Interplay of space and time in Spacetime invariant interval

1. Feb 24, 2015

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.

Thanks

Franco

Last edited: Feb 24, 2015
2. Feb 24, 2015

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.

3. Feb 24, 2015

Fredrik

Staff Emeritus
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).

4. Feb 24, 2015

Ittiandro

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

5. Feb 24, 2015

Staff: Mentor

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.

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.

6. Feb 24, 2015

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.

. .

7. Feb 24, 2015

Staff: Mentor

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).

8. Feb 24, 2015

Staff: Mentor

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.)

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).

9. Feb 24, 2015

Ittiandro

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

10. Feb 24, 2015

Staff: Mentor

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).

11. Feb 24, 2015

Staff: Mentor

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.

12. Feb 24, 2015

Staff: Mentor

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).

13. Feb 25, 2015

stevendaryl

Staff Emeritus
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.

14. Feb 25, 2015

Ittiandro

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
.

15. Feb 26, 2015

Staff: Mentor

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 $1000000 \; 2^{-6.8/1.56} = 49000$

Last edited: Feb 27, 2015
16. Feb 27, 2015

Ittiandro

Thanks for your input

On 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’=.0000415

So T=.000294 and T’=.0000415. The relationship between T and T’ is nowhere near Gamma=5

Can you clarify this for me please?

Thanks

Franco

17. Feb 27, 2015

Staff: Mentor

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

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.

Last edited: Feb 27, 2015
18. Mar 1, 2015

Ittiandro

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

19. Mar 1, 2015

Staff: Mentor

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.

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.

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

20. Mar 1, 2015

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

21. Mar 1, 2015

Staff: Mentor

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 $b^2-a^2$.

22. Mar 2, 2015

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 [Broken] 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?

Thanks

Franco

Last edited by a moderator: May 7, 2017
23. Dec 12, 2015

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

24. Dec 12, 2015

Staff: Mentor

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.

25. Dec 12, 2015

Fredrik

Staff Emeritus
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.