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I have been trying to understand a more or less geometric derivation of the Lorentz transformation, and I'm getting stuck at one spot. The wikipedia article for the Lorentz transformation for frames in standard configuration lists the following equations:

$x^{\prime} = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$

$y^{\prime} = y$

$z^{\prime} = z$

$t^{\prime} = \frac{t-(v/c^2)x}{\sqrt{1-\frac{v^2}{c^2}}}$

I've been able to work everything out except for $-(v/c^2)x$ in the $t^{\prime}$ equation. I haven't seen any explanations for this, which makes me feel like I'm missing something simple. Where does this part of the equation come from? Shouldn't $t^{\prime} = \gamma \cdot t$?

EDIT: Ok, so I reviewed the idea I was using to derive the Lorentz factor and thus the transformation for $t^{\prime}$. Suppose you have the two frames I've described, and you have a light wave moving perpendicular to the X axis in the second ($\prime$) frame.

Light Path Diagram

Using basic trig with the diagram, you can derive:

$t^{\prime}=t\cdot\sqrt{1 - \frac{v^2}{c^2}}$

Obviously this would contradict the transformation provided by wikipedia. What step am I missing here? I don't really want a proof that I'm wrong or that the equation I've derived is incorrect - I'm already pretty convinced of that. What I would really like is an intuitive explanation as to why mine is invalid and how I would go about deriving the correct equation through similar means.

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Marek's answer is fine, although possibly not at the right level for the OP. Note that if the time transformation were $t'=\gamma t$, then all observers would agree on simultaneity, but there are straightforward and well-known ways of showing that this can't happen, e.g., Einstein's famous thought experiment with the train and the two flashes of lightning: bartleby.com/173/9.html –  Ben Crowell Jul 27 '11 at 17:14
The other thing to understand here is that length contraction and time dilation are both different things from what the Lorentz transformations describe. Length contraction and time dilation describe the properties of clocks and rulers in frames where they're not at rest, compared to frames in which they are at rest. To recover time dilation as a special case of the Lorentz transformations, you have to pick two events, $(t_1,x)$ and $(t_2,x)$, and substitute them into the Lorentz transformations. Then the $\gamma v x/c^2$ terms cancel. –  Ben Crowell Jul 27 '11 at 17:17
Thanks @Ben-Crowell, that's really helpful. I'm thinking that I may have underestimated the complexity involved :) –  Jake Jul 28 '11 at 4:31

2 Answers 2

I'll not derive the transformation (that has been done in countless books and articles, I am sure you can find them yourself) but instead will try to explain why the formula you propose can't be correct.

For starters, observe that since you don't touch $y$ and $z$, we might as well work in 1+1 dimensions. Also, let $c=1$ so that we aren't bothered by unimportant constants (you can restore it in the end by requiring that formulas have the right units). Then it's useful to reparametrize the transformation in the following way $$x' = \gamma(x - vt) = \cosh \eta x - \sinh \eta t$$ $$t' = \gamma(t - vx) = -\sinh \eta x + \cosh \eta t$$ where we introduced rapidity $\eta$ by $\tanh \eta = v$ and this by standard (hyperbolic) trigonometric identities implies $\cosh \eta = \gamma = {1 \over \sqrt{1 - v^2}}$ and $v \gamma = \sinh \eta$, so that this reparametrization is indeed correct.

Now, hopefully this reminds you a little of something. In two-dimensional Euclidean plane we have that rotations around the origin have the form $$x' = \cos \phi x + \sin \phi y$$ $$y' = -\sin \phi y + \cos \phi x$$ and this is indeed no coincidence. Rotations preserve a length of vector in Euclidean plane $x'^2 + y'^2 = x^2 + y^2$ and similarly, Lorentz transformations preserve space-time interval (which is a notian of length in Minkowski space-time) $x'^2 - t'^2 = x^2 - t^2.$ You can check for yourself that only the stated transformation with hyperbolic sines and cosines can preserve it and consequently the change you introduced will spoil this important property. Also, if you are familiar with phenomena like relativity of simultaineity, one could also argue on physical grounds that your proposed change can't lead to physical results.

Incidently, there has recently been asked similar question to yours, namely how to derive that the transformation is linear purely because of the preservation of space-time interval. You might want to check it out too.

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You should look at this answer, because it derives the term you want right away. Einstein's postulates <==> Minkowski space. (In layman's terms)

The reason it's $t'=(t-vx)/\sqrt{1-v^2}$ and not $t'=t/\sqrt{1-v^2}$ (you must set c=1 to follow anything in relativity) is simple--- it's failure of simultaneity at a distance. The coordinate lines t=constant can't stay horizontal in a space-time diagram--- they have to get tilted up by the same amount that the time axis is tilted left. The remaining factors can be understood by reproducing time-dilation and length-contraction arguments, but failure of simultaneity is the most important nonintuitive effect, and it is the first discussed by Einstein in his paper, for this reason.

The form of the Lorentz transformation should be constrasted with the form of a rotation of the x and y coordinates, so that the x coordinate gets a slope of m:

$$x' = { x+my \over\sqrt{1+m^2}}$$ $$y' = { y-mx \over\sqrt{1+m^2}}$$

or if you use different units for x and y, say x in inches and y in centimeters,

$$x' = { x + my \over \sqrt{1+{m^2\over c^2}} }$$ $$y' = {y - {mx\over c^2}\over \sqrt{1+{m^2\over c^2}} }$$

Where c is a universal constant of nature: the isoceles slope of right, which is the slope of an isoceles right triangle with legs along the x and y axis. It's magnitude is 2.54 cm/inch.

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protected by Qmechanic Jan 19 at 18:58

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