Spacetime intervals again - still figuring out the formula

[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:
[tex]c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2[/tex]
Since [tex]ds=cd\tau[/tex]
this equals:
[tex]c^2=(cd\tau/dt)^2+v_{space}^2[/tex] (1)

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

The [tex]c^2[/tex] 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 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:
[tex]cd\tau/dt[/tex]
should actually be written as:
[tex]c\frac{d\tau/dx_5}{dt/dx_5}[/tex]
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
[tex]cd\tau/dx_5[/tex]
(because [tex]dt/dx_5=1[/tex] always)

[tex]t[/tex] 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:
[tex]c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2[/tex]
Since [tex]ds=cd\tau[/tex]
this equals:
[tex]c^2=(cd\tau/dt)^2+v_{space}^2[/tex] (1)

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

The [tex]c^2[/tex] in (1) is invariant by assumption.

OK, I agree that if you choose to define "velocity through time" as [tex](cd\tau/dt)[/tex] 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), [tex]d(ct)[/tex] 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 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:
[tex]cd\tau/dt[/tex]
should actually be written as:
[tex]c\frac{d\tau/dx_5}{dt/dx_5}[/tex]
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
[tex]cd\tau/dx_5[/tex]
(because [tex]dt/dx_5=1[/tex] always)

[tex]t[/tex] is the time as measured by an observer in his own rest frame.

If you define "velocity through time" as [tex](cd\tau/dt)[/tex] 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 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 https://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), [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.

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 [tex]ds[/tex] and [tex]d\tau[/tex] 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:

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

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 [tex]cd\tau/dt[/tex] for the velocity in the time dimension.

OK, I agree that if you choose to define "velocity through time" as [tex]cd\tau/dt[/tex] 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), [tex]d(ct)^2[/tex] is the invariant"?

For the same reason as [tex]d(c\tau)^2[/tex] is the invariant in the Minkowski style of this equation.
When I rewrote the Minkowski equation
[tex]ds^2=d(ct)^2-dx^2-dy^2-dz^2[/tex] (1)
into the Euclidean equation
[tex]d(ct)^2=ds^2+dx^2+dy^2+dz^2[/tex], (2)
I switched the role of the invariant with the time component in the 4-vector. Indirectly this also switches the role of [tex]t[/tex] and [tex]\tau[/tex] (because [tex]ds=cd\tau[/tex]). 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
[tex]c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2[/tex]
by differentiation with respect to [tex]t[/tex] (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 [tex]\tau[/tex] instead of [tex]t[/tex]. The reason for this is historically determined: differentiation with respect to [tex]t[/tex] results in the 4-vector [tex](c, v_1, v_2, v_3)[/tex] but this 4-vector is not invariant under a Lorentz transformation in Minkowski space. Therefor, the 4-vector was multiplied with the factor [tex]\gamma[/tex] to artificially make it invariant. This [tex]\gamma[/tex] leads to the transformation of [tex]t[/tex] into [tex]\tau[/tex]. 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:

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

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:

[tex]dS^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2[/tex]

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.

[tex] dS^2[/tex] 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 [tex]dx^2+dy^2+dz^2-c^2dt^2[/tex], which is the interval using the -+++ metric instead of the +--- metric.

 

[JesseM]

[tex] dS^2[/tex] 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 [tex]d\tau[/tex] between events with a timelike separation:

[tex]dS^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2[/tex]

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

So, [tex]dS = i*c*d\tau[/tex]

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 [tex]d\tau[/tex], 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 [tex]d\tau[/tex] 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 [tex]dx^2+dy^2+dz^2-c^2dt^2[/tex], 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 [tex]c^2 dt^2 - dx^2 - dy^2 - dz^2[/tex], 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 [tex]ds[/tex] becomes complex (because [tex]ds^2<0[/tex]). 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. [tex]ds[/tex]) 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 across 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]

Measuring displacements in Minkowski spacetime

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 across 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 [tex]t_P[/tex], [tex]t_S[/tex], [tex]t_R[/tex] be the times read off that observer's clock. (Obviously, [tex]t_R \geq t_S[/tex].)

According to this observer, the spatial-displacement from P to Q is
[tex]\Delta x_{Q\mbox{ from }P}=c(t_R-t_S)/2[/tex] ,
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 [tex]t_Q= (t_R+t_S)/2 [/tex],
that is, the average of the send and receive clock-readings.

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

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

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
[tex] c^2( t_{receive} - t_P)( t_{send} - t_P) [/tex] is invariant.
Indeed,
[tex]
\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*}
[/tex]
(This calculation is simpler if we assume [tex]t_P=0[/tex].
To keep to the spirit of emphasizing the time-measurements, one should start at the bottom with [tex](t_R-t_P)(t_S-t_P)[/tex] and obtain the expression [tex](c\Delta t)^2- \Delta x^2 [/tex].)


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

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

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

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

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

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

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

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

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

 

[jtbell]

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

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

Now let's proceed:

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

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

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

 

[JesseM]

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

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

Well, do you understand how to factor variables out of equations in algebra? For example, do you understand why if you factor [tex]x^2[/tex] out of [tex](x^3 + 2x^5 + 5)[/tex] you'd get [tex](x^2)*(x + 2x^3 + 5/x^2)[/tex]? If so, just factor [tex](1/dt^2)[/tex] out of the equation [tex](1 - (1/dt^2*c^2)(dx^2 + dy^2 + dz^2))[/tex] 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 [tex]\frac{dx^2 + dy^2 + dz^2}{c^2 dt^2}[/tex] by dt^2, you'll get [tex]\frac{dx^2 + dy^2 + dz^2}{c^2}[/tex]?

 

[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 https://www.physicsforums.com/showpost.php?p=541106&postcount=21
make sense?

How about the physical interpretation that I described in post #56
https://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 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

[tex]ds[/tex] is now no longer the 4D displacement but just the displacement in the time dimension. The factor [tex]cdt[/tex] 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
https://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.