why negative

[swansont]

No, I asked why [math] s^2=x^2+y^2+z^2-c^2t^2 [/math]

 

 

I am patient. You will say I am stubborn. My question is not about the meaning of "interval", it about zero.

 

"spacetime interval" has a meaning.

 

[math] s^2=0 [/math]

 

and of course

 

[math] ds^2=0 [/math]

 

Why do you call zero "spacetime interval"? How did that come out?

 

When the result of an operation is zero, It is usually called "null result", not "something". Is that more clear as a question?

 

 

I answered this already. S^2 = 0 only for light. S^2 is not zero in general.

[D H]

"spacetime interval" has a meaning.

The problem is that you appear to be misinterpreting that meaning. [math]s^2=x^2+y^2+z^2-(ct)^2[/math] is the "distance", aka "interval", between the origin and some point in four dimensional spacetime. This can easily be extended to get the "distance" between two points in spacetime: [math]s^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2-(c(t_2-t_1))^2[/math], or more simply, [math]s^2 = \Delta r^2 - c^2\Delta t^2[/math]. This is very similar to a distance, but that it can take on negative values means it does not satisfy one of the key mathematical concepts that characterize "distance". Nonetheless this is an extremely useful concept, so it is handy to give it a name. That name is the spacetime interval.

 

Your misunderstanding stems from your thinking that a light-like interval, one in which [math]s^2=0[/math], is *the* spacetime interval. It is but one example of a spacetime interval.

[michel123456]

Thank you. I understand quickly but you have to explain for a long time.

 

Now I will tell you what I understand.

 

_Distance (spacelike interval in spacetime) is positive.

_Duration (timelike interval in spacetime) is negative.

_Distance for light (lightlike interval in spacetime) is null.

 

Is it correct?

Edited December 28, 2010 by michel123456

[IM Egdall]

Thank you. I understand quickly but you have to explain for a long time.

 

Now I will tell you what I understand.

 

_Distance (spacelike interval in spacetime) is positive.

_Duration (timelike interval in spacetime) is negative.

_Distance for light (lightlike interval in spacetime) is null.

 

Is it correct?

 

No. Space interval is the distance between two events.

Time interval is the elapsed time between two events

Both the space interval and time interval are affected by uniform motion (if you are moving with respect to me, you measure a different space interval and different time interval between the two events than I do.)

 

The spacetime interval squared equals the difference between the square of the space interval and the square of c times the time interval. The spacetime interval is NOT affected by uniform motion. (For example, if you are moving with respect to me, you calculate the same spacetime interval between two events that I do).

 

Time-like: The time interval squared is greater than the space interval squared.

 

We write the spacetime interval as S^2 = (ct)^2 - (x^2 +y^2 + z^2)

 

Space-like: The space interval squared is greater than the time interval squared

 

Here we write the spacetime interval as S^2 = (x^2 + y^2 +z^2) - (ct)^2

 

Light-like: The time interval and space interval are equal, so the spacetime interval is zero.

Edited December 28, 2010 by I ME

[michel123456]

Sorry for misunderstanding. I am tired of discussing moving bodies, and since I have difficulties on understanding simple things, I focus on objects standing at rest in the same FOR. For those 2 objects:

1.the distance between them is positive. I guess everyone agree on this.

2.the time between them is negative. That is the result I understand from the precedent discussion. Do you agree?

[D H]

The spacetime interval squared equals the difference between the square of the space interval and the square of c times the time interval. The spacetime interval is NOT affected by uniform motion. (For example, if you are moving with respect to me, you calculate the same spacetime interval between two events that I do).

 

Time-like: The time interval squared is greater than the space interval squared.

 

We write the spacetime interval as S^2 = (ct)^2 - (x^2 +y^2 + z^2)

 

Space-like: The space interval squared is greater than the time interval squared

 

Here we write the spacetime interval as S^2 = (x^2 + y^2 +z^2) - (ct)^2

 

Light-like: The time interval and space interval are equal, so the spacetime interval is zero.

Yech. While some people do use s²=(ct)²-(x²+y²+z²) rather than s²=x²+y²+z²-(ct)², it is a very very bad idea to mix conventions. Stick with one.

 

 

 

Thank you. I understand quickly but you have to explain for a long time.

 

Now I will tell you what I understand.

 

_Distance (spacelike interval in spacetime) is positive.

_Duration (timelike interval in spacetime) is negative.

_Distance for light (lightlike interval in spacetime) is null.

 

Is it correct?

Assuming you are using the convention s²=x²+y²+z²-(ct)², that is kind of right. Remember that this is an expression for the square of the interval. This makes a time-like interval imaginary rather than negative.

Edited December 28, 2010 by D H

[michel123456]

(...)

Assuming you are using the convention s²=x²+y²+z²-(ct)², that is kind of right. Remember that this is an expression for the square of the interval. This makes a time-like interval imaginary rather than negative.[/size][/font]

 

Interesting comment.

That make my statements completely wrong. I should have stated: (aouch its harder than i first thought)

 

_Distance (spacelike interval in spacetime) is positive or negative, when squared is positive.

_Duration (timelike interval in spacetime) is imaginary, when squared is negative.

_Distance for light (lightlike interval in spacetime) is null.

 

That makes time perpendicular to space, isn't it?

 

---------------------------------------------------

 

for a single object standing still we have:

[math] x=0 [/math], [math] y=0 [/math], [math] z=0 [/math]

 

and thus

 

[math] s^2= -c^2t^2 [/math] (1)

 

 

taking the square root we get

 

[math] s = c t i[/math] (2) or [math] s = (ci) t[/math] (3)

 

equation (3) is convenient because it gives a comprehensive interpretation of the timelike interval s.

 

One would expect that for a non moving object, the interval s would be equal to t. That is not the case. What we see is that s is equal to t multiplied by a coefficient equal to ci.

 

Well, there is a way to make the equation ressemble what was expected, but the price to pay is inacceptable:

if and only if [math] c= i [/math] then equation (3) becomes [math] s = -t [/math] (4)

 

But [math] c= i [/math] is wrong.

Edited December 28, 2010 by michel123456

[Celeritas]

Study the figure posted in → Post 240 → of the Clocks, Rulers … and an issue for relativity thread, then read what follows here. Just a quick note wrt that figure, where it says “spacetime system” should have been stated “2-space system (time implied)”.

 

In the figure of Post 240, the light ray always travels the hypthenuse of a right triangle, which is always the longest leg per Pythagorus’ theorem. However, an observer at rest in the moving x’,y’ system records the photon to travel directly along +y’ during time t’. The photon must travel ct’ = y’ (= y) in the primed system. So why then is the polarity of ct of opposite polarity wrt x,y,z in the eqn for the length of the invariant spacetime interval? …

 

s² = (ct)² - (x²+y²+z²)

 

The reason …

 

In the unprimed system …

 

(ct)² = x² + y² … which for the given lightpath, and motion only along x/x’ is … (ct)² = (vt)² + y²

 

but y = y’ = ct’ … as shown in the referenced figure, so

 

(ct)² = x² + y²

(ct)² = x² + (ct’)²

 

Now, we have 2 light paths here, and so which is then the hypothenuse? The equation models the Pathagorus’ eqn per the unprimed POV, using ct’ for y as they are numerically identical, and so (ct)² is the longest leg of the right triangle. That then gives us …

 

(ct)² = x² + (ct’)² ... where ct > ct' since t = γt'

(ct’)² = (ct)² - x²

 

as the above eqn is technically …

 

y² = (ct)² - x²

 

and so the eqn for the length of the invariant spacetime interval …

 

(ct’)² = (ct)² - x²

 

The spacetime interval is a relation between variables of 2 differing frames, that are related by a euclidean-like metric (the Minkowski metric), not a purely euclidean metric (that Galileo used). The minus sign on the right side of the equality is the result of that.

 

Best regards,

Celeritas

Edited October 3, 2016 by Celeritas

[KipIngram]

Here, let me try. This is like using c as an agreed upon "zero point." That's sloppy wording, if you want to find something that everyone will agree on the speed of, regardless of their state of motion, it has to be moving at c. Something moving at speed 0 for me isn't moving at speed 0 for you, if you and I are moving w.r.t. one another. If two events are connected by a light beam, we will all agree on their interval: zero. So we're using that "automatically agreed upon" case as the reference point in our interval measurement.

 

I have mixed feelings about that explanation. It feels to me like there's something worthwhile in it, but also something that a little hand-wavy. So don't take it as gospel, but if it sheds any light at all for you then that's great.

 

The only other thing I can think of to say about this is "because." Each user will have his or her own dx, dy, dz, and dt values for the two events, based on states of motion. But the math of relativity is such that they'll all agree on x^2+y^2+z^2-(c*t)^2. That's why it's useful: because every one agrees. We defined it thusly because it's useful.

[<Veronica>]

Michel, your questions remind me of the good old philosopher Socrates!  👊