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I have been reading a few books lately and browsing the internet for an explanation on the concept of "four-dimensional" space (or, space-time, as some of them have been calling it). But, what those books and sources did majorly was confuse me.

In some places, I see the author calling it "four dimensional space-time", where the fourth dimension is Time, and at other places, another person says, "Time is not any dimension at all. The fourth dimension is a dimension perpendicular to the cube". About the tesseract/Hypercube etc. I tried reading about the Hypercube and I was almost convinced that Time is not really a dimension when a few other books I read (on Einstein's General Relativity) started calling it the "curved 4-Dimensional space-time", to further confuse me.

Would be glad if anybody could explain these concepts and the difference, if any, between them. Or could anyone recommend any books or articles on them? Thanks in advance.

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  • $\begingroup$ The term "four-dimensional spacetime" means that spacetime is locally homeomorphic to $\mathbb{R}^4$ where $\mathbb{R}^n$ is the standard $n$-tuple of real numbers. $\endgroup$ – Ryan Unger Dec 19 '15 at 9:46
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    $\begingroup$ @0celo7 True, but utterly useless for someone trying to get a intuitive understanding. (And very helpful, compact notation for someone who already has an intuitive understanding). $\endgroup$ – Sebastian Riese Dec 19 '15 at 17:30
  • $\begingroup$ You know, coordinates are personal names(usually real nos.) that you assign to events. Now, it turned out, that the name 'time', that we all agreed upon wasn't really something we should have agreed upon according to einstein. Hence, we started calling time coordinate by different name too, someone says this event happened at 2s, someone said 1s, even those who agreed upon starting their clocks together. So, time turned out to be a relative coordinate. Hence, we included it in the 4-d spacetime as a whole to signify their names are all relative and (1/2) $\endgroup$ – Viesr Dec 20 '15 at 10:06
  • $\begingroup$ it is the space-time structure as a whole that is fundamental, just like space was in gallilean days.(2/2) $\endgroup$ – Viesr Dec 20 '15 at 10:06
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One can mathematically assign a vector dimension to the three space coordinates. Vector spaces have strict mathematical consequences.For euclidean vectors, which are the ones used in classical mechanics, the "length" of the vector $V = (x,y,z)$ is equal to $\left|l\right|= \sqrt{x^2+y^2+z^2}$ and characterizes it uniquely.

One can hypothesize a four vector in space, i.e. that there exists a fourth space dimension which we cannot see similar to ants walking on a two dimensional plane who cannot see the third dimension. One could have a four dimensional "cube" then which is your tesseract, which can be projected to our three dimensions.

These are euclidean fourvectors $V = (x, y, z, u)$ obeying the length that characterizes vectors having an extra variable under the square root sign: $\left|l\right| = \sqrt{x^2+y^2+z^2+u^2}$.

The space time of the physics of special relativity is different. It is not a euclidean vector space, Where the forth dimension, time enters:

Define the event to have spacetime coordinates $(t,x,y,z)$ in system $S$ and $(t′,x′,y′,z′)$ in $S′$. Then the Lorentz transformation specifies that these coordinates are related in the following way:

relativity

where

gamma

and $c$ is the velocity of light, and $v$ the relative velocity of $S$ to $S'$.

The "length" of this four "vector" is not equal to the square root of the squared difference of the variables for $S$ and $S'$. (it is how one ends up with three space length contraction in special relativity)

It is simpler to think of the four vector of energy and momentum. Here momentum is a euclidean three vector, but the length of the four vector is (setting $c=1$) $$m= \sqrt{ E^2 - \vec p \cdot \vec p},$$ where p is a normal euclidean three vector. So it is not the usual dot product of a euclidean four dimensional space, but of a pseudoeuclidean one.

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  • $\begingroup$ Nomenclature nitpicking: The Minkowski space is a vector space in the mathematical sense (this basically just means that it is closed under addition and multiplication with a scalar and those operations have some nice properties). It is not a scalar product space in the mathematical sense, because it only has a (not positive definite) pseudo-scalar product. (And even a scalar product space can have a scalar product $g$ which is given by a non-diagonal matrix $g(X, Y) = g_{ij} X^i Y^j$. Of course you can always choose a basis where $g_{ij} = \delta_{ij}$.) +1 anyway. $\endgroup$ – Sebastian Riese Dec 19 '15 at 17:23
  • $\begingroup$ @SebastianRiese would qualifying in the last sentence with "euclidean vector space" make it better? $\endgroup$ – anna v Dec 19 '15 at 17:37
  • $\begingroup$ The last sentence of the post? That would make it worse (as Minkowski space is, indeed, a pseudo-euclidean vector space, not a euclidean vector space). Qualifying the "vector space" in the second sentence of the post with "Euclidean", however, would make it a lot better (as Euclidean is usually used to mean a $\mathbb R^n$ with the standard scalar product and then the following statements are absolutely correct). $\endgroup$ – Sebastian Riese Dec 19 '15 at 17:55
  • $\begingroup$ @SebastianRiese I meant as above $\endgroup$ – anna v Dec 19 '15 at 18:33
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The concept of a "dimension" is a mathematical term used to denote how many coordinates a given system we are considering has. For example, a point has no dimensions. A line is 1 dimensional. A sheet of paper is 2 dimensional. Physical objects are 3 dimensional. These are intuitive spatial concepts of dimensionality.

When physicists talk about 4-dimensional spacetime, they are not using dimension in this spatial intuitive sense. They are using it as a term in relation to what we call a vector space.

A vector space is a mathematical concept and is anything that has these properties. An example of a vector would be (1,5,7). This vector lives in a 3-dimensional world because it has 3 coordinates.

When physicists say spacetime has 4 dimensions, they just mean it has 4 coordinates. That is what the fourth dimension is, an extra coordinate $t$. So (1,8,7,15) could be a spacetime vector. Spacetime points are usually denoted as $(t,x,y,z)$

I would consult Wheeler and Taylor's Spacetime Physics as a starting point.

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