dt² (1 - v² * c^-2) = t₀²

From here, solving should be pretty easy. Let's just skip to the answer, which is:

dt = (t₀² * (1 - v² * c^-2)^-1)^1/2

Or, alternatively:

dt = t₀ * (1 - v² * c^-2)^-1/2

So yes. That's the way this works. There, you have a mathematical expression for the way time changes relative to a difference in velocity between observers.

We can plug in some numbers here and get a general idea of how this equation behaves. Going back to our example with things on the moon, a radio transmission from Earth to Moon takes 1.2822 seconds for observers on either body. But for observers moving on a rocket going 5,000 m/s, it's still 1.2822 seconds, because 5,000 m/s is such an insignificant speed. But what if the speed of the shuttle is 90% of the speed of light? In that case, it's 2.9416 seconds that pass on the moon, for those 1.2822 seconds on the shuttle.

Another way to analyze this is to observe what happens at the extremes. If v is infinite, then dt becomes 0. If v is 0, then dt becomes t₀. Larger v means a larger difference in time perception from stationary and non-stationary reference frames. This will be important later.

What's interesting, is that just as time is not constant, neither is mass. A similar equation times an object's rest mass, m₀, is the mass of an object traveling at that speed. So, in essence, we have:

(1 - v² * c^-2)^-1/2 * m₀ = M

Where M is the mass accounting for relativistic effects at high speeds.

Now, let's look at what exactly energy is. Kinetic energy, at least, is defined by the integral of a force multiplied by a distance. Because of newton's laws, this force is d(M*v)/dt. So, the integral looks something like:

d(M*v)/dt * ds, where the ds is the distance that the equation is being integrated over.

Now, since ds/dt is just simply v, this means the form can be reduced to v * d(M*v). We know what M is based on the prior equation, and this becomes v * d((1 - v² * c^-2)^-1/2 * m₀*v). This is not an easy integral to evaluate, but it can be evaluated using integration by parts.

The solution becomes:

KE = M * c² - m₀ * c²

Where KE is the kinetic energy. Rearranging this, we can write:

KE + m₀ * c² = M * c²

Or, in terms of energies:

KE + E₀ = E

Here, we're basically saying the E₀ term is the energy in which the reference frame is stationary, and E is the relative energy. Now, assume the mass is stationary in a certain frame, such that kinetic energy is zero. We still have that the energy is E₀, and we know that E₀is m₀ * c². So, in essence, this really just comes down to E = m * c², which is the nice, cute and simple formula that everypony and their grandmother remembers.

So, the answer is yes. I do believe the idiot that was Einstein. I believe this is completely true, and I do attempt to understand the mathematics behind it to the best of my abilities. My understanding leads me to believe that this is entirely correct.