GREPhysics.NET: GR9277 #94

GR9277 #94

Problem

GREPhysics.NET Official Solution

\prob{94}
Which of the following is a Lorentz transformation? (Assume a system of units such that the velocity of light is 1.)

$ x^'&=&4x\\ y^'&=&y\\ z^'&=&z\\ t^'&=&.25t $
$ x^'&=&x-0.75t\\ y^'&=&y\\ z^'&=&z\\ t^'&=&t $
$ x^'&=&1.25x-0.75t\\ y^'&=&y\\ z^'&=&z\\ t^'&=&1.25t-0.75x $
$ x^'&=&1.25x-0.75t\\ y^'&=&y\\ z^'&=&z\\ t^'&=&0.75t-1.25x $
None of the above

Special Relativity $\Rightarrow$ }Lorentz Transformation

Lorentz transformations are given by

$x^{'} = \gamma (x-vt)$

$t^{'} = \gamma (t - vx/c^2)$

Factoring out the terms, choice (C) is $x=5/4(x-3/5t)$ , and thus $\gamma=5/4$ and $v=3/5$ . Since the equation for t fits the form above, this is a valid Lorentz Transformation.

Alternate Solutions

student2008
2008-10-16 13:54:16

LOL! The simplest way here is to calculate the interval $S^2=(t')^2-(x')^2$ , it should be equal to $t^2-x^2$ . This can't be true for (A) & (B), and in (D) $S\equiv 0$ . But, it is correct for (C).

In general situation (i.e. when y's and z's also change) this is the only practical method. You would get stuck using the general form of transformations.

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Comments

tinytoon
2008-11-07 02:18:56

I think Yosun's solution is valid, but the easiest way to do it is to recognize that Lorentz transformations are the following:

$x' = \gamma(x-vt)$

and

$t' = \gamma (t-vx/c^2)$

In units of c = 1, which the problem kindly gives us, the second transformation becomes:

$t' = \gamma (t-vx)$

Now we can clearly see that the $\gamma$ 's and the $v$ 's are symmetrical in the two transformations and out pops (C).

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student2008
2008-10-16 13:54:16

neon37
2008-10-29 01:52:17

Note: ds=0 for light.

$ds^2 \gt 0$ space-like
$ds^2 = 0$ null
$ds^2 \lt 0$ time-like

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FortranMan
2008-10-12 08:58:48

you can eliminate D by remembering that $\gamma$ must be greater than 1. To absolutely eliminate E, note that here $\frac{v}{c^2} = v$ , then double check $\gamma$ .

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evanb
2008-06-26 19:37:47

Actually, that's not enough to satisfy that it's a Lorentz transformation.

Consider:

x' = 10x - 1t
t' = 10t - x

According to the official solution, this would be a Lorentz transformation with $\gamma=10$ and $\beta=0.1$ , but we can check that such a transformation is unphysical: $\gamma$ and $\beta$ are not independent, and because

$\frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-0.1^2}} = \frac{1}{\sqrt{0.99}} << 10 = \gamma$, this is not a legit Lorentz transform.

However, since the numbers here --happen-- to be 5/4 and 3/4, we're in the clear.

Maxwells_Demon
2008-09-21 17:12:38

Wait, how are we in the clear??????? If gamma really is 5/4 then beta should be 3/5. Beta is clearly 3/4 = 0.75

So how is it a valid transformation?

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