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This question makes no sense so I think it must be a misprint. It says:

Frame $Σ$ moves at speed v with respect to frame $Σ'$ in the common $x, x'$ directions. The two frames are aligned at time $t = t_0 = 0$.

(a) Write down the Galilean transformation as a matrix transforming the $Σ$-frame coordinates $(ct, x)$ to the $Σ'$ coordinates $(ct' , x' )$. You can ignore the transverse directions and work with $2 × 2$ matrices.

(b) Write down the Lorentz transformation matrix. Take the limit $c → ∞$, and show that this fails to reproduce the Galiliean transformation matrix.

My answer:

a) Is found easy enough $x'=x-vt$ and $t'=t$ so the transformation matrix is

$$ \begin{pmatrix} 1 & 0 \\ -v & 1 \\ \end{pmatrix} $$

b) It says take the limit as $c\rightarrow∞ $ but if we do this then $x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}} \rightarrow \frac{x-vt}{\sqrt{1-0}}=x-vt$ and $t'=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}} \rightarrow \frac{t-0}{\sqrt{1-0}}=t$ which is the exact same as the Galilean transformation.

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  • $\begingroup$ Maybe the problem writer doesn't think $vx/c^2$ Goes to zero? You could always pick an $x$ big enough where it doesn't. $\endgroup$ – user1379857 Mar 29 '18 at 1:15
  • $\begingroup$ @user1379857 if $c \rightarrow ∞ $ the only way to make sure that x is bigger would be to make x=∞ and $\frac {∞}{∞}$ is undefined . So I don't see how that would work ? $\endgroup$ – bhapi Mar 29 '18 at 1:18
  • $\begingroup$ Is this a question from a particular book, or is it from an unofficial problem sheet? I'd agree that it seems to be a misprint, but the wording of the question makes it seem rather deliberate. $\endgroup$ – J. Murray Mar 29 '18 at 1:23
  • $\begingroup$ @J.Murray It's from a problem sheet. the third part of the question goes . (c) The reason for the failure above is that in the c → ∞ limit, the variables ct and ct' lose meaning. So, write down the Galilean transformation as a matrix transforming the Σ- frame coordinates (t, x) to the Σ0 coordinates (t ' , x' ). i.e., the first coordinate does not have the c factor. Write down the Lorentz transformation the same... $\endgroup$ – bhapi Mar 29 '18 at 1:28
  • $\begingroup$ ...way, as a matrix transforming (t, x) to (t ' , x' ). Now show that the Lorentz transformation matrix reduces to the Galiliean transformation matrix in the limit c → ∞. $\endgroup$ – bhapi Mar 29 '18 at 1:28
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With the added details in the comments, this question makes sense. It's not talking about whether the Lorentz transformation itself reduces to the Galilean transformation in the limit as $c\rightarrow \infty$ (it does). Instead, it's talking about the actual matrix $$\frac{1}{\sqrt{1-v^2/c^2}}\pmatrix{1 & -\frac{v}{c} \\ -\frac{v}{c} &1}$$

which transforms $\pmatrix{ct \\ x}$ into $\pmatrix{ct' \\ x'}$.

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  • $\begingroup$ ah yes thank you , that does make a lot more sense :) one thing though in your matrix I think you meant to put $\beta=\frac{v}{c}$ instead of v, no ? $\endgroup$ – bhapi Mar 29 '18 at 1:51
  • $\begingroup$ Ah yes, whoops. Too used to natural units ;) $\endgroup$ – J. Murray Mar 29 '18 at 1:52
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This isn't a misprint, and the problem is valid.

Your mistake is that you're expressing the transformation matrices in terms of $t$ and $t'$, instead of in terms of $ct$ and $ct'$ as the problem tells you to. If you express the matrices in terms of $ct$ and $ct'$, the bottom left corner of the Galilean transformation matrix will be $-v/c$ instead of $-v$, and the bottom left corner of the Lorentz transformation matrix will be

$$\frac{-v/c}{\sqrt{1-v^2/c^2}}\ \ .$$

The limit of that expression as $c\to\infty$ is $0$, not $-v/c$, so the limit of the Lorentz matrix as $c\to\infty$ is different from the Galilean matrix. (Note that the problem doesn't have you take the limit of the Galilean matrix as $c\to\infty$.)

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  • $\begingroup$ thank you I see now that I'd been making a silly mistake . $\endgroup$ – bhapi Mar 29 '18 at 1:58

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