# Do contracting objects show red shift?

by Yashbhatt
Tags: contracting, objects, shift
 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?
 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 $$f=\frac{c+v_r}{c+v_s}f_o$$ c=velocity of waves in a medium $$v_r$$ is the velocity measured by the source using the source’s own proper-time clock(positive if moving toward the source $$v_s$$ is the velocity measured by the receiver using the source’s own proper-time clock(positive if moving away from the receiver) Gravitational redshift $$\frac{\lambda}{\lambda_o}=\frac{1}{\sqrt{(1 - \frac{2GM}{r c^2})}}$$ 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) $$1+Z=\frac{\lambda-\lambda_o}{\lambda_o}$$
 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?
 Mentor P: 11,884 Do contracting objects show red shift? Yes, the light emitted from the center will be red shifted. Why do you ask?
 P: 1,857 I see there was a post which Drakkith answered while I was working on the latex,
 P: 181 I am asking if the red shift at the center will be relatively more than towards the sides.
P: 1,857
 Quote by Yashbhatt 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?
a lower relative velocity between emitter and observer would result in a lower redshift so yes. see the formulas I posted in my first post you can work out the relations from those
P: 1,857
 Quote by Yashbhatt I am asking if the red shift at the center will be relatively more than towards the sides.
There is two types of shifts to deal with from the sides, from the center only one, see above. Visually the doppler shift and gravitational shift would combine. To separate them you need to identify the cause

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
 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. Attached Thumbnails
 P: 1,857 gotcha, your talking from the surface on both points, then yes. A will move away from us faster than B
 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
Mentor
P: 11,884
 Quote by Mordred 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
Yes, the sides of the sphere are technically slightly further away than the center point of the sphere, so when the sphere contracts, the sides move towards the center and there is a very slight blueshift since they are getting closer to us.

This is in addition to any redshift experienced by the time dilation/relative velocity of the sides as they contract. A little complicated, huh?
 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
P: 181
 Yes, the sides of the sphere are technically slightly further away than the center point of the sphere, so when the sphere contracts, the sides move towards the center and there is a very slight blueshift since they are getting closer to us.
Closer to us? Can you please explain how?

 edit forgot to add redshift and blueshift of spinning bodies, lol the combinations are endless
In that case, we should see red shift for the part which moves away and blueshift for the part which moves towards us, right?
That would be quite fascinating if we also consider the difference in rotational speeds at the equator and at the poles.
P: 1,857
 Quote by Yashbhatt Closer to us? Can you please explain how?.
see the image on this link I posted above.

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

 Quote by Yashbhatt In that case, we should see red shift for the part which moves away and blueshift for the part which moves towards us, right?.
correct
Mentor
P: 11,884
 Quote by Yashbhatt Closer to us? Can you please explain how?
Consider a circle. Every point on the circle is equidistant from the center. Now imagine you are at the center of a circle which has a radius starting at you and ending at the center point of the sphere. Since the circle curves towards you, it passes through the edges of the sphere closer to you than a straight line drawn from the edges of the sphere through the center. So when the edges of the sphere contract, they get closer to the center point and thereby closer to you.