goodabouthood said:
It says that x' = gamma (x-vt)
Now what exactly does that formula mean?
That formula is one half of the Lorentz Transform. The purpose of the LT is to convert coordinates of an event in one Frame of Reference to the coordinates of the same event in another Frame of Reference moving at speed v with respect to the first one.
goodabouthood said:
I take it that x is a point in my coordinate system, let's say 2 meters.
You should use the term "event" rather than point because "point" implies just the location part of the event which includes three components of location (x, y & z) and one component of time (t). You must always include the time components although in many cases we leave y and z out of the nomenclature when they are always equal to zero but it is always understood that they are really still there.
goodabouthood said:
I take it that v is the speed of the other reference frame or am I not correct in saying that? What exactly does that V refer to?
You are correct, v is the speed along the x-axis that the second reference frame is traveling with respect to the first frame.
goodabouthood said:
I take it that the t = time in my coordinate system.
Yes, it is the time of an event, the fourth coordinate.
goodabouthood said:
So would this make sense? In my coordinate system an event happens at distance of 2 meters and at 2 seconds.
Yes, if you mean, the event happened 2 meters from the origin along the x-axis at 2 seconds after the origin. It's important to understand that the two frames share a common origin where all four coordinates equal zero; that is, all the location components equal zero and the time component equals zero.
goodabouthood said:
The following train is moving at .5c relative to me.
Is this following equation right?
2-.5(2) x gamma = the distance in the other coordinate system.
It depends on what you really meant, but let me make some comments.
First off, events are relative to the origin of the frame, not to you, even if you were located at the origin of the frame at the beginning of the scenario because one second later, you are no longer located at the origin because the event describing you now has a time value of 1 second.
Secondly, you have to use consistent units. If your unit for distance is meters and your unit for time is seconds, then you cannot use .5 for the speed if you meant .5c which is a whole lot faster, so I'm going to assume that the train is traveling at .5c.
Thirdly, things can move in a frame and you describe their motion by a series of events, all within the one frame. So if the train were at the origin at the start of your scenario, its coordinates would be t=0 and x=0 (considering just the front of the train). Two seconds later, it will have moved one half the speed of light along the x-axis. So if we use the unit for time of seconds and the unit for distance of light-seconds, then the coordinates would be t=2 and x = 1. (But remember, I have changed the distance unit from meters to light-seconds.) If you leave the two numbers the same and you're talking about something that moved from the origin to that event, then you are talking about light because it is the only thing that can maintain both the x and t coordinates equal to each other.
goodabouthood said:
It appears according to the Interactive Minkowski Diagram when the t and d are the same number in my coordinate system, they will be different but equal numbers in the other coordinate system.
That's true. Let's work it out. First we have to calculate gamma. I'm assuming you know how to do this and you will get a value of 1.1547. Now we can continue calculating for light with both coordinates equal to 2. You will note that I am now using β instead of v because I am using compatible units and I want to make things easier when I do the calculation for t'. But first x':
x' = gamma (x-βt)
x' = 1.1547 (2-.5(2))
x' = 1.1547 (2-1)
x' = 1.1547 (1)
x' = 1.1547
Now we have to calculate t':
t' = gamma (t-βx)
t' = 1.1547 (2-.5(2))
t' = 1.1547 (2-1)
t' = 1.1547 (1)
t' = 1.1547
So it's true and obvious if you look at the calculation that when t and x are the same in one frame, they will be the same in any other frame. But this is just demonstrating that the speed of light is the same in all frames.
Now I think it is much more interesting to do the calculation for the event of an object (a train) moving at .5c. Remember, at t=2, x=1. Now we have:
x' = gamma (x-βt)
x' = 1.1547 (1-.5(2))
x' = 1.1547 (1-1)
x' = 1.1547 (0)
x' = 0
Now we have to calculate t':
t' = gamma (t-βx)
t' = 1.1547 (2-.5(1))
t' = 1.1547 (2-.5)
t' = 1.1547 (1.5)
t' = 1.732
Now what is this telling us? First off, whenever you calculate the events for a moving object that starts out at the origin and you transform them to a frame that is moving at the same speed as the object, you find that the x' coordinates of the event go to zero. This shows us that the object is at rest in this comoving frame.
Secondly, we see the time has gone from 2 seconds to 1.732 seconds which is the time divided by gamma. Does this sound familiar? It's showing time dilation.
