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It is very long time ago that I took a physics lesson, so I want to refresh my memory. I think I learned that there is only one inertial frame in Minkowski spacetime (or special relativity time) that $ct$ and $x$ are orthogonal. (inertial frames are assumed as Lorentz-invariant, and we assume one space axis, $x$.) So, why is it?

($ct$ is in vertical axis, $x$ is in horizontal axis.)

To avoid confusion: I think I found a question on my old textbook :)

Edit: orthogonality in Euclidean term is assumed.

Show that the $S'$ axes, $x'$ and $ct'$, are nonorthogonal in a spacetime diagram. Assume that the $S'$ frame moves at the speed of $v$ relative to the $S$ frame ($S'$ is moving away from $S$ to the side of $+x$) and that $t = t' = 0$ when $x = x' = 0$. (The $S$ axes, $x$ axis and $ct$ axis, are defined orthogonal in a spacetime diagram.)

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Your textbook uses misleading notation. The direction of the $S'$ axes in an $S$ spacetime diagram look non-orthogonal in the euclidean sense, which is the wrong way to look at it: the geometry of spacetime uses the Lorentz inner product, with respect to which both the $S$ and $S'$ are indeed orthogonal. – Emilio Pisanty Jun 20 '12 at 14:07
@EmilioPisanty Right.. let us just assume Euclidean sense. – Bill Fritzgard Jun 20 '12 at 14:21
That's just the thing. To your euclidean eyes, the axes don't look orthogonal, but as far as spacetime goes then they're perfectly fine. You might wish to call Lorentz-inner-product orthogonality something different, say, "good-angle-nality", but then the result you are asking about simply says that for every particle worldline there exists a unique spatial axis that is good-angle-nal to it. After a while, you'll get tired, and realize that it's easier to call it orthogonal while keeping in mind that it means something different from the (euclidean!) usual. – Emilio Pisanty Jun 20 '12 at 15:11
@EmilioPisanty Thanks. – Bill Fritzgard Jun 20 '12 at 15:15

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This is false.

Each inertial frame has spatial and temporal axes that are orthogonal to each other (in the Lorentz inner product, of course). Of course, different frames have different axes: temporal axes differ (to reflect relatively-moving origins, which also happens in galilean relativity) and spatial axes also differ (to reflect relativity of simultaneity, which does not happen in galilean relativity).

You may be thinking of the fact that if we concentrate on some inertially-moving particle, then there is a unique spatial axis that is orthogonal to the particle's worldline. This axis is of course the spatial axis of the rest frame of the particle: the Lorentz frame that has its temporal axis along the particle's worldline.

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You're right. Sorry for my comment. – Bill Fritzgard Jun 20 '12 at 14:19
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There is a way to formulate your question that it has a nontrivial answer-- suppose you have both a euclidean and Lorentzian notion of perpendicularity. So that the dot product of two vectors has two meanings:

$$ A\cdot_E B = A_t B_t + A_x B_x$$ $$ A\cdot_L B = A_t B_t - A_x B_x$$

Then if both $A\cdot_E B = 0$ and $A \cdot_L B = 0$, then by adding and subtracting the equations you learn that $A_t B_t = A_x B_x=0$, so that exactly one of the vectors has a component along the t axis, and exactly one of the vectors have a component along the x-axis, and this proves the result.

The reason is that when you do a Lorentz rotation (a boost), you tilt the time axis toward the velocity vector, and you tilt the space axis toward the velocity vector too, so that they hug the 45-degree line of light propagation more closely, so the result is obvious from the picture--- the only time they are Euclidean perpendicular is when you are at rest.

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