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Assume the Lorentz transformations obey the relationship $$g_{uv}\Lambda^u_{p}\Lambda^v_\sigma = g_{p\sigma},$$ where $g_{uv}$ is the metric tensor of special relativity.

How can one show, under that assumption, that the Lorentz matrix $\Lambda^a_b$ has an inverse?

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5 Answers

up vote 3 down vote accepted

You may order the matrices like this: $$ \Lambda_\rho^\mu g_{\mu\nu} \Lambda^\nu_\sigma = g_{\rho\sigma} $$ I suppose all the letters should have been Greek. They're called mu, nu, rho, sigma, good to learn them.

In my form, one may view $\mu$ as the summed over index in the first product on the left hand side and $\nu$ as the summed over index in the second product. So making a convention for a matrix $\Lambda$ so that its components are $\Lambda^\mu_\rho$ where $\rho$ is the row and $\mu$ is the column, the equation above is the matrix equation $$ \Lambda \cdot g \cdot \Lambda^T = g $$ where $T$ means transposition. The matrix on the right hand side is nonsingular, i.e. it has a nonzero determinant, so the factors on the left hand side must also have a nonzero determinant i.e. be invertible.

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Assume the Lorentz transformation $\Lambda$ is not invertible. Then it is in particular not injective and there exists $0\not=u\in\ker\Lambda$.

The inner product $g$ is nondegenerate so there's a vector $v$ with $g(u,v)\not=0$ and we end up with the contradiction $$ 0\not=g(u,v)=g(\Lambda u, \Lambda v)=g(0,\Lambda v)=0 $$ where we have used the fact that Lorentz transfomations leave the inner product invariant, which is exactly your starting equation.

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There is typo: 0≠u∈kerΛ. instead of u≠0∈kerΛ –  Shaktyai Sep 21 '12 at 16:39
    
@Shaktyai: that's intentional as the important thing is that $u\in\ker\Lambda$, and not $0\in\ker\Lambda$ - read it as $(0\not=u) \wedge (u\in\ker\Lambda)$, same as $a=b=c$ is $(a=b)\wedge(b=c)$; I lost a $\Lambda$, though, which is now fixed (the equation was still correct by accident) –  Christoph Sep 21 '12 at 18:12
    
0∈kerΛ is always true for any linear operator... –  Shaktyai Sep 21 '12 at 19:50
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Starting from $$ g_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma = g_{\rho\sigma} $$ we contract with $g^{\sigma\tau}$ $$ g_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma g^{\sigma\tau} = g_{\rho\sigma} g^{\sigma\tau} = \delta_\rho^\tau $$ and reorder the factors $$ g_{\mu\nu} g^{\sigma\tau} \Lambda^\nu{}_\sigma \cdot \Lambda^\mu{}_\rho = \delta_\rho^\tau $$ which shows that $\Lambda$ has a left-inverse and is thus injective.

As $\Lambda$ is an endomorphism, this is sufficient to show it's invertible and we have $$ (\Lambda^{-1})^\tau{}_\mu = g_{\mu\nu} g^{\sigma\tau} \Lambda^\nu{}_\sigma = \Lambda_\mu{}^\tau $$

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this is essentially Chris' answer –  Christoph Sep 21 '12 at 17:23
    
Much appreciated! –  Mew Sep 21 '12 at 22:33
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I have come up with the following proof:

Begin with the relationship,

$g_{vu}\Lambda^u_{p}\Lambda^v_\sigma = g_{p\sigma}$

Which is the same as,

$\Lambda_{pv}\Lambda^v_\sigma = g_{p\sigma}$

Now multiply both sides of the equation by $g^{ap}$ to yield,

$g^{ap}\Lambda_{pv}\Lambda^v_\sigma = g^{ap}g_{p\sigma}$

This simplifies to:

$\Lambda^a_{v}\Lambda^v_\sigma = \delta^a_\sigma$

Also, we know that

$(\Lambda_v^a)^{-1}\Lambda^v_\sigma = \delta^a_\sigma$ By definition of inverse.

Therefore, $(\Lambda_v^a)^{-1} = \Lambda^a_{v}$

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+1: this is a correct answer, though it misses some subleties due to $\Lambda^\mu{}_\nu\not=\Lambda_\nu{}^\mu$ - see my 2nd answer –  Christoph Sep 21 '12 at 17:29
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$$g_{uv}\Lambda^u_{p}\Lambda^v_\sigma=g_{p\sigma} \Longleftrightarrow \Lambda^Tg\Lambda=g$$ where g is the matrix whose entries are $g_{uv}$ $$ \det(\Lambda^Tg\Lambda)=\det(g) $$ $$ \det(\Lambda^T)\det(g)\det(\Lambda)=\det(g) $$ Obviously, $\det(g)\neq0$ and $\det(\Lambda^T)=\det(\Lambda)$

Then $$\det(\Lambda)^2=1$$

Since $\det(\Lambda)$ never vanishes, the matrix $\Lambda$ is always invertible.

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