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For S and S' in standard configuration, the Galilean transformations are:

x' = x - vt, y' = y, z' = z, t' = t

From the Lorentz transformations for v << c:

x' = x - vt, y' = y, z' = z, t' = t - vx/c^2

So it looks as if the Galilean transformations become increasingly accurate for:

vx -> 0, v << c

And exact for v = 0 for all x.

Yet, all text books I've come across state that the Galilean transformatons become more accurate for the condition v << c only.

So what are the conditions under which the Galilean transformations become more accurate and why?

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You have already written it. For $v < < c$ it is most accurate. And $t\prime = t$ for this case. For $v = 0$ you have nothing to transform!! What exactly is your question? – user1355 Feb 6 '11 at 16:42
More accurate than what? If you mean more accurate than the Lorentz transform the answer is "never", though there are certainly conditions when it is adequate to the available precision for any given measurement. – dmckee Feb 6 '11 at 21:25
@sb1 my question is as in the title, but I wanted to make sure that people didn't just state the condition v << c stated in all text books and not explain why vx -> 0 isn't included, as you've done in your comment. – John McVirgo Feb 6 '11 at 22:53
up vote 4 down vote accepted

If I understand correctly, the question is about whether, in deriving the Galilean transformation as an approximate limiting case of the Lorentz transformation, it is necessary to impose the requirement that $vx$ is small, in addition to the "obvious" requirement that $v$ is small. The answer is yes, it is necessary. Of course, since $vx$ is not a dimensionless quantity, we have to specify what we mean when requiring that it is small. The specific requirement is that $vx/c^2\ll \Delta t$, where $\Delta t$ is the precision with which we wish to calculate time intervals.

Suppose that, in the inertial reference frame in which you are at rest, two stars explode simultaneously, one here and one in the Andromeda galaxy. Consider the same two events in the reference frame of someone walking at a liesurely pace. The Lorentz transformation indicates that they will be separated by a time interval of about 2 days. If you care about levels of precision less than that, you can't use the Galilean transformation, even though $v/c$ is small.

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I have not quite understood your problem. For $v << c$, $vx/c^2$ is negligible unless $x$ is huge. Hence $t\prime = t$ We get the Galilean transformation back. I guess, you want to get back the Galilean transformation "exactly". No John, that never happens. All we can say is Galilean transformation is sufficiently accurate for description of transformations from one inertial frame to another for the $v << c$ case. If the separation between two events is huge then yes, even for ordinary velocities you can't use Galilean transformations, you have to use Lorentz transformations.

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It means that even though v << c, using the GT to transform large accelerations as 4-vectors, for example, won't give the expected accuracy. Which is why vx -> 0 is important to keep in mind. – John McVirgo Feb 6 '11 at 23:13

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