Interplay of space and time in Spacetime invariant interval

[Ittiandro]

Did you get the algebra to work out? Because the scheme you outlined in Post #22 is not what Nugatory described in #21 and then more recently in #27. The equations for length contraction and time dilation that you describe in #22 as the translation according to Lorentz are not the Lorentz transformation equations.
It's best to do exactly the same thing. Mark it off in segments of 100 km, too. Make the distance between the 100 km tick marks the same on both axes.

You are confusing yourself because you are thinking that this represents a light beam traveling a distance of 100 km when it instead corresponds to the the time it would take a light beam to travel a distance of 100 km. Lots of other things can happen in that time interval. A light beam traveling a distance of 100 km is just one of them, even though in the situation you happen to be considering there is no light beam doing that.
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Maybe I can better describe my problem by referring to

http://demonstrations.wolfram.com/MinkowskiSpacetime/ Here we have 4 snapshots each showing graphically and algebraically how the invariant interval remains the same by shifting from one frame of reference to another starting from given values of x and ct in the 1st frame ( black)and a given v/c value .

Let’s look at the 1st snapshot , where x=0.584, ct=0.00 and v/c=.50.

We use the Lorentz transformation to translate the x and ct coordinates from frame X to frame X’ ( red) and the

Delta S^2=Delta ct^2- Deltax^2 formula to arrive at the same invariant value of -0.341 in both frames.It would help if both x and ct axes were graded in equal segments showing the spatial values of x and the spacetime values of ct..

In the black frame x =0.584 and ct=0.00. Translated into the second frame ( red) these coordinates become x=0.674 and

Ct=-0.337.

What I do not understand is how in each frame the coordinate x relates to the coordinate ct .

From what I understand, x and ct should be linked: there is a value of ct for each value of x. Could ct=0.00 in this snapshot be the same even if x changed? I don’t think so, but how does x translate into ct?

Also, I don’t understand how the x and ct coordinate would change in the graphic by changing v/c which is set to .50 in this case. There seems to be a sliding bar at the top to do this, but it is inactive,

Thank you

Ittiandro

 

[Dale]

From what I understand, x and ct should be linked

x and ct are not linked in general, any more than x and y are linked in general.

Now, if you have a line, then x and y are linked on that line. But that is only on that specific line and does not imply any sort of other linkage.

 

[Mister T]

Also, I don’t understand how the x and ct coordinate would change in the graphic by changing v/c which is set to .50 in this case. There seems to be a sliding bar at the top to do this, but it is inactive,

Changing v/c changes the angle between the x-axis and the x'-axis (and also the angle between the ct-axis and the ct'-axis, since those angles are always equal).

I think that to make that slider active you need to purchase the software. The link you provided appears to be an advertisement for that software.

It would help if both x and ct axes were graded in equal segments showing the spatial values of x and the spacetime values of ct..

There are no numbers shown because there are no numbers. It's assumed that the same scale is used on each axis so that a light line makes a 45-degree angle.

You would benefit from a few concrete examples with numbers. What is your primary source for information? Do you have one or are you just flipping through web pages hoping that you'll see something of value? I suggest you order a copy of Spacetime Physics, 2nd edition, by Taylor and Wheeler. It's very readable.

 

[Ittiandro]

x and ct are not linked in general, any more than x and y are linked in general.

Now, if you have a line, then x and y are linked on that line. But that is only on that specific line and does not imply any sort of other linkage.

In my first post I asked
quote
Do I understand correctly that the more x ( the spatial distance between 2 events) increases( or decreases) in this formula, the more ct decreases (or increases) and viceversa, but in the end s^2 will remain the same for different combinations of x and ct. In other words every time x or ct increases( or decreases) this will be compensated by an equal decrease( or increase) of the other dimension, so that in the end s^2 remains the same. Is this correct?
unquote

The answer I received from Khashishi was

quote
Yes that seems to be correct unquote

I thought so, too. Now I'm confused

Ittiandro

 

[Mister T]

In my first post I asked
quote
Do I understand correctly that the more x ( the spatial distance between 2 events) increases( or decreases) in this formula, the more ct decreases (or increases) and viceversa, but in the end s^2 will remain the same for different combinations of x and ct. In other words every time x or ct increases( or decreases) this will be compensated by an equal decrease( or increase) of the other dimension, so that in the end s^2 remains the same. Is this correct?
unquote

The answer I received from Khashishi was

quote
Yes that seems to be correct unquote

I thought so, too. Now I'm confused

Read what Fredrik wrote in Post #3. Following that you indicated that you understood that it was true under the condition of constant velocity motion. In that case you'd have $\Delta x=v \Delta t$ relating the coordinates.

 

[Mister T]

@Ittiandro Here's an excerpt from a lesson I'm working on. Perhaps it will help answer the question you're asking.

On a long stretch of abandoned road, Sam marks out an x-axis, and he has placed high-precision electronic clocks along it.

Each of the numbers on the x-axis is separated by a distance of 300 m, which is the distance traveled by light in a time of 1 μs. Sam carefully synchronizes all of the clocks. A rocket zooms along this x-axis at a steady speed. Tess is aboard the rocket and she wants to signal Sam using two events separated in time by just a few microseconds. The plan is to use a circuit that shoots sparks. The rocket is zooming along, an antenna protrudes from the bottom of the rocket and its tip is just a millimeter above Sam’s roadway. The first spark is created, it leaves the rocket antenna and hits Sam’s x-axis at x = 3. When the second spark is fired it hits at x = 6. Sam’s clocks are set up to read the time when hit by a spark. According to these clocks, the first spark hit at a time t = 2 μs and the second at a time t = 7 μs.

Here's the spacetime diagram:

The spacetime interval is 4 μs because $$(\Delta t)^2-(\Delta x)^2=(7-2)^2-(6-3)^2=(5)^2-(3)^2=(4)^2.$$
In Tess's frame the two events occur at the same location (the tip of her rocket antenna) so $\Delta x'=0$. Since the spacetime interval is invariant, it must also equal 4 μs in her frame of reference. $$(\Delta t')^2-(\Delta x')^2=(4)^2$$ $$(\Delta t')^2-(0)^2=(4)^2$$ Therefore $\Delta t'=4$, meaning 4 μs of time elapses in between the events in Tess's frame.

 

[Ittiandro]

@Ittiandro Here's an excerpt from a lesson I'm working on. Perhaps it will help answer the question you're asking.

/QUOTE]

Great! Getting there!

Your diagram is very clear. I only wish that the other sources I visited on the Internet were this clear. You guys do a terrific job.

Now I see where I got confused

First, I had wrongly understood that the x -axis should represent a spatial distance in units of length and that the time values should go on the y axis( by converting them in the same units of length as the x-axis using c as a conversion factor).. This is where my confusion arose.

I see instead that in your diagram , the spatial distance on the x-axis, too, is expressed in time, each consecutive number along the x-axis representing 1 μs increments or, spatially, a distance of 300 m at speed c.

As I said, I thought that this should happen on the y-axis ( ct) only. Now my problem is solved. I even did some drills to test the Lorentz transformations from one frame to another for given x , ct and c values.

I notice that in your diagram the speed is set at c because the line is slanted at 45 deg. I imagine you can set the speed at any subluminal value . In this case the slope of the line would increase to less than 45 deg , until it becomes vertical, parallel to the Y-axis if there is no spatial movement.

I also guess that if the speed goes down, suppose fro 1 to .50 c the space intervals between each consecutive μs value on the x-axis become shorter, so instead of being 300 m, they will become correspondingly shorter, in this case 150 m instead of 300 m. Do I have to assume that also the distances between each μs increment on the the Y-axis will become graphically shorter, to equal those on the x-axis?

Now there is one more point I want to clarify: why for calculating the Invariant Interval S^2 the Δx^2 value is subtracted from the Δct^2 value instead of being added, like in the Pythagorean formula?

I ‘ll see if I can get hold of J.A. Wheeler book “ Spacetime physics”. I already have his other book called “ Gravitation” which deals wit this issue, but I found it a bit forbidding, at my level of layman, albeit educated, I guess, with not much of a background in maths. Maybe this my limit. But sometimes it is true that students are as good or bad in their leraning curve as their teachers. A good teacher can do wonders!Thanks for your helpIttiandro

 

[Mister T]

First, I had wrongly understood that the x -axis should represent a spatial distance in units of length and that the time values should go on the y axis( by converting them in the same units of length as the x-axis using c as a conversion factor).. This is where my confusion arose.

I see instead that in your diagram , the spatial distance on the x-axis, too, is expressed in time, each consecutive number along the x-axis representing 1 μs increments or, spatially, a distance of 300 m at speed c.

The two approaches are equivalent. You measure distance and time in the same units, they can be either units of distance or units of time. I wish I had instead used units of distance for this post, but I was doing it the other way in the lesson so I just left it that way.

I notice that in your diagram the speed is set at c because the line is slanted at 45 deg.

No, it's not. The speed is $\frac{\Delta x}{\Delta t}$.

Now there is one more point I want to clarify: why for calculating the Invariant Interval S^2 the Δx^2 value is subtracted from the Δct^2 value instead of being added, like in the Pythagorean formula?

Because the speed of light is the same in all reference frames.

 

[Ittiandro]

Mister T

1. How would the x-axis and ct-axis of your diagram look if represented in units of distance?

2. If the speed is Δx/Δt, then the speed here is .60c, Ι guess?

3. .60c is the speed of what? Assumedly, the sparks emitted by the rocket's antenna travel at c.Is it the rocket's speed ( Tess)? Bottom line, if the conversion standard is c, how can we speak of speeds lower than c, like, in the Wolfram diagram, .20c or, in your example .60c,?

4. In the 1st Wolfram diagram at the top of the page http://demonstrations.wolfram.com/MinkowskiSpacetime/ the speed is given as .2 c. If the speed is Δx/Δt, Ι am unable to calculate it from the given x value (.500) and ct.

5. Shouldn't the speed always be given at the onset, because it shapes the slope of the line and ultimately contributes to define the invariant interval? In fact the Wolfram diagrams are articulated according to different pre-set speeds.
In your example, instead, the speed is not given as a constitutive pre-set datum, but it emerges indirectly as Δx/Δt. I'm sure there is a reason, but I can't see it.

Thanks for your help

Ittiandro

 

[Mister T]

How would the x-axis and ct-axis of your diagram look if represented in units of distance?

You would just multiply every number on each axis by 300 and change the units from microseconds to meters.

.60c is the speed of what?

The rocket.