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Do contracting objects show red shift? 
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#1
May114, 11:11 PM

P: 181

If we consider a point on the central part of a contracting object as observed from earth, the point is continuously moving away from us as the object contracts. Moreover, light emerges from a deeper gravitational well as the object contracts. So, shouldn't contracting objects show a red shift? And if so it should be less towards the edges than towards the center, right?



#2
May114, 11:51 PM

P: 1,857

Interesting question, Ok lets assume nothing interferes with the mean free path of light from the center of the contracting object. So in this case you would only have a gravitational redshift, but not a doppler shift, as this source is not contracting. Now if the emitter source is the surface of the object (the contracting part). Then yes this would have both a doppler shift due to movement away from the observer and a gravitational redshift. Climbing out of a higher gravity well. Not sure what your describing from the edge statement
"And if so it should be less towards the edges than towards the center, right?" However simply put the amount of redshift due to gravity, is due to the difference in gravity between the emitter and source. Where doppler shift is due to rate of relative velocity from emitter and source. You should be able to pick your corresponding relations from there. Doppler shift [tex] f=\frac{c+v_r}{c+v_s}f_o[/tex] c=velocity of waves in a medium [tex]v_r[/tex] is the velocity measured by the source using the source’s own propertime clock(positive if moving toward the source [tex]v_s[/tex] is the velocity measured by the receiver using the source’s own propertime clock(positive if moving away from the receiver) Gravitational redshift [tex] \frac{\lambda}{\lambda_o}=\frac{1}{\sqrt{(1  \frac{2GM}{r c^2})}} [/tex] G=gravitational constant c=speed of light M=mass of gravitational body r= the radial coordinate (measured as the circumference, divided by 2pi, of a sphere centered around the massive body) and just for completeness Cosmological redshift (due to expansion and contraction of the universe) [tex]1+Z=\frac{\lambda\lambda_o}{\lambda_o}[/tex] 


#3
May214, 12:13 AM

P: 181

I mean that if we observe some spherical object from the earth, it would appear as a circle. So, the center of that apparent circle will be moving directly away from as the object contracts while the edge of that circle will be moving obliquely and hence should show lesser Doppler shift.
Is that correct? 


#4
May214, 12:19 AM

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P: 11,884

Do contracting objects show red shift?
Yes, the light emitted from the center will be red shifted. Why do you ask?



#5
May214, 12:21 AM

P: 1,857

I see there was a post which Drakkith answered while I was working on the latex,



#6
May214, 12:27 AM

P: 181

I am asking if the red shift at the center will be relatively more than towards the sides.



#7
May214, 12:31 AM

P: 1,857




#8
May214, 12:33 AM

P: 1,857

edit: In other words a side emitter could have a combined higher redshift if the rate of contraction is high enough, that combined redshift could be higher than the center. The center has no contraction so only has a gravitational redshift, where as a side emitter has doppler+gravitational 


#9
May214, 12:55 AM

P: 181

I don't mean the center of the spherical body. In the diagram below, should the red shift be more at point A as compared to point B because A will be moving directly away from us as the object contracts while B will move rather diagonally.



#10
May214, 12:58 AM

P: 1,857

gotcha, your talking from the surface on both points, then yes. A will move away from us faster than B



#11
May214, 01:03 AM

P: 1,857

however there is also a transverse doppler lol unfortunately I don't know that formula. off hand, the redhshift/blueshift effect due to transverse doppler can be seen here.
http://www.conspiracyoflight.com/Tra...er_Effect.html formula for transverse is on this page, see transverse doppler halfway down http://en.wikipedia.org/wiki/Relativ...Doppler_effect 


#12
May214, 01:18 AM

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P: 11,884

This is in addition to any redshift experienced by the time dilation/relative velocity of the sides as they contract. A little complicated, huh? 


#13
May214, 01:25 AM

P: 1,857

yep angled motion, with other forms of shifts such as varying gravity well potential, and cosmological redshift, can get extremely complex REAL fast lol
edit forgot to add redshift and blueshift of spinning bodies, lol the combinations are endless 


#14
May214, 11:05 AM

P: 181

That would be quite fascinating if we also consider the difference in rotational speeds at the equator and at the poles. 


#15
May214, 01:09 PM

P: 1,857

http://www.conspiracyoflight.com/Tra...er_Effect.html 


#16
May214, 01:50 PM

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#17
May214, 02:01 PM

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