# Do contracting objects show red shift?

1. May 1, 2014

### Yashbhatt

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. May 1, 2014

### Mordred

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)

$$1+Z=\frac{\lambda-\lambda_o}{\lambda_o}$$

Last edited: May 2, 2014
3. May 2, 2014

### 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?

4. May 2, 2014

### Staff: Mentor

Yes, the light emitted from the center will be red shifted. Why do you ask?

5. May 2, 2014

### Mordred

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

6. May 2, 2014

### Yashbhatt

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

7. May 2, 2014

### Mordred

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

8. May 2, 2014

### Mordred

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

Last edited: May 2, 2014
9. May 2, 2014

### Yashbhatt

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.

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10. May 2, 2014

### Mordred

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

11. May 2, 2014

### Mordred

Last edited: May 2, 2014
12. May 2, 2014

### Staff: Mentor

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?

13. May 2, 2014

### Mordred

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

Last edited: May 2, 2014
14. May 2, 2014

### Yashbhatt

Closer to us? Can you please explain how?

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.

15. May 2, 2014

### Mordred

16. May 2, 2014

### Staff: Mentor

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.

17. May 2, 2014

### Staff: Mentor

I think you have the geometry wrong here (unless I'm misunderstanding, also a possibility). The contraction of the sphere moves any point on the intersection of the circle and the surface of the sphere to somewhere outside the circle... Although it does also move that point closer to the line between the center of the sphere and the circle.

18. May 2, 2014

### Staff: Mentor

True, but I'm not talking about the intersection of the circle and the sphere. Just the way the edges of the sphere (aka the "horizon") get closer to the center.

19. May 2, 2014

### Staff: Mentor

No, I have to take what I said back. Now that I see what you're saying, you're right.

If the distance between me and the central point of the contracting object is $R$ and at some moment the radius of the contracting object is $r$, then the points on the edge of the disk of the object will be at a distance $\sqrt{R^2+r^2}$ from me, and this will become smaller (the point is moving towards me) as $r$ diminishes with the contraction.

Of course this effect is well and thoroughly negligible if $R\gt\gt{r}$, whereas the effect on the points that are directly on the line between the two centers is independent of $R$.

Last edited: May 3, 2014
20. May 2, 2014

### Yashbhatt

Sorry, but I still din't get you. If any point on the sphere is moving towards us we won't be able to observe it since it will be on the other side of the sphere.

21. May 2, 2014

### Staff: Mentor

I'm sorry, I'm not sure how to explain it any better than I already have. Maybe someone else can?

22. May 2, 2014

### Mordred

lets try this your observer C on earth, take the image you posted on post number 9. with a and b. Now that circle you posted is 3D, a planet. So as B moves towards point A on that 3D image. the distance between B and C decreases, hence its moving towards you=blueshift.

23. May 2, 2014

### Mordred

In Drakkiths example observer A is the center of a circle, B is a point on the surface. As the surface contracts B moves towards A =blueshift

24. May 2, 2014

### Chronos

It depends on the observer. I remain convinced there is no privileged observer in the universe.

25. May 2, 2014

### Mordred

were setting the observer points in our examples, so privileged observer isn't even an issue, not sure where your coming from in this thread?

"It depends on the observer" 100% correct

"I remain convinced there is no privileged observer in the universe" 100% unrelated to the context and examples in this thread

Last edited: May 3, 2014