# Line element in Euclidean Space

1. Jun 27, 2015

### Tony Stark

The line element is defined as

How is dx2+dy2+dz2 be written as gijdqidqj.
Is some sort of notation used??

2. Jun 27, 2015

### Mentz114

3. Jun 27, 2015

### Noctisdark

The metric tensor gij is defined as gij = Ei*Ej, you can see that in Euclidean (flat) space that gij is to 0 whenever i is not equal to j but gij = 1 when i=j, now you may want to simplify the whole equation!

4. Jun 27, 2015

### Orodruin

Staff Emeritus
This is only true in a Cartesian coordinate system. There are several possible coordinate systems on Euclidean space which are neither orthogonal nor normalised. Generally, the metric tensor defines the inner product instead of the other way around.

5. Jun 28, 2015

### Tony Stark

6. Jun 28, 2015

### Mentz114

Like this
$dx^2+dy^2+dz^2=g_{ij}dx^i dx^j$

Remember that $x^1\equiv x, x^2\equiv y, x^3 \equiv z$. $i,j$ are spatial indexes.

7. Jun 29, 2015

### Tony Stark

Thanks Mentz. Is there an online link which has elaborate description of Einstein's Notation?? Please mention

8. Jun 29, 2015

### Mentz114

That's all I got. You'll need to understand tensor notation. It is well explained in lots of books and online articles and courses.

9. Jun 29, 2015

### pervect

Staff Emeritus
What's wrong with the wiki link Mentz gave? Was it unclear? Too technical?

Basically, the Einstein convention is that you sum over repeated indices. In your example, $g_{ij} dx^i dx^j$ i and j are repeated, so you sum over i,j having all possible values. The Wiki link explains this - or tries to, so have several posters.

There is some ambiguity - what are all possible values? There are conventions for that too. But while textbooks will generally explain their conventions and follow them rigorously, you won't always see all posters here (including me) follow suit, leaving it up to the reader to sort out the minor ambiguities.

Wiki says:
So using the above convention $g_{ij} dx^i dx^j$ in cartesian coordinates, where by modern convention $t= x^0 \quad x = x^1 \quad y = x^2 \quad z=x^3$ would be just dx^2 + dy^2 + dz^2, while $g_{\mu\nu} dx^\mu dx^\nu$ would be the space-time interval. Depending on more sign conventions, the space-time interval might be interpreted either as dx^2 + dy^2 + dz^2 - dt^2 , or possibly dt^2 - dx^2 - dy^2 - dz^2.

In cylindrical coordinates, we might have coordinates $r, \phi, z$ rather than x,y,z. If we identify $x^1=r \quad x^2 = \phi \quad x^3=z$ (the exact assignment here may vary even more between writers) then we would still write $g_{ij} dx^i dx^j$, but this corresponds to $dr^2 + r^2d\phi^2 + dz^2$, which you will hopefully recognize as the formula for the line element for distance in cylindrical coordinates. So in cylindrical coordinates with the given assignments $g_{11}=1 \quad g_{22}=r^2 \quad g_{33}=1$. The result, though, gives you the distance in whatever coordinate system you use.

There are a lot of ambiguities in the notation, either a fuller explanation is given (in a well-written textbook), or the reader is expected to fill in the details from context.