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GR- Energy conservation, effective potential graph sketch

  • Thread starter binbagsss
  • Start date
  • #1
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Homework Statement





vrq.png


I would like to ask about parts c) and d) , the graph sketching bits.


2. Homework Equations


##V(r) = ( \frac{J^2}{r^2}+\epsilon)(r-\frac{1}{r}) ##

where the value of ## \epsilon ## is set according to whether time-like or null etc.

The Attempt at a Solution



Q1)for part c)

##V(r) = ( \frac{J^2}{r^2})(r-\frac{1}{r}) ##

I am able to deduce the shape of the curve from the fact that the potential ##V'(r)=0 ## is a quadratic and solveable, and so i can check whether this is a stable or unstable point. So to sketch I:
a) ## \lim r \to \infty ## and ## \lim r\to 0 ##
b) to deduce the cts shape consider the critical point and check whether it is stable or unstable.
and that gives me the following which is fine.

(giving me the following:
vr null.png
)

however for part d) instead we have ##V(r) = ( \frac{J^2}{r^2}+1)(r-\frac{1}{r}) ## and now ##\lim r \to \infty = r ## but ##\lim r \to 0 = -1/r^3 ##. from this I can draw the start and end , but am unsure how to connect in the middle- curving upward or downward. compared to the above, null example, where the critical point is solvable and i can check stability to deduce the shape, here the critical point is unsolvable so I am unsure how to do this.

I have the following so far:

vr me.png


the answer is:
vr tl.png

I AM UNSURE WHY, how to deduce the shape on the 'jointy' bit.

Many thanks in advance.

(I believe there may be an error in the solution since for a time-like geodesic instead of ##\epsilon=1 ## it should be ##\epsilon=-1## however I am interested in the principal of the graph sketching, so, assuming the graph sketch is consistent with the value of epsilon chosen for the solution, I'd like to ignore this please and focus on the graph sketching. Thanks).
 

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Answers and Replies

  • #2
1,206
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bump
many thanks
 

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