Spacetime intervals again - still figuring out the formula

In summary: I don't remember. Anyway, this space is very similar to our own spacetime, just that it has a negative time component.But what does that have to do with relativity?In summary, JesseM is still trying to figure out what relativity is, and is looking for an explanation of the formula for spacetime. In relativity, we deal with events that occur at a particular point in space (x, y, z) at a particular time (t). The spacetime interval between two events is \Delta s = \sqrt {c^2 (t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2
  • #1
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ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2

I'm still trying to figure it out. What I was thinking, you're always moving through spacetime at c, right? Through time, but that will be less when traveling in spatial demensions. (If any of this is wrong, let me know, I read this somewhere but can't remember where and it's some time ago so my memory might've messed it up or I may simply have read something that isn't true)

So wouldn't that make the distance traveled in spacetime only dependant on the amount of time? d = ct? I can't think of anything else that would follow the above so the spacetime formula must've come from somewhere else. Where?
 
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  • #2
I'm not sure what you are "trying to figure out". The formula, just written down by itself, seems to me to be a lot clearer than your explanation of it.
 
  • #3
In relativity we deal with events that occur at a particular point in space (x, y, z) at a particular time (t). The spacetime interval between two events is

[tex]\Delta s = \sqrt {c^2 (t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 - (z_2 - z_1)^2}[/tex]

That's basically all your equation means. It has nothing to do with motion, specifically. It's written with ds, dt, dx, etc. because we often work with infinitesimal differences using calculus.

The spacetime interval is important because it has the same numeric value in any reference frame, for a given pair of events. We say it is invariant under Lorentz transformations between reference frames.
 
  • #4
NanakiXIII said:
I'm still trying to figure it out. What I was thinking, you're always moving through spacetime at c, right?
I'd say the "everything always moves through spacetime at the speed of light" is just a bad explanation of relativity which has been made popular by Brian Greene, but which is more confusing than helpful (the math works out, but the way he defines 'speed through spacetime' is very counter-intuitive). If you want more details this was discussed at length on this thread.
 
  • #5
JesseM said:
I'd say the "everything always moves through spacetime at the speed of light" is just a bad explanation of relativity which has been made popular by Brian Greene, but which is more confusing than helpful...

JesseM,

Amen!
 
  • #6
jtbell said:
[tex]\Delta s = \sqrt {c^2 (t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 - (z_2 - z_1)^2}[/tex]

That's basically all your equation means. It has nothing to do with motion, specifically. It's written with ds, dt, dx, etc. because we often work with infinitesimal differences using calculus.

I don't get where the formula using differentials comes from. If I calulate the total differential

[tex]ds = \frac{\partial{s}}{\partial{t}}dt + \frac{\partial{s}}{\partial{x}}dx [/tex]

of the function

[tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

and square it, I get the ugly

[tex]ds^2=\frac{c^4t^2dt^2-x^2dx^2-2c^2txdtdx}{c^2t^2-x^2}[/tex]

So what am I missing?
 
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  • #7
He meant the very famous

[tex] ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu} [/tex]


Daniel.
 
  • #8
quasar987 said:
If I calulate the total differential

[tex]ds = \frac{\partial{s}}{\partial{t}}dt + \frac{\partial{s}}{\partial{x}}dx [/tex]

of the function

[tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

and square it, [...]

Where did you get the function

[tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

from?
 
  • #9
jtbell said:
Where did you get the function

[tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

from?

I figured

[tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

is kinf of

[tex]\Delta s = \sqrt{(c\Delta t)^2 - (\Delta x)^2}[/tex]

where t1 = x1 = 0...

How do you get the expression

[tex](ds)^2 = (cdt)^2 - (dx)^2 - (dy)^2 - (dz)^2[/tex]

anyway?
 
  • #10
By definition.That's the definition of the relativistic interval between 2 inifnitesimally close points in a flat Minkowski space...


Daniel.
 
  • #11
Another thing to notice is that the spacetime interval defined above is just c times the proper time (time as measured on the clock of someone moving along that worldline), [tex]d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)[/tex]

Since all frames must agree on how much time ticks on a clock between two points on its worldline, this shows why it's necessary that [tex]d\tau[/tex], and hence [tex]ds = cd\tau[/tex], must be a frame-invariant quantity.
 
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  • #12
I'm trying to figure out why the formula looks like that. I don't understand. Why are we multiplying dt^2 by c^2? Why are we substracting dx through z^2?
 
  • #13
NanakiXIII said:
I'm trying to figure out why the formula looks like that. I don't understand. Why are we multiplying dt^2 by c^2? Why are we substracting dx through z^2?
It´s like pythagoras in a weird space.
time can be considered with some right to be very closely related to space. Some would say that you could measure time in meters and space in seconds - that meter and second are merely different units for basically the same thing. The conversion factor between those units would then be: 1m = 300000000 m/s * 1s.
And like pythagoras gives you a distance in normal space, ds gives you a distance in spacetime. Mostly you express this distance in units of meters, so when you have time given in seconds you first have to transform it to meters -> c*t = time in meters.
The only difference to normal space is that the time component counts as a negative space component, that´s why it´s called Minkowski space instead of euclidean space.
However this distance has a physical meaning: it equals the time between two events in the frame where they happen at the same place, or the spatial distance in the frame where they happen simultaneous, whichever is appropriate.
 
  • #14
NanakiXIII said:
I'm trying to figure out why the formula looks like that. I don't understand. Why are we multiplying dt^2 by c^2? Why are we substracting dx through z^2?
Like I said, what you're really calculating is the proper time. If you see a clock moving past you at velocity v, you know that if the time elapsed on your own clock between two points is [tex]dt[/tex], and if the time elapsed on that clock between those same two points is [tex]d\tau[/tex], then the two time intervals are related by the time dilation formula:

[tex]d\tau = dt/\gamma[/tex]
where [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex]

And in your coordinate system, v is equal to the distance the clock covered during the time interval [tex]dt[/tex], giving

[tex]v = \sqrt{dx^2 + dy^2 + dz^2}/dt[/tex]

So, plugging that into the time dilation formula:

[tex]d\tau = dt \sqrt{1 - v^2/c^2}[/tex]
[tex]d\tau = dt \sqrt{1 - (dx^2 + dy^2 + dz^2)/(c^2dt^2)}[/tex]
[tex]d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}[/tex]
[tex]d\tau = \sqrt{dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)}[/tex]

which gives the proper time formula I posted above:

[tex]d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)[/tex]

And as I said, if you define [tex]ds = cd\tau[/tex], you get the formula for the spacetime interval that you posted, they are pretty much interchangeable (although proper time [tex]d\tau[/tex] has a more obvious physical meaning than [tex]ds[/tex], I'm not sure what the motivation is for ever using [tex]ds[/tex]).
 
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  • #15
quasar,

In your post #6, I think there are some mistakes in your expressin for ds^2. Doesn't the first term needs a factor of t, the second term a factor of x and the cross term an x and a t? And didn't you forget to square the denominator?

Or am I missing something?
 
  • #16
No, that's true.

I'll edit it.
 
  • #17
time can be considered with some right to be very closely related to space. Some would say that you could measure time in meters and space in seconds - that meter and second are merely different units for basically the same thing. The conversion factor between those units would then be: 1m = 300000000 m/s * 1s.

Thanks, that helps. The first part of the formula (c^2dt^2) is immediately more clear. I still don't understand, however, why we're substracting the spatial parts. In Pythagoras, you use addition.
I'm just looking for a geometrical (if one's allowed to call it that) explenation to the formula. Such complex (well, to me, anyway) maths like that of JesseM is no doubt correct and might lead me to better understanding, but it would take me a lot of time. I'm not especially gifted at physics and maths and I haven't studiedthem extensively (I'm still in high school, or whatever it's called in your part of the world).
 
  • #18
NanakiXIII said:
I'm just looking for a geometrical (if one's allowed to call it that) explenation to the formula.

Since c is the speed of light, and dt is the time interval between the events, then c*dt is the distance traveled by light between the time the first event hapened and the time the second event hapened.

And dx²+dy²+dz² = D², according to pythagoras, is just the spatial distance (squared) separating the position of hapening of the two events.

We conclude from the substraction of D² from (c*dt)² that if the result is negative (ds² < 0), then it means that (c*dt) < D. That is to say, the distance traveled by light during the event is smaller than the actual distance separating the event. So not even light could have gone from A to B between their hapening. And since nothing can travel faster than light, then nothing in event A could have caused event B.

If the result is positive (ds² > 0), it means that (c*dt) > D. Then A might be the cause of B.

If, finally, the result is 0 (ds² = 0), then (c*dt) = D and the only thing in A that could have caused B is the light that was emited from A.
 
  • #19
quasar,

I just saw the corrections you made to your expression for ds^2 in post #6. They look right.

Now remember you defined your spatial and time intervals as being from 0 to x and from 0 to t, respectively. But when you calculate the differential ds^2, those intervals become differentials as well. That is x -> dx and t -> dt. if you make those substitutions in your expression, you'll see that your numerator is just the square of your denominator. So your whole expression is equal to what you now have in the denominator. And that's the right answer.

This one really had me wrapped around the axle for a while; the solution finally hit me while I was jogging yesterday! The way you got into trouble was by using the variables x and t to represent not only VALUES of the domain of S but also INTERVALS in the domain of S. That works fine in algebra and analytic geometry, but in calculus you need different variables for x and dx or you'll always be calculating derivatives at x=0, because the calculation automatically sends dx->0.

In this case, since there's nothing physically significant about assigning the coordinates (0,0) to your first event, you don't lose any physics as long as you remember that once the derivative is taken x = dx and t = dt.

Did that help at all, or was it confusing?
 
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  • #20
Thanks jdavel. I will look seriously into this after my exams, but I do see your point, I just need to really see it for myself. Thanks again!
 
  • #21
NanakiXIII said:
Thanks, that helps. The first part of the formula (c^2dt^2) is immediately more clear. I still don't understand, however, why we're substracting the spatial parts. In Pythagoras, you use addition.
I'm just looking for a geometrical (if one's allowed to call it that) explenation to the formula. Such complex (well, to me, anyway) maths like that of JesseM is no doubt correct and might lead me to better understanding, but it would take me a lot of time. I'm not especially gifted at physics and maths and I haven't studiedthem extensively (I'm still in high school, or whatever it's called in your part of the world).

Here's a geometric interpretation:
Given the [Minkowski-distance] formula ds2=(cdt)2-dx2, consider the set of [events] points with a constant value for ds2 (say, 1) from given [event] point. Those [events] points "equidistant from the given [event] point" lie on a hyperbola on the xt-plane.

When you apply the Lorentz transformations to an [event] point on that hyperbola, you get mapped to another [event] point on that hyperbola.

The Euclidean analogy, of course, has a circle (instead of the hyperbola) and rotations (instead of the Lorentz Transformations).
 
  • #22
Ah, for some reason I didn't realize that with a third dimension, the formula was simply a^2+b^2+c^2 (I never used it with anything but two dimensions). Ok, so we're substracting spatial distance from temporal distance converted to spatial distance? But if t has been converted to a 'fourth spatial dimension' (one probably wouldn't call it that, but bear with me), shouldn't we be adding, using 'four dimensional Pythagoras' (another phrase I just made up because I have no idea what to call it)?
I understand what you're saying, quasar987, but then if c^2dt^2=D^2, ds^2 would be 0. That would mean there is no interval, right? That seems a bit odd.
 
  • #23
NanakiXIII said:
...But if t has been converted to a 'fourth spatial dimension' (one probably wouldn't call it that, but bear with me), shouldn't we be adding, using 'four dimensional Pythagoras' (another phrase I just made up because I have no idea what to call it)?...
I

That would be nice, but alas, (cdt)^2 + dr^2 isn't invariant under a Lorentz transform. (cdt)^2 - dr^2 is, and it's that invariance that makes this definition of the interval useful.

It's sort of like (but not exactly!) the way total mechanical energy is conserved if you define KE as 1/2mv^2. It would be simpler if you could leave the 1/2 off. But then "energy" wouldn't be conserved.

You have to take what nature gives you, not what makes your calculations easy.
 
  • #24
NanakiXIII said:
Ah, for some reason I didn't realize that with a third dimension, the formula was simply a^2+b^2+c^2 (I never used it with anything but two dimensions). Ok, so we're substracting spatial distance from temporal distance converted to spatial distance? But if t has been converted to a 'fourth spatial dimension' (one probably wouldn't call it that, but bear with me), shouldn't we be adding, using 'four dimensional Pythagoras' (another phrase I just made up because I have no idea what to call it)?
I understand what you're saying, quasar987, but then if c^2dt^2=D^2, ds^2 would be 0. That would mean there is no interval, right? That seems a bit odd.

When you say, "shouldn't we", what exactly do you mean? In order to achieve what goal should we add those quantities? The formula of the space-time interval wasn't man-made for a purpose (like you seem to believe). Nobody said one day, "um, it would be kinda cool to define something fancy like the SPACE-TIME INTERVAL as (cdt)² - dx² - dy² - dz². Yeah, in fact, I'M cool. :cool:" No, it just so happens that since space and time coordinates transform from one inertial frame to another according to the Lorentz transformations, the quantity [itex]\Delta s^2[/itex] gives out the same number in every inertial frames.

And we like that because it eases our computations. And it also has a nice interpretation in terms of causality. So we gave it a name for matter of "we're lazy". We said let's call it the "interval" instead of always referring to it has "the invariant obtained from the interval four-vector".

At least that's my interpretation, exagerated of course, but it shouldn't be too far from the thruth. :tongue2:
 
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  • #25
(cdt)^2 + dr^2 isn't invariant under a Lorentz transform.

What does that mean? I don't really know what Lorentz transformations are, they're kind of difficult to understand, at least from what I've read.

When you say, "shouldn't we", what exactly do you mean?

I mean that it seems more logical to me.

The formula of the space-time interval wasn't man-made for a purpose (like you seem to believe). Nobody said one day, "um, it would be kinda cool to define something fancy like the SPACE-TIME INTERVAL as (cdt)² - dx² - dy² - dz². Yeah, in fact, I'M cool. " No, it just so happens that since space and time coordinates transform from one inertial frame to another according to the Lorentz transformations, the <couldn't copy and past this> gives out the same number in every inertial frames.

It's not man-made (umm well it must be, right?) or it wasn't made for a purpose (it was just thought up for no apparent reason?)? Also...gives out the same number in every intertial frame? You mean ds^2 always equals the same? I'm guessing I'm misreading.
 
  • #26
NanakiXIII said:
What does that mean? I don't really know what Lorentz transformations are, they're kind of difficult to understand, at least from what I've read.

The Lorentz transformations are pretty much all there is to relativity. Given two events (let's say a firecracker exploding and somebody shouting "Weee!") and two observers in relative moving to each other (let's say you in a car, and me standing on a sidewalk), if one observer says that the two events hapened a distance [itex]\Delta d[/itex] appart and a time interval [itex]\Delta t[/itex] appart, then, when you plug the numbers [itex]\Delta d[/itex] and [itex]\Delta t[/itex] in the equations we call the Lorentz transformation, what you get out of those equation is the distance [itex]\Delta d^*[/itex] and the time interval [itex]\Delta t^*[/itex] that the other observer perceive has separating the same two events. *pant pant*


Also...gives out the same number in every intertial frame? You mean ds^2 always equals the same? I'm guessing I'm misreading.
Yes, ds^2 always equal the same in any inertial frame for a given pair of events. But I'm guessing you don't know what an inertial frame is.

It just means a system of coordinates (x-y-z) moving at constant velocity relative to another coordinate system (x*-y*-z*). In the above exemple where you're in a car and I'm on the sidewalk. Since we move at constant velocity relative to each other, we are two distinct inertial frames.
 
  • #27
So Lorentz transformations are the transformation between different points of view?

Then what would be a non-inertial frame? Also, if ds^2 is always the same, doesn't that kind of defeat the purpose of having that formula?
 
  • #28
NanakiXIII said:
So Lorentz transformations are the transformation between different points of view?

Exactly. They allow you to calculate what one observer observes, given what another observer observes.

if ds^2 is always the same, doesn't that kind of defeat the purpose of having that formula?

Fundamental physical reality should not depend on who is observing it, according to relativity theory. Therefore, quantities that do not vary from one observer to another (that is, quantities that are Lorentz invariant) are more fundamental than quantities that depend on the observer.

Another example of a Lorentz-invariant quantity is

[tex] E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2 = (m c^2)^2 [/tex]

where [itex]m[/itex] is what some people call the "rest mass" of an object, and what most physicists call simply "the mass" of the object.

Notice the similarity between this formula and the one for [itex]ds^2[/itex]. That's not a coincidence. In relativity, time and position together form a four-vector [itex](ct, x, y, z)[/itex], and energy and momentum form another four-vector, [itex](E, p_x c, p_y c, p_z c)[/itex]. All the fundamental laws of physics can be written in a way so that they have the same form for all inertial observers, by using four-vectors. We say that these laws are Lorentz covariant.
 
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  • #29
Ah, thanks.

But I meant that if ds^2 is always the same, then there's no need of calculating any interval, because all intervals in the universe are the same. But I think I misunderstood. ds^2 of one interval is the same for different observers, that's what was meant, no? I interpreted it as 'every interval is of equal size'.

Now, I think I understand that, but I still don't see why we're substracting. I don't even see the link between that and your explenations. Could we start over from where I asked 'why substract D^2 from c^2dt^2?' ?
 
  • #30
NanakiXIII's original questions on why the c is involved in [tex]cdt[/tex] and why we are subtracting the [tex]dx_i[/tex] instead of adding it is essentially answered by stating that this is inherent to the traditional Minkowski interpretation of SR.

Euclidean interpretations of relativity are slowly gaining support among physicists. It was probably first proposed by Robert d'E Atkinson in his 1963 paper "General relativity in Euclidean terms". Hans Montanus has brought the subject to live again in his paper "Proper time formulation of relativistic dynamics" after which it was further elaborated by amongst others prof. Alexander Gersten and prof. Jose Almeida. The topic is studied by some amateur physicists as well, that I count myself to.

The core elements of Euclidean special relativity are the universal velocity c for all objects in 4D space-time and the ++++ metric (i.e., the [tex]dx_i[/tex] is added instead of subtracted). It allows a much more intuitive approach of relativity and gives natural explanations for time dilation and length contraction. Much of the discussion in this thread would be put in a completely different context if approached via Euclidean relativity. In fact, there was already a hardly noticed reference to it in the answer of RobPhy where he states:
The Euclidean analogy, of course, has a circle (instead of the hyperbola) and rotations (instead of the Lorentz Transformations).

For all who struggle with the counter-intuitive Minkowski interpretation I would highly recommend to study Euclidean relativity for a while to see how it eases your understanding of relativity. A good starting point may be my own website http://www.rfjvanlinden171.freeler.nl where I maintain a collection of links on Euclidean relativity and where you can also find my own writings on the topic.
 
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  • #31
NanakiXIII said:
Now, I think I understand that, but I still don't see why we're substracting. I don't even see the link between that and your explenations. Could we start over from where I asked 'why substract D^2 from c^2dt^2?' ?
Did you read my earlier post #14 where I derived the formula [tex]d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)[/tex] from the formula for time dilation in relativity, [tex]d\tau = dt/\gamma[/tex]? If you substitute [tex]d\tau^2 = ds^2/c^2[/tex] into that you get [tex]ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2[/tex].
 
  • #32
Besides the fact that I only understand half of the mathemathics you used (mainly step three and four of those four lines of latex are a bit blurry), I'm really looking for a more concrete explenation. Surely there's some logic behind it that one could understand?
 
  • #33
Well, how about this: think of a rectangular triangle. If you calculate the length of one of the rectangular sides (A) from the hypotenuse (H) and the other rectangular side (B), you get: [tex]A^2=H^2-B^2[/tex]
This reflects the Minkowski way of doing relativity.

The Euclidean way of doing relativity simply says: [tex]H^2=A^2+B^2[/tex]
Both formulas essentially contain the same information. but the roles of the variables have switched.

In the Minkowski formula, A is the invariant while in the Euclidean formula H is the invariant. The latter requires all objects to have velocity c in 4D Euclidean space which will become clear from the examples I will give below:

Instead of
[tex]ds^2=d(ct)^2-dx^2-dy^2-dz^2[/tex] (1)
you can also say:
[tex]d(ct)^2=ds^2+dx^2+dy^2+dz^2[/tex] (2)
In (1), [tex]ds[/tex] is the invariant. In (2), [tex]d(ct)[/tex] is the invariant. We must now make plausible that we can switch this role just like that.

For this to work out, we must make the assumption that the vectorsum of velocities in all 4 dimensions always equals c. That means that the velocity in the time dimension must be:
[tex]v_{time}=\sqrt{c^2-v_{space}^2}[/tex] (3)

For a photon [tex]ds^2=0[/tex] so equation (2) reduces to
[tex]d(ct)^2=dx^2+dy^2+dz^2[/tex],
which is simply true. [tex]dct[/tex] is indeed Lorentz-invariant and the time-component vanishes in this case because the speed of the photon equals c.

A moving mass particle cannot reach lightspeed so [tex]ds^2>0[/tex].
The factor [tex]dct[/tex] in equation (2) is now still Lorentz invariant as result of our assumption that the object moves with velocity c in 4D. After all, this is supposed to represent the 4-dimensional distance traveled in a time duration [tex]dt[/tex] which should always equal [tex]c[/tex] times [tex]dt[/tex].

[tex]ds[/tex] in this case should then represent the distance traveled in the time dimension in a timeduration [tex]dt[/tex]. N.B: this is not [tex]dt[/tex] but [tex]v_{time}dt[/tex]. So the total distance traveled is
[tex]\sqrt{(v_{time}dt)^2+dx^2+dy^2+dz^2)}
= \sqrt{dt^2(\sqrt{c^2-v_{space}^2})^2+(v_{space}dt)^2}[/tex]
[tex]= dct [/tex]

So the Euclidean way does make sense as long as the assumption about the velocity c in 4D is maintained. And it gets rid of the minus sign that you have been struggling with all the time! Bottom line is that this minus sign is merely a result of the awkward way of doing relativity with the Minkowski model of space-time.
 
  • #34
NanakiXIII said:
Besides the fact that I only understand half of the mathemathics you used (mainly step three and four of those four lines of latex are a bit blurry)
You mean [tex]d\tau = dt/\gamma[/tex] and [tex]d\tau^2 = ds^2/c^2[/tex]? The first is just the standard formula for time dilation in relativity, it means that if I see a clock traveling at velocity v for some period of time dt in my frame (say, 1 hour), then the clock itself will appear slowed down in my frame, and I will only observe it to tick forward by [tex]d\tau = dt/\gamma = dt\sqrt{1 - v^2/c^2}[/tex] during that time period. This time dilation formula can itself be derived from the requirement that light must be measured to travel at the same speed in all reference frames--see the "light clock" explanation http://www.kineticbooks.com/physics/17467/17486/sp.html [Broken].

The second formula should just be thought of as the definition of ds in terms of the proper time [tex]d\tau[/tex]. As far as I know ds doesn't have any independent physical meaning, I think people just came up with ds because they wanted an invariant measure of "distance" in spacetime that had units of length, instead of units of time. Since the proper time is a frame-invariant type of "distance" between two events in spacetime that has units of time, multiplying it by a constant with units of length/time will give a frame-invariant measure of distance with units of length.
NanakiXIII said:
I'm really looking for a more concrete explenation. Surely there's some logic behind it that one could understand?
Well, I think the light-clock explanation gives a pretty concrete explanation for the relation between time experienced in my frame and the proper time elapsed on a clock that's moving in my frame, and deriving [tex]d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)[/tex] from that is pretty straightforward. Also, once you understand the meaning of proper time, it's clear that it must be a frame-invariant quantity, since different frames can't disagree on how much time elapses on a physical clock as it passes two given points in spacetime (with 'points in spacetime' marked by particular physical events, like 'clock departs earth' and 'clock arrives at mars'). But if that's not concrete enough I don't know of any simpler explanation; I suppose you could just trust that the formula follows from the Lorentz transformation, and the Lorentz transformation follows from the assumption that the laws of physics should work the same way in all reference frames and the assumption speed of light should be the same in all reference frames.
 
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  • #35
Mortimer said:
Instead of
[tex]ds^2=d(ct)^2-dx^2-dy^2-dz^2[/tex] (1)
you can also say:
[tex]d(ct)^2=ds^2+dx^2+dy^2+dz^2[/tex] (2)
In (1), [tex]ds[/tex] is the invariant. In (2), [tex]d(ct)[/tex] is the invariant. We must now make plausible that we can switch this role just like that.
I don't see how dt could be an invariant unless you change the definition of dt so it's no longer what's measured on physical clocks. Surely if you pick two physical events and ask observers in different frames to measure the time between them using their own clocks, they will not all get the same answer.
Mortimer said:
For this to work out, we must make the assumption that the vectorsum of velocities in all 4 dimensions always equals c.
How do you define "velocity" in the time dimension? What physical procedure would you use to measure it?
 
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