I have heard relativistics only very compressed during my student time. Now I looked up the definitions again and a question comes into my mind:

A contravariant vector is transformed like this: $(a^\mu)'=L_{\mu\lambda}a^\lambda$ wherein $L_{\mu\lambda}$ is the Lorentz matrix. A covariant vector is transformed like this: $(a_\mu)'=L^{-1}_{\lambda\mu}a_\lambda=L^{T}_{\lambda\mu}a_\lambda=L_{\mu\lambda}a_\lambda$. Does this mean no matter if I have a co- or contravariant four vector I use the same matrix to transform them from one coordinate system to the other?

And one general question: Why do I need co- AND contravariant four vectors? Isn't one enough?

EDIT: My definitions

I think I found the solution. We do not have the common definitions of the four vectors! I have a definition like this:


The normal definition is:


Then, regarding the answers, it doesn't matter where I put the indices of the Lorentz matrix, does it? Within the new definition, the Lorentz matrix becomes non-symmetric

$L=\left(\begin{array}{ccc}\cosh \theta &0&0&\mathrm{i}\sinh \theta \\ 0&1&0&0\\0&0&1&0\\ -\mathrm{i}\sinh&0&0&\cosh \theta\end{array}\right)$

In contrast, the normal Lorentz matrix is

$L=\left(\begin{array}{ccc}\cosh \theta &0&0&\sinh \theta \\ 0&1&0&0\\0&0&1&0\\ \sinh&0&0&\cosh \theta\end{array}\right)$

Now we see that indeed $(a_\mu)'=L^{-1}_{\lambda\mu}a_\lambda=L^{T}_{\lambda\mu}a_\lambda=L_{\mu\lambda}a_\lambda$. So in this representation, the transformation of co- and contravariant four vectors are done by the same matrix. Can anyone confirm?


3 Answers 3

  1. As others have pointed out, the indices in your expression is wrong. It should be $a'^\mu=L^\mu{}_\lambda a^{\lambda}$. Covariant vectors transform like $a'_\mu=L_\mu{}^\lambda a_\lambda$.
  2. Covariant and contravariant vectors are dual to each other. Their distinction simplifies the expression of inner product.
  3. $a^\mu=(a_1,a_2,a_3,\mathrm{i}ct)$ is what most elementary relativity teaches. In this expression, there is no distinction of covariant and contravariant vectors, so upper and lower indices are all the same. Lorentz transform matrix is (complex) orthogonal (not unitary) and is the same for either vectors. This convention simplifies life, which is why it is taught to first learners.
  4. The distinction between covariant and contravariant vectors are very important in general relativity, so in order to be generalizable, it is better to distinguish covariant and contravariant vectors even in special relativity.
  • $\begingroup$ In my book I just found the definitions I used and I can't clearly see why this should be wrong and why it is necessary to write it like you did under 1. $\endgroup$
    – DaP
    Commented Feb 1, 2013 at 9:52
  • $\begingroup$ @DaPhil: Is your book using imaginary time? In that case as I said, there is no distinction between upper and lower indices. $\endgroup$
    – Siyuan Ren
    Commented Feb 1, 2013 at 10:30
  • $\begingroup$ Actually, the book does not use imaginary time... $\endgroup$
    – DaP
    Commented Feb 1, 2013 at 10:35
  • $\begingroup$ @DaPhil: Then your book is wrong. The indices must match each other. $\endgroup$
    – Siyuan Ren
    Commented Feb 1, 2013 at 11:36
  • $\begingroup$ I would be surprised if it is wrong. It's some kind of the standard text book about theoretical physics in Germany. I guess I am understanding something wrong. Could you explain where is the difference between $L^\mu_{\hphantom{\mu}\nu}$ and $L_\mu^{\hphantom{\mu}\nu}$? What do you mean by the indices must match? I can't see matching indices in your answer either. $\endgroup$
    – DaP
    Commented Feb 1, 2013 at 11:43

A Lorentz matrix should not turn a covariant vector to a contravariant vector, only the metric does. So you see that you have various errors in your index notation. The second sign for an error: You should only sum over one upper and one lower index! Why? The dual space is the space of linear functions that assigns a vector a number. In most cases the dual of the dual space is the space itself, so what I try to say is: Only a covector * contravector or vice versa can give a number.

So in my opinion, the first line should look like: $ (a^{\mu})' = \Lambda^{\mu}_{\nu}a^{\nu} $

The second one: $ (a_{\mu})' = \Lambda_{\mu}^{\nu}a_{\nu} $

The inverse transform is the same Lorentz boost with negative velocity, if you perform the boost and then the inverse one you should be back in the original system.

I hope this awnsers your questions.

  • $\begingroup$ btw can sombeday show me how i can seperate the lower and upper indices so that they are not on top of each other? $\endgroup$
    – Noldig
    Commented Jan 31, 2013 at 14:05
  • $\begingroup$ You can use \hphantom{/mu} to do so. $\endgroup$
    – DaP
    Commented Jan 31, 2013 at 14:25
  • 1
    $\begingroup$ or just write something like \Lambda_{a}{}^{b} $\endgroup$ Commented Jan 31, 2013 at 15:06

It helps if the matrix is written $L^{\mu}_{\ \ \lambda}$. Contravariant vectors transform as, $$ a'^{\mu}=L^{\mu}_{\ \ \lambda}a^{\lambda} $$ and covariant vectors transform as, $$ a'_{\mu}=[L^{-T}]_{\mu}^{\ \ \lambda}a_{\lambda}=a_{\lambda}[L^{-1}]^{\lambda}_{\ \ \mu} \ . $$ The Lorentz transformation matrices leave the Minkowski metric $\eta_{\mu\lambda}=diag[1,-1,-1,-1]$ invariant. $$ \eta'_{\mu\lambda}=[L^{-T}]_{\mu}^{\ \ \rho}[L^{-T}]_{\lambda}^{\ \ \sigma}\eta_{\rho\sigma}=\eta_{\mu\lambda} $$ Transposing the second Lorentz matrix gives, $$ \eta_{\mu\lambda}=[L^{-T}]_{\mu}^{\ \ \rho}\eta_{\rho\sigma}[L^{-1}]^{\sigma}_{\ \ \lambda} $$ which is $\eta=L^{-T}\eta L^{-1}$ in matrix form. Since $\eta$ is not a unit matrix the inverse $L^{-1}$ is not equal to its transpose $L^{T}$ and so $L^{-T}\neq L$ and hence contravariant vectors transform differently from covariant vectors.

The reason for the two sorts is because different physical quantities transform in different ways. For example, coordinates $x^{\mu}$ are contravariant and momenta $p_{\mu}$ are covariant.


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