Schutz GR Book, Question about World line.

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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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.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?

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