Red-shift question from another thread.

  • Thread starter Jufro
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  • #1
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Homework Statement


This is not a HW questions but from another thread.
https://www.physicsforums.com/showthread.php?p=4495692&posted=1#post4495692

The statement I made was that if z increases for a source, then it is accelerating away. Or, it could that if z is constant then the sources moves with constant velocity away from an observer.

I just need to know where my logic is breaking down and what direction I can try to find the right answer.


Homework Equations



I took the equation:
1+z = [itex]\sqrt{\frac{1+v/c}{1-v/c}}[/itex]

The Attempt at a Solution



So taking d/dt on both sides I end up with:

dz/dt = 1/2 [itex]\frac{1+v/c}{1-v/c}-1/2[/itex]*[itex]\frac{(1/c dv/dt)*(1-v/c)-(1/c dv/dt)*(1+v/c)}{(1-v/c)2}[/itex]

Re-writting this:

dz/dt= [itex]\frac{-v* dv/dt}{c2*(z+1)*(1-v/c)2}[/itex]

Since the first term is negative (from the minus sign) and v is positive, then a positive dz/dt would require that dv/dt is negative or that the sources acceleration is radially inward when I had suspected outward.

This may be a result of me using the flat-spacetime (Minkowski metric) but I am not sure. Can anyone point me in the right direction.

Please and thank you.
 

Answers and Replies

  • #2
vela
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Homework Statement


This is not a HW questions but from another thread.
https://www.physicsforums.com/showthread.php?p=4495692&posted=1#post4495692

The statement I made was that if z increases for a source, then it is accelerating away. Or, it could that if z is constant then the sources moves with constant velocity away from an observer.

I just need to know where my logic is breaking down and what direction I can try to find the right answer.


Homework Equations



I took the equation:
$$1+z = \sqrt{\frac{1+v/c}{1-v/c}}$$

The Attempt at a Solution



So taking d/dt on both sides I end up with:
$$\frac{dz}{dt} = \frac{1}{2} \left(\frac{1+v/c}{1-v/c}\right)^{-1/2} \frac{(1/c\ dv/dt)(1-v/c)-(1/c\ dv/dt)(1+v/c)}{(1-v/c)^2}.$$ Re-writing this:
$$\frac{dz}{dt}= \frac{-v\ dv/dt}{c^2(z+1)(1-v/c)^2}.$$ Since the first term is negative (from the minus sign) and v is positive, then a positive dz/dt would require that dv/dt is negative or that the sources acceleration is radially inward when I had suspected outward.

This may be a result of me using the flat-spacetime (Minkowski metric) but I am not sure. Can anyone point me in the right direction.

Please and thank you.
You dropped a sign when you applied the quotient rule. The numerator should end up as a sum, not a difference.
 
  • #3
92
8
Ah, I used the wrong derivative in the second half of my quotient rule. Thanks for that.
 

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