Basically how do we know that since it is not possible to see the 4th dimension, is it for simplicity ?
That's true, but there is more to it than that.It is orthogonal because you can fix your position in space and measure time independently.
Similarly x is orthogonal to y and z because you can fix your position in y and z and measure x independently.
Yes, just as you can define coordinate systems in which x and y are not orthogonal. The precise meaning of that must refer to the math.Just to add to DrGreg's post, note that you can define coordinate systems in which the timelike and spacelike directions are not orthogonal. It just makes the maths nasty and hard to interpret and, consequently, rather error prone.
In any given inertial frame, ##I_1##, the measured position in space and time are independent of each other. An external inertial frame, ##I_2##, might see a connection between ##I_1##'s space and time, but that is irrelevant to the issue of orthogonality of ##I_1##'s space and time.But SR treats moving frames then space and time are linked ?
The speed and the angle are essentially the same thing. (Technically the speed is the slope, but it is not much effort to convert a slope to an angle and each uniquely determines the other)What I mean is that the transform depends on v the speed and α the angle of the time axis.
Suppose one has a flat plane. One can parameterize it with coordinates so that x and y are orthogonal, or one could parameterize it so that they're not orthogonal.I thought the inclination of the time axis could be seen as an extra parameter giving a slightly different Lorentz transform. So that any discrepancy in an accelerator experiment for example could be explained by that ?
I'd like to give you a slightly different answer. They are orthogonal by definition of the inner product. The same is true of the x and y axes. They are only orthogonal if you define the angle between them to be a right angle.
This can't be right, because there are curved spacetimes that have a metric tensor which is everywhere diagonal (the best known example is Schwarzschild spacetime), which means that the four coordinate axes are orthogonal at every point.That the x, y, z and t axes are orthogonal is essentially the definition of "flat" spacetime.
In curved spacetime the timelike direction is also orthogonal to spacelike directions. We can see this in the tetrad formalism. https://en.wikipedia.org/wiki/Frame_fields_in_general_relativityI'd like to give you a slightly different answer. They are orthogonal by definition of the inner product. The same is true of the x and y axes. They are only orthogonal if you define the angle between them to be a right angle.
That the x, y, z and t axes are orthogonal is essentially the definition of "flat" spacetime.
That definition is then compatible with experimental results.
You could postulate any other geometry for gravity-free spacetime, but that wouldn't be compatible with experiment.
You could define a different inner product where the x, g and t axes are not orthogonal. That would, in general, give you curved spacetime.
More precisely, in curved spacetime we can always pick, at a particular point, a set of one timelike direction (vector) and three spacelike directions (vectors) that are all mutually orthogonal; this is called a tetrad and a mapping of tetrads to points in a spacetime is called a frame field. But in a general curved spacetime there does not need to be any single global coordinate chart that matches up with a frame field (in the sense that at every point the vectors in the tetrad mapped to that point by the frame field also point in the same directions as the coordinate basis vectors at that point). The example I gave (Schwarzschild spacetime) is a curved spacetime in which there is a single global coordinate chart that matches up with a frame field.In curved spacetime the timelike direction is also orthogonal to spacelike directions. We can see this in the tetrad formalism.
This orthogonality also holds in Galilean spacetime.In other words,From Minkowski's "Space and Time"... [bolding by me]
(e.g. http://www.minkowskiinstitute.org/mip/MinkowskiFreemiumMIP2012.pdf , p47)
We decompose any vector, such as that from [itex] O[/itex] to [itex] x, y, z, t[/itex] into four components [itex]x, y, z, t [/itex]. If the directions of two vectors are, respectively, that of a radius vector [itex]OR[/itex] from [itex]O[/itex] to one of the surfaces [itex]\mp F = 1[/itex], and that of a tangent [itex]RS [/itex] at the point [itex] R[/itex] on the same surface, the vectors are called normal to each other. Accordingly, [tex]𝑐^2𝑡𝑡_1−𝑥𝑥_1−𝑦𝑦_1−𝑧𝑧_1=0[/tex] is the condition for the vectors with components [itex]x, y, z, t [/itex] and [itex] x_1, y_1, z_1, t_1[/itex] to be normal to each other.
locate the intersection of an observer's 4-velocity with the unit-hyperbola (the Minkowksi circle) centered at the tail of the observer's 4-velocity.
The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity. That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.
The "intuition" to have is that
the tangent to the "circle" in that geometry
orthogonal to the radius vector.