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The Lorentz transformation is a mapping of the $x', y', z',$ and $t'$ coordinates relative to $x, y, z, t$ that preserves the speed of light:

(1) $x^2 + y^2 + z^2 - c^2t^2 = x'^2 + y'^2 + z'^2 - c^2t'^2 = 0$

If $K'$ is moving with speed $v$ in the $x$ direction relative to $K$ (so that $K$ is moving with speed $-v$ in the $x$ direction relative to $K'$), and we use the Galilean transformation, we get an erroneous

(2) $x'^2 + y'^2 + z'^2 - c^2t'^2 = -2vxt + v^2t^2$

which is NOT equal to zero, and hence does not describe a sphere of radius $ct$ emanating from the origin of $K'$, but something else (I'm pretty sure it's a sphere that's been shifted negatively in the $x$ direction by a small amount, relative to $K'$). The Lorentz transformation says we modify both the $x' = x - vt$ and the $t' = t$ relationships in order to ensure that we end up getting 0.

However, as a naive mathematician, I look at equation (2) and think, "Ok, so we've got to modify the rule for $x'$, or $t'$, or both, to get rid of that nasty $-2xvt + v^2t^2$ term in the $x'$ rule. Why not just set $t'$ up so that instead of coming out to $-c^2t^2$, it actually comes out to $-c^2t^2 + 2vxt - v^2t^2$, so that the extra terms cancel out?" I naively went ahead with this idea and I got the following coordinate transformation:

$$ x' = x - vt\\ y' = y\\ z' = z\\ t' = \frac{1}{c} \sqrt{(v^2 + c^2)t^2 - 2vxt} $$

If you plug this into (1), you get 0 like you're supposed to.

Now, I haven't been able to find an explanation as to why this transformation is unsatisfactory. Of course, it's no longer linear - but is there some basic reason why the transformation must be linear? Obviously this transformation satisfies Einstein's postulate about the constancy of the speed of light for all inertial observers.

Presumably, then, it must violate the first postulate, regarding the equivalence of all physical phenomena for inertial observers - otherwise we'd need more information in order to derive the Lorentz transformation, which I think is supposed to be derivable from the two postulates alone.

So, in the end, I have one question I'm trying to answer:

  1. Does this transformation violate equivalence for inertial observers? Why? (ideally, someone could give me an example of a force that suddenly appears in my coordinate system that didn't already exist in the other coordinate system)
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  • $\begingroup$ @BenSelfridge In intuitive terms, beyond preserving the interval: A physically acceptable transformation is expected to transform any linear uniform motion into a linear uniform motion. Your transformation maps every linear uniform motion that passes through the origin into a similar one, but does not do so for arbitrary linear motions that do not intercept the origin. Interesting example anyway. $\endgroup$ – udrv Aug 6 '16 at 13:19
  • $\begingroup$ Lorentz transformations in special relativity is not about conservation of some "$x^2-t^2$" form, with x and t being coordinates. They're about conserving the metric at every point in spacetime. The former is not really geometric, while the latter is. $\endgroup$ – user12029 Aug 6 '16 at 21:54
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Flat spacetime in special relativity is the same everywhere, so we must have translational symmetry in time and space. If we start with two events separated by $\Delta x^\mu$ in one frame, the separation $\Delta x'^\mu$ in a boosted frame should only depend on $\Delta x^\mu$, and not on $x^\mu$ itself. This is only satisfied for linear transformations.

Using your transformations, observers would be able to identify a privileged point. This violates the equivalence principle even for observers at rest relative to each other, since it depends on where they put their origins.

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  • $\begingroup$ I'm sweeping some details under the rug. This only works for SR in Cartesian coordinates, and translational symmetry isn't really a direct postulate of SR but a consequence of other things. But this is the cleanest way to show why transformations like the one above are unsuitable. $\endgroup$ – knzhou Aug 5 '16 at 19:15
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    $\begingroup$ See, I can make sense out of this mathematically, but it's hard to get a physical intuition. Is there any way you can describe an experiment whereby the observer at the origin of K' would be able to tell he was moving? $\endgroup$ – Ben Selfridge Aug 5 '16 at 19:19
  • $\begingroup$ I think I may have put it in terms I can understand. $\endgroup$ – Ben Selfridge Aug 6 '16 at 15:21
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1. Response to your specific example: Suppose you and I are both located at a point we choose to call the origin. You are traveling at velocity $3/5$ with respect to me. (I take $c=1$).

Our friend, one light year away according to me, flashes a light at time $t=1/2$.

According to you, that light flashes at time $i\sqrt{13/50}$.

So I see the light flashing at a real time, while you see it flashing at an imaginary time, whatever that means. Presumably it means you don't see it flashing at all. That's one way you know you're moving.

2. Response to the more general question of "why linear?": More fundamentally: If we both watch a third person moving away from the origin at some other velocity, and if no forces are acting on that person, and if he passes through both $(x_1,t_1)$ and $(x_2,t_2)$ (according to me) then we must have $x_1/t_1=x_2/t_2$. And if you must agree that he has no forces acting on him, then we must also have $x_1'/t_1'=x_2'/t_2'$. These considerations will force the implication $$(x_1 t_2=x_2 t_1) \Rightarrow (x_1' t_2'=x_2' t_1')$$ which very much restricts the allowable transformations.

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  • $\begingroup$ Ok, this may be the answer for me. Here's how I interpret it. If x'(t') = at' is the position of some mass moving at a constant velocity in K', then we would expect x(t), the position of the mass in terms of K, to be some linear function in t. But if you do the derivative of x for this mass, you don't get a constant. I can't do the math myself, can someone help me? $\endgroup$ – Ben Selfridge Aug 6 '16 at 15:27
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This is where it is good to distinguish (x,y,z,t) as the coordinates on Minkowski space vs in a tangent space at any point. The Minkowski space comes to you as a manifold, but the tangent space comes to you as a vector space (so it makes sense to ask for linearity). It is then a happy case that you can identify them. This is what the $\Delta x$ of @knzhou answer is trying to say by moving you to the vector space instead of the manifold. If you look at what categories your objects are in, what you can do to them becomes no choice at all.

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  • $\begingroup$ I agree that we should be working with the tangent spaces, not the manifold. But I don't think that's enough to conclude that the transformations must be linear. There are lots of categories whose objects are vector spaces. The question comes down to: Why are we looking at the category of vector spaces and linear maps, as opposed to, say, the category of (topological) vector spaces and homomorphisms? $\endgroup$ – WillO Aug 5 '16 at 22:38
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    $\begingroup$ @WillO I think the key physical point is that the maps we're considering here are not maps between spacetime points. Lorentz transformations are passive; they map observers' coordinates, not the points the coordinates refer to. $\endgroup$ – knzhou Aug 6 '16 at 19:25
  • $\begingroup$ This is still not enough to get linearity, since you could, say, switch between Cartesian and polar coordinates. But the second thing is that in SR, we explicitly care about coordinate systems where the coordinates have a direct physical interpretation in terms of networks of clocks and rulers. That restricts us to 'Cartesian-like' coordinates and forces us to use linear maps. You can have "polar" Lorentz transformations but of course they're not linear in the coordinates. $\endgroup$ – knzhou Aug 6 '16 at 19:26
  • $\begingroup$ @knzhou: I don't object to anything you say, but I do not think it addresses the question. You've shown that if we impose certain requirements, we can deduce linearity. The OP asked whether, if we impose certain other requirements, we can deduce linearity. So I think you're giving a perfectly good answer to a question that wasn't asked. $\endgroup$ – WillO Aug 6 '16 at 19:51
  • $\begingroup$ It's the fact that it's on the tangent space/passive+the fact that you must agree on light. But that gives you orthogonal transformation of the vielbien. To just get linear you want that the people can coherently match up what they mean by adding and scaling vectors. $\endgroup$ – AHusain Aug 6 '16 at 21:35

protected by Qmechanic Aug 5 '16 at 23:03

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