Interplay of space and time in Spacetime invariant interval

[Ittiandro]

According to Special Relativity, the same event could have a different time duration and a different space extension for different observers, depending on their frame of reference. Relativity subsequently introduced the the notions of Spacetime as a continuum ( as opposed to the classical view of Space and Time as separate entities) and invariant interval in order to allow for the same event to be perceived by different observers univocally, i.e. with the same spatial-temporal dimension.

The formula is s^2=dx^2-dct^2.

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?

If I understand correctly, is it possible to represent this mathematically in a set of equations s^2=dx^2-dct^2 for different values of x and ct?

I am not physicist, just trying to have a basic conceptual grasp of this issue. So may be my question is off, but I had to ask.ThanksFranco

 

[Khashishi]

Yes, that seems correct. Usually, the term event is used to refer to something that happens at a single instant in time and space, and we would use two events (a start event and an end event) to describe something that happens extended over an interval. But I don't think there's anything wrong with your usage of the term.

 

[Fredrik]

The formula is s^2=dx^2-dct^2.

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?

Only if you chose to only consider constant-velocity motion and forgot to mention it. You can change d(ct) without changing dx, if you just change your velocity (which is equal to c dx/d(ct)). Obviously, if you do this, you will have to replace one of the two events by another, but you have to do that even when you change dx to compensate for the change in d(ct).

 

[Ittiandro]

Only if you chose to only consider constant-velocity motion and forgot to mention it. You can change d(ct) without changing dx, if you just change your velocity (which is equal to c dx/d(ct)). Obviously, if you do this, you will have to replace one of the two events by another, but you have to do that even when you change dx to compensate for the change in d(ct).

Yes, of course, I meant constant velocity, because , I think, if you change either ct or cdx you no longer have the same events.
One more question, though: once Special Relativity destroyed the absolute character of Space and Time, why was it necessary to reintroduce, with SPACETIME, an invariant interval, which in a way brings back the idea of some absoluteness?
The farthest I can go to explain this to myself is that science strives for unity and univocality and therefore the idea that Space and Time are relative and depend on the frame of reference does not sit well with such a search for univocality .
Certainly the S.R. GAMMA factor and, in G.R., the impact of gravity on time are so true and of such a practical importance that modern technology must take them into account. If not, even the GPS and satellite communications data would be off. Still I cannot see the practical and predictive value of such a thing as a SPACETIME CONTINUUM and its notion of INVARIANT interval.
It may sound like a philosophical question, but behind the seeming abstractness of maths and the speculations of theoretical physics, there must be some purposes or reasons, amenable to conceptual expression and having perhaps a practical import.

Thanks

Franco

 

[PeterDonis]

once Special Relativity destroyed the absolute character of Space and Time, why was it necessary to reintroduce, with SPACETIME, an invariant interval, which in a way brings back the idea of some absoluteness?

Because that's how the universe works. Experiments have shown that spacetime intervals are, in fact, invariant, so that's how we represent them in our theory.

I cannot see the practical and predictive value of such a thing as a SPACETIME CONTINUUM and its notion of INVARIANT interval.

You don't think SR and GR have practical and predictive value? Even though you gave a specific example (GPS) where they obviously do? That practical and predictive value comes from modeling the universe as a spacetime continuum with invariant intervals.

 

[Ittiandro]

Experiments may well have shown that Spacetime intervals are invariant and I am not here to argue against, since I am not a physicist , but I am curious to know beyond mere BELIEF what kind of empirical evidence corroborates the invariance of spacetime intervals, just in the same way that the MUONS experiments, for ex, or the clock-on-the-airplane experiment have shown EXPERIMENTALLY that time and space expand or contract depending on the speed at which objects travel..

Your question quote You don't think SR and GR have practical and predictive value? unquote implies " How can you not think so? ". Well, there is no such a thing as a general theory distinct from its postulates and corollaries and not all of them may necessarily have the same strength or predictive value a priori, without specific empirical evidence( this is why science progresses by discarding certain hypotheses ( or certains aspects of them) and replacing them with new hypotheses having a higher predictive value).

Indeed, I have already acknowledged the practical and predictive value of both SR and GR in what concerns the relativity of time and space and the application of this idea to modern technology. I can't see, though, how the spacetime invariance follows from the Relativity of Space and Time ( even more so that the conclusions are opposite, indeed, even taking into account that Space and Time are not the same entity as SPACETIME).
If spacetime invariance is true, it must have 1) its own specific EMPIRICAL evidence and 2) its practical technological applications. Until then, we cannot unequivocally say that this is the way the universe works...WE may like to think so, or maybe the idea is mathematically consistent, but.it may not be the whole story.

. .

 

[Nugatory]

Experiments may well have shown that Spacetime intervals are invariant and I am not here to argue against, since I am not a physicist , but I am curious to know beyond mere BELIEF what kind of empirical evidence corroborates the invariance of spacetime intervals, just in the same way that the MUONS experiments, for ex, or the clock-on-the-airplane experiment have shown EXPERIMENTALLY that time and space expand or contract depending on the speed at which objects travel..

It's the same experiments. The Minkowskian formulation based on space-time intervals is just a different (cleaner, more elegant, generally easier to use) way of expressing the math of special relativity, so any experiment that supports relativity supports the invariance of the space-time interval. Indeed, the invariance of the space-time interval is required by the Lorentz transformations of Einstein's original approach; it's just a different way of expressing that the speed of light is $c$ in all frames.

(The part of your post that I bolded above is not correct - you would have to work in the essential point about relative motion to make it correct. This may seem like it doesn't matter here, but in fact it is essential to understanding how Minkowski's formulation is equivalent to Einstein's).

 

[PeterDonis]

I have already acknowledged the practical and predictive value of both SR and GR in what concerns the relativity of time and space and the application of this idea to modern technology.

But the underlined part is not what gives SR and GR their practical and predictive value. The relativity of time and space is not a fundamental concept of the theories; it's just a consequence that comes out. The fundamental concept of the theories as they are formulated and used today is spacetime and its invariant intervals.

Even in Einstein's original formulation (before Minkowski showed how to formulate the theory much more cleanly using spacetime), the relativity of time and space was not a fundamental concept. Einstein did not develop SR by asking himself "what would happen if time and space were relative?" He developed SR by asking himself "how would mechanics work if the speed of light were invariant in all inertial frames?" So even in Einstein's formulation, the fundamental concept is the invariance of the speed of light (and also the principle that physical laws must be the same in all inertial frames). Relativity of time and space is a consequence of the theory; it's not what the theory is based on.

(And, as Nugatory noted, the invariance of the spacetime interval is just an alternate way of expressing the fact that the speed of light is the same in all inertial frames; so the two different formulations are only different on the surface, they actually express the same thing.)

If spacetime invariance is true, it must have 1) its own specific EMPIRICAL evidence and 2) its practical technological applications.

No. Empirical evidence validates a theory; it doesn't validate a particular way of formulating it but not others. As Nugatory said, all the evidence that validates SR validates spacetime invariance, since spacetime invariance is a valid way to formulate SR. It also validates Einstein's original formulation (which is equivalent to the spacetime formulation, as above).

 

[Ittiandro]

It's the same experiments. The Minkowskian formulation based on space-time intervals is just a different (cleaner, more elegant, generally easier to use) way of expressing the math of special relativity, so any experiment that supports relativity supports the invariance of the space-time interval. Indeed, the invariance of the space-time interval is required by the Lorentz transformations of Einstein's original approach; it's just a different way of expressing that the speed of light is $c$ in all frames.

(The part of your post that I bolded above is not correct - you would have to work in the essential point about relative motion to make it correct. This may seem like it doesn't matter here, but in fact it is essential to understanding how Minkowski's formulation is equivalent to Einstein's).

A lot of food for thought!
I am not up to advanced maths, though, and even less to Minkowski's maths. So, mathematically, there is quite a bit which may escape me.
I tend to view maths, though, not as a reality in itself, but as a language which expresses the underlying reality of our world , in a infinitely more precise way than concepts,, but still a language.. So, I believe that not knowing maths should not be an impediment to a basic understanding.

Coming to the point, that part of my post which you bolded and which seems incorrect to you is the essential conceptual version of the t' and t relationship in the light of the the GAMMA factor. The formula essentially states that the time measured by the observer in motion (t') is equal to the time of the stationary observer (t) multiplied by the GAMMA factor, from which follows that the more velocity increases, the shorter the elapsed time becomes for for the observer in motion. This is no more no less than what the formula says and which I said.
I am aware, though, that in relativistic terms, t' would also be right in saying that he is not moving and that t is moving instead..

Also, when you say that quote any experiment that supports relativity supports the invariance of the space-time interval.unquote, I' be interested to know how and in which sense an experiment like, for instance, the one with the MUONS shows both the relativity of Time and Space and at the same time the invariance of the SPACETIME interval.
Could you shed some light for me in basic conceptual terms or with a minimum of maths, if it is possible?

Thanks

Franco

 

[PeterDonis]

I' be interested to know how and in which sense an experiment like, for instance, the one with the MUONS shows both the relativity of Time and Space and at the same time the invariance of the SPACETIME interval.

What you are calling the "relativity of time and space" is a consequence of the invariance of the spacetime interval. You can't have one without the other. The same measurements that show that the muons have much less elapsed time than clocks on Earth, also show that the spacetime interval is invariant between the two events in question (a given muon being created in the upper atmosphere, and the same muon being detected at the Earth's surface).

 

[Dale]

what kind of empirical evidence corroborates the invariance of spacetime intervals

All of this empirical evidence:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

If you start with the Lorentz transform then you can derive the fact that the spacetime interval is invariant. Alternatively, you can start with the invariance of the spacetime interval and derive the Lorentz transform. They are logically equivalent, so any evidence which supports one necessarily supports the other.

 

[Dale]

I' be interested to know how and in which sense an experiment like, for instance, the one with the MUONS shows both the relativity of Time and Space and at the same time the invariance of the SPACETIME interval.

You can calculate how many muons reach the surface of the Earth using any of the following approaches:

In the Earth's frame you can use the time dilation of the muons
In the muon's frame you can use the length contraction of the earth
In any frame you can use the invariant interval along the muon's worldline

All three computations yield the same answer regarding the number of muons reaching the surface. So any measurement which supports one will support the other two (since they are all the same).

 

[stevendaryl]

One more question, though: once Special Relativity destroyed the absolute character of Space and Time, why was it necessary to reintroduce, with SPACETIME, an invariant interval, which in a way brings back the idea of some absoluteness?

It doesn't explain everything about relativity, but to me, the geometric view of spacetime helps to understand a lot of the peculiarities of Special Relativity.

Think of a plain white piece of paper. You can draw any straight line on a piece of paper and call that "horizontal". Draw a line perpendicular to the first, and call it "vertical". Now draw a third line segment in any direction whatsoever.

You can characterize the third line segment by two numbers: \delta H, how far the line segment extends in the horizontal direction, and \delta V, how far the line extends in the vertical direction. Note that these two numbers are completely dependent on how you chose your vertical axis and horizontal axis. There is a "relativity" of directions on the piece of paper; you can choose any line you want, and call that "horizontal". However, not everything is relative. The combination

L = \sqrt{|\delta H^2 + \delta V^2|}

is invariant. No matter what you chose for your horizontal and vertical axis, as long as they are perpendicular, you will find that L has the same value.

Special Relativity in two spacetime dimensions, x and t is very similar. The separation between two events can be characterized by two numbers: \delta x and \delta t. You have a choice of any inertial frame to use as the basis for your t axis and your x axis. So the values of \delta x and \delta t are relative to that choice. However, the combination S = \sqrt{|\delta x^2 - c^2 \delta t^2|} is invariant, it has the same value, no matter what reference frame you choose.

 

[Ittiandro]

You can calculate how many muons reach the surface of the Earth using any of the following approaches:

In the Earth's frame you can use the time dilation of the muons
In the muon's frame you can use the length contraction of the earth
In any frame you can use the invariant interval along the muon's worldline

All three computations yield the same answer regarding the number of muons reaching the surface. So any measurement which supports one will support the other two (since they are all the same).

Good! Your answer is the one that addresses my question closest.
It would be interesting though:if you could elaborate a bit more how the invariant interval along the muon's worldline can be used within either of the two frames ( Earth's frame and muons' frame) to yield the same number of muons reaching the Earth. My maths are not very hot, though. Perhaps a graphic illustration would be more appropriate or a combination of the two, keeping the math, though, as simple as the nature of the issue allows.
Thanks

Ittiandro
.

 

[Dale]

Here is a graphic illustration:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html#c2

That link also shows the time dilation calculation in the Earth frame. The key is in the red box, where they calculate the "moving half-life" (7.8 us) from the standard half-life (1.56 us) times the time dilation factor (5).

The spacetime interval calculation would be $\Delta\tau = \sqrt{\Delta t^2 - \Delta x^2/c^2} =\sqrt{ (34.04 us)^2 - (10 km/c)^2 }= 6.8 us$ and $1000000 \; 2^{-6.8/1.56} = 49000$

 

[Ittiandro]

Here is a graphic illustration:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html#c2

That link also shows the time dilation calculation in the Earth frame. The key is in the red box, where they calculate the "moving half-life" (7.8 us) from the standard half-life (1.56 us) times the time dilation factor (5).

The spacetime interval calculation would be $\Delta\tau = \sqrt{\Delta t^2 - \Delta x^2/c^2} =\sqrt{ (34.04 us)^2 - (10 km/c)^2 }= 6.8 us$ and $1000000 \; 2^{-6.8/1.56} = 49000$

Thanks for your inputOn the link http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html#c2 I am getting bogged down on some of the maths:

1.For the N0N-relativistic Frame , the Relativistic Earth- Frame and Relativistic Muon-Frame, the survival rate is expressed as 2^-21.8, 2^-4.36 and 2^-4.36 respectively. I can’t understand, mathematically, how these three survival rates are computed.2. How can Gamma =5 ?

Supposedly gamma is 1/sqrt(1-[v2/c^2]).

Then for v=.98c, Gamma=1/0.1414. Why Gamma=5?

Let T= time for Earth-bound observer and T’=time for the Muon

Let T=10^4 meters/(.98)(3x10^8 ms)=.000294 ms. for both the Non-Relativistic and Relativistic Earth Frames

Let v=.98c

If T’=Tx Gamma and Gamma = 1/sqrt(1-[v2/c^2),

then T’=.0000415So T=.000294 and T’=.0000415. The relationship between T and T’ is nowhere near Gamma=5Can you clarify this for me please?ThanksFranco

 

[Dale]

For the N0N-relativistic Frame , the Relativistic Earth- Frame and Relativistic Muon-Frame, the survival rate is expressed as 2^-21.8, 2^-4.36 and 2^-4.36 respectively. I can’t understand, mathematically, how these three survival rates are computed.

The 21.8 and 4.36 are the number of half lives. After one half life, by definition, half (2^-1) of the muons have decayed. After another half life then half of the remaining muons have decayed, leaving a total of only one fourth (2^-2) of the original muons. After n half lives only (2^-n) of the original muons remain.

http://en.wikipedia.org/wiki/Half-life

How can Gamma =5 ?

Supposedly gamma is 1/sqrt(1-[v2/c^2]).

Then for v=.98c, Gamma=1/0.1414. Why Gamma=5?

You are forgetting to square the velocity. For $v/c=0.98$ we find that $1/\sqrt{1-v/c}=1/0.1414$ but $\gamma=1/\sqrt{1-v^2/c^2}=1/0.1999$.

Also, don't get too hung up on the rounding in the Hyperphysics example. If you get it close but your numbers are slightly off it is probably just because you rounded off in a different spot than they did.

 

[Ittiandro]

Here is a graphic illustration:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html#c2

That link also shows the time dilation calculation in the Earth frame. The key is in the red box, where they calculate the "moving half-life" (7.8 us) from the standard half-life (1.56 us) times the time dilation factor (5).

The spacetime interval calculation would be $\Delta\tau = \sqrt{\Delta t^2 - \Delta x^2/c^2} =\sqrt{ (34.04 us)^2 - (10 km/c)^2 }= 6.8 us$ and $1000000 \; 2^{-6.8/1.56} = 49000$

Thanks
Getting there!

I have gone over again the Muon experiment link. Its skeleton is mathematically clearer now, but there are a couple of things I’d like to clarify.I can see that:

1) the surviving muons are much more than they were expected in the pre-relativistic context and

2) the number of surviving Muons is the same in both frames of reference , due to the Time Dilation factor( Gamma) for the R.E.F.O. observer and the Space Contraction factor for the R.M.F.O. observer

I suppose this is one way to put ( and demonstrate) the Invariance of Spacetime for different frames of reference. What is still not clear, though, is how

1) they come up with a number of halflives of 4.36( as opposed to 21.8) and

2) how the number of halflives ( 4.36 ) is the same in both frames even though the distances are not the same . In fact, for T=10^4m/ (.98)(3x10^8)[ REFO frame] and T=2000 m/(.98)(3x10^8)[ RMFO frame] the number of halflives is always 4.36.

3) Why only relatively few muons survive, even in the relativistic context? What happens to the others?

Thanks

Franco

 

[Dale]

1) they come up with a number of halflives of 4.36( as opposed to 21.8)

In the muon's frame the muon is at rest and therefore not time dilated, so the halflife is just the standard 1.56 us. In the muon's frame the trip lasts 6.77 us, which gives 6.77/1.56 = 4.34 halflives. (don't worry about the rounding)

In the Earth's frame the muon is moving at .98 c and is therefore time dilated by a factor of 5.03, so the coordinate halflife is 5.03*1.56 us = 7.84 us. In the Earth's frame the trip lasts 34.0 us, which gives 34.0/7.84 = 4.34 halflives.

In any frame the spacetime interval is 6.77 us, so in any frame you get 6.77/1.56 halflives.

2) how the number of halflives ( 4.36 ) is the same in both frames even though the distances are not the same . In fact, for T=10^4m/ (.98)(3x10^8)[ REFO frame] and T=2000 m/(.98)(3x10^8)[ RMFO frame] the number of halflives is always 4.36.

I would say that the number of halflives is the same because the number of halflives is always equal to the spacetime interval divided by the resting halflife, both of which are the same in all frames.

3) Why only relatively few muons survive, even in the relativistic context? What happens to the others?

Muons are unstable particles. They usually decay into an electron and some neutrinos.

 

[Ittiandro]

Thanks Dalespam
I got it. Great!
Now another related question on this:
based on the DS^2=DX^2-DcT^2.equation, is it possible to show mathematically how DS^2 remains the same for the same speed of .98 c and distance in both frames?
Can this be represented graphically as well on the x/y cartesian coordinates, if this can be done using the lay-out of the forum, maybe by an attachment? Thanks for your clarity

Franco

 

[Nugatory]

Now another related question on this:
based on the DS^2=DX^2-DcT^2.equation, is it possible to show mathematically how DS^2 remains the same for the same speed of .98 c and distance in both frames?
Can this be represented graphically as well on the x/y cartesian coordinates, if this can be done using the lay-out of the forum, maybe by an attachment?

It's hard to demonstrate this graphically because a graph is has be drawn on a piece of paper or a computer screen, and both of those surfaces obey the Cartesian coordinate rule $ds^2=dx^2+dt^2$ (if we call the vertical axis the "t axis" instead of the "y axis") from the Pythagorean theorem. There's a plus sign in there where we need a minus sign, which just goes to show that space-time doesn't work like ordinary space.

It's easy enough to demonstrate it with a bit of algebra though. Put one end of your interval at the origin $(x=0, t=0)$ and the other end at the event $(x=a, t=b)$ so $\Delta{x}=a-0=a$, $\Delta{t}=t-0=t$, and $\Delta{s}^2=\Delta{t}^2-\Delta{x}^2=b^2-a^2$. Use the Lorentz transforms to calculate the coordinates of the point $(x=a, t=b)$ in the moving frame and use these to calculate $\Delta{s}^2$ in that frame. When you do, all the $v$ and $c$ and $\gamma$ stuff will miraculously cancel out and you'll be left with the same $b^2-a^2$.

 

[Ittiandro]

Thank you, I need some more help
In order to demonstrate algebraically that the Spacetime interval is invariant in all frames I should be able( I think) to obtain two equations :

Ds^2=dct^2-dX^2 ( frame X, stationary) and

Ds’^2=dct’^2-dX’^2 ( moving frame X’)

where Ds^2= Ds’^2, because the time-dilation factor from the stationary X frame is compensated by a corresponding space contraction from the moving muon frame . The Time dilation in one frame is offset by a space contraction in the other.I know that, according to the Lorentz transformation, I have to translate the spatial-temporal coordinates of the moving frame X’( the MUON frame) into the coordinates of the “ stationary” frame, i.e. the Earth-bound observer.

This translation according to Lorentz should be based on the equations

L’=L/Gamma and T’=T*Gamma, where L is the distance and T the time in the “ stationary” frame X and L’,T’ are the distance and time, respectively, in the moving frame X’The values I’d be using are those of the Muon experiment at the http://hyperphysics.phyastr.gsu.edu/hbase/relativ/muon.html#c2 link:

v=.98c and L=10^4 mts I was about to try and translate algebraically the MUON Experiment by taking as a point of departure the equation ds^2=dct^2-dx^2 and using the Lorentz transformations. I tried and tried, but I got a roadblock.

Can you help please?ThanksFranco

 

[Ittiandro]

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

What I understand is that, if the spatial coordinate X is in METERS ( or KM), the TIME coordinate Y must also be expressed in METERS or Km ( based on the invariant “c”) in order to provide a common denominator with the spatial coordinate X expressed in MT or KM .

The underlying assumption in the relativistic model , from what I understand, is that objects always MOVE even when they are stationary, because they MOVE through TIME. These objects are always represented as WORLDLINES within the Cartesian coordinates, even when they do not move spatially, (in which case the worldline is vertical and parallel to the Y axis.)

Now how do we represent all this with the Cartesian coordinates?

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?

In theory, I know that TIME, too, has to be expressed in METERS ( or KM) based on the speed of “c” which is invariant. So, for ex., 1 sec of “c” time would appear on the Y coordinate as 300,000,000 mt ( or 300,000 Km) .

But what is this time which we have to express in MT or Km on the Y coordinate, subdivided in equal Mt or KM segments?

Can anybody explain or refer me to books or to the Web? I have searched the Internet for one week, though, unsuccessfully.

Thanks

Franco

 

[Dale]

It would really help if you used proper SI. The SI unit of length is the meter, not the METER, and it is abbreviated m, not mt or Mt or MT. 1000 meters is a kilometer which is abbreviated km, not KM or Km.

 

[Fredrik]

You don't have to measure time and distance in the same units. You can if you want to. That would be one way to describe what we're doing when we're choosing units such that c=1. But in SI units, we have c=299792458 m/s. If t is in seconds (s), and c is in meters per second (m/s), then ct is in meters. It's common to draw diagrams with x on one axis and ct on the other. This way you get the same units on both axes, and the world line of a ray of light makes a 45 degree angle with each axis.

The book "Spacetime physics" by Taylor and Wheeler is one of the standard recommendations for people who are looking for an introduction to SR. I often recommend the first chapter of "A first course in general relativity" by Schutz, but it requires that you know some math.

 

[Dale]

I know that TIME, too, has to be expressed in METERS ( or KM) based on the speed of “c” which is invariant. So, for ex., 1 sec of “c” time would appear on the Y coordinate as 300,000,000 mt ( or 300,000 Km) .

But what is this time which we have to express in MT or Km on the Y coordinate, subdivided in equal Mt or KM segments?

One way to think of this is to consider how ships would navigate. They would use nautical miles to measure horizontal distance and fathoms to measure vertical distances. If you wanted to express distances that were not purely horizontal or vertical then you would need to use some conversion factor.

Similarly in spacetime. The SI system is designed using s for time measurements and m for distance measurements, and c is used as a conversion factor between the two. It doesn't matter if you convert your distances to s or your time to m.

Writing the time axis as ct is a way to remind people not to forget the conversion factor. But a lot of times you can just use units where c=1 and then drop the c entirely.

 

[Nugatory]

based on the DS^2=DX^2-DcT^2.equation, is it possible to show mathematically how DS^2 remains the same for the same speed of .98 c and distance in both frames

That's easy to do. You have $\Delta{x}=x_2-x_1$ and $\Delta{t}=t_2-t_1$ in one frame. Use the Lorentz transforms to find what these look like in the second frame and then you just have to grind through the algebra to find that $\Delta{s}^2=\Delta{t}^2-\Delta{x}^2=\Delta{t}'^2-\Delta{x}'^2=\Delta{s}'^2$.

Can this be represented graphically as well on the x/y cartesian coordinates,

That's not so easy to do. Any graphical representation is going to have to be drawn on a sheet of paper or the screen of a computer, and both of those surfaces obey the Euclidean geometry instead of the Minkowski geometry. You can draw an x-axis on a sheet of paper, and you can relabel the y-axis the t-axis, but the distance between two points on the sheet of paper is going to be the Pythagorean $\Delta{s}^2=\Delta{t}^2+\Delta{x}^2$ instead of the Minkowski $\Delta{s}^2=\Delta{t}^2-\Delta{x}^2$.

 

[Herman Trivilino]

Can anybody explain or refer me to books or to the Web?

I agree with Fredrik's recommendation of Spacetime Physics by Taylor and Wheeler. The opening passage will quickly convince you that the authors are addressing the same geometrical issues you are asking about. You can easily order it from amazon or the like. There's a 2^nd edition that I haven't read yet, so I can't make a recommendation that it's any better. Both are available.

 

[Ittiandro]

That's not so easy to do. Any graphical representation is going to have to be drawn on a sheet of paper or the screen of a computer, and both of those surfaces obey the Euclidean geometry instead of the Minkowski geometry. You can draw an x-axis on a sheet of paper, and you can relabel the y-axis the t-axis, but the distance between two points on the sheet of paper is going to be the Pythagorean $\Delta{s}^2=\Delta{t}^2+\Delta{x}^2$ instead of the Minkowski $\Delta{s}^2=\Delta{t}^2-\Delta{x}^2$.

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? ..

In other words.suppose that we have 3 worldlines to represent on the Cartesian coordinates framework : WL a, ( no spatial motion) is vertical, parallel to the ct axis, WL b: slopes to the right along X , WL c: sloping further down .as spatial movement along X increases. For any given X values, how do we represent the corresponding Y(ct) values on the Y axis? I'd say it is more a question of methodology than of proving the validitity of the invariant formula, because I already understood that from the algebraic formula.
May be somebody can take it from here.

Thanks
Ittiandro

 

[Dale]

For any given X values, how do we represent the corresponding Y(ct) values on the Y axis?

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?

 

[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.