
#1
Jan2413, 06:21 PM

P: 28

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 3d 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.




#2
Jan2413, 06:49 PM

PF Gold
P: 4,081

This
ds^s=dx^2+dz^2+dy^2(cdt)^2 is also called the properinterval, and for points (4D points) that can be joined by a lightbeam the properdistance 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 c^{2}, the unit is time. See this thread http://www.physicsforums.com/showthread.php?t=666591 



#3
Jan2413, 07:59 PM

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#4
Jan2413, 08:04 PM

C. Spirit
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P: 4,916

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.




#5
Jan2413, 09:11 PM

P: 28

thanks




#6
Jan2413, 09:33 PM

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



#7
Jan2513, 10:40 AM

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P: 4,081




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