¿can someone explain me the phisical meaning of space-time distance?

Question

you know, from the theory of relativity we obtain:

(dx)^2+(dy)^2+(dz)^2+(c*dt*i)^2=(ds)^2

where dx, dy and dz are distances measured on their respective axes x, y and z.

i is an imaginary number

and c*dt*i is a temporary dimension

and applying the theorem of pythagoras to each coordinate you obtain a "space-time" distance.

I understand that, what I don't understand is the phisical meaning of a "space time distance"

¿what is exactly a space-time distance?

sorry if you find a mistake in the text, I don't speak english very well, I speak spanish.

Answer 1

If you take a point in spacetime, you have something which exists in 0D (zero dimentions).
If you extend this point from itself orthogonally, you have a line which exists in 1D (one dimention).
If you extend this line from itself orthogonally, you have an area which exists in 2D (two dimentions).
If you extend this area from itself orthogonally, you have a volume which exists in 3D (three dimentions).
If you extend this volume from itself orthogonally, you have a *shape* which exists in 4D (four dimentions).

As you rightly point out, the first three dimentions of space are usually labelled (in cartesian co-ordinates) x, y and z. To maintain constant units, the temporal fourth dimention is usually stated as being ct (the units of ct being that of space).

In this way, non-euclidean geometry can be done in 4D (or even higher dimentions if you so wish).

You can do vector analysis in 3D easy enough, right? Well it is near-enough the same in 4D as long as, rather than labelling the fourth dimention just t, you label it ct.

In this way, pythagoras theorem can be applied and, taking the projection of the vector on the ct axis, the time displacement can be calculated (i.e. the time from point A to point B on the vector).

Note: The vector can not have a gradient steeper than the 'light line' if it does represent displacement as this would violate the theory of special relativity. Also, this sort of analysis can be done in euclidean geometry too, but doesn't show too much because spacetime is not flat, it's curved.

Answer 2

Because it has no extension, time cannot be a dimension. If a distance lacked the qualification time does, the road from one city to another would only exist as you drive on it, and only at the spot you were driving. So Einstein's Nobel Prize would have been as worthless as Yassir Arafat's if he had gotten it for this theory.

The Fourth Spatial Dimension was proposed a few decades before Einstein, but had degenerated into Hollywood-type fantasies by his time, so he was embarrassed to propose it. The extension of 3D into 4D is dependent on the excitation of energy, Zero extension is actually an interface--something on the box affects the box, even though it is not inside the box. The reason why dishonest physicists normalize infinities is that they don't understand the factors that should be in the equation.

Answer 3

Let me explain the utility of the spacetime interval in another way. It's frequently pointed out that light travels the path of shortest distance or shortest time. For example, in going from A to B, bouncing off a lot of mirrors inbetween, it takes the path of least time. However, if we extend this principle to mass particles, it takes a more complicated form, which is that it will "take the path of least action". Action is a mathematical concept that's hard for most to understand, but that's the way it comes out of Lagrangian equations of motion, and it's a very valuable tool. But when we generalize the Lagrangian equations to include relativistic effects, the principle is restated in this form, "the path of shortest spacetime interval". They are mathematically equivalent. We've regained the simplicity of the original statement of "path of least time", but now we have something that works for mass particles. This power of generalization was made possible by putting time on equal footing with space, so that relativists think of geometry in Minkowski space, which is a 4D space that has x, y, z, and t, rather than a 3D space which is changing in time. The path of any particle in Minkowski space is a geodestic, which is the shortest distance between any two points, just like how commericial jets travel from city A to city B along the great circle, which is a geodestic on the spherical earth.

Answer 4

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Answer 5

modern physics describes reality (the universe) as existing in four dimensions: space (x, y, and z) and time (t or ct). since time is one of the dimensions of the universe, it is referred to as "space time". when calculating the distance between two points in a four-dimensional space, taking into account only three dimensions is not sufficient. the "metric" of a four dimensional space must take into account all four dimensions. some thing may exist in the same x, y, z location as another but at a different time. although these two things may have three coordinates in common, there is a distance between them in the fourth dimension (time).

Answer 6

*Hick-up* What the heck was the question? And why are you typing upside down question marks? Your computer Spanish too?