You would probably be a bit less confused if you used the standard terminology. “Temporal displacement” is ambiguous and non standard, discard it. The standard terms are “coordinate time” and “proper time”.Yet this is confusing if τ is the proper time, it is the temporal displacement along the rotated worldline which must be ct - the temporal displacement...
Please help me I am confused again...
Yes, sorry about that but I was responding to this quote:You would probably be a bit less confused if you used the standard terminology. “Temporal displacement” is ambiguous and non standard, discard it. The standard terms are “coordinate time” and “proper time”.
so I was trying to use the same terminology; it is part of what makes things confusing when different terms are used as it becomes impossible to be sure what is being referred to...This is Minkowski geometry, not Euclidean geometry. Increasing the spatial displacement (the difference between starting and ending x coordinates) while holding the temporal displacement (the difference between starting and ending t coordinates) constant reduces the length of the (timelike) interval between the end points.
OK. Tyring to look at the maths rather than picturing it, I have a difficulty seeing why a hyperbolaRegarding dτ2=dt2−dx2. The quantity that is invariant is dτ. Both dt and dx are frame variant.
In spacetime, if you do a whole series of boosts you will find that after each boost you have a different dx and dt and the same dτ. You will also find that the set of all points with the same dτ traces out a hyperbola, meaning that in Minkowski geometry timelike distance (proper time) is defined by hyperbolas and hyperbolas are unchanged under boosts.
No. The “hypotenuse” is 0. In fact, the coordinate time axis is usually labeled ct explicitly just to make it clear that coordinate time is being treated geometrically and considered as a coordinate distance.So isn't ct a measurement from the origin rather than a measurement along the t axis?
I am unsure what you mean. As I understand it:No. The “hypotenuse” is 0.
So how does that make any sense?wikipedia said:
Note the scare quotes on @Dale's use of the word "hypotenuse". This is hyperbolic trigonometry we are talking about. The length of the "hypotenuse" is the square root of the difference of the squares of the sides, not the sum.Unless you mean that x and ct are not the sides of a right angle triangle... in which case there can be no hypotenuse?
The formula for calculating the spacetime interval in units where c=1 is ##ds^2=-dt^2+dx^2+dy^2+dz^2##. Notice the - sign in front of the ##dt^2## term. That makes it so that the “hypotenuse” is not the longest side.I am unsure what you mean. As I understand it:
For visualizations, I really like the approach Rob uses in post #23. To be able to interpret the diagram, one needs to be familiar with the concept of light clocks.It's not irrelevant to me. I have to look at the Euclidian representation of the graph. Knowing that longer Euclidian length = shorter worldline length is useful to me. In fact, what I was trying to point out is how the Euclidian length on a minkowski diagram has, in effect, the opposite meaning to what one's intuition would suggest.
Calculate the length of what?So, specifically for light, if we are plotting the world line for a pulse of light that was released at the origin and traveled for a coordinate time of 1 year then it will have traveled a coordinate distance of 1 light year. This will form a 45 degree slope from the origin, that is the “hypotenuse”. Now, as an exercise for you, calculate the length using the Minkowski formula.
The spacetime interval between the two events. One event is the emission of the light flash and the other is the reception. Note you correctly calculated the value to be zero. The events have a lightlike separation. You seem to be confusing the spacetime interval with the proper length. The value of the spacetime interval equals the proper length only for events with a spacelike separation.Calculate the length of what?
Nowhere can I find any suggestion that it involves a different metric (if that is the correct term?)Wikipedia said:Minkowski diagrams are two-dimensional graphs that depict events as happening in a universe consisting of one space dimension and one time dimension. Unlike a regular distance-time graph, the distance is displayed on the horizontal axis and time on the vertical axis. Additionally, the time and space units of measurement are chosen in such a way that an object moving at the speed of light is depicted as following a 45° angle to the diagram's axes.
It is not the same quantity, but there are some similarities. In both cases it is an invariant measure of distance in the space. The differences are that in space it is called distance and there is only one kind of distance, while in spacetime it is called the spacetime interval and there are three different kinds of spacetime intervals (space like, time like, and null).YEs, but one is the aggregate length in 3 dimensions and the other in 4, so does ds represent the same quantity in each case?
It makes little sense to use the term "invariant" to refer to ##a^2+b^2+c^2## in the context of Minkowski Spacetime.Surely a2+b2+c2 is still the aggregate length in Minkowski Spacetime; while -ct2+a2+b2+c2 is the Spacetime interval and both are invariant intervals.