# Intro: Visualise, VISUALISE! (Or are we just manifesting it?) Often one of the greatest roadblocks to learning relativity is the ability for one to visualise the fabric of space itself. Surely it isn't that much of a tall task, is it! Well, even Einstein was purported to have struggled with the bane of all mathematicians - differential geometry. And guess what you need to understand relativity to the fullest? Therefore, we come here today to learn about **Spacetime Diagrams**, which allowed me and will allow you to sidestep the "Optional Rememberance" Boss that is the hyperboloid shape carved out by a gravitational well. Emotions will run HIGH as we each retrace the steps of the greats that walked this Earth and carved out a place for themselves in the scientific hall of fame, all whilst not breaking a single sweat! Or maybe a couple sweats if you're reading this the night before exams. YOU DO YOU! I'm no person to judge. ![[spacetimepunchdrunk.png]] # Finding the Speed of a Frame First, let's imagine a frame $u$. This will be our [[Galilean Relativity (Relativity)#Reference Frames|reference frame]], so we can first define the grid as shown: ![[BETTERspacetimegrid.png]] *Fig 1: A Spacetime Grid!* The y axis is denoted with a 'ct' and the x axis with a 't'. This way, we can adjust for relativistic effects by using the y-axis as our guideline. It's kind of like when you have an observer car travelling faster than any other in a race in order to set a guideline. Now, let's introduce another frame, this time moving at a speed $v$ relative to the frame $u$. We'll call it $u'$. ![[spacetimediag2.png]] *Fig 2: The grid, but with the new frame!* As per the [[Lorentz Transformations]], when an object moves at an apparent speed $v$ to the observer, time will dilate. Therefore, for each unit of time that the stationary object experiences (the $ct'$ frame), the moving observer will record a shorter time elapsed. Here, let me draw another grid for you so you can better visualise it. This time, I'll show the two grids and how each 'unit time' will look for them. When I overlap it, do you see that one unit of '$ct' time is actually greater than the value of the $ct$ axis? ![[colourcodedspacetimediagoverlap.png]] *Fig 3: the overlapped grid! Blue corresponds to the ct', x' axes.* So when $ct'$ is 1, the observer at $ct$ sees a value of 1.3. Spooky! There's one more property of the moving frame $u'$ that we can deduce without drawing any extra lines - the speed of the frame itself! To find the speed $v$ in terms of the speed of light $c$, we only have to take the tangent of the angle from the original $ct$ axis to the $ct'$ axis. In this case, $v$ would be $4/8$ = $0.5c$. Think about it - if $x$ were in light years, then the diagram basically tells us that for every eight years that pass, the frame will move 4 light years from the stationary observer's perspective. >[!Danger]- Using the Invariant Hyperbola to prove this! >TBA, check back when I learn it. # Enter Worldline. You see the grids? Those not on the axes would be known as **worldlines** - means to describe the motions of objects in each frame! Let's first erase everything on the 'grid' to be left with just the axes. We'll leave one perfectly vertical line at a displacement of $x=5$ from the origin of the observer: ![[worldline1.png]] *Fig 4: Worldline 1* This worldline would show that the object is stationary relative to the observer. As it is parallel to the ct axis, its displacement from the observer at x=0 does not change. Conversely, a worldline directly parallel to the x axis would be impossible - that'd imply teleportation! This would also apply to different grids - for example, a frame moving at a speed $v$ would have a stationary worldline as such: ![[worldline2.png]] *Fig 5: Now imagine a worldline moving!* An object can also move within the frame of the observer; for example, an object can accelerate or decelerate as it moves over time, depicted through this diagram: ![[decelworldline.png]] *Fig 6: A worldline of an object decelerating according to the frame of the observer!* As time goes on, the object's displacement first increases relative to the observer O then decreases, signifying that it has decelerated and that its velocity is now negative, facing towards the direction of negative displacement. ![[accelworldline.png]] *Fig 7: The accelerating worldline!* If an object accelerates relative to the observer, then the rate to which the displacement from the origin changes per unit time will increase, corresponding to a **decrease** in the apparent gradient of the worldline (as seen above!). Remember to detach the axes from the worldlines! The axes determine the motion of the frames themselves, whereas worldlines just refer to objects moving within those frames. # Extrapolating Intervals *Short chapter, but important anyways!* Now, say we have an event that we see takes place at the coordinates of (3, 4). We can use a spacetime diagram to better visualise this: ![[spacetimepoint.png]] So, this is essentially saying that at a distance of 4 from the observer at the origin of the frame, the event will appear to have occurred ct = 3 units of time before. Now, remember that our perception of an 'event' will follow the speed of light. In a spacetime diagram, the speed of light is reflected by a 45-degree angle. So, for this event, we draw said line towards the origin to find it's actual distance from the frame: ![[spacetimeeventorigin.png]] Perfect! # Spacetime Intervals - Conserving Quantities Now, for the spacetime diagram to work, as both displacement and time are not constant between frames, we have to use something known as the *spacetime interval*. To define it, we use its formula: $(\Delta s)^2 = (\Delta x)^2 - (c \Delta t)^2$ Where $\Delta x$ and $c \Delta t$ represent the displacement from the origin of the event in the $x$ and $ct$ axes respectively. Where it is actually $(\Delta s)^2$ that constitutes the spacetime interval. If the particle is moving in two or more directions, just add the $(\Delta y)^2$ and $(\Delta z)^2$ components to the spacetime interval! >[!Success]- Example of a conserved Spacetime Interval > >So, let's consider the point above (again). In the frame of the stationary observer, its spacetime interval would be: >$\Delta s = 9 - 16 = -7$ >Now, let's assume that another observer is moving away from the first frame, say, with a speed of 0.6c. Let's use the Lorentz transforms to find the resultant $\Delta x$ and $c \Delta t$! >$\Delta x' = \frac{1}{\sqrt{ 1 - 0.6^2 }} (3 - 0.6 \times 4) = 0.75$ >$c \Delta t' = \frac{1}{\sqrt{ 1 - 0.6^2 }} (4 - \frac{0.6 \times 3}{c^2}) = 2.75$ >And if we plug these values into our expression for the spacetime interval, we get: >$(\Delta s)^2 = 0.75^2 - 2.75^2 = -7$ ## EXT: The Invariant Hyperbola TBA, MOST LIKELY will learn this over the summer. # Intro to Minkowski Space TBA, just putting this here until I get to take the course :) # What's Next? Congratulations! This is the end of the Special Relativity series. From here on out, it's gonna be General Relativity - and it's gonna get TOUGH! Tuff. Pah! I'm no geologist. Whatever - you get the gist. 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