4th DIMENSION EQUATIONS CALCULUS HELP

[bazucajoe10]

4th DIMENSION EQUATIONS! CALCULUS HELP

I have to explain the 4th dimension and hyper-space using for my High School calculus AP final project. IF someone can give me some insight or some equations to help me see and learn more about the 4th dimension it will be greatly appreciated.

Some stuff i know
*(2x+1)^4 gives you the dimentions of a Hyper-Cube.
* ds^2=dx1^2 + dx2^2 + dx3^2 + dx4^2 for a point in 4D space

P.S. what does time have to do with the 4th dimension?

 

[Alkatran]

Time is considered to be the 4th dimension. But you should be considering the 4th dimension as an extension to the gemometrical ones we observe more easily...

Here is an excellent site for info on the 4th dimension. (though it may not be what you're looking for).

 

[robphy]

Visit some of the pages describing the work done by Thomas Banchoff
http://www.geom.uiuc.edu/~banchoff/projects.html

Here is Abbott's Flatland
http://www.geom.uiuc.edu/~banchoff/Flatland/

Another resource:
http://www.geom.uiuc.edu/docs/outreach/4-cube/

 

[jcsd]

I have to explain the 4th dimension and hyper-space using for my High School calculus AP final project. IF someone can give me some insight or some equations to help me see and learn more about the 4th dimension it will be greatly appreciated.

Some stuff i know
*(2x+1)^4 gives you the dimentions of a Hyper-Cube.
* ds^2=dx1^2 + dx2^2 + dx3^2 + dx4^2 for a point in 4D space

P.S. what does time have to do with the 4th dimension?

Wasn't the second equation one that I gave you? Here's an explanation of it and how it relates to time (Actually I'm pretty sure that I've posted a very simliar explanation a while ago):


Pythagoras's theorum states that for a right-angled triangle of legs a and b and hypotneuse of c:

$$c^2 = a^2 + b^2$$

Using Pythagoras's theorum we can find the length of the shortest line (ds)between two points on a piec of graph paper (x₁, y₁) and (x₂,y₂)

$$ds^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$$

Now it should be clear that wherever we decide to put the origin and whichever directions we decide for the x and y-axis (with the obvious conditon thta they are orthogonal) the length of the shortest libne between the two points will always be the same. If we call the 'change' in the x direction 'dx' and the 'change' in the y direction 'dy' we can say that:

$$ds^2 = dx^2 + dy^2$$

Now let's say we want to find the distance between two points in 3 dimensions:

by Pythagoras's theorum we can detrmine that (I don't too much about the excat level of 'AP Physics' not being from the US but I imagine that this would be familair to you as finding the length of a diagonal on a cuboid):

$$ds^2 = (\sqrt{dx^2 + dy^2})^2 + dz^2 = dx^2 + dy^2 + dz^2$$

Unsuprisingly if we go into 4 dimensions (x₁, x₂, x₃ and x₄) we find that the distance between two points is given by:

$$ds^2 = {dx_1}^2 + {dx_2}^2 + {dx_3}^2 + {dx_4}^2$$

So that is the significance of that equation, a word of warning though the equation is only true in Euclidian spaces (that is 'flat' spaces).

In my next post I'll relate 4 dimensions to time.

 

[jcsd]

As I mentioned in my last post, the quantity 'ds' is unchanged by basic trnaformations such as rotation and moving the origin and as said this is failry obvious as the diatnce between two points doens't change if you rotate your viewpoint or move. But as I'm sure you know in special relativity there is something called 'length contarction described by the equations which make up a tranformation called a 'Lorentz transformation'. This transformation describes how objects of differing relative speeds see things, they are:

$$x' = \gamma(x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma(t - \frac{vx}{c^2})$$
where
$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
(note: that v is in the x direction, that's the reason that x changes why the other stay the same under the transformation))
If we apply this transformation to our 3-dimensional equation we find that:

$$ds'^2 = \gamma^2(dx - vt)^2 + y^2 + z^2$$

Therefore

$$ds = ds'$$

if and only if v = 0, so the distance between two points (note innthos example the two point's rest frame is in the unprimed rest frame) is no longer constant as it was before, so our equation no longer does a good job of describing space as it is frame dependent. So how can we define soemthing that is absolute like ds was until we brought in the Lorentz transformation?

You may of noticed that we left out one part of the Lorentz transformation, the 'time part'. As it happens if we include the time part we can find a quantity that isn't frame dependent.

If we now define 'ds' as the 'spacetime interval' given by:

$$ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2$$

When we apply a Lorentz tranformation we find that:

$$ds = ds'$$

no matter what 'v' is.

So that is why we use the conspt of four-dimensional spacetime in special relativity as it is inavariant under a Lorentz transformation (I did prove that it was Lorentz invariant last time, but I'll leave that to you), .i.e it will be the same whatever the speeds of the observers.

One thing to note the equation for our inetrval isn't the same as our equatiuon for the distance between two points in four-dimensions, we can ignore the factor of c^2, but as you can see that the temporal dimension has a different sign to the spatial dimensions (whether it is plus or minus is a matter of convention).

 

[bazucajoe10]

Thank You

thanks so much for your help. Your explanation was exactly what i needed. you just saved my ass from failing my project. thank you so much

 

[Alkatran]

I can't believe I didn't realize you were talking about distances.