Why all the different ways of mapping in Special Relativity?

[Rainer]

Hello everybody,

I've got a really abstruse and half-baked question for the experts. I'm not proud of my half-baked question, but after months of diligent wikipedia surfing and armchair pondering this is the best I could do.

So, Lorentz Transformations. I get them. They're great. But then you have metric tensor covariant transformations, I don't get them. Not only do I not get them, I don't get why Einstein needed another way to map four dimensional events ("four-vectors"?) from one rest frame to another if he already had the Lorentz Transformation at his service. Isn't it kind of redundant?

If someone can shed any light on that, that would be great. And my follow-up question would be:

Tensors of rank zero are said to be scalars. Fine that's easy. To map a vector from one rest frame to another I just multiply the matrix through by the scalar. But then tensors of rank 1, 2, 3, 4? What sort of operation does that translate into?

Thanks

[sol invictus]

So, Lorentz Transformations. I get them. They're great. But then you have metric tensor covariant transformations, I don't get them.


I'm not completely sure what you mean by "metric tensor covariant transformations".

Lorentz transformations are the special class of coordinate transforms that leave the Minkowski metric (ds^2=-c^2 dt^2+dx^2+dy^2+dz^2) invariant. Any other transformation (apart from rigid translations) will change the form of the metric. So I'm going to assume you mean coordinate transformations that do not leave the metric unchanged.


Not only do I not get them, I don't get why Einstein needed another way to map four dimensional events ("four-vectors"?) from one rest frame to another if he already had the Lorentz Transformation at his service. Isn't it kind of redundant?

Well, no, it's not redundant. For example if you want to express the laws of physics in spherical coordinates, or in a rotating or accelerating frame, you need something more general than Lorentz transforms.


Tensors of rank zero are said to be scalars. Fine that's easy. To map a vector from one rest frame to another I just multiply the matrix through by the scalar. But then tensors of rank 1, 2, 3, 4? What sort of operation does that translate into?


A vector is a rank 1 tensor. The rank of the tensor is (in physics lingo) the number of spacetime indices it has. So a rank n tensor has n indices and 4^n components. Each index transforms with one transformation matrix (and depending on whether the index is contravariant or covariant it's a slightly different matrix, but that's a detail).

Perhaps I've misunderstood you, and by "metric tensor covariant transformations" you just mean the matrix here (http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form)? If so, that's just the matrix you use to transform tensor indices, in the special case of a Lorentz transformation. It's not redundant with Lorentz transformations, it is a Lorentz transformation. Try multiplying it by a 4-vector of position.

[W.D.Clinger]

I've got a really abstruse and half-baked question for the experts.
I'm not an expert. I'm not even a physicist, so I'm not going to answer your main question. I'll answer your follow-up question, though:
And my follow-up question would be:

Tensors of rank zero are said to be scalars. Fine that's easy. To map a vector from one rest frame to another I just multiply the matrix through by the scalar. But then tensors of rank 1, 2, 3, 4? What sort of operation does that translate into?
I'm going to give you the mathematical answer first; physicists might give you a slightly different and more complicated answer (see below).

Tensors are multilinear forms: functions that map some number of vectors to some number of vectors. Because tensors are multilinear, they can be described by a structured set of numerical coefficients in some coordinate system. A rank zero tensor corresponds to a scalar. When written down, the coefficients of a rank 1 tensor are a sequence of n numbers (where n is the dimension of the vectors on which the tensor operates); the coefficients look like a vector but actually describe the dual of a vector. The coefficients of a rank 2 tensor form a square array of numbers. The coefficients of a rank 3 tensor form a cubic array of numbers. The coefficients of a rank 4 tensor form a 4-dimensional hypercube of numbers. And so on.

If you associate a tensor with every point of some space, you get a tensor field. In physics, the word "tensor" often means a tensor field, which is confusing until someone warns you about it.

ETA: If you feed enough vectors to a tensor, its output will be a scalar. The rank of a tensor is the number of vectors you have to feed it before its output will be a scalar. A rank 0 tensor is already a scalar. A rank 1 tensor operates on 1 vector by computing the inner product of its coefficients (regarded as a vector) with the components of the vector; that produces a scalar. A rank 2 tensor can operate on 2 vectors, in which case it produces a scalar, or it can operate on just 1 vector, in which case it produces a rank 1 tensor that can be fed the other vector to produce a scalar. And so on.

[kalen]

Hello everybody,

I've got a really abstruse and half-baked question for the experts. I'm not proud of my half-baked question, but after months of diligent wikipedia surfing and armchair pondering this is the best I could do.

So, Lorentz Transformations. I get them. They're great. But then you have metric tensor covariant transformations, I don't get them. Not only do I not get them, I don't get why Einstein needed another way to map four dimensional events ("four-vectors"?) from one rest frame to another if he already had the Lorentz Transformation at his service. Isn't it kind of redundant?



Thanks

I don't think that Einstein came up with "another way" to do Lorentz transformations. When you see the lamba_mu_nu notation this is just a more compact way of expressing a Lorentz transform. Apparently, he liked to write down very little and, in fact, the commonplace convention of summing over indicies with the same Greek symbol is named after him.

The Lorentz transformations also form a mathematical "group" which tells you, amongst other things, that any "alternative" formulation of the Lorentz transformations you may come across is mathematically equivalent to the one you are familiar with. Sometimes it is more convenient, depending on the problem at hand, to use spherical coordinates in a covariant tensor formulation. Very sometimes.

[Rainer]

Thank you all for your clarifications. It has been of great help.

I'm not completely sure what you mean by "metric tensor covariant transformations".

I don't think that Einstein came up with "another way" to do Lorentz transformations.

Okay I see now that what I am separating into two things is really just one and the same: a Lorentz Transformation is a Lorentz Transformation is a Lorentz Transformation.

I guess I am grappling with the two ways or guises in which this transformation appears. On the one hand you have a very simple manipulation of pythagoras' theorem. On the other hand you have tensor matrix algebra.



ETA: If you feed enough vectors to a tensor, its output will be a scalar. The rank of a tensor is the number of vectors you have to feed it before its output will be a scalar. A rank 0 tensor is already a scalar. A rank 1 tensor operates on 1 vector by computing the inner product of its coefficients (regarded as a vector) with the components of the vector; that produces a scalar. A rank 2 tensor can operate on 2 vectors, in which case it produces a scalar, or it can operate on just 1 vector, in which case it produces a rank 1 tensor that can be fed the other vector to produce a scalar. And so on.

Hmmmm... So If I have a 4x1 matrix of 4 components (i.e. 1 four-vector), and I want to perform a Lorentz Transformation on it, I would be multiplying a 4x1 matrix by a rank four tensor. So you're saying that the transformed four-vector would also constitute a rank 3 tensor, and that if I wanted to subsequently map another four-vector by the same Lorentz factor, I could then use this rank 3 tensor instead of the original rank 4 tensor?

Is there a way to do a Lorentz Transformation for a greater-than-four dimensional vector?

[W.D.Clinger]

Hmmmm... So If I have a 4x1 matrix of 4 components (i.e. 1 four-vector), and I want to perform a Lorentz Transformation on it, I would be multiplying a 4x1 matrix by a rank four tensor. So you're saying that the transformed four-vector would also constitute a rank 3 tensor, and that if I wanted to subsequently map another four-vector by the same Lorentz factor, I could then use this rank 3 tensor instead of the original rank 4 tensor?
No. I'm not saying anything like that.

You may be confusing dimension with rank. The Lorentz transformation is a rank 2 tensor.

What's more, I responded to you as though tensors were covariant in all indexes, so applying a rank 2 tensor to a vector would yield a rank 1 covariant tensor (which is dual to a vector). The situation is actually more complicated: tensors can be covariant in some indexes but contravariant in others (http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors). A vector is a rank 1 contravariant tensor (dual to a rank 1 covariant tensor).

The Lorentz transformation is a mixed tensor: covariant in one index but contravariant in the other. That means you can think of it as a 4x4 square matrix that, applied to a 4-vector (by multiplying the two), yields another 4-vector. To get a scalar, you could then take the dot product of some 4-vector with that result.

The reason we talk about rank 2 tensors instead of matrices is that we want to discuss the geometric object independently of any coordinate system. The matrix representation depends on the coordinate system, and must be transformed when the coordinate system changes; the "covariant" and "contravariant" stuff has to do with how that transformation is done.

Is there a way to do a Lorentz Transformation for a greater-than-four dimensional vector?
The Lorentz transformation can be generalized to more than 4 dimensions, but I don't know whether it would be considered proper to refer to the generalization as a Lorentz transformation.

[sol invictus]

I guess I am grappling with the two ways or guises in which this transformation appears. On the one hand you have a very simple manipulation of pythagoras' theorem. On the other hand you have tensor matrix algebra.

Well, ask yourself the following question: what transformations preserve the hypotenuse length that appears in Pythagoras' theorem? That is, start with a triangle in the 2D plane. What can you do to the plane that doesn't change the length of the hypotenuse? Better, what can you do that won't change any length?

It turns out there are three fundamental such operations; "fundamental" means any other is some combination of those three. Can you guess what they are?

Two are translations - you can "slide" everything up or down (that's one), or left and right (that's two). The third is a rigid rotation (that makes three) - you just rotate everything by some angle around some central point. Let's forget about the translations, because they're fairly trivial and not part of the Lorentz group. So, how does that rotation act on a vector in the plane? Like this:

cos(w) sin(w)
-sin(w) cos(w)

where t is the angle of rotation, and that's supposed to be a 2x2 matrix that multiplies a vector like (x y). Notice that it bears a very close resemblance to the 2x2 part of the Lorentz transformation matrix on the wiki I linked to.

Now, same question in 3D? The answer is three translations, and three independent rotations (in the xy plane, in the yz plane, in the xz plane). To represent the action of rotations on a 3D vector we need a 3x3 matrix, but for a rotation in the xy plane it will look just like the above, plus one extra row and column 001. It will act on (x y z).

4D Euclidean space (no time)? Same thing, but two extra rows and columns.

Now, you're not interested so much in rotations - although they form part of the Lorentz group, properly speaking - but in "boosts", changes of velocity. It turns out you can think of those as "rotations" in the "xt plane" (or yt or zt). So there are three independent boosts. If we make the first two rows and columns of our matrix t and x, the Lorentz transformation that boosts in the x direction is almost identical to the rotation matrix we started with. The "only" difference is that due to the fact that t is time and not space, the rotation angle w becomes imaginary (which turns cos into cosh, hyperbolic cosine, and sin into sinh), and the first two rows and columns of the matrix turn into

cosh(b) sinh(b)
-sinh(b) cosh(b)

You can check that the action of this matrix on (x t) preserves the Lorentzian "length" -t^2+x^2, just as the rotation matrix acting on (x y) preserves x^2+y^2 (the Pythagorean, or Euclidean, length).

If you're interested in more (or fewer) dimensions, the above procedure should make it relatively obvious how to generalize to them.

Does that help?


The Lorentz transformation can be generalized to more than 4 dimensions, but I don't know whether it would be considered proper to refer to the generalization as a Lorentz transformation.

People routinely refer to the Lorentz group in d dimensions, it's fairly standard usage.

[Rainer]

No. I'm not saying anything like that.
You may be confusing dimension with rank. The Lorentz transformation is a rank 2 tensor.


Yes, I was. Thanks.

The Lorentz transformation is a mixed tensor: covariant in one index but contravariant in the other. That means you can think of it as a 4x4 square matrix that, applied to a 4-vector (by multiplying the two), yields another 4-vector. To get a scalar, you could then take the dot product of some 4-vector with that result.


Great.


...The "only" difference is that due to the fact that t is time and not space, the rotation angle w becomes imaginary (which turns cos into cosh, hyperbolic cosine, and sin into sinh), and the first two rows and columns of the matrix turn into

cosh(b) sinh(b)
-sinh(b) cosh(b)


Whoa this is deep.

One slight detail--this wikipedia page has both sinh(b) being negative:

http://en.wikipedia.org/wiki/Lorentz_transformation


Does that help?


Yes. Thanks very much.

[Vorpal]

If in the (x,y)-plane, we have the rotation matrix [cos(w) sin(w); -sin(w) cos(w)], and we let {y = it, w = ib}, then
[ x'] = [ cos(ib) sin(ib) ][ x] = [ cosh(b) i.sinh(b) ][ x]
[it'] = [ -sin(ib) cos(ib) ][it] = [-i.sinh(b) cosh(b) ][it]
So in (x,t)-space, we have the hyperbolic rotation {x' = (cosh b)x + (-sinh b)t, t' = (-sinh b)x + (cosh b)t}, just as we should have, with both sinh's negative.

[sol invictus]

One slight detail--this wikipedia page has both sinh(b) being negative:


That was a typo - sorry! The wiki, and Vorpal, are correct.

[Rainer]

In a Lorentz boost the y axis value remains unchanged. Does that make it an "eigenvector"?

[W.D.Clinger]

In a Lorentz boost the y axis value remains unchanged. Does that make it an "eigenvector"?
No. An eigenvector is a vector whose direction is unchanged by the transformation while its components are multiplied by the eigenvalue.

Using Wikipedia's notation for a boost in the x-direction (http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form), the eigenvectors of the boost are the vectors whose first (ct) component is the additive inverse (negative) of the x component, and whose y and z components are zero. The eigenvalue is \gamma+\beta\gamma.

[sol invictus]

In a Lorentz boost the y axis value remains unchanged. Does that make it an "eigenvector"?

I assume you mean a Lorentz boost in the x direction? If so, the answer is yes.

No. An eigenvector is a vector whose direction is unchanged by the transformation while its components are multiplied by the eigenvalue.

You mean, like a vector pointed in the y (or z, or any linear combination) direction? All of those are eigenvectors, with eigenvalue 1. That's a case where you have a 2D "degenerate" eigenspace (i.e. two eigenvectors with the same eigenvalue, so that any linear combination is an eigenvector).

Wikipedia's notation for a boost in the x-direction (http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form), the eigenvectors of the boost are the vectors whose first (ct) component is the additive inverse (negative) of the x component, and whose y and z components are zero. The eigenvalue is \gamma+\beta\gamma.

There are two eigenvectors in the x,t "plane", with two eigenvalues - the one you identified, and \gamma-\beta\gamma. The two eigenvectors are null rays, they are tangent to a ray of light moving in the +x or -x direction. They're eigenvectors because a Lorentz boost in a direction parallel to a ray of light doesn't change its "direction" in spacetime, only its frequency (and energy).

Here's a question - what are the eigenvectors and eigenvalues for a rotation (say in the xy plane) in 3D space? One is easy to find; the other two a little more tricky.

[W.D.Clinger]

You mean, like a vector pointed in the y (or z, or any linear combination) direction? All of those are eigenvectors, with eigenvalue 1. That's a case where you have a 2D "degenerate" eigenspace (i.e. two eigenvectors with the same eigenvalue, so that any linear combination is an eigenvector).
Thank you for correcting me. What I should have said is that it isn't the unchanging value of the y component that makes any linear combination of [0 0 1 0] and [0 0 0 1] into eigenvectors, but the fact that a Lorentz boost in the x-direction leaves their direction unchanged.

Here's a question - what are the eigenvectors and eigenvalues for a rotation (say in the xy plane) in 3D space? One is easy to find; the other two a little more tricky.
Unless the rotation is a multiple of 180 degrees, I believe the other two will require some imagination.

[sol invictus]

Thank you for correcting me. What I should have said is that it isn't the unchanging value of the y component that makes any linear combination of [0 0 1 0] and [0 0 0 1] into eigenvectors, but the fact that a Lorentz boost in the x-direction leaves their direction unchanged.

Re-reading Rainer's question, I see now why you said what you did. It's not completely clear what s/he meant.

Unless the rotation is a multiple of 180 degrees, I believe the other two will require some imagination.

:)

[Floyt]

(
Now, you're not interested so much in rotations - although they form part of the Lorentz group, properly speaking - but in "boosts", changes of velocity. It turns out you can think of those as "rotations" in the "xt plane" (or yt or zt).

[math nincompoop] So if a rotation in the t planes is a change in velocity, what would a translation in these planes equate to?
)

[sol invictus]

(


[math nincompoop] So if a rotation in the t planes is a change in velocity, what would a translation in these planes equate to?
)

Just a time or space translation (or both if you translate at an angle). It just moves whatever you act on from one time and location to a different time and different location.

The group consisting of the Lorentz group (boosts and rotations) plus translations in space and time is called the Poincare group. It's the full* symmetry group of flat spacetime (the spacetime of special relativity).


*Actually there are some discrete symmetries as well, like parity.

[sol invictus]

Another interesting physics brain teaser:

space translation invariance gives you conservation of momentum
time translation invariance gives you conservation of energy
rotation invariance gives you conservation of angular momentum

What does boost invariance give you?

[Floyt]

Another interesting physics brain teaser:

space translation invariance gives you conservation of momentum
time translation invariance gives you conservation of energy
rotation invariance gives you conservation of angular momentum

What does boost invariance give you?

Wild stab - it shows that the observers are identical? (as the "boost", to my 30mins-on-the-web understanding, is the conversion between two frames with different observers)

[Vorpal]

Another interesting physics brain teaser:
...
What does boost invariance give you?
A boost along the x direction has the infinitesimal generator x∂t+t∂x, which corresponds to tpx - xE being constant, which in turn looks like x - tv except for a factor of E. So this is something like "conservation of the initial position of center of momentum" or somesuch, at least whenever E is also a constant of motion.

[sol invictus]

A boost along the x direction has the infinitesimal generator x∂t+t∂x, which corresponds to tpx - xE being constant, which in turn looks like x - tv except for a factor of E. So this is something like "conservation of the initial position of center of momentum" or somesuch, at least whenever E is also a constant of motion.

Bingo.

It looks funny - and isn't very useful - because boost generator doesn't commute with the usual Hamiltonian (i.e time translations). That's why you've never heard of it as a conservation law or quantum number.

Interestingly, you can actually use that boost generator as the Hamiltonian (that's equivalent to using Rindler time for time; the generator is timelike in part of the spacetime). All sorts of bizarre and interesting things happen when you do so - there are horizons, "Hawking radiation", etc. Of course the physics is identical, but it certainly doesn't look it.

[Rainer]

Unless the rotation is a multiple of 180 degrees, I believe the other two will require some imagination.

Yeah, got the hint.

After five years of trying, however, I still cannot explain what a complex number is, nor understand anyone's explanation.

Maybe in another five years.

Thanks