# How to describe the 4th dimension?

#### Rico L

what is 4th dimension? is it just 1 dimensional space? how come time is described as the 4th dimensional path?

thanks..

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#### Dale

Mentor
The 4th dimension is time.

#### Rico L

how to describe it? in a mathematical way plzz... :)

#### Nabeshin

Depends on how you define the 4th. In relativity we say fourth simply because this theory (in its most basic version) only includes 3 spatial dimensions so the natural addendum is to put time as the 4th. In reality, we treat it more like its the zeroth dimension, which is probably more correct. To deal with this time mathematically, here is a snippit:

In flat 3-dimensional space, the infinitesimal distance between any two points is given (in cartesian coordinates) by:
$$ds^2=dx^2+dy^2+dz^2$$
Which you might recognize as simply the infinitesimal form of the familiar Pythagorean theorem. In a flat 4-dimensional spacetime, the new expression is:
$$ds^2=-dt^2+dx^2+dy^2+dz^2$$
Note how this is not the simple Pythagorean theorem because of the sign change! (The negative on the time part is merely convention. What is important is the time component has a different sign from the spatial component) So time enters into the calculation of the distance between any two points, so that the resulting path is made up of the temporal and spatial displacements according to the above form.

#### Rico L

Depends on how you define the 4th. In relativity we say fourth simply because this theory (in its most basic version) only includes 3 spatial dimensions so the natural addendum is to put time as the 4th. In reality, we treat it more like its the zeroth dimension, which is probably more correct. To deal with this time mathematically, here is a snippit:

In flat 3-dimensional space, the infinitesimal distance between any two points is given (in cartesian coordinates) by:
$$ds^2=dx^2+dy^2+dz^2$$
Which you might recognize as simply the infinitesimal form of the familiar Pythagorean theorem. In a flat 4-dimensional spacetime, the new expression is:
$$ds^2=-dt^2+dx^2+dy^2+dz^2$$
Note how this is not the simple Pythagorean theorem because of the sign change! (The negative on the time part is merely convention. What is important is the time component has a different sign from the spatial component) So time enters into the calculation of the distance between any two points, so that the resulting path is made up of the temporal and spatial displacements according to the above form.

but why x, y,z and t has to be squared? i take this chance to ask another similar question as well.. why C has to be squared in E=MC^2 ... and when you calculate kinetic energy.. E=half MV^2...... why squared?

#### diazona

Homework Helper
As far as the velocities: for one thing, it makes the units work out. For another thing, those particular formulas can be successfully used to calculate experimental results. Why, did you think it should be something else?

The reason that the x, y, z, and t are squared in the line element equation Nabeshin posted is completely different. That has to do with the Pythagorean theorem. In 3D space, we observe that the formula
$$\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$$
allows us to calculate the distance between two points, and similarly in 4D space the formula
$$\Delta s^2 = -\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2$$
allows us to calculate the "interval" between two points, which is a useful quantity (it tells you whether a particle could get from one point to the other).

#### Tac-Tics

You just use R^4 instead of R^3.

#### Simon43254

$$ds^2=-dt^2+dx^2+dy^2+dz^2$$
Note how this is not the simple Pythagorean theorem because of the sign change! (The negative on the time part is merely convention. What is important is the time component has a different sign from the spatial component) So time enters into the calculation of the distance between any two points, so that the resulting path is made up of the temporal and spatial displacements according to the above form.

Sorry, I know you said it's convention but why? I mean I understand why in newtons law of gravity, there is a negative in front of GMm/r^2, but you haven't really explained why this convention and what the importance of having a negative in front of the dt^2 is.

#### Nabeshin

Ok one thing I forgot to mention in my first post that I will say now. I implicitly was using units where c=1. If you do not set c=1, the equation is:
$$ds^2=-c^2 dt^2 +dx^2 +dy^2+dz^2$$
Now you can see the units all match up (they're all units of length squared). Sorry if that confused anyone.

As for why the time component has an opposite sign, do you want a rigorous derivation? A hand-wavy argument is that we really want ds2 to be an invariant quantity. If the signs were all positive, then the line element would not be invariant. Perhaps you can come up with an example to convince yourself of this.

It's importance, though, cannot be understated. The fact that the sign is different shows us that this spacetime is NOT simply 4-dimensional euclidian space, where we would expect:
$$ds^2=dw^2+dx^2+dy^2+dz^2$$
Also, in the euclidian scheme the interval is always greater than or equal to zero. With minkowski space (the -dt^2 one has this name), however, we can have an interval which is positive, negative, or zero. Perhaps you have heard of this, as these intervals correspond to the three different classes of events. With the convention I am using (-,+,+,+) it goes like this:
ds2>0: This means the spatial component (dx2 +dy2 +dz2) is larger than the time component. What this physically means is that these two points in spacetime are causally disconnected. That is, one could not possibly have caused the other, as to do so a signal would have had to have traveled faster than the speed of light.
ds2=0: This is the world line for light, where the spatial component equals the time component. Objects with this interval are only causally connected for interactions with massless particles (which necessarily travel at the speed of light).
ds2<0: These are time-like world lines, and indicate causal events. People, footballs, and anything with mass will follow a world line like this.

So you see, a lot of structure comes out of that one simple minus sign!

Hopefully I have answered your question okay, sorry I can't think of a slicker way of explaining the origins of the sign difference though. Perhaps someone else can help me out there.

#### russ_watters

Mentor
i take this chance to ask another similar question as well.. why C has to be squared in E=MC^2 ... and when you calculate kinetic energy.. E=half MV^2...... why squared?
I'm not sure where you want to go with this line of questioning, but you are asking about something that is simply a word we use to describe a mathematically useful relationship. It's just a definition.

But for the relationship itself, work is force times distance and kinetic energy is easily derivable from work: Consider the work required to create some potential energy by lifting a rock, then converting it to kinetic energy by dropping it.

#### Simon43254

It took a couple of reads to understand till it made perfect sense. The only thing that I don't quite see the importance of is:

As for why the time component has an opposite sign, do you want a rigorous derivation? A hand-wavy argument is that we really want ds2 to be an invariant quantity.
Why is it important that ds2 be invariant? And also why the c2 is attached to the dt2. As effectively the product of this is a speed and a time, so hence a distance. Which does make sense in it's self. But why not just right this product as a distance?

Si

PS. Can you direct me to where you have all the nifty codes for writing squared and other signs? Much appreciated.

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#### Dale

Mentor
Why is it important that ds2 be invariant?
What we are looking for is something in spacetime geometry (aka Minkowski geometry) which is analogous to distance in Euclidean geometry. In Euclidean geometry distance is invariant under rotations and translations. The equivalent class of operations in spacetime is spatial translations, spatial rotations, time translations, and boosts (relative velocity). So we are looking for something which is invariant under those operations, this thing will be the spacetime analog to spatial distance.

#### Simon43254

Oh ok, that makes sense. Thanks for clearing that up Dale.

#### Nabeshin

And also why the c2 is attached to the dt2. As effectively the product of this is a speed and a time, so hence a distance. Which does make sense in it's self. But why not just right this product as a distance?
I'm not quite sure what you mean "why not just write this product as a distance", but the point is it makes no sense to add quantities with different units. When you actually derive the form of the interval you see the c2 there, but in relativity we generally use units where c=G=1, so I tend to forget about factors of c and G and insert them back so at the end. Can seem a bit haphazard to someone not familiar with using geometric units, so I apologize for that.
PS. Can you direct me to where you have all the nifty codes for writing squared and other signs? Much appreciated.

Cheers!

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#### Rico L

this is amazing!!

thanks a lot people! :P

#### Rico L

just 1 last question.. what does the s of ds^2 stands 4?

and the t stands 4 time ?
thanks

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#### Dale

Mentor
The s stands for arc length or distance.

oh.. cheers