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

CaspianXI
2009-03-19 15:59:11

If you're low on time, here's an easy way to solve this problem.

$x' = \gamma - \gamma v t$
$t' = \gamma - \gamma v x$

Thus, the first terms in both x' and t' must be the same. And the coefficients of the second terms for both x' and t' must be the same. (C) is the only one which satisfies this, so you're done.

CaspianXI
2009-03-20 20:55:39

Oops... typo. I *meant* to list the equations as:

$x' = \gamma x - \gamma v t$
$t' = \gamma t - \gamma v x$

The rest of the logic still applies.

WARNING: $t' \neq \gamma t - \gamma v x$ in general. This is only true because we've been given that we're using some funky units where c = 1.

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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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