# Lorentz transformation as a higher dimensional rotation

1. May 3, 2010

### closet mathemetician

In the section entitled "Invariance of Length: The Euclidean Picture" the article discusses how rotations within an n-dimensional space keep length invariant. However, if you rotate and object into a higher, n+1 dimension, then the length, as defined in n dimensions, is NOT invariant under that rotation.

So I started thinking about this. Consider a two-dimensional spacetime, in which the graph of a reference frame would look like a two-dimensional plane. Now suppose that we rotate the "moving" reference frame about its diagonal axis into the third dimension.

Could this rotation, when viewed from the perspective of the original inertial reference frame, produce a coordinate transformation that is equivalent to the results obtained by performing a Lorentz transformation?

Also, if the axis of rotation of the plane were the diagonal, lengths along the diagonal would not change under the rotation, but lengths on any other portion of the plane would. Could this be why the speed of light is invariant, because the diagonal axis length of both frames remains the same under the rotation?

I'd like to come up with a rotation matrix to try to prove this mathematically, but I'm having trouble on how to proceed.

Last edited by a moderator: May 4, 2017
2. May 4, 2010

### Ich

No. Lorentz boosts are hyperbolic rotations in that plane, they behave differently.

3. May 4, 2010

### Rasalhague

Rotating an n-dimensional Euclidean space, En, in some higher dimensional embedding space doesn't have any effect on the intrinsic geometry of your original En. Length is invariant, as are all other intrinsic properties. It's only your way of representing it that's changed. The difference between circular rotation (the kind we're familiar with) and hyperbolic rotation (what you're thinking of as a Lorentz transformation) isn't due to one of them taking place in a higher dimensional space; for a given n-dimensional space, you can have circular rotations and hyperbolic rotations.

If the original reference frame is 2-d, embedding that 2-d space in a higher dimensional space and rotating it in the higher dimensional space will mean nothing to the original 2-d space.

Let's stick to a single number of dimensions, say four. Lorentz transformations include circular rotations, which mix two spatial coordinates, e.g.

$$\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & -\sin \theta & \cos \theta & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

and hyperbolic rotations, also called boosts, which mix the time coordinate with one spatial coordinate, e.g.

$$\begin{bmatrix} \cosh \phi & -\sinh \phi & 0 & 0\\ -\sinh \phi & \cosh \phi & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

where phi, the argument of the hyperbolic functions in the latter matrix, is the rapidity: http://en.wikipedia.org/wiki/Rapidity

What these matrices have in common is that they preserve the spacetime interval, which is the spacetime analogue of length in Euclidean space:

ds2=-dt2+(dx2+dy2+dz2)/c2.

Last edited by a moderator: May 4, 2017
4. May 4, 2010

### Phrak

If I understand you, a vector rotated out of the 2 dimensional plane, say, will have a projected length into the plane of a different than the length of the three dimensional vector. You might keep in mind that all normal projections of the vector will be less than or equal to the rotated vector in the Euclidean space you seem to have in mind.

Despite Lorentz transforms being referred to as rotations they are a combination of stretches (and compressions) and shear transformations of the coordinates basis in (x,y,z,t). With a proper choice of coordinate axes the transformation of the coordinate system is purely stretches and compressions. These transformations are called scaling.

Anyway, the length of some vectors will be stretched under a Lorentz transform. But rotating the vector into a fifth dimension and then projecting it back into spacetime will only produce a shorter vector.

However, there is always some clever way to get around any obstacle such as above, but would probably require something much more involved compared to a relatively simpler Lorentz transformation of spacetime. Replacing vectors with their duals and assuming your fifth dimension is non orthogonal will get you longer projected vectors, but it probably will not all hang together properly.

We could drop all this coordinate business, and examine affine constructions only, where lengths and time intervals only get shorter...

Last edited by a moderator: May 4, 2017
5. May 4, 2010

### Ich

What are you talking about? Boosts are hyperbolic rotations, and they leave the length of vectors unaffected.

6. May 4, 2010

### Phrak

OK, I see the problem. We can talk about the drawn length of a vector as I have done, or the norm under the metric.

Last edited: May 4, 2010
7. May 4, 2010

### closet mathemetician

I understand that the rotation into three dimensions leaves lengths on the plane intact. The point is that when this rotation is viewed from a 2-dimensional perspective, lengths become skewed. This is due to perspective.

Lengths defined using two dimensions are invariant under two-dimensional rotations. Lengths defined using two dimensions are NOT invariant under three-dimensional rotations.

8. May 4, 2010

### DrGreg

For the benefit of other readers of this thread who don't seem to have grasped the point of this question, I believe you are proposing that the Lorentz transformation, in 2 dimensions, might be viewed as the 2D projection of an object undergoing a Euclidean rotation in 3 dimensions.

Well, no, that doesn't work. If you check the maths, you will find that the (Euclidean) "length of the diagonal" actually does change under Lorentz transformation, although the direction of the diagonal does not. In 2D, the two diagonals are proportional to (xct) and (x + ct), and one of them gets bigger and the other gets smaller. (In fact the rescale factors are the doppler red- and blue- shift factors, which are reciprocal.)

As has been mentioned by others, the reason that Lorentz transformations are referred to as "rotations" has nothing to do with any 3rd dimension. A rotation is something that leaves "length" unchanged. In spacetime, we don't use Euclidean length given by Pythagoras by s2 = x2 + y2. Instead, we use the "spacetime interval" given by s2 = x2c2t2. The Lorentz transform leaves the spacetime interval unchanged, so by analogy we describe it as a "rotation" (or, better, as an isometry). Rasalhague's post shows the similarity between Euclidean rotations described via trig functions and Lorentz transforms that can be described using hyperbolic functions.

Last edited by a moderator: May 4, 2017
9. May 4, 2010

### Rasalhague

Yeah, sorry, closet, I didn't realise you were talking about projections.

10. May 5, 2010

### closet mathemetician

Yes, DrGreg, that is exactly the question. And no problem, Rasalhague, I probably didn't phrase the question well enough.

I need to think about the diagonals changing length.