# Special Relativity - Need4Speed. Picture this. You're in a car, making your way to work, and some flashy sports coupe just happens to overtake you. From the corner of your eye, you can just make out the shape of a rude gesture and a cheeky smirk past the UV-reflective windows. The gall! You thus make it your goal to catch up to him, and to show him who's boss. The problem is, you two are on the interstellar hyper-relay to 51 Pegasi (they're running an interstellar tourism campaign there), and there isn't really a hope in hell you can catch up to the space Ferrari without breaking speed limits. So, as the Ferrari takes the next exit onto another highway with a speed limit that's lower than yours, yet still parallels your highway, you take this as your chance to put your pedal to the metal to determine who's boss, once and for all. Is it you, or does the car appear to 'shrink' when you accelerate? (It also looks a little 'bluer' than usual, but we can ignore that for now.) This, well, is an example of **special relativity** - modifications we make to our interpretation of relativity for objects at speeds at a non-negligible proportion of the speed of light. %%comEcON%% # Time Dilation and the Lorentz Gamma Alright, so the *core theory* for this is actually **finding the spatial displacement as a function of the time displacement.** Just basic problem solving here, expressing each variable as a function of another so that you can make accurate comparisons. Now let's interpret the 'time' dimension (the fourth dimension) as a 'timeline' - so a linear axis that just travels along the page, from the left to the right. For this example, let's imagine a light source AND a mirror that are moving, relative to us, at Now, light will hit a mirror and reflect back to us. In the frame of the light source dimension, **looking at this beam of light**, it hasn't moved AT ALL. back-and-forth motion, no displacement. But in the time dimension, what actually happens is that the mirror moves along the 'timeline' denoted by the $ct$ axis. Here's a diagram: ![[specialrelativitylightdiagram.png]] When we define these frames, it's important to consider the **proper time** of the light. This is defined as being measured from the frame where the object is completely stationary (or the inertial frame of the object/event we're talking about), so in this case, the proper time will be measured from the frame of the light! So, here's what time dilation actually is. If we look at the geometry of the light rays and the distances they have to travel, we can see: - For $t_0$, the total time to travel the distance D is $\frac{2D}{c}$. - Whereas for $t$, the total time $t$ to travel the distance $L$ is $\frac{2L}{c}$ - So therefore the time dilation is $t - t_0 = \Delta t = \frac{2L}{c} - \frac{2D}{c}$ Now we find $L$ in terms of $d$, where we get: $L = \sqrt{D^2 + \frac{1}{2}(v^2t^2)}$ Then substitute the equation for $D$ when you look at the straight line graph, that being $2D = ct_{0}$, $t_0$ being the proper time: $L = \sqrt{ \frac{1}{2}(c^2t_{0}^2) + \frac{1}{2} (v^2t^2) }$ Now we can solve for $t$ since $\frac{2L}{c} = t$ as well: $\frac{2\left( \sqrt{ \frac{1}{2}(c^2t_{0}^2) + \frac{1}{2}(v^2t^2)} \right)}{c} = t$ $ct = 2\left( \sqrt{ \frac{1}{2}(c^2t_{0}^2)+\frac{1}{2}(v^2t^2)} \right)$ $c^2t^2 = 4\left( \frac{1}{2}\left( c^2t_{0}^2 + \frac{1}{2}(v^2t^2) \right) \right)$ $c^2t^2 = 2c^2t_{0}^2 + 2v^2t^2$ $t^2 = 2t_{0}^2 + \frac{2v^2}{c^2} t^2$ $t^2\left( 1-\frac{2v^2}{c^2} \right) = 2t_{0}^2$ $t^2 = \frac{2t_{0}^2}{1-\frac{2v^2}{c^2}}$ So therefore: $t = \frac{t_{0}}{\sqrt{ 1- (v / c)^2 }}$ Of which $(v/c)^2$ is often written as $\beta^2$ and the constant $\frac{1}{\sqrt{1-(v/c)^2}}$ is the **LORENTZ GAMMA** $\gamma$. For our case, it'd be $\frac{1}{\sqrt{ 1- \frac{0.6^2}{1} }} = 1.25$, meaning that time in our 'observer' frame is actually dilated! # Example Problems # What's Next? And there goes the foundational stuff. Terr-i-fic!