Building a 4 dimensional spatial model (forgetting time for now)

[geordief]

The space that we are used to is 3D and we construct it  (the model) by ensuring  that the 3 planes (x=0:y=0:z=0) are orthogonal to each other.

 

How do we move on from there if we want to build a model with 4 such planes ,all presumably orthogonal to each other in the same way?

 

To my untrained eye it seems    that the new( 4th) plane will   have "nowhere to go"

ie ,if it is orthogonal to the  x=0 plane then  it feels like it should be identical to the  y=0 or the z=o plane which already occupy those areas("area" is a poor choice of word ,but I hope the meaning is clear)

 

How do we "shoehorn " this extra plane into the model when it feel like there is no space available.

 

PS I am obviously thinking about the spacetime model  but the difficulty I have seems to apply to any 4d  model with 4 spatial axes (which is what the spacetime model can be viewed as sine "ct"  is a spatial distance) 

[swansont]

Mathematically it's not a problem; you just won't be able to visualize it. You define the next dimension so that the dot products of a unit vector with a unit vector of the other dimensions is zero. (probably need a cross product definition as well)

Math has many example where functions or basis vectors are orthogonal, and have more than three dimensions. Some are infinite. A Hilbert space is an example, commonly found in physics.

[geordief]

19 minutes ago, swansont said:

Mathematically it's not a problem; you just won't be able to visualize it. You define the next dimension so that the dot products of a unit vector with a unit vector of the other dimensions is zero. (probably need a cross product definition as well)

Math has many example where functions or basis vectors are orthogonal, and have more than three dimensions. Some are infinite. A Hilbert space is an example, commonly found in physics.

So ,if we start with any unit vector there are an infinite number of ways that one can define a new unit vector so that its dot product with the  original vector is zero?

 

Is this dot product a more general  requirement for a new spatial dimension than the right angle requirement which I would be more used to?

 

 

 

 

[swansont]

4 hours ago, geordief said:

So ,if we start with any unit vector there are an infinite number of ways that one can define a new unit vector so that its dot product with the  original vector is zero?

 

Is this dot product a more general  requirement for a new spatial dimension than the right angle requirement which I would be more used to?

They're basically the same thing. The dot product is zero when the vectors are orthogonal.

[geordief]

1 hour ago, swansont said:

They're basically the same thing. The dot product is zero when the vectors are orthogonal.

Is it fairly easy to show  when two vectors in a space with n degrees of freedom  are orthogonal?

Suppose the model had 10  dimensions would there be 2^10 orthogonal axes?

 

 

[swansont]

4 hours ago, geordief said:

Is it fairly easy to show  when two vectors in a space with n degrees of freedom  are orthogonal?

You take the dot product. If it's zero, they are orthogonal. 

4 hours ago, geordief said:

Suppose the model had 10  dimensions would there be 2^10 orthogonal axes?

AFAIK 10 dimensions means 10 axes. 

[geordief]

6 minutes ago, swansont said:

 

AFAIK 10 dimensions means 10 axes. 

My confusion .I was (double) counting the positive and negative axes separately.