Interplay of space and time in Spacetime invariant interval

[Ittiandro]

I guess I don't understand your confusion here. If the object is traveling from the origin at 0.6 c then the worldline goes from the origin in a straight line through X=0.6 and Y=1.0. Is that somehow unclear in any way?

Thank you for trying to answer my question

I guess by x=0.6 and y= 1.0 you mean 60% of c and c, respectively. Then I take you to mean that these two values can be graphically represented ( in the absence of acceleration) as a straight line starting from the origin of the coordinates ( the intersection of the x, y axes ) and sloping upward through the intersection point of x(0.6) and Y(1)..

This still does not address my question. In addition it raises a new one.

Let me start from the new one, In your answer you express the values on both the X and Y axes in the same unit of measurement( the speed of light, or a fraction of it) ,whereas from what I understood of the invariant interval formula, the X axis refers to the SPATIAL distance and should be in UNITS of LENGTH, while the Y axis ( which in the classical notion of Space and Time expressed the TIME component) now should express TIME converted in units of length by using the speed of light as a conversion factor. I can’t see any such conversion when you represent ct on the Y axis as 1 ,( supposedly c.) and the X axis value as 0 .6 c,, which is not in Units of length.

If my questions are not based on some fundamental misunderstanding of the issue, which is possible, then you can perhaps help me to understand how the X axis values in units of length (as they should be by definition) generate the ct values on the Y axis expressed as meters of Time or other equivalent units of length) of Time..
I understand that 1 sec of time can be represented on the Y axis, for instance, as 300,000,000 Meters by using c as a conversion factor, but why 1 second and not 2 seconds, 5 seconds or 3600 seconds? .I think this is in function of the X axis SPATIAL value, but how does this tie in with the Time length on the Y axis? .

 

[Nugatory]

Just measure distance in light-seconds and time in seconds or distance in meters and time in units of 1/300000000 of a second; plot the numerical value of the time on according to an observer at rest at $x=0$ on the y-axis and the numerical value of the position relative to that observer on the x-axis. Now when we say "x=0.6, y=1.0" (or "t=1.0" - "t" or "y"is just a matter of the label we write next to the axis) we're talking about a single point on the Cartesian plane.

Nothing awful happens if you instead measure distance in meters (or feet, or furlongs) or time in seconds (or weeks, or fortnights)... the worldline of a flash of light will no longer be a 45-degree angle so your diagrams will be harder to read and your equations will be cluttered up by factors of c everywhere, so you're doing things the hard way... But if you're careful to keep track of the units you'll still get the physics right.

 

[Nugatory]

I realize this, but even if if the surface used obeys to the Euclidean geometry,still won't it give at least an approximate visual appreciation of how ct shifts on the Y axis in function of the spatial movement of the object/events along the X axis? ..

Yes. Draw the worldline of an moving object. Now pick some point on that worldline. That point has an x-coordinate and a y coordinate in the Cartesian plane. You can interpret that pair (x,y) as saying "the object is at point x at time t according to an observer who is at rest (worldline is vertical) in these coordinates".

The relationship between the numerical value of the time t and the value on the y-axis is just a matter of the units we choose; c is the conversion factor between the x-axis scale and the y-axis scale. It's convenient to choose units in which c=1 because then we can use ordinary graph paper with square cells - a 45-degree line represents the worldline of a flash of light moving at a speed of one light-second per second.

However, none of this has anything to do with visualizing the invariance of the spacetime interval. That's a lot harder to see in these diagrams.

 

[Dale]

I guess by x=0.6 and y= 1.0 you mean 60% of c and c, respectively.

No most definitely not. If I use units of years for y (time) and units of light years for x (space), then x=0.6 and y=1.0 is the event that is a distance of 0.6 ly away and 1.0 y after the origin.

Drawing a line through that point and through the origin gives a worldline of an object traveling at 0.6 c.

Let me start from the new one, In your answer you express the values on both the X and Y axes in the same unit of measurement( the speed of light, or a fraction of it)

Sorry about being unclear. No, I did not express either X or Y in units of c. I didn't specify the units at all. All I did was specify the speed of the object, which corresponds to the slope of the worldline. Knowing the slope of the worldline you know that the point X=0.6 and Y=1.0 is on the line regardless of the units.

 

[Herman Trivilino]

Hi
I understand the algebraic explanation of the invariant interval, but I’m unable to translate it graphically with the Cartesian coordinates. I expected this to be more intuitive than the algebraic formulation, but it is not

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.

For the X coordinate, it is clear: here, if the distance between events A and B is, say, 1000 Km ( or I traveled 1000 Km from point A to point B ), the X coordinate will be subdivided in equal segments expressed in Km, for instance 10 segments of 100 km).

But how about the Y coordinate?

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.
.

 

[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.
.

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.

 

[Herman Trivilino]

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

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

@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

 

[Herman Trivilino]

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

 

[Herman Trivilino]

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.