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

• zepp0814
In summary, the distance between two points in 3-dimensional space is measured in terms of the proper-interval. This interval is invariant under the Poincare group of transformations, which includes 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.
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.

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.

zepp0814 said:
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.

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.

thanks

Mentz114 said:
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.

robphy said:
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.

Last edited:

1. What is the formula for finding the distance between two points in 3D space?

The formula for finding the distance between two points, (x1, y1, z1) and (x2, y2, z2), in 3D space is: Distance = √((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

2. How do you incorporate time dimension into finding the distance between two points in 3D space?

To incorporate time dimension, also known as the fourth dimension, into finding the distance between two points in 3D space, you can use the formula: Distance = √((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 + (t2-t1)^2). This takes into account the time difference between the two points in addition to the spatial dimensions.

3. Can the Pythagorean theorem be applied to finding the distance in 3D space?

Yes, the Pythagorean theorem can be applied to finding the distance in 3D space. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This can be extended to 3D space by taking the square root of the sum of the squares of the differences in each dimension as shown in the formula in question 1.

4. What units are typically used to measure distance in 3D space?

The units used to measure distance in 3D space can vary depending on the context and application. Common units include meters, feet, kilometers, and miles. In scientific research, the units may be more specialized, such as nanometers or astronomical units.

5. Can the formula for finding the distance between two points in 3D space be extended to more than four dimensions?

Yes, the formula for finding the distance between two points in 3D space can be extended to more than four dimensions. It follows the same pattern of taking the square root of the sum of the squares of the differences in each dimension. For example, in 5D space, the formula would be: Distance = √((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 + (t2-t1)^2 + (u2-u1)^2).

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