Spacetime intervals again - still figuring out the formula

[NanakiXIII]

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?

[pervect]

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.

[jtbell]

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 - (z_2 - z_1)^2}

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.

[JesseM]

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 (http://www.physicsforums.com/showthread.php?t=59901).

[jdavel]

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!

[quasar987]

\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}

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

ds = \frac{\partial{s}}{\partial{t}}dt + \frac{\partial{s}}{\partial{x}}dx

of the function

s(x,t) = \sqrt{(ct)^2-x^2}

and square it, I get the ugly

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

So what am I missing?

[dextercioby]

He meant the very famous

ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}


Daniel.

[jtbell]

If I calulate the total differential

ds = \frac{\partial{s}}{\partial{t}}dt + \frac{\partial{s}}{\partial{x}}dx

of the function

s(x,t) = \sqrt{(ct)^2-x^2}

and square it, [...]

Where did you get the function

s(x,t) = \sqrt{(ct)^2-x^2}

from?

[quasar987]

Where did you get the function

s(x,t) = \sqrt{(ct)^2-x^2}

from?

I figured

s(x,t) = \sqrt{(ct)^2-x^2}

is kinf of

\Delta s = \sqrt{(c\Delta t)^2 - (\Delta x)^2}

where t1 = x1 = 0...

How do you get the expression

(ds)^2 = (cdt)^2 - (dx)^2 - (dy)^2 - (dz)^2

anyway?

[dextercioby]

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


Daniel.

[JesseM]

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

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 d\tau, and hence ds = cd\tau, must be a frame-invariant quantity.

[NanakiXIII]

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?

[Ich]

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.

[JesseM]

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 dt, and if the time elapsed on that clock between those same two points is d\tau, then the two time intervals are related by the time dilation formula:

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

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

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

So, plugging that into the time dilation formula:

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

which gives the proper time formula I posted above:

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

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

[jdavel]

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?

[quasar987]

No, that's true.

I'll edit it.

[NanakiXIII]

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).

[quasar987]

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 travelled 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 spacial 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.

[jdavel]

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?

[quasar987]

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!

[robphy]

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).

[NanakiXIII]

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.

[jdavel]

....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.

[quasar987]

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 \Delta s^2 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:

[NanakiXIII]

(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 apparant 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.

[quasar987]

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 \Delta d appart and a time interval \Delta t appart, then, when you plug the numbers \Delta d and \Delta t in the equations we call the Lorentz transformation, what you get out of those equation is the distance \Delta d^* and the time interval \Delta t^* 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.

[NanakiXIII]

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?

[jtbell]

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

E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2 = (m c^2)^2

where m 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 ds^2. That's not a coincidence. In relativity, time and position together form a four-vector (ct, x, y, z), and energy and momentum form another four-vector, (E, p_x c, p_y c, p_z c). 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.

[NanakiXIII]

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?' ?

[Mortimer]

NanakiXIII's original questions on why the c is involved in cdt and why we are subtracting the dx_i 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 dx_i 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.

[JesseM]

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 (http://www.physicsforums.com/showpost.php?p=539900&postcount=14) where I derived the formula d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2) from the formula for time dilation in relativity, d\tau = dt/\gamma? If you substitute d\tau^2 = ds^2/c^2 into that you get ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2.

[NanakiXIII]

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?

[Mortimer]

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: A^2=H^2-B^2
This reflects the Minkowski way of doing relativity.

The Euclidean way of doing relativity simply says: H^2=A^2+B^2
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
ds^2=d(ct)^2-dx^2-dy^2-dz^2 (1)
you can also say:
d(ct)^2=ds^2+dx^2+dy^2+dz^2 (2)
In (1), ds is the invariant. In (2), d(ct) 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:
v_{time}=\sqrt{c^2-v_{space}^2} (3)

For a photon ds^2=0 so equation (2) reduces to
d(ct)^2=dx^2+dy^2+dz^2,
which is simply true. dct 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 ds^2>0.
The factor dct 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 dt which should always equal c times dt.

ds in this case should then represent the distance traveled in the time dimension in a timeduration dt. N.B: this is not dt but v_{time}dt. So the total distance traveled is
\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}
= dct

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.

[JesseM]

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 d\tau = dt/\gamma and d\tau^2 = ds^2/c^2? The first is just the standard formula for time dilation in relativity, it means that if I see a clock travelling 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 d\tau = dt/\gamma = dt\sqrt{1 - v^2/c^2} 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 here (http://www.kineticbooks.com/physics/17467/17486/sp.html).

The second formula should just be thought of as the definition of ds in terms of the proper time d\tau. 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. 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 d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2) 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.

[JesseM]

Instead of
ds^2=d(ct)^2-dx^2-dy^2-dz^2 (1)
you can also say:
d(ct)^2=ds^2+dx^2+dy^2+dz^2 (2)
In (1), ds is the invariant. In (2), d(ct) 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. 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?

[Mortimer]

JesseM said:
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.

Perhaps its easier to see if written like this:
c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
Since ds=cd\tau
this equals:
c^2=(cd\tau/dt)^2+v_{space}^2 (1)

cd\tau/dt is the same as v_{time}
which follows from the relations
v_{time}=\sqrt{c^2-v_{space}^2}=c/\gamma
and
d\tau=dt/\gamma

The c^2 in (1) is invariant by assumption.
As to your question how to measure the velocity in the time dimension, you see here one possible physical procedure but I prefer another one which is explained here (http://www.rfjvanlinden171.freeler.nl/dimensionshtml/node2.html). (that saves me a lot of typing in this post)
That text essentially explains that the expression:
cd\tau/dt
should actually be written as:
c\frac{d\tau/dx_5}{dt/dx_5}
which is the ratio between the proper-time-velocity and the time-velocity in a 5D space and generally defines the velocity in the time dimension as
cd\tau/dx_5
(because dt/dx_5=1 always)

t is the time as measured by an observer in his own rest frame.

[jdavel]

NanakiXIII,

Try this.

A theory of relativity gives the relation between measurements made in each of two frames of reference S and S', where S' is moving at a speed v with respect to S.

Specifically, it relates the results of measurements made in each frame of the distance between events (space intervals). And it relates the results of measurements made in each frame of the elapsed time between events (time intervals).

If the measured value for an interval (time or space) is the same in both frames, it's called absolute. If not, it's called relative. Speed is called absolute if the ratio of the space interval to the time interval (for which the speed is defined) is the same in both frames. If not, it's called relative.

According to classical relativity (what Galileo, Newton and everyone else believed until 1905):

1) Time intervals are absolute
2) Space intervals (in general) are relative
As a result of 1) and 2),
3) All measurements of speed (including the speed of light) must be relative.

According to Einstein's relativity (what everyone has believed since 1905):

1) Measurements of the speed of light are absolute.
2) Measurements of any speed less than the speed of light are relative
As a result of 1) and 2)
3) Time intervals AND space intervals must be relative.
As a result of the specific way that time intervals and space intervals are relative,
4) The time interval squared times the speed of light squared minus the space interval squared is absolute.

That's the logic of why the space time interval is what it is. To see mathematically why the spacetime interval is what it is, you have to do the math.

[JesseM]

Perhaps its easier to see if written like this:
c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
Since ds=cd\tau
this equals:
c^2=(cd\tau/dt)^2+v_{space}^2 (1)

cd\tau/dt is the same as v_{time}
which follows from the relations
v_{time}=\sqrt{c^2-v_{space}^2}=c/\gamma
and
d\tau=dt/\gamma

The c^2 in (1) is invariant by assumption. OK, I agree that if you choose to define "velocity through time" as (cd\tau/dt) then this math works out. But your equation only has c^2 as an invariant, not dt; why did you earlier say that "In (2), d(ct) is the invariant"? As to your question how to measure the velocity in the time dimension, you see here one possible physical procedure but I prefer another one which is explained here (http://www.rfjvanlinden171.freeler.nl/dimensionshtml/node2.html). (that saves me a lot of typing in this post)
That text essentially explains that the expression:
cd\tau/dt
should actually be written as:
c\frac{d\tau/dx_5}{dt/dx_5}
which is the ratio between the proper-time-velocity and the time-velocity in a 5D space and generally defines the velocity in the time dimension as
cd\tau/dx_5
(because dt/dx_5=1 always)

t is the time as measured by an observer in his own rest frame. If you define "velocity through time" as (cd\tau/dt) then of course all you have to do to measure it is to measure both the elapsed time in your reference frame and the elapsed proper time on the moving clock, and plug them into that formula. But above you seem to suggest that your concept of "velocity through time" is not just a definition in terms of existing ideas in relativity, but is actual part of some new theory you have devised which involves an extra dimension (or two extra dimensions?) beyond the three space and one time dimensions we are accustomed to. If you're discussing a new theory of your own, then that really shouldn't go in this forum--as said in the IMPORTANT! Read before posting (http://www.physicsforums.com/showthread.php?t=17355) sticky thread at the top of this forum, this forum is just meant for discussing the theory of relativity, and "is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories". If you want to discuss a new theory of your own, that's what the Theory Development (http://www.physicsforums.com/forumdisplay.php?f=12) forum is for.

[PeteSF]

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

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 d\tau, and hence ds = cd\tau, must be a frame-invariant quantity.

Thinking about spacelike intervals...

The interval would be the proper length of a rod with position and (inertial) motion such that the two ends of the rod coincided with the two events in question simultaneously in the rod's rest frame...

Right? And is there a simpler way of saying it?

[JesseM]

Thinking about spacelike intervals...

The interval would be the proper length of a rod with position and (inertial) motion such that the two ends of the rod coincided with the two events in question simultaneously in the rod's rest frame...

Right? And is there a simpler way of saying it? That can't be right, because the separation between two events which are simultaneous in some inertial frame must be spacelike, and both ds and d\tau are imaginary for spacelike intervals.

[PeteSF]

You're right... My understanding grows a little, and I'd better do the maths.

***

OK, it seems the interval is i times the proper length:

\Delta s^2 = c.\Delta t^2 - (\Delta x^2 + \Delta y^2 + \Delta z^2)
\Delta s = i\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}

Is this result just a fun fact, or does it lead to any insights?

[Mortimer]

JesseM said:
But above you seem to suggest that your concept of "velocity through time"
is not just a definition in terms of existing ideas in relativity, but is
actual part of some new theory you have devised which involves an extra
dimension (or two extra dimensions?) beyond the three space and one time
dimensions we are accustomed to. If you're discussing a new theory of your
own, then that really shouldn't go in this forum
Point taken. My fault. In this thread I'll stick to the definition cd\tau/dt for the velocity in the time dimension.
OK, I agree that if you choose to define "velocity through time" as cd\tau/dt then this math works out. But your equation only has c^2 as an invariant, not dt; why did you earlier say that "In (2), d(ct)^2 is the invariant"?
For the same reason as d(c\tau)^2 is the invariant in the Minkowski style of this equation.
When I rewrote the Minkowski equation
ds^2=d(ct)^2-dx^2-dy^2-dz^2 (1)
into the Euclidean equation
d(ct)^2=ds^2+dx^2+dy^2+dz^2, (2)
I switched the role of the invariant with the time component in the 4-vector. Indirectly this also switches the role of t and \tau (because ds=cd\tau). The switch was mathematically justified by the assumption of the universal velocity c for objects in 4D space-time.
Equations (1) and (2) describe the Minkowski and Euclidean 4-vectors for position.
Equation (2) is transformed into the Euclidean velocity 4-vector
c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
by differentiation with respect to t (which is what we are supposed to do if we calculate velocities).
The Minkowski 4-vector for velocity is derived from the 4-vector for position by differentiation with respect to \tau instead of t. The reason for this is historically determined: differentiation with respect to t results in the 4-vector (c, v_1, v_2, v_3) but this 4-vector is not invariant under a Lorentz transformation in Minkowski space. Therefor, the 4-vector was multiplied with the factor \gamma to artificially make it invariant. This \gamma leads to the transformation of t into \tau. The Euclidean 4-vector is invariant as is but requires the assumption of the universal velocity c in 4D space-time.

As I indicated earlier in this thread, Euclidean relativity is not new. It is at least 40 years old. A very clean description (i.e., restricting itself to the core) is published by prof. Alexander Gersten (Ben-Gurion univ., Israel) and can be found here: www.bgu.ac.il/~gersten/papers/euclirev.pdf
Other articles on Euclidean relativity (including mine) often stack a number of speculations on top of it.

[JesseM]

You're right... My understanding grows a little, and I'd better do the maths.

***

OK, it seems the interval is i times the proper length:

\Delta s^2 = c.\Delta t^2 - (\Delta x^2 + \Delta y^2 + \Delta z^2)
\Delta s = i\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}

Is this result just a fun fact, or does it lead to any insights? Yeah, that works, I hadn't thought of it like that before. It'd be even simpler if you forgot about ds and instead used dS, defined like this:

dS^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2

Then for two events with a spacelike separation, dS would simply represent the distance between the events as measured in the frame where they happened simultaneously. Obviously in that frame, dt=0, so dS is just the ordinary distance formula. But this seems more like a "fun fact" rather than something that leads to any big insights.

[PeteSF]

Thanks Jesse!

[Ich]

But this seems more like a "fun fact" rather than something that leads to any big insights.
dS^2 is interpreted either as a proper time or as a proper length, depending on the sign. Both are equally valid, so I don´t see why one of those interpretations should be a fun fact.

[Mortimer]

I agree with "Ich". I can see what JesseM tries to say but its a bit confusing to use the term "space-like" in this context as well as the expression dx^2+dy^2+dz^2-c^2dt^2, which is the interval using the -+++ metric instead of the +--- metric.

[JesseM]

dS^2 is interpreted either as a proper time or as a proper length, depending on the sign. What do you mean by "depending on the sign"? You have to change more than just the sign to go from the proper length dS between events with a spacelike separation to the proper time d\tau between events with a timelike separation:

dS^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2

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

So, dS = i*c*d\tau Both are equally valid, so I don´t see why one of those interpretations should be a fun fact. I agree both are equally valid, I just meant that if you already understand how to get the equation for d\tau, you don't really gain any big new insight by figuring out the derivation for dS...but the reverse is true as well I suppose (although the expression for proper time is useful in more problems since you can integrate d\tau over a curved path to find the total proper time, while I don't think there's any situation where you'd want to integrate dS).

[JesseM]

I agree with "Ich". I can see what JesseM tries to say but its a bit confusing to use the term "space-like" in this context Why is it confusing? The concept of the distance between two events in the frame where they are simultaneous only makes sense if there is a spacelike separation between them, if they are timelike separated there is no frame where they are simultaneous. as well as the expression dx^2+dy^2+dz^2-c^2dt^2, which is the interval using the -+++ metric instead of the +--- metric. Again, if you want to find the distance between two events in the frame where they are simultaneous, you have to write it that way. If you used c^2 dt^2 - dx^2 - dy^2 - dz^2, that would give i times the distance.

[Mortimer]

No big deal. Strictly speaking you are not saying anything wrong.
I see no physical difference between the +--- metric and the -+++ metric. Although there is a mathematical difference (the i) I find this hardly relevant because the Minkowski "distance" between events is not a real distance in the sense as we use it in space anyway.

[NanakiXIII]

JesseM, no, I meant those four lines you used to get to the final formula.


3) Time intervals AND space intervals must be relative.
As a result of the specific way that time intervals and space intervals are relative,
4) The time interval squared times the speed of light squared minus the space interval squared is absolute.

That's the logic of why the space time interval is what it is. To see mathematically why the spacetime interval is what it is, you have to do the math.


I'm going to assume 3 is correct, I don't fully understand. But I don't see how 3 leads to 4.


My apologies for being away for a bit, I've been and am ill. Also, sorry for not replying to all your replies, I just picked out what seemed important.

[PeteSF]

I find this hardly relevant because the Minkowski "distance" between events is not a real distance in the sense as we use it in space anyway.
...but it is for events with spacelike separation, yes?

[Mortimer]

The point is that with a spacelike separation the invariant ds becomes complex (because ds^2<0). If you don't like that you can multiply the whole interval with i and get a real number again. That's really just a matter of convention. But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.
For your understanding, it is probably important to remember that the distance between events (i.e. ds) in Minkowski space-time is just a "gadget" that happens to be invariant under a Lorentz transformation. It has no real physical meaning like for instance a distance in space or a timelapse on a clock. The math is merely a tool that helps making the correct calculations (just as complex numbers are for instance).

[PeteSF]

But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.

What?
Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes?
FTL speeds are required for information transfer between such events, but is that relevant?

[Mortimer]

Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes? FTL speeds are required for information transfer between such events
Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").

but is that relevant
Well, depends on what you want to do with it, doesn't it? If i will never ever be able to come accross the situation, what's the point in discussing it like it is real? It's like speculating about tachyons and so.

[Ich]

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").


Spacetime has to be measured with RODS and CLOCKS, according to Einstein.
Either a length, measured (at least symbolically) with rods, or a time difference, measured with clocks, are physically meaningful and well defined. There exists a prescription how to measure proper length, and it is invariant. What more do you want?

But JesseM is right, as world lines are necessarily timelike, ds=dtau will be used more often in calculations and is therefore more useful in a way. But it´s not the whole truth about the spacetime interval. You need both interbpretations posslible to be consistent.

[robphy]

But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.
For your understanding, it is probably important to remember that the distance between events (i.e. ) in Minkowski space-time is just a "gadget" that happens to be invariant under a Lorentz transformation. It has no real physical meaning like for instance a distance in space or a timelapse on a clock. The math is merely a tool that helps making the correct calculations (just as complex numbers are for instance).




Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes? FTL speeds are required for information transfer between such events

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").


Well, depends on what you want to do with it, doesn't it? If i will never ever be able to come accross the situation, what's the point in discussing it like it is real? It's like speculating about tachyons and so.

This may be helpful.


Here's a procedure to measure displacements in [Minkowski] spacetime using a clock, a light source, and detector. (This is the "radar method".)
Given two events, P and Q, what is the square-interval?

Consider an inertial observer meeting P.
One his worldline, there is
an event S when at a light ray can be sent to Q, and
an event R when that light ray's reflection is received from Q.

Let t_P, t_S, t_R be the times read off that observer's clock. (Obviously, t_R \geq t_S.)

According to this observer, the spatial-displacement from P to Q is
\Delta x_{Q\mbox{ from }P}=c(t_R-t_S)/2 ,
that is, half of the measured round-trip time multiplied by the speed of light.

According to this observer, the time-coordinate of the distant-event Q
is DEFINED by t_Q= (t_R+t_S)/2 ,
that is, the average of the send and receive clock-readings.

So,
according to this observer, the time-displacement from P to Q
is \Delta t_{Q\mbox{ from }P}=t_Q-t_P,
or
\Delta t_{Q\mbox{ from }P}=(t_R+t_S )/2 -t_P.
that is, the average of the send and receive clock-readings minus the clock-reading t_P.

The following quantity can tell us about the causal relationship of Q from P.
( t_R - t_P)(t_S - t_P).

If the events on this worldline happen in the sequence S-then-P-then-R,
then Q is spacelike-related to P. This quantity is negative.

If the events on this worldline happen in the sequence P-then-S-then-R,
then Q is in the timelike-future of P. This quantity is positive (since it's the product of two positive numbers).

If the events on this worldline happen in the sequence S-then-R-then-P,
then Q is in the timelike-past of P. This quantity is positive (since it's the product of two negative numbers).

If any two events coincide, then that quantity is zero.
If it's S and P that coincide, then Q is in the lightlike-future of P.
If it's R and P that coincide, then Q is in the lightlike-past of P.
If S, R, and P coincide, then Q coincides with P.


Now, consider any two inertial observers that meet P and
perform this procedure to make measurements of Q.
So, each observer will have a different set of send and receive events and
thus a different set of clock-readings for send and for receive.

According to special relativity, the quantity
c^2( t_{receive} - t_P)( t_{send} - t_P) is invariant.
Indeed,

\begin{align*}
(c\Delta t_{Q\mbox{ from }P})^2-(\Delta x_{Q\mbox{ from }P})^2
&=c^2\left(\frac{t_R+t_S}{2} -t_P\right)^2-\left(c\frac{t_R-t_S}{2} \right)^2\\
&=c^2\left(\frac{(t_R-t_P)+(t_S-t_P)}{2}\right)^2-\left(c\frac{(t_R-t_P)-(t_S-t_P)}{2} \right)^2\\
&=c^2\left( \left(\frac{(t_R-t_P)+(t_S-t_P)}{2}\right)^2-\left(\frac{(t_R-t_P)-(t_S-t_P)}{2} \right)^2\right) \\
&=c^2\left( \frac{2(t_R-t_P)(t_S-t_P)}{4} - \frac{-2(t_R-t_P)(t_S-t_P)}{4} \right) \\
&=c^2 (t_R-t_P)(t_S-t_P) \\
\end{align*}

(This calculation is simpler if we assume t_P=0.
To keep to the spirit of emphasizing the time-measurements, one should start at the bottom with (t_R-t_P)(t_S-t_P) and obtain the expression (c\Delta t)^2- \Delta x^2 .)


So, to comment on the sections I quoted above,
there is a way to measure (with a physical setup) the separation of two spacelike-related events.

[JesseM]

JesseM, no, I meant those four lines you used to get to the final formula.OK, if you're comfortable with these formulas:

Time dilation:
d\tau = dt/\gamma = dt\sqrt{1 - v^2/c^2} (1)

Velocity = distance/time:
v = \sqrt{dx^2 + dy^2 + dz^2}/dt (2)

Then those last four lines are just a matter of algebra:

First, start with equation (1):
d\tau = dt \sqrt{1 - v^2/c^2}

Then square both sides of equation (2) and substitute in for v^2:
d\tau = dt \sqrt{1 - (dx^2 + dy^2 + dz^2)/(c^2dt^2)}

Then factor out 1/dt^2 from the expression inside the square root:
d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

Then do \sqrt{(1/dt^2)*stuff} = (1/dt)\sqrt{stuff}, and the 1/dt cancels out with the dt that was already outside the square root:
d\tau = \sqrt{dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)}

If there's still any of this algebra you're not clear on, let me know which step is giving you trouble.

[NanakiXIII]

I don't understand how you got to the third formula

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

[jtbell]

It may be a little bit more obvious if we first rewrite the preceding equation in a slightly different format:

d\tau = dt \sqrt{1 - \frac {dx^2 + dy^2 + dz^2}{c^2dt^2}}

Now let's proceed:

d\tau = dt \sqrt{\frac {dt^2}{dt^2} - \frac {dx^2 + dy^2 + dz^2}{c^2dt^2}}

d\tau = dt \sqrt{\frac {1}{dt^2} dt^2 - \frac {1}{dt^2} \frac {dx^2 + dy^2 + dz^2}{c^2}}

d\tau = dt \sqrt{\frac {1}{dt^2} \left( dt^2 - \frac {dx^2 + dy^2 + dz^2}{c^2}\right)}

[JesseM]

I don't understand how you got to the third formula

d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))} Well, do you understand how to factor variables out of equations in algebra? For example, do you understand why if you factor x^2 out of (x^3 + 2x^5 + 5) you'd get (x^2)*(x + 2x^3 + 5/x^2)? If so, just factor (1/dt^2) out of the equation (1 - (1/dt^2*c^2)(dx^2 + dy^2 + dz^2)) in the same way.

[NanakiXIII]

Yes, but if you factor out 1/dt^2, aren't you just multiplying everything by dt^2?

Maybe I'm doing something wrong, I'm not that great at algebra, but if I factor out 1/dt^2....

dt^2-c^2*dt^2(dx^2+dy^2+dz^2)

I know, that looks kind of messy, but I don't know how to use Latex.

[JesseM]

Yes, but if you factor out 1/dt^2, aren't you just multiplying everything by dt^2? Yes, exactly, although you have to include that (1/dt^2) on the outside, just like if you factor x out of (x^2 + 2x), you have to divide the whole thing by x and then include x on the outside, like (x)*(x + 2). Maybe I'm doing something wrong, I'm not that great at algebra, but if I factor out 1/dt^2....

dt^2-c^2*dt^2(dx^2+dy^2+dz^2) That shouldn't be a c^2*dt^2 multiplying the space interval, it should be 1/c^2. Remember, the original thing in the square root was:

1 - (dx^2+dy^2+dz^2)/(c^2*dt^2)

So, if you multiply everything by dt^2, it will cancel out with the dt^2 in the denominator of (dx^2+dy^2+dz^2)/(c^2*dt^2), giving:

dt^2 - (dx^2+dy^2+dz^2)/(c^2)

Maybe it'd help if I put the fraction in Latex--do you see why if you multiply \frac{dx^2 + dy^2 + dz^2}{c^2 dt^2} by dt^2, you'll get \frac{dx^2 + dy^2 + dz^2}{c^2}?

[NanakiXIII]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

And yeah, the Latex is a lot easier to read.

[JesseM]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense. As I said earlier, I can't think of any conceptual explanations that don't involve any math--if someone else can think of one hopefully they'll jump in, but it may just be that there is no such purely conceptual explanation.

[robphy]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

And yeah, the Latex is a lot easier to read.

So, does the geometric interpretation using a hyperbola that I described in post #21 http://www.physicsforums.com/showpost.php?p=541106&postcount=21
make sense?

How about the physical interpretation that I described in post #56
http://www.physicsforums.com/showpost.php?p=549398&postcount=56 ?

[Mortimer]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

In a last attempt to try and make you understand it (and promote Euclidean special relativity) I have put my remarks of posts #33, 36 and 42 in a web page that you find here (http://www.rfjvanlinden171.freeler.nl/4vectors). Maybe it helps.

[NanakiXIII]

robphy, the first post just makes little sense to me. The second isn't much better, I've forgotten what it's about by the time I'm halfway. It just seems to advanced.

Ok, I just thought of something when looking at that site. ds^2=c^2dt^2-dA^2, so if you turn that around, c^2dt^2=ds^2+dA^2. This has been said before, I think. But I just realized, no idea why that happened just now, that that means that c^2dt^2 is the 'long side of the triangle', of which I can never remember the proper name. This makes no sense to me. Why is the temporal dimension bigger or equal to the entire interval, when it is but one of four dimensions used to calculate that interval? I suppose this is exactly the same as what I asked before, only worded differently. But maybe it'll help understand what I don't understand, if anyone didn't.

[Mortimer]

But I just realized, no idea why that happened just now, that that means that c^2dt^2 is the 'long side of the triangle'

I think you are getting it now! (it's hypothenuse b.t.w.)
I repeat two pieces of the webpage that are important in this respect:

the components of the Minkowski 4-vector can have no physical meaning. Their function is purely mathematical

and

ds is now no longer the 4D displacement but just the displacement in the time dimension. The factor cdt that plays this role in the Minkowski style of this 4-vector has become the actual 4D displacement

That's it. Simply do not try to seek a physical meaning behind the Minkowski 4-vector. That meaning is only obvious in the Euclidean 4-vector.

[robphy]

That's it. Simply do not try to seek a physical meaning behind the Minkowski 4-vector. That meaning is only obvious in the Euclidean 4-vector.

I invite the readers of this thread to read this discussion I had with Mortimer
http://www.physicsforums.com/showthread.php?t=73582 in which I elaborate on the physical interpretation of the velocity 4-vector and its components.

[quasar987]

From post #19, in relation to post #9:

But when you calculate the differential ds^2, those intervals become differentials as well. That is x -> dx and t -> dt.

Would you mind writting mathematically what you mean by this statement? Thx.