1
$\begingroup$

I am revising special relativity introducing more matrix form in the equation. Currently I am reading book in which transformation matrix is defined as $${\Lambda= \begin{bmatrix} \gamma & -v\gamma & 0 & 0 \\ -v\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} }$$ and the four-velocity as $${U^{\alpha}=\Lambda^{\alpha}_{\overline{\beta}}(\vec e_{\overline{0}})^{\overline{\beta}}=\Lambda^{\alpha}_{\overline{0}}}$$ The first thing that bothers me is that for positive ${v}$ I get negative component of the velocity vector $${\vec U= \begin{bmatrix} \gamma \\ -v\gamma \\ 0 \\ 0 \end{bmatrix} }$$ And the second thing is that when I apply this rule of transformation twice (for example when I have the following task:Ship is travelling with 0.6c relative to the Earth and another ship is travelling with 0.6c with respect to the first shipt. Find the four-velocity of the second ship relative to the Earth.(it is not homework)) For the four-velocity I get $${ \vec U= \begin{bmatrix} \gamma^{2}+v^{2}\gamma^{2} \\ -v\gamma^{2}-v\gamma^{2} \\ 0 \\ 0\\ \end{bmatrix} }$$ The second component is negative which definitely differs from the real vector (I checked this using Lorentz transformation, getting the equations, building the t''-x'' axes of the second ship's frame with respect to Earth). Where am I wrong? (c=1)

$\endgroup$
2
$\begingroup$

Timaeus's answer could be correct. The $\Lambda$ matrix from your book may have been intended as a passive transformation (one that acts on the coordinate system) and you mistakenly used it as an active transformation (one that acts on the object). Alternatively, the $\Lambda$ matrix from your book may have been intended as the active transformation of a co-vector and my answer below addresses that.

Because boosts are not orthogonal matrices, you have to think about the different way co-vectors and contra-vectors transform. The $\Lambda$ matrix you wrote down is the +v boost for a co-vector. You then applied it to a vector and incorrectly interpreted the result as if it were a contra-vector. The co-vector $U_\alpha$ with +v velocity really does have $-v\gamma$ in it as you found. You must raise U's index with the metric to see what the contra-variant $U^\beta$ is, and you see the +v that you expected.

$$U^\beta = \eta^{\beta \alpha}U_\alpha$$ $$\begin{bmatrix}\gamma \\v\gamma \\0 \\0\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 &-1 & 0 \\ 0 & 0 & 0 &-1\end{bmatrix} \begin{bmatrix}\gamma \\-v\gamma \\ 0 \\ 0\end{bmatrix}$$

A contra-vector transforms as $$Unew^\alpha = \Lambda^\alpha _{-\beta} Uold^\beta$$ Then a co-vector transforms like $$Unew_\alpha = Uold_\beta(\Lambda^{-1})^\beta _{-\alpha}$$

This matrix boosts contra-vectors by +v. I have redefined your $\Lambda$ to be the matrix that gives a +v boost to a contra-vector. $${\Lambda=\begin{bmatrix}\gamma & v\gamma & 0 & 0 \\ v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}}$$

$$\begin{bmatrix}\gamma \\v\gamma \\0 \\0\end{bmatrix} =\begin{bmatrix}\gamma & v\gamma & 0 & 0 \\ v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

This matrix boosts co-vectors by +v. $${(\Lambda^{-1})^T=\begin{bmatrix}\gamma & -v\gamma & 0 & 0 \\ -v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}}$$

$$\begin{bmatrix}\gamma \\-v\gamma \\0 \\0\end{bmatrix} =\begin{bmatrix}\gamma & -v\gamma & 0 & 0 \\ -v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

Notice that rotations R are orthogonal which means $(R^{-1})^T=R$ and the transformations for co-vectors and contra-vectors are the same. Hence we don't talk about co-vectors and contra-vectors when doing rotations.

$\endgroup$
  • $\begingroup$ I got the idea but have one question. Why do we have to change it to contra-vector? What if we use its co-vector components and when do we have to do that? $\endgroup$ – Blake Aug 20 '15 at 6:48
  • $\begingroup$ And what if the metric matrix is -1 1 1 1? $\endgroup$ – Blake Aug 20 '15 at 6:52
  • $\begingroup$ 1) You don't have to use the contra-variant components. It is perfectly okay to specify the 4-velocity by its co-varient components. Just be sure and boost it by $(\Lambda^{-1})^T$ as shown above, and realize that $-v\gamma$ is then the correct second component of the co-vector. $\endgroup$ – Gary Godfrey Aug 25 '15 at 4:45
  • $\begingroup$ 2) I think the sign of my metric is just my convention. I like time like vectors $|TimeComponent|>|SpaceComponent|$ to have $X_\alpha X^\alpha > 0$. If you choose to multiply the metric by -1, you make the convention $X_\alpha X^\alpha < 0$. $\endgroup$ – Gary Godfrey Aug 25 '15 at 5:07
1
$\begingroup$

The first thing that bothers me is that for positive ${v}$ I get negative component of the velocity vector

If your object was at rest then to someone moving to the right the object at rest is moving to the left in their frame.

As for the second question it is same issue. Someone at rest will be moving to the left relative to someone moving faster than 0.6c to the right.

$\endgroup$
  • $\begingroup$ No, they are moving to the right in the Earth frame and still we have minus signs. $\endgroup$ – Blake Aug 19 '15 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.