Find Dist. b/w 2 Points in 3D Space incl. Time Dim.

[zepp0814]

so i am hoping that all of you are familiar with the fact that a basic way of finding the distance between two points in 3d space is a^2+b^2+z^2=c^2. anyway i wanted to know in the equation ds^s=dx^2+dz^2+dy^2-(cdt)^2 does this show the distance between two points in in 3-d space including 1 time Dimension or does it show something different. also is this is the distance between 2 point then is there a unit of measurement of is the distance really something like -3.45E17.

 

[Mentz114]

This

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

is also called the proper-interval, and for points (4D points) that can be joined by a light-beam the proper-distance is negative ( using your metric equation).

It is important because we identify it with the time ticked on a clock while traveling along a curve.

It is invariant under the Poincare group of transformations - i.e. Lorentz boosts, 3d rotations and 4d translations.

As you've written it, the unit of ds is length, but if you divide through by c², the unit is time.

See this thread https://www.physicsforums.com/showthread.php?t=666591

 

[Chestermiller]

so i am hoping that all of you are familiar with the fact that a basic way of finding the distance between two points in 3d space is a^2+b^2+z^2=c^2. anyway i wanted to know in the equation ds^s=dx^2+dz^2+dy^2-(cdt)^2 does this show the distance between two points in in 3-d space including 1 time Dimension or does it show something different. also is this is the distance between 2 point then is there a unit of measurement of is the distance really something like -3.45E17.

If you are talking about calculating the distance between 2 points in 3D space using the 4D spacetime equation, then you are implicitly assuming that the two points lie within the same rest frame, such that dt = 0.

 

[WannabeNewton]

The term distance should be used carefully here. The line element (called first fundamental form in classical differential geometry) measures the arc length of a segment between two points on the curve; in the way you have written it we have an infinitesimal arc length of neighboring points on a curve. This is not the same thing as the distance between two points in 3 - space for which we use a metric. The line element involves the metric tensor.

 

[zepp0814]

thanks

 

[robphy]

This

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

is also called the proper-interval, and for points (4D points) that can be joined by a light-beam the proper-distance is negative ( using your metric equation).

ds^2 = ...

I think you mean "joined by a _timelike_ curve"... "the [square-]interval is negative".
When joined by a light-beam, the square-interval is zero.

It is important because we identify it with the time ticked on a clock while traveling along a curve.

..."[minus c^2 multiplied by the square of the] time ticked on a clock while traveling along an _inertial_ curve" between two nearby events.

 

[Mentz114]

ds^2 = ...

I think you mean "joined by a _timelike_ curve"... "the [square-]interval is negative".
When joined by a light-beam, the square-interval is zero.

..."[minus c^2 multiplied by the square of the] time ticked on a clock while traveling along an _inertial_ curve" between two nearby events.

Yes, I expressed my self very badly. I meant to say in the light-cone. Sigh. My apologies to the OP for making such a blue of my answer.