Time dilation, length contraction, but velocity invariant

[Stephanus]

Does the above diagram make sense to you based on the specification of the scenario?

Absolutely, 200% makes sense. B >>------------------------>> R and B <<---------------------------------<< R
Just like that! Because we are ALWAYS moving in spaceTIME, so the light makes 45⁰ angle from B to R instead of horizontal lines.

There's no before and after B moves or before and after its worldline moves. B is at rest in our defining stationary frame so its worldline is shown as a vertical line because its x-coordinate is a constant 0. Later on, when we transform to the frame moving at 0.6c with respect to the stationary frame, B will be moving at -0.6c all the time. B does not change its motion. We are merely describing its state of motion, either constantly at rest or constantly moving, in two different frames but never changing its state of motion in either frame.

I always consider B and R are always moving. But this does not register in my unconscious mind. Next time, I should have to picture it instinctively rather than mathematically.

You got the right answer but I don't know how you did it.

It's time for you to see how the Lorentz Transformation works. It's real simple and you don't have to think about what you are doing. You can take the (x,t) coordinates of any event (dot) from the stationary frame and transform to the (x',t') coordinates of the moving frame. You repeat this process for all the events (dots) or at least some strategic ones so that you can see how the new worldlines appear.

Here at the three equations which apply for units where c=1 as in our case:

γ = 1/√(1-v2)
x' = γ(x-vt)
t' = γ(t-vx)

I think
t' = γ(t-vx) is simpler than
$t' = \frac{(t-x)*\sqrt{\frac{1-v}{1+v}} - x*\sqrt{1-v^2}}{1+v}$

You've already determined gamma to be 1.25 but I'll go through the motions here for anyone else that might be looking on. Remember, v=0.6:

γ = 1/√(1-v²) = 1/√(1-0.6²) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

Now we calculate the new time coordinate, t', for the new frame from the old coordinates, x=12 and t=0:

t' = γ(t-vx) = 1.25(12-0.6(0)) = 1.25(12) = 15

As you can see, we got the same answer you got but with a lot less effort, or at least, a lot less thinking.

Now it's just as important to obtain the x-coordinate which you did not do:

x' = γ(x-vt) = 1.25(0-0.6(12)) = 1.25(-7.2) = -9

Forgot that. I would have calculated x from my graphic equation $x = vt = -0.6 * 15 = -9$ But later, I'll do it with Lorentz boost. in x direction.

Can you please start using the Lorentz Transformation?

I wouldn't be in Physics Forum if I had known Lorentz before. Thanks, I'll use them later. But I think this is important for me. Because it is geomatrically correct. I'm doing it in cartesian system. Because I have to be sure, how to calculate any points based on simpler rule.
1. Hypotenuse in 4D $\sqrt{(ct)^2-x^2-y^2-z^2}$
2. Lorentz factor $\gamma = \frac{1}{\sqrt{1-v^2}}$

This doesn't make any sense. The Blue and Red lines lie in all frames.

This, I don't understand. Perhaps later.

I've asked you several times, where is W? is he stationary in one of the frames? If so, which one and what are his coordinates and state of motion? Why have you introduced him anyway?

W is at rest while B and R are moving 0.6c to the "West"
W I think is in 0,0. W is the green line.

Whatever B and R see in one frame, they see in all other frames. For example, B sees the light pulse being emitted when his clock is at zero in both frames. B sees the reflected light pulse when his clock is at 12 in both frames. R sees the reflection occur when his clock reads 6 in both frames. Different frames do not change anything that any observer sees or measures.The Red clock starts first in the moving frame. Both clocks start at the same time in the stationary frame.

I'll contemplate it.
Thanks gwhellsjr for your invaluable effort and insight.

 

[Stephanus]

Dear PF Forum,
Can someone explain to me?

Let's take a scenario where an observer sends out a light pulse to a reflector 6 feet away and measures with his clock how long it takes for the reflection to get back to him. Since light travels at 1 foot per nanosecond, it will take 12 nsecs for him to see the reflection and he will validate that the light is traveling at 1 foot per nsec for the 12 feet of the round trip that the light takes. If he is following the precepts of Special Relativity, he will define the time at which the reflection took place at 6 nsecs according to his clock and I have made that dot black:

In this diagram, the observer is shown as the thick blue line with dots marking off one-nanosecond increments of time according to his clock. Since he is at rest in this Inertial Reference Frame (IRF), the horizontal grid lines align with his clock and the vertical grid lines align with his measurement of distance.

The reflector is shown as the thick red line with similar dots. The thin blue line represents the pulse of light that the observer emits and the thin red line represents the reflection.

Now let's transform the coordinates of all the dots in the above diagram to a new IRF moving at 0.6c with respect to the original IRF:

At 0.6c, gamma is 1.25 and you will note that the dots are placed 1.25 nsecs apart as defined by the gird lines. Also, the distance to the reflector is contracted to 4.8 feet (6/1.25) as defined by the grid lines. You will also note that the light pulse and its reflection continue to travel at 1 foot per nanosecond but it takes less time in this IRF (3 nsecs) for the light to reach the reflector and more time (12 nsecs) for the reflection to get back to the observer. The total Coordinate Time is 15 nsecs and the total Coordinate Distance the light traveled is 15 feet so the velocity is maintained at 1 foot per nanosecond.

Here is a diagram that depicts the parameters that would apply if we just took the Length Contraction and Time Dilation factors into account:

Note that we would be using the length at the start of the scenario and the time as defined by the observer as when the reflection occurred to establish where the light reached at the time of the reflection which is way off base and would result in a contracted velocity as shown by the extra thin blue line. Instead, we need to use the Coordinate Distance of where the reflector has moved to when the light pulse reaches it and the Coordinate Time at which this occurred, not the time that the observer defines the reflection to occur according to his clock.

BR is the frame where Blue line and Red line lies.

This doesn't make any sense. The Blue and Red lines lie in all frames.

Why it doesn't make sense?
What does "The Blue and Red lines lie in all frames" mean?
Thanks.

 

[Ibix]

Each of the diagrams is a representation of a frame of reference. The red and blue lines are manifestly in both. You could transform to another frame and draw a third diagram, and the red and blue lines would be there too.

The point is that a frame of reference is just a different choice of point of view. It's closely analagous to turning a map upside down because you happen to be facing southwards. Things on the map don't appear and disappear as you do it, they just change place.

I'm not sure of the context of your comment, but you probably meant something like "BR is the frame where the red and blue lines are vertical". That's the rest frame of the red and blue observers.

 

[Stephanus]

Each of the diagrams is a representation of a frame of reference. The red and blue lines are manifestly in both. You could transform to another frame and draw a third diagram, and the red and blue lines would be there too.

The point is that a frame of reference is just a different choice of point of view. It's closely analagous to turning a map upside down because you happen to be facing southwards. Things on the map don't appear and disappear as you do it, they just change place.

I'm not sure of the context of your comment, but you probably meant something like "BR is the frame where the red and blue lines are vertical". That's the rest frame of the red and blue observers.

You mean BR is in TD,LC,ConstantC1.PNG, TD,LC,ConstantC2.PNG and TD,LC,ConstantC3.PNG. Yes they are.
It's my English. I tought ghwellsjr meant, the frame exists in space time diagram, Up and down, left - right, far left - far right, everywhere in the space diagram. Okay,.. it's an English problem. Thanks for making it clear for me to understand ghwellsjr answer.

 

[Ibix]

Langauge is very important in relativity, mainly because it's a complex topic and quite subtle errors of language can mean something completely different. Having less-than-perfect English isn't a problem, but you will need to use the language precisely.

A frame of reference is just a choice of point of view. Once you've chosen a frame, there are a natural set of coordinates to use (at least in SR), and you can draw a space-time diagram from that perspective. You can use the Lorentz transforms to switch to another frame's natural set of coordinates, and draw the diagram from a that perspective.

For example, when drawing maps we typically choose the frame of reference in which North appears as straight up. But there's nothing stopping us picking a frame where West is straight up, or where 037° is straight up. Here's a map of mainland Britain drawn from each of those three reference frames:

I marked London with a red dot; it has different coordinates in each map - something like (1.5,-4.25), (-4.25,-1.25) and (4.9,-1.5). I could write down a mathematical relationship between the coordinates of any point in one map and its coordinates in another; that would be the way to transform coordinates between those two frames.

Similarly, ghwellsjr has picked two different frames of reference and drawn a map of the x-t part of space-time as seen from each one. He could write down a mathematical relationship between the coordinates of any event in one diagram and its coordinates in another - and that would be the Lorentz transforms (in fact, he knows the transforms and used them to generate the maps, but working from the maps to deduce the transforms is - conceptually, at least - what Einstein did).

NB: Map is CC attribution/share-alike licensed - please see here for the original, with details of licencing and attribution.

 

[Stephanus]

I marked London with a red dot; it has different coordinates in each map - something like (1.5,-4.25), (-4.25,-1.25) and (4.9,-1.5). I could write down a mathematical relationship between the coordinates of any point in one map and its coordinates in another; that would be the way to transform coordinates between those two frames.

Thank you very much for you effort to help me understanding Space Time diagram.