Schutz GR Book, Question about World line.

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The discussion centers on a question regarding a spacetime diagram from B. Schutz's book "A First Course in General Relativity." It clarifies that a world line represents the path of a particle in spacetime, with the vertical axis as time (t) and the horizontal axis as position (x). For velocities less than the speed of light (|v| < 1), the world line has a negative slope, indicating that it moves from the upper left to the lower right. This occurs because there are two possible values for v that satisfy |v| = 0.5, allowing for both positive and negative slopes. The omission of this explanation by Schutz is noted as a potential point of confusion for readers.
Lou Arnold
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This isn't homework, nor is it an exercise problem; merely a question about a diagram.

Re: B.Schutz book "A First Course in General Relativity" 2nd Edition, (Asian print version), page 5, Figure 1.1 "A spacetime diagram in natural units".

From section 1.4 Spacetime diagrams:
A world line is the locus of events observed by the reference frame as a particle moves with velocity v.
The vertical axis is t. The horizontal axis is x.
The equation is: slope=dt/dx = 1/v where v=1 is the speed of light.
When v=1, the slope of the world line is positive at 45 degrees. When v>1, the slope is positive but less than 45 degrees.

Q: The World line for |v|<1 has a negative slope - goes from upper left to lower right. Can someone explain why this slope is negative and why not positive greater than 45 degrees?

Lou.
 
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Welcome to PF.

Suppose that you wish to draw a world line corresponding to |v| = .5, which is an example of |v| < 1. There are two possible values of v that satisfy |v| = .5; namely, v = +.5 and v = -.5. So, if you wanted to draw a world line that satisfies |v| = .5, you have a choice of drawing a line with a positive slope of 2 or a line with a negative slope of -2. Either line would make an angle with respect to the x-axis that has a magnitude greater than 45o. Schutz chose to draw a line corresponding to the negative value of v that satisfies |v| < 1.
 
I understand your logic. Not very kind of Schutz to omit that small explanation
Thanks for your help.
I see dizzying notation coming in the next sections. I may need more help.
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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