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What is a Lorentz reflection of $\mathbb R^3$? Is there a way to visualize it? Suppose I have a plane, P, what would (Lorentz) reflecting in it differ from (Euclid) reflecting in it?

I know that the Lorentzian metric is one where the inner product $(x,y)=x_1y_1+x_2y_2-x_3y_3$ but is the reflection along the (Euclidean) orthogonal to P but ending up at points with distance wrt the given metric?

In the Euclidean case, for $n\in \mathbb R^3$ such that $n$ is a unit vector normal to a plane P, the reflection in P is given by $R(x)=x-2(x,n)n$ , where $x\in \mathbb R^3$.

Is the equation for the Lorentzian case the same except the inner product $(.,.)$ is the Lorentzian one?

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I don't think this is commonly used terminology in physics. Certain authors may define it, but if you use the term, probably other people will not understand you. Physicists usually define a time reflection operator T that does $t\rightarrow -t$, and a spatial reflection operator P (for parity) according to $(x,y,z)\rightarrow(-x,-y,-z)$. –  Ben Crowell Feb 27 '12 at 1:36
    
Using the Lorentz metric, not all nonzero vectors $v$ can be normalized to have $(v,v) = 1$, but rather to $\pm 1$ if $(v,v) \not= 0$. In ${\mathbf R}^3$, for a nonzero vector $v$ the Euclidean reflection across the plane perp. to $v$ is given by $R(x) = x - 2((x,v)/(v,v))v$. This formula works regardless of whether or not $(v,v)=1$. The same exact formula defines a reflection when $(\cdot,\cdot)$ is the Lorentz metric, provided $(v,v) \not= 0$. In Euclidean geometry the only vector whose norm is 0 is the vector 0, but in Lorentz geometry more vectors have norm 0 than the zero vector. –  KCd Jan 29 '13 at 12:52
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1 Answer 1

Expanding the comment by KCd: if $(\cdot , \cdot )$ is a symmetric bilinear form on $\mathbb R^n$ and $v$ is a vector such that $(v,v)\ne 0$, then the map $$ R(x)=x-2\frac{(x,v)}{(v,v)}v \tag1 $$ defines an involution on $\mathbb R^n$ that preserves $(\cdot , \cdot )$.

Involution: since $(R(x),v)=-(x,v)$, it follows that $$R(R(x)) = R(x)+2\frac{(x,v)}{(v,v)}v = x \tag2$$

Form-preserving: $$ \begin{split} (R(x),R(y))&=(x,y)-2\frac{(x,v)}{(v,v)}(v,y)-2\frac{(y,v)}{(v,v)}(v,x)+4\frac{(x,v)(y,v)}{(v,v)^2}(v,v) \\ & = (x,y) \end{split} \tag3 $$

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No pictures? But it's tagged [intuition]! :P –  Mario Carneiro Jun 13 '13 at 14:49
    
@MarioCarneiro If the OP (who put the tag) requests a picture, I'll try to come up with one. But s/he was last seen Feb 27 '12. –  ˈjuː.zɚ79365 Jun 13 '13 at 14:55
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