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I am not overly versed in special relativity, so do not be afraid to correct anything here that is erroneous. $\newcommand{\b}[1]{\mathbf{#1}} \newcommand{\V}[4]{ \begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \end{pmatrix} } \newcommand{\v}[3]{ \begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \newcommand{\vn}[1]{ \begin{pmatrix} #1 ^1 \\ #1 ^2 \\ #1 ^3 \end{pmatrix} } \newcommand{\Vn}[1]{ \begin{pmatrix} #1 ^0 \\ #1 ^1 \\ #1 ^2 \\ #1 ^3 \end{pmatrix} } \newcommand{\LT}{ \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} }$


At relativistic speeds, length is contracted as $$L’=\frac L\gamma\tag1$$ time is dilated as $$T’=\gamma T\tag2$$ and the effective mass of matter is increased as $$m’=\gamma m\tag3$$


The Lorentz transform of 4-vectors can be written as a single matrix: $$\LT$$ which can be applied to the space-time 4-vector—$$\V{ct’}{x’}{y’}{z’} = \LT\V{ct}xyz\tag{A}$$—and to the energy-momentum 4-vector—$$\V{E’}{p_x’c}{p_y’c}{p_z’c} = \LT\V{E}{p_xc}{p_yc}{p_zc}\tag{B}$$


Are the following true or false?

  • Equation $(\text{A})$ has embedded within it all the information communicated by equations $(1)$ and $(2)$.
  • Equation $(\text{B})$ has embedded within it all the information communicated by $(1)$, $(2)$ and $(3)$.

I do not know enough about this math to expand them and arrive at an answer myself. However, my research suggest that the matrix does in fact concisely state the same as $(1)$ through $(3)$.

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Short answer

  • True
  • True

Long answer

Let's set $c = 1$ first.

Time Dilation

Consider two events $\mathcal{E}_1$ and $\mathcal{E}_2$ measured in reference frame $K$ that happen at the same location, denoting them with the coordiantes $(t_1,x,y,z)$ and $(t_2,x,y,z)$. Your equation (A) tells us that for an observer $K'$ boosted in the $x$-direction,

$$t_2' - t_1' = \gamma(t_2 - t_1) - \gamma \beta (x-x) = \gamma (t_2 - t_1)$$

which is the correct expression for the time dilation of a clock stationary in $K$.

Length Contraction

Consider two events $\mathcal{E}_1$ and $\mathcal{E}_2$ measured in reference frame $K$ that happen simultaneously but are separated in space, denoting them with the coordinates $(t,x_1,y_1,z_1)$ and $(t,x_2,y_2,z_2)$; additionally, for simplicity, assume that these events correspond to the endpoints of a meterstick that is comoving with observer $K'$ and aligned with the $x$-axis (i.e $y_1 = y_2$ and $z_1 = z_2$). I have highlighted the word "simultaneously" because it is important to understand that in order to use a meterstick as a ruler to measure length in your own reference frame, the ruler must either be stationary, or you must measure the position of the endpoints according to your own planes of simultaneity.

Since the ruler is stationary in $K'$, the observer in $K'$ does not have to measure the endpoints of the ruler simultaneously to determine its length $L'$, and your equation (A) tells us that for an observer $K'$ boosted in the $x$-direction,

$$L' = x_2' - x_1' = - \gamma \beta (t-t) + \gamma(x_2 - x_1) = \gamma (x_2 - x_1) = \gamma L$$

which is the correct expression for the length contraction of a ruler stationary in $K'$.

Effective Mass

Consider a massive object stationary in $K$. Its four-momentum is given in $K$ by $(m,0,0,0)$ and in $K'$ by $(m',p_x',p_y',p_z')$ (here I have used the relationship $E = mc^2 = m$ since we have set $c = 1$). Your equation (B) tells us that for an observer $K'$ boosted in the $x$-direction,

$$m' = \gamma m$$

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  • $\begingroup$ It should be maybe mentioned that (B) follows from (A) as well, so (A) has all the information in it needed to derive (B), (1), (2), (3). $\endgroup$ – Photon Dec 17 '17 at 10:23
  • $\begingroup$ That is correct. More generally, for a given transformation, all four-vectors will transform in the same way, and all of the relativistic phenomena mentioned by OP are direct consequences of the type of transformation specified in the question (a Lorentz boost). $\endgroup$ – JM1 Dec 17 '17 at 10:38

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