space time intervals


by zepp0814
Tags: intervals, space, time
zepp0814
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#1
Jan24-13, 06:21 PM
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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.
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Mentz114
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Jan24-13, 06:49 PM
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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 travelling 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 http://www.physicsforums.com/showthread.php?t=666591
Chestermiller
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Jan24-13, 07:59 PM
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Quote Quote by zepp0814 View Post
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
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Jan24-13, 08:04 PM
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space time intervals


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
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Jan24-13, 09:11 PM
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Jan24-13, 09:33 PM
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Quote Quote by Mentz114 View Post
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 travelling 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
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Jan25-13, 10:40 AM
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Quote Quote by robphy View Post
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.


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