A Lot of Basic Question About Lorentz-Minkowski Geometry

  • Thread starter nomather1471
  • Start date
  • Tags
    Geometry
In summary, the author is discussing the concept of subspaces within a vector space and how they are related to one another. He states that there are three-dimensional subspaces of time-like vectors, and that subspaces with only space-like vectors are one-dimensional. He also discusses the concept of a null vector, and proves that E1+E2 is a null vector. Finally, he asks a question about whether a subspace containing a null vector must be a lightlike subspace. Thanks for your time.
  • #1
nomather1471
19
1
Hi there, i have a lot of question about Lorentz-Minkowski geometry:

1)
Is Lorentz metric degenere or non-degenere? Why?

2) In spacelike subspaces only spacelike vectors live in it there is not problem here but how can

we say that timelike subspaces include null and spacelike vectors and lightlike subspaces include

spacelike vector?

3) Let E1 spacelike and E2 timelike vector then E1+E2 is null vector, proof...?

4) If Sp{E1,E2} spacelike, Sp{E1,E3} and Sp{E2,E3}

are timelike planes Sp{E1,E2+E3} is lightlike plane?

5) E1+E2+E3 is spacelike vector but

Sp{E1,E1+E2+E3} is lightlike plane?

6) E2+E3 is lightlike vector but Sp{E2+E3,E3} is timelike plane?
 
Physics news on Phys.org
  • #2
That is a lot of questions.
nomather1471 said:
3) Let E1 spacelike and E2 timelike vector then E1+E2 is null vector, proof...?
This is not true in general. In units where c=1 and using the (-+++) signature, then the sum of
E1=(0,1,0,0) and E2=(1,0,0,0) is null
E1=(0,1,0,0) and E2=(5,0,0,0) is timelike
E1=(0,5,0,0) and E2=(1,0,0,0) is spacelike

Hopefully other people will address the other questions.
 
  • #3
DaleSpam said:
That is a lot of questions.This is not true in general. In units where c=1 and using the (-+++) signature, then the sum of
E1=(0,1,0,0) and E2=(1,0,0,0) is null
E1=(0,1,0,0) and E2=(5,0,0,0) is timelike
E1=(0,5,0,0) and E2=(1,0,0,0) is spacelike

Hopefully other people will address the other questions.

Thanks, i agree with you, these questions are from the Rafael Lopez Minkowski Lectures, about question 3 the author able to think the unit vectors, the original is E2+E3 is null so if we take E2=(0,1,0) and E3=(0,0,1) then E2+E3=(0,1,1) which is a null vector, here the signature is (+,+,-) in E13...
 
  • #4
nomather1471 said:
Hi there, i have a lot of question about Lorentz-Minkowski geometry:

1)
Is Lorentz metric degenere or non-degenere? Why?

It's non-degenerate, because the representing matrix [itex](\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)[/itex] (sorry for changing the convention within the thread, I'm used to the west-coast convention; if I'd change it, most probably I'd get confused myself).

nomather1471 said:
2) In spacelike subspaces only spacelike vectors live in it there is not problem here but how can we say that timelike subspaces include null and spacelike vectors and lightlike subspaces include spacelike vector?

I don't know, what you mean by "subspace". In the usual sense of subspaces of a vector space, indeed there exist three-dimensional subspaces of time-like vectors. They can be defined as to be 3D subspaces Minkowski-perpendicular to an arbitrary time-like vector.

Proof: Let [itex]n[/itex] a time-like unit vector, i.e., [itex]n \cdot n=\eta_{\mu \nu} n^{\mu} n^{\mu}=1[/itex]. Then you can build the following Lorentz transformation
[tex]({\Lambda^{\mu}}_{\nu})=\begin{pmatrix}
n^0 & \vec{n}^t \\
\vec{n} & 1+\frac{n^0-1}{\vec{n}^2} \vec{n} \otimes \vec{n}
\end{pmatrix}
[/tex]
in obvious notation. Then
[tex]E_0'=n=\Lambda E_0, \quad E_j'=\Lambda E_j[/tex]
build a new Minkowski-orthonormal system, and [itex]\text{span}(E_1',E_2',E_3')[/itex] is a 3D subspace containing only space-like vectors.

On the other hand subspaces with only time-like vectors are necessarily one-dimensional, because suppose you have two linearly independent space-like vectors [itex]n_1[/itex] and [itex]n_2[/itex] you can construct the vector
[tex] a=n_1-\frac{(n_1 \cdot n_2)}{n_2 \cdot n_2} n_2 \neq 0[/tex]
within the subspace. It's Minkowski-perpendicular to the space-like vector [itex]n_2[/itex], but then it must be either spacelike or lightlike.

To prove this you just take [itex]a \cdot n_2=0[/itex] and write it out in components
[tex]a^0 \cdot n_2^0-\vec{a} \cdot \vec{n}=0 \; \Rightarrow \; |a^0|=\frac{|\vec{a} \cdot \vec{n}|}{n^0} \geq \frac{|\vec{a} \cdot \vec{n}|}{|\vec{n}|} \geq |\vec{a}|,[/tex]
and thus
[tex]a \cdot a=(a^0)^2-\vec{a}^2 \leq 0.[/tex]

In contradiction to the assumption the subspace necessarily contains at least one spacelike or lightlike vector. In conclusion, subspaces that contain only time-like vectors are one-dimensional.

nomather1471 said:
3) Let E1 spacelike and E2 timelike vector then E1+E2 is null vector, proof...?
is wrong as shown by the counter example by DaleSpam as are the other claims, or are their additional restrictions on the vectors [itex]E_j[/itex]?
 
  • #5
Sorry but i can't understand anything with physics notation, i think Lorentz metric be a non-degenerate because of the definition of degenere metric...I mean subspaces by subset of Lorentz space, timelike subspaces are space which has a non-degenerate metric on it and lightlike subspace are space which has a degenerate metric and different from {0}, but how can we guarantiate the lightlike space has a spacelike vector, our teach told this with tangent plane to the timeconi but i am not sure of the truth, the question is if any subspace contain null vector is this set must be a lightlike subspace?
 

1. What is Lorentz-Minkowski geometry?

Lorentz-Minkowski geometry, also known as Minkowski space, is a mathematical model used to describe the geometry of special relativity. It is a four-dimensional space that combines three dimensions of space and one dimension of time, and it allows for the measurement of distances and intervals between events in spacetime.

2. How is Lorentz-Minkowski geometry different from Euclidean geometry?

Unlike Euclidean geometry, which is based on the concept of absolute space and time, Lorentz-Minkowski geometry takes into account the effects of special relativity, such as time dilation and length contraction. This means that the measurements of distances and intervals in Lorentz-Minkowski geometry are relative and can vary depending on the observer's frame of reference.

3. What is the significance of Lorentz-Minkowski geometry?

Lorentz-Minkowski geometry plays a crucial role in the understanding and application of special relativity. It allows us to accurately describe and predict the behavior of objects moving at high speeds and in different frames of reference. It also forms the basis of many important concepts in modern physics, such as spacetime diagrams and the Lorentz transformation.

4. Are there any practical applications of Lorentz-Minkowski geometry?

Yes, Lorentz-Minkowski geometry has several practical applications in fields such as particle physics, cosmology, and GPS technology. In particle accelerators, it is used to calculate the trajectories of high-speed particles. In cosmology, it is used to describe the expansion of the universe. In GPS technology, it is used to correct for the effects of relativity on satellite signals.

5. Can Lorentz-Minkowski geometry be visualized?

Yes, although it is a four-dimensional space, Lorentz-Minkowski geometry can be visualized using mathematical tools such as spacetime diagrams and the Minkowski space plane. These visualizations allow us to understand and interpret the geometric concepts of special relativity in a more intuitive way.

Similar threads

  • Special and General Relativity
Replies
6
Views
2K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
12
Views
1K
Replies
10
Views
1K
  • Special and General Relativity
Replies
14
Views
772
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
3
Views
2K
Replies
6
Views
2K
  • Special and General Relativity
Replies
7
Views
2K
  • Special and General Relativity
2
Replies
42
Views
5K
Back
Top