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A Lorentz boost along the x-axis conserves area in the x-t plane. Can anyone think of an argument to prove this fact that's simpler than the one below? I don't want to assume the form of the Lorentz transformation, because I want to use the conservation of area as part of a derivation of the Lorentz transformation. I also don't want to assume that c is the same in all frames. The only assumptions are supposed to be:
1 Spacetime is homogeneous and isotropic. No point has special properties that make it distinguishable from other points, nor is one direction distinguishable from another.
2 Inertial frames of reference exist. These are frames in which particles move at constant velocity if not subject to any forces. We can construct such a frame by using a particular particle, which is not subject to any forces, as a reference point.
3 Equivalence of inertial frames: If a frame is in constant-velocity translational motion relative to an inertial frame, then it is also an inertial frame. No experiment can distinguish one preferred inertial frame from all the others.
4 Causality: Observers in different inertial frames agree on the time-ordering of events.
5 No simultaneity: The experimental evidence shows that observers in different inertial frames do not agree on the simultaneity of events.
Start with a rectangular area in the x-t plane, and for simplicity, let this rectangle have unit area. Then the area of the new parallelogram is still 1, by the following argument. Let the new area be A, which is a function of v. By isotropy of spacetime (assumption 1), A(v)=A(-v). Furthermore, the function A(v) must have some universal form for all geometrical figures, not just for a figure that is initially a particular rectangle; this follows because of the definition of affine area in terms of a dissection by a two-dimensional lattice, which we can choose to be a lattice of squares. Applying boosts +v and -v one after another results in a transformation back into our original frame of reference, and since A is universal for all shapes, it doesn't matter that the second transformation starts from a parallelogram rather than a square. Scaling the area once by A(v) and again by A(-v) must therefore give back the original square with its original unit area, A(v)A(-v)=1, and since A(v)=A(-v), A(v)=[itex]\pm1[/itex] for any value of v. Since A(0)=1, we must have A(v)=1 for all v. The argument is independent of the shape of the region, so we conclude that all areas are preserved by Lorentz boosts. The argument is also purely one about affine geometry (it would apply equally well to a Euclidean space), so there is no reason to expect the area A in the (t,x) plane to have any special physical significance in relativity; it is simply a useful mathematical tool in the present discussion.
For anyone who wants to see the whole derivation of the Lorentz transformation from assumptions 1-5, it's here: http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.7 . I'm basically looking to see if I can make the argument simpler and easier to understand for people with less math and physics background.
Thanks in advance!
-Ben
1 Spacetime is homogeneous and isotropic. No point has special properties that make it distinguishable from other points, nor is one direction distinguishable from another.
2 Inertial frames of reference exist. These are frames in which particles move at constant velocity if not subject to any forces. We can construct such a frame by using a particular particle, which is not subject to any forces, as a reference point.
3 Equivalence of inertial frames: If a frame is in constant-velocity translational motion relative to an inertial frame, then it is also an inertial frame. No experiment can distinguish one preferred inertial frame from all the others.
4 Causality: Observers in different inertial frames agree on the time-ordering of events.
5 No simultaneity: The experimental evidence shows that observers in different inertial frames do not agree on the simultaneity of events.
Start with a rectangular area in the x-t plane, and for simplicity, let this rectangle have unit area. Then the area of the new parallelogram is still 1, by the following argument. Let the new area be A, which is a function of v. By isotropy of spacetime (assumption 1), A(v)=A(-v). Furthermore, the function A(v) must have some universal form for all geometrical figures, not just for a figure that is initially a particular rectangle; this follows because of the definition of affine area in terms of a dissection by a two-dimensional lattice, which we can choose to be a lattice of squares. Applying boosts +v and -v one after another results in a transformation back into our original frame of reference, and since A is universal for all shapes, it doesn't matter that the second transformation starts from a parallelogram rather than a square. Scaling the area once by A(v) and again by A(-v) must therefore give back the original square with its original unit area, A(v)A(-v)=1, and since A(v)=A(-v), A(v)=[itex]\pm1[/itex] for any value of v. Since A(0)=1, we must have A(v)=1 for all v. The argument is independent of the shape of the region, so we conclude that all areas are preserved by Lorentz boosts. The argument is also purely one about affine geometry (it would apply equally well to a Euclidean space), so there is no reason to expect the area A in the (t,x) plane to have any special physical significance in relativity; it is simply a useful mathematical tool in the present discussion.
For anyone who wants to see the whole derivation of the Lorentz transformation from assumptions 1-5, it's here: http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.7 . I'm basically looking to see if I can make the argument simpler and easier to understand for people with less math and physics background.
Thanks in advance!
-Ben
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