Understanding Spacetime Diagrams

[Pencilvester]

I assume you are not saying that the equation $ds^2=dt^2−dx^2$ is another way of writing one of the Lorentz transformations?

Correct (I’m not saying that). It is an expression of the line element, or the metric, for the Minkowski plane, i.e. what makes the Minkowski plane fundamentally different from the Euclidean plane. The quantity $ds^2$ is invariant—for any two points on the plane, regardless of the frame of reference or the choice of coordinates, $ds^2$ will be the same. The analogous version of this in Euclidean space is the distance between two points (squared). It doesn’t matter how you rotate the space or move it around or assign coordinates to the points—the distance between the points stays the same.

 

[robphy]

Here are two 3-4-5 triangles in Minkowski spacetime.

I tried to read through the diagram, but, in general, because of my level, I could not understand what is going on in it. There is a lot of detail, and I wasn't sure which lines represent the world lines of which frames, whose frame of reference the diagram was drawn in, etc. Also, all the lines are diagonal and there is no set of vertical lines and horizontal lines for the time and space axis of the main frame. Things need to be greatly simplified for me:)

The diagrams below are based on the standard spacetime diagram that was drawn by @Mister T above .

I have reproduced my diagram with usual [unrotated] graph paper
along with the rotated graph paper version.

Note that the observer's frames of reference are determined by the joining opposite corners of the diamonds (that is, the diagonals of the diamonds). So, it's drawn in the frame where the red-observer (along the vertical OP) is at rest. The space-direction for red is along the horizontal PQ... just like the ordinary spacetime diagrams like what you are drawing. To get the grid for red, just join the corresponding corners parallel to the red diamond diagonals.)

(The rotated grid lines represent light-cones, which in my opinion is at the heart of special relativity...
and it's this grid that allows calculations to be done easier, when the method is followed... because it's easier to count little diamonds in the grid to compute the area.
One advantage of the rotated graph paper is that
it can display many reference frames on the same diagram.
If you want the grid for blue, just join the corresponding corners parallel to blue diamond diagonals. )

 

[Herman Trivilino]

Regarding the hypotenuse not being the longest side of the triangle, if you take a literal measurement of this diagram with a rule and measure the length of the red line segment between the two sparks, the length will come out to be 5.83, which would be the longest side of the triangle. So, I must be missing something?

Yes, you are missing the fact that the length of the hypotenuse can't be measured with a ruler on a spacetime diagram. In this case (see Post #26) the length is 4 seconds.

You are writing $t'^2=t^2-x^2$, so in this case we have $t'^2=5^2-3^2$, so $t'= 4$.

A more general expression is $t'^2-x'^2=t^2-x^2$, but in this case $x'=0$ because both events happened at the same place in the rocket frame. When two events happen at the same place, the time that elapses between them is called the proper time $\tau$. Therefore $\tau^2=t^2-x^2$. It's a relativistic invariant.

 

[Pencilvester]

I tried to draw a line of simultaneity for the rocket at the point of the second explosion (the explosions are marked with asterisks). That looks like this:

This shows that from the rocket's perspective, the tree frame's clock read 3.2 seconds during the time when the second explosion occurred.

Just a helpful tip: in spacetime diagrams where light traces out lines at 45 degree angles (which is the case here as well as in most spacetime diagrams you’ll see), the worldline of a physical object and that object’s lines of simultaneity will always be symmetrical with respect to light rays—i.e. light rays bisect the angles made between a worldline and it’s line of simultaneity. I see in your otherwise well drawn diagram that the rocket’s worldline goes over 2 units for every 3 units up, which means the lines of simultaneity should go over 3 units for every 2 units up, but it looks like yours is going over 4. The rocket should say the tree’s clock read 3.0 at the time of the second explosion.

 

[Pencilvester]

the rocket’s worldline goes over 2 units for every 3 units up

Never mind, I was looking at too small a portion of your lines. It does look like on average your slopes are 5/3 and 3/5.

 

[NoahsArk]

I see in your otherwise well drawn diagram that the rocket’s worldline goes over 2 units for every 3 units up, which means the lines of simultaneity should go over 3 units for every 2 units up

This is a very helpful tip! So, to get the slope of the line of simultaneity given the X and Y coordinates of the worldline, we just flip the numbers. E.g. if the car is moving at .1 c, then for every movement of .1 to the right, the y movement is 1 (.1, 1). The line of simultaneity, then, would have coordinates of (1, .1) and the slope is .1?

The rotated grid lines represent light-cones, which in my opinion is at the heart of special relativity

It was through watching a video about light cones that I first started to understand the rule about why two events can never occur in reverse order, no matter which frame of reference is observing them, when the first event had caused the second event to occur:

 

[Herman Trivilino]

It was through watching a video about light cones that I first started to understand the rule about why two events can never occur in reverse order, no matter which frame of reference is observing them, when the first event had caused the second event to occur:

If the first event caused the second event then their separation is timelike and what you say about the order is true. If the events have a spacelike separation then the order of the events varies with different reference frames. In particular, there will be a frame where the events are simultaneous.

 

[Pencilvester]

the slope is .1?

Yes, $\frac{dt}{dx}=0.1$ for the lines of simultaneity of an object moving with $\frac{dx}{dt}=0.1$.