# Spacetime Invariant interval and gamma

• Ittiandro
In summary: The two events are different spacetime events.Also, I don't know what you mean by "square1".In summary, the conversation revolves around the invariant interval spacetime issue and the correct calculation for it based on data given in the book "Spacetime Physics" by E.F. Taylor and A. Wheeler. The conversation also raises questions about the velocity of the rocket and its impact on the time interval between two events, as well as the purpose of considering time as a 4th dimension in spacetime. However, it is clarified that the purpose is not to avoid reading different time and space values in the same event, but to relate measurements of events from different observers.
Ittiandro
Member advised to use the formatting template for all homework help requests
I am revisiting the invariant interval spacetime issue , as explained in the book Spacetime Physics by E.F. Taylor and A. Wheeler. The explanation is clear and the invariant interval is correct, based on the data given and this is the whole point I am making, as explained below.

In a nutshell, the invariant interval is calculated by the authors by taking as an example a rocket flying over a lab. The rocket fires from its antenna two sparks into the lab, in sequence, with a time interval of 33.69 ns between them and a space interval of 2 m. relative to the stationary framework of the lab.
Table 1-4 (attached ) explains the calculation, yielding an invariant interval of 9.9 which is the same for the rocket framework, based on a 33.03 ns time interval between them and 0 space interval , because the two sparks are fired from the same place ( the rocket’s antenna) .

I wonder, though, why :

1) the data do not include the velocity of the rocket and

2) why is the time interval between the two sparks 33.03 ns for the rocket and not a different one ?
How is it arrived at? Shouldn’t the velocity of the rocket ( as a % of c) be required , too, to calculate gamma , hence the time dilation ?

IN other words, not only must the time interval between the two sparks be shorter for the moving rocket ( which it is , in the example) but it doesn’t have to be 33.03 ns, because it depends on the velocity of the rocket, which is not given..Why?

In this case, the interval between the two events ( the sparks) would not be the same for the stationary framework and the rocket. We would then be back to square1!
I’m sure I have missed something, because the whole purpose of considering time as a 4th dimension of a spacetime grid, is precisely to avoid reading different time and space values in the same event(s) depending on the observer's framework.
Can anybody clarify this for me?ThanksItttiandro

#### Attachments

• Invariant interval.docx
275.9 KB · Views: 220
Ittiandro said:
IN other words, not only must the time interval between the two sparks be shorter for the moving rocket ( which it is , in the example) but it doesn’t have to be 33.03 ns, because it depends on the velocity of the rocket, which is not given..Why?
Although the speed of the rocket relative to the lab is not given explicitly, you can find it directly from the data given.

Dale
Ittiandro said:
In a nutshell, the invariant interval is calculated by the authors by taking as an example a rocket flying over a lab. The rocket fires from its antenna two sparks into the lab, in sequence, with a time interval of 33.69 ns between them and a space interval of 2 m. relative to the stationary framework of the lab.
Table 1-4 (attached ) explains the calculation, yielding an invariant interval of 9.9 which is the same for the rocket framework, based on a 33.03 ns time interval between them and 0 space interval , because the two sparks are fired from the same place ( the rocket’s antenna) .

I wonder, though, why :

1) the data do not include the velocity of the rocket and
The rocket speed can be determined from ##\gamma## which is the ratio of the times: ##\gamma = t'/t##
2) why is the time interval between the two sparks 33.03 ns for the rocket and not a different one ?
It is the square root of the space-time interval which is measured in the lab frame - this is the proper time for the events.
I have missed something, because the whole purpose of considering time as a 4th dimension of a spacetime grid, is precisely to avoid reading different time and space values in the same event(s) depending on the observer's framework.
The purpose is to relate the spatial and time measurements of events of differently moving observers. The differences are real differences an cannot be avoided.

AM

Ittiandro said:
1) the data do not include the velocity of the rocket
This is untrue, although not stated explicitly. Unless the antenna moves relative to the rocket, you have its position at two times and can directly compute the velocity as (2 m)/(33.69 ns).

Ittiandro said:
2) why is the time interval between the two sparks 33.03 ns for the rocket and not a different one ?
How is it arrived at? Shouldn’t the velocity of the rocket ( as a % of c) be required , too, to calculate gamma , hence the time dilation ?
In the rocket's frame, ##\Delta x^2 = 0## and therefore the spacetime interval is just ##c^2 \Delta t^2##. You never need to compute the gamma factor.

@Ittiandro please do not post work in attachments. Post it directly in the thread. Thanks.

Ittiandro said:
the whole purpose of considering time as a 4th dimension of a spacetime grid, is precisely to avoid reading different time and space values in the same event(s) depending on the observer's framework.

No, it isn't. I don't know where you got this idea from, but it's wrong. The invariant is the spacetime interval, not space or time separately.

Ittiandro said:
In other words, not only must the time interval between the two sparks be shorter for the moving rocket (which it is, in the example) but it doesn’t have to be 33.03 ns, because it depends on the velocity of the rocket, which is not given. Why?
You can look at it this way. You start with five quantities to mess around with: ##\Delta t##, ##\Delta x##, ##\Delta t'##, ##\Delta x'##, and ##\beta##, but they aren't completely independent. First, you must have that ##(c\Delta t)^2 - (\Delta x)^2 = (c\Delta t')^2 - (\Delta x')^2##. Second, because of the way the problem is set up, you have ##\Delta x'=0## and the relationship ##\beta = \Delta x /(c\Delta t)##. So there are only two degrees of freedom. Once you choose values for two of the quantities, the remaining two are determined.

In the rocket frame from which the signals are emmited the space time interval is c22=c2dt2-dx2
dΓ=time measured in the rocket frame between the events (proper time.)
dt=time measured between the events from the lab-frame.
dx=spatial distance between the two events as measured from the lab frame.
from this we get dt=ϑdΓ,
Where ϑ=1/√(1-v2/c2),
Thus relative velocity ( v ) betwee the two frame (rocket and lab) frame is related.

Apashanka said:
In the rocket frame from which the signals are emitted the space time interval is c22=c2dt2-dx2
This is not incorrect but it unnecessary and possibly confusing to say "in the rocket frame". The space-time interval between two arbitrary events e1(x,y,z,t) and e2(x+Δx, y, z, t+Δt) is always ΔS2 = (cΔt)2 - (Δx)2, the spatial and temporal co-ordinates being measured in the inertial observer's reference frame. In this case, the observer in the rocket frame measures the space-time interval as: ΔS2 = (cdΓ)2 since the events occur at the same spatial location (Δx=0).

The space-time interval for these same events in the lab frame is ΔS2 = (cΔt)2 - (Δx)2 where Δt and Δx are measured in the lab frame. This space-time interval measured in the lab frame must be the same as that measured in the rocket frame, which is where we get c22=c2dt2-dx2 (where dΓ is the proper (rocket) time interval between the events and dt and dx are the time and space intervals between the same events as measured in the lab frame.

AM

PeterDonis said:
No, it isn't. I don't know where you got this idea from, but it's wrong. The invariant is the spacetime interval, not space or time separately.

Peter, my vocabulary may not have been as precise as it should be, because I am neither a scientist nor a student of physics, but I never implied that space and time separately can be be invariant!

In fact, I know full well that, with the advent of the S.R, they have been shown to be all but invariant, depending on the observers' framework, whether stationary or in uniform motion. What I meant, is that by combining space AND time as SPACETIME and by converting time into spatial units of distance ( meters of time..) observers both from a stationary framework and from a moving one will no longer have different spatial/temporal measurements for the same events and the distance between them , which will be INVARIANT for all the frameworks. This is, I think , not only true, but also a substantial step forward in our understanding of the Universe. If this is not the main idea behind SPACETIMEI , as you seem to imply, what is it then?

Thanks for your input and corrections

Ittiandro

Thanks for your clarifications. They carry me one step further.
Until now I always took Gamma as the SQRT(1- v^2). This is right, but I didn’t realize that gamma can also be expressed as the ratio t/t’. Based on this:1. t’ then represents the Lorentz transformation of t, because of the time dilation caused by the speed of the rocket

2. the time lapse of 33.0228 ns given in table 1-4 of the textbook in regard to the rocket framework is the only one that makes the Invariant Interval the same for both frameworks.3. Since V=d/t and we know d=2 m and t= 33.690 ns( t), then the rocket’s speed V=0.05936 m/ns or 0.19802 c4. 4. I also figured out V another way, starting from gammaGamma=0.980195903If 0.980195903= sqrt( 1-v^2) extract V by squaring both termsi.e. 0.960784009=1-v^2, hence solving for v^2 , v^2=0.039215991 hencev=0.198030278 ct’=33.690 x sqrt ( 1-0.198030278^2)=33.0227 ns . Q.E.D.
I still have a problem, however, in finding the Invariant interval for t = 16.678 ns and the spatial distance 4.0000 m. , as laid out in the text drill..By relating the spatial distance ( 4 m) to the time meters ( 5 m.time ) , the text extracts V as 80% C and from then it extracts the same invariant interval of 3 m. for both frameworks ( as it should be) . I took a different route, but it gives me an invariant interval for 4.47 m. for t’ ( rocket framework) and not 3 m. I am sure I have missed something. Here is my reasoning. Please let me know where I went wrong.Rocket framework:a. t’ is not given. I thought I would have to find the rocket speed V, first . Knowing that it took 16.678 ns to cover 4 m spatial distance , than by the formula V=d/t we get V= .44506 c ( 4 m/.299792458 ns=13.34 m/ns =. 44506 c )b. If v=.44506c, then we find Gamma as sqrt(1- .44506^2)= .8955c. From Gamma .8955 we get t’=16.678 ns x.8955 =14.935 nsd. Invariant interval for t’: 14.935 x .299782458 =4.477 time meters . Not the same! Where did I go wrong?Thanks
Ittiandro

Ittiandro said:
by combining space AND time as SPACETIME and by converting time into spatial units of distance ( meters of time..) observers both from a stationary framework and from a moving one will no longer have different spatial/temporal measurements for the same events

"Different spatial/temporal measurements" means "different measurements of times and distances separately". Those are still different for different observers in SR, so your statement above is wrong.

Ittiandro said:
and the distance between them , which will be INVARIANT for all the frameworks

The spacetime interval between the events will be invariant, i.e., the same for all observers. But the word "distance" is not usually used to mean the spacetime interval, because there is too much potential for confusion with spatial distance, which is not invariant.

Ittiandro said:
Thanks for your clarifications. They carry me one step further.
Until now I always took Gamma as the SQRT(1- v^2).
##\gamma = \frac{1}{\sqrt{1- v^2}}## where v is really v/c.
4. 4. I also figured out V another way, starting from gamma

Gamma=0.980195903
This is 1/γ. But you used 1/γ in your calculation below so you get the right answer.

If 0.980195903= sqrt( 1-v^2) extract V by squaring both terms

i.e. 0.960784009=1-v^2, hence solving for v^2 , v^2=0.039215991 hence

v=0.198030278 c

t’=33.690 x sqrt ( 1-0.198030278^2)=33.0227 ns . Q.E.D.
Correct (using 1/γ). See my post #3 above: ##\gamma## = (lab time)/(proper time).. You used (proper time)/(lab time) = .980195003 which is 1/γ.

I still have a problem, however, in finding the Invariant interval for t = 16.678 ns and the spatial distance 4.0000 m. , as laid out in the text drill..
The space-time interval for these two events in question is: ##\Delta S^2 = (c\Delta t)^2 - (\Delta x)^2## so plug in the numbers:

##\Delta S^2 = (3*10^8*16.678*10^{-9} t)^2 - (4)^2 = 34.034##

By relating the spatial distance ( 4 m) to the time meters ( 5 m.time ) , the text extracts V as 80% C and from then it extracts the same invariant interval of 3 m. for both frameworks ( as it should be) . I took a different route, but it gives me an invariant interval for 4.47 m. for t’ ( rocket framework) and not 3 m.
The space-time interval is ##\Delta S^2## and dimensions of distance2. It works out to 34.034 m2.
Rocket framework:

a. t’ is not given. I thought I would have to find the rocket speed V, first . Knowing that it took 16.678 ns to cover 4 m spatial distance , than by the formula V=d/t we get V= .44506 c ( 4 m/.299792458 ns=13.34 m/ns =. 44506 c )
I thought you said the time interval in the lab frame was 16.678 ns.

This means that the proper time is ##t = \sqrt{\Delta S^2}/c = \sqrt{34.034}/3*10^8 = 19.446 ns.##
and the relative speed is:
##v = d/t = 4/(16.678*10^{-9}) = 2.4*10^8 m/sec = .8c##

AM

PeterDonis said:
"Different spatial/temporal measurements" means "different measurements of times and distances separately". Those are still different for different observers in SR, so your statement above is wrong.

Peter, I am afraid I lost you here, because when I said that by combining space AND time as SPACETIME and by converting time into spatial units of distance ( meters of time..) observers both from a stationary framework and from a moving one will no longer have different spatial/temporal measurements for the same events , I meant exactly what you said in your rebuttal above. So, how can I be wrong if you say exactly the same?

I know that for most of us it can be difficult and confusing to translate maths into a conceptual language or, like in my case, to speak of mathematical concepts without knowing maths in depth , not only because the mathematical precision is inevitably lost when conceptualizing, but also because in order to effectively communicate certain mathematical concepts one would have to have a command of the language which is rarely found.

I am always amazed at the clarity with which Albert Einstein laid out the principles of his S.R. in a short paper!

Anyway, the gamma issue will explain what I meant by “ different spatial/temporal measurements”

I’ll use the same scenario as in the textbook, already detailed in my post:

Two sparks emitted by a rocket flying over the lab hit the room at two consecutive points spaced by 4 m and with a time lapse of 33. 690 ns between them.

If we don’t calculate from within the SPACETIME perspective, the stationary lab observer and the moving observer in the rocket will indeed clock the respective time lapses between the sparks differently:

In fact, the text example translates the 33.360 ns reading of the stationary observer to 33.02 ns , because of the gamma conversion. I assume that the same can be said for the distance : for the rocket observer the distance of 4 m. in the lab measured by the lab technician will be shorter because of the length contraction caused by the speed.

This no longer happens once space and time are brought together as SPACETIME and both observers will then measure the SAME SPACETIME INTERVAL, by using C as a common denominator. Isn’t this what you mean? I thought there were no two ways to express the essence of the SPACETIME concept !ThanksIttiandro

Ittiandro said:
If we don’t calculate from within the SPACETIME perspective, the stationary lab observer and the moving observer in the rocket will indeed clock the respective time lapses between the sparks differently:
We are having difficulty understanding what you mean by the above. One always has to make measurements of space and time intervals between events. That is how the space-time interval is determined. So perhaps you could explain what you mean by "calculate from within the spacetime perspective" and why you think it is different than each observer measuring space and time differently.

This no longer happens once space and time are brought together as SPACETIME and both observers will then measure the SAME SPACETIME INTERVAL, by using C as a common denominator. Isn’t this what you mean? I thought there were no two ways to express the essence of the SPACETIME concept !
Why does it no longer happen i.e. different inertial observers getting different measurements? The space-time interval between two events is derived from the measurements of the distance interval and time interval between those events by an inertial observer.

AM

Ittiandro said:
I meant exactly what you said in your rebuttal above.

You may have meant it but you didn't say it. I am trying to correct the way you say what you're trying to say, because the way you're saying it now is wrong: it doesn't say what you are trying to say.

Specifically, this is the problem:

Ittiandro said:
will no longer have different spatial/temporal measurements for the same events

This is wrong. I don't care if you want it to mean the same thing that I said. It doesn't mean that. It means that observers in different states of motion will measure the same spatial distances and the same times. It doesn't mean that observers in different states of motion will agree on spacetime intervals. "Spatial/temporal measurements" does not mean "spacetime intervals". It means the measurements of spatial distances and times that each individual observer makes.

Ittiandro said:
This no longer happens once space and time are brought together as SPACETIME and both observers will then measure the SAME SPACETIME INTERVAL

No, they will not. It's impossible for them both to measure the same spacetime interval directly. All they can measure is spatial distances and times using their own frames. They then both calculate the same spacetime interval between the two events.

If one observer happens to be present at both events (for example, the observer riding in the rocket in your example), then their measurement of the elapsed time between the two events will be the same as the spacetime interval between those events. (He can obviously confirm this by calculation since in his frame the spatial distance between the events is zero.) But that is a special case. An observer at rest in the lab cannot be present at both events: they are spatially separated in his frame. So he can't directly measure the spacetime interval between them. He can only measure the time between them and the distance between them and calculate the spacetime interval from those measurements.

I expect that you will be tempted to respond that what I said above is the same as what you said. Don't. You might have meant to say what I said above, but you didn't say it. You applauded Einstein's clarity of language. Part of that clarity was precision: he picked precise words to express his precise concepts. That's what I'm trying to get you to do.

PeterDonis said:
You may have meant it but you didn't say it. I am trying to correct the way you say what you're trying to say, because the way you're saying it now is wrong: it doesn't say what you are trying to say.

Peter,
I think I understand you now.

Most of my exchanges revolve around the idea that by combining space AND time as SPACETIME and by converting time into spatial units of distance ( meters of time..) observers both from a stationary framework and from a moving one will no longer have different spatial/temporal measurements for the same events.
I now realize where I went wrong : I believe you say that Spacetime and and spacetime intervals are mathematical concept that observers from different frameworks neither will be able to experience nor to measure as they do with Space and Time separately .

So the introduction of the SPACETIME concept still won’t prevent observers from different frameworks in different states of motion relative to one another to measure Space and Time differently.

Thanks again

Ittiandro

Peter
Regarding my last reply, where I came to agree with your observations ( and corrections) I still have a new, more general question: if , as I understood you to mean) Spacetime and spacetime invariant intervals are mathematical constructions which, however mathematically true, are beyond our sense experience and cannot be measured by observers ( as I initially implied throughout my posts ) in the same way that they can measure space and time separately, within their respective frameworks , what is then the practical import of the notion of Spacetime?

Please understand, mine is not a rhetorical questions, implying perhaps that it is only a useless speculation...Far from this!

Indeed, I am not saying that the truth and significance of physical theories and of knowledge in general should be measured by the extent to which they can be applied to technology. Aristotle's opening statement in his Magnum Opus, The Metaphisics , was that " Men desire knowledge for the sake of knowledge..". I fully agree with him..
There is an anecdote about the famous Greek astronomer Aristarchus of Samos, who , true to the popular stereotype of the " absent minded professor " ..fell into a ditch while looking upward, contemplating the night sky, , ..At which point, the story goes, a down-to-Earth ( and probably less intellectually endowed girl) passing by, poked fun at this " weirdo" who misses what is really important why looking at useless things. I don't certainly think like this " savvy" maid..

I am also aware of ( and not unsympathetic to) a modern Platonic/Pythagorean trend among reputed scientists, physicists and mathematicians ( Roger Penrose, Mario Livio..), holding that mathematics and mathematical relationships, in other words NUMBER, in Pythgorean terms. ARE the Universe..

To conclude, my question, in essence, is :
Modern technology , like the GPS technology and probably other applications, could not function as they do without taking into account the Special Relativity, essentially the Lorentz transformations and the Gamma conversion...
Is ( or can) the notion of Spacetime be put to use in the same way by our technology?

THanks

Ittiandro

Ittiandro said:
are mathematical constructions
This is a rather irrelevant notion. All mathematical tools that we use to do physics are mathematical constructs.

Ittiandro said:
are beyond our sense experience and cannot be measured by observers
Physics has very little to do with ”our sense”. Contrary to what you seem to suggest, invariants are arguably more physical than arbitrary coordinate assignments.

Ittiandro said:
ARE the Universe..
This is purely philosophical and has little to do with the actual physics.

Ittiandro said:
Spacetime and spacetime invariant intervals are mathematical constructions which, however mathematically true, are beyond our sense experience

Not really. As I said in post #16, an observer who is present at both of a pair of events can directly measure the spacetime interval between them: it's just the elapsed time on his clock. And in principle, there will be some observer who can be present at both of any pair of events that are timelike separated. So timelike intervals, at any rate, can be directly measured. The only limitation is that it is impossible for observers in relative motion to both measure the same timelike interval directly; at most, only one of them can. The other has to calculate it.

So it's not true that spacetime intervals, or at least timelike intervals, are only mathematical constructions.

Ittiandro said:
Is ( or can) the notion of Spacetime be put to use in the same way by our technology?

It already is. As you say, GPS technology uses SR (and also GR). That is using spacetime.

Ittiandro said:
Peter
Regarding my last reply, where I came to agree with your observations ( and corrections) I still have a new, more general question: if , as I understood you to mean) Spacetime and spacetime invariant intervals are mathematical constructions which, however mathematically true, are beyond our sense experience and cannot be measured by observers ( as I initially implied throughout my posts ) in the same way that they can measure space and time separately, within their respective frameworks , what is then the practical import of the notion of Spacetime?
Many people said the same thing about the mathematical construct we call "energy" until 150 years after Newton. The space-time interval between two events is "physical" in the same sense that energy is physical. Its magnitude is related to the space and time intervals between two events. It is also very useful because it is the same in all inertial reference frames and its sign tells us whether the events were time-like, or space-like (ie. whether there could be a frame of reference in which the events occurred in the same place separated only by time, or at the same time separated only by space) or light-like (separated by a time and distance such that Δx = cΔt for all observers)

AM

## 1. What is the Spacetime Invariant interval?

The Spacetime Invariant interval is a measure of the distance between two events in the fabric of spacetime. It takes into account both the spatial and temporal components of the events and is used to determine the interval's invariance under different reference frames.

## 2. How is the Spacetime Invariant interval calculated?

The Spacetime Invariant interval is calculated using the Minkowski metric, which is a mathematical representation of the fabric of spacetime. It involves taking the difference in the squared spacetime coordinates between the two events and then taking the square root of the result.

## 3. What is gamma in the context of Spacetime Invariant interval?

Gamma, also known as the Lorentz factor, is a mathematical term used in special relativity to describe the relationship between an object's observed time and its proper time. It is represented by the symbol γ and is calculated as the reciprocal of the square root of 1 minus the square of the object's velocity relative to the speed of light.

## 4. How does gamma affect the Spacetime Invariant interval?

Gamma is a crucial factor in the calculation of the Spacetime Invariant interval. It is used to account for the effects of time dilation and length contraction in special relativity. As an object's velocity approaches the speed of light, gamma increases, resulting in a larger Spacetime Invariant interval.

## 5. Why is the Spacetime Invariant interval important in physics?

The Spacetime Invariant interval is important in physics because it provides a fundamental measure of the distance between two events in the fabric of spacetime. It is a key concept in special relativity and is used to understand the effects of time and space on objects moving at high speeds. It also helps to explain the concept of causality and the limitations of information transfer in the universe.

Replies
10
Views
1K
Replies
4
Views
1K
Replies
6
Views
568
Replies
2
Views
529
Replies
14
Views
619
Replies
1
Views
2K
Replies
37
Views
3K
Replies
7
Views
702
Replies
35
Views
3K
Replies
2
Views
2K