Why Is dz/dt Negative in Cosmology?

In summary, the question is asking for the sign and physical explanation of dz/dt, which represents the change in redshift over time. The equation for dz/dt is (1+z)H_o[1-(1+z)^(3/2)], and it is negative for z>0 due to the decelerating nature of the model. This is supported by a graph and is physically explained by the decreasing redshift with time.
  • #1
Sirius24
14
0

Homework Statement


I am working with cosmology and expansion.
Obtain dz/dt as a function of z and H_o only. What is the sign for dz/dt and why?
z is the redshift and H_o is the Hubble constant.

Homework Equations



The Attempt at a Solution



So I found the equation for dz/dt = (1+z)H_o[1-(1+z)^(3/2)] based on previous sections of this problem. I can see that for z>0, dz/dt is negative. I also have a graph that confirms this. I don't understand why this is true physically.

Why is dz/dt negative?
 
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  • #2
If I recall correctly, that's because the model you're using is actually decelerating, which means that the redshift will be decreasing with time, making [tex] \frac{dz}{dt} [/tex] negative.
 

FAQ: Why Is dz/dt Negative in Cosmology?

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