# Gravitational blueshift

1. Feb 11, 2004

### Sammywu

It's known and experimented when you shoot a light beam upward from Erath surface to a light detector at higher altitude, the light detector will detect a redshift. This is called gravitational redshift.

Both my friend and I agreed from how this was deduced, we shall see a blueshift when the light beam is shot from ahigher altitude down to the surface. Apparently this is so easy to perform, but everyone seems to keep silent on this.

Was this blueshift so simple and nobody care, or the experimental result showed opposite and so everyone just try to pretend blind on this issue?

I also noticed Andrewgray put this as one topic in his new theory. Maybe more people than us have noticed this and just kept silent for the time being.

2. Feb 11, 2004

### pmb_phy

Experimental evidence only shows that the clocks run at different rates. The term "redshift" has its origns historically and pertaining to light recieved by the Earth which came from the Sun which will be redshifted. But the only real thing going on here is a difference in frequencies when detected locally.

Keep in mind that as measured by any single observer, the light moving through a stationary gravitational field does not change. The only change comes into play when there is a comparison of locally measured frequencies.

Last edited: Feb 11, 2004
3. Feb 11, 2004

### Staff: Mentor

I think the reason this sort of blueshift isn't discussed is we don't see it much in nature. We see gravitational redshift when looking at black holes and such.

4. Feb 11, 2004

### Peterdevis

This is not true. The frequency of light is changing when it travels in a gravitational field. Ofcours you need two measurements to measure changes (by definition).
The gravitational redshift ( or blueshift)is a direct consequence of EEP ( Einsteins Eqiuvalence Principle) and not of the details of general relativity

5. Feb 11, 2004

### Sammywu

6. Feb 11, 2004

### pmb_phy

7. Feb 11, 2004

### Sammywu

PMB, Thanks. I read the paper written by Okun except the section 6, because I am not familiar with the grating experiment. It will take me some times to read that.

What I got here is:
1. The gravitational redshift is really due to the energy of the atom in a gravity field is lower than one outside of the gravity field. Does this also mean its rest mass is really lower than the rest mass of one free of gravity?

Any way, this does not exclude my question. A light beam shot from a tower down to the surface shall be observed as blueshift by the observer. Strange! Why was this experiemnt performed? People seem to consider that is an maeningless experiment.

2. The concept of comoving local inertial frame is useful to me. In trying to resolving questions of which clock is truly slower in two world line intersecting at two event points, I seem to need to llok for somethings like that. This probably also shows a possibility to help explaining the " H & K experiment" that I posted in another thread. Still, the kinetical energy time dilationc formula shall be corrected, because that describes an absolute effect not a relative effect.

3. I think it made clear that using gravitational work to explain the gravitational redshift is completely wrong is good. I might have read that in a textbook and I think I was quite confused when it does make a double effect issue.

One worm in the can is spotted though. This document explained the photon energy actually was not changed. How about the Hubble redshift coming from the stars in the expanding universe? How do we explain that the photon has different energy seen from the source and the observer?

8. Feb 12, 2004

### Peterdevis

I take back my words.

Yes I was

So when i have it right, the freqency of light is coördinate independent (in the sence like tensors)but the components (that what we measures) change by coördinate transformations.

9. Feb 12, 2004

### pmb_phy

It is not quite precise to that components change by coordinate transformations. What happens is that you have a new set of axes and the projects of the vector on the new axes are different than the projects on the old axes. If R is the position vector of a point with respect to the origin then

$$x = \mathbf{R}\bullet\mathbf{e}_{x}$$

x is the scalar product of two vectors and the scalar product remains unchanged by a change in coordinates.

http://www.geocities.com/physics_world/ma/invariant.htm

Last edited: Feb 12, 2004
10. Feb 12, 2004

### Peterdevis

PMB,

I was not thinking on scalars but on tensors (ofcourse is a scalar a tensor of rank 0). These are geometrical objects who are independent of coördinates and specified in a point (you may not think that they are stretching from one point to another point).
When you choose a basis for this tensors, you can describe the tensor with his components. When you change from coördinates, the components changes but also the basis in such away that the tensor stays the same (invariant).

So, now my question is the freqency of light such an object (i suppose a tensor of rank 0 = scalar)

11. Feb 13, 2004

### Sammywu

Did I hit a jackpot here? I found the argument of what caused Gravitational redshift is a central debate of two schloars. I just saw another website posting the redshift is caused by the gravitational work performed to the photon which L. B. Okun disputed.

12. Feb 13, 2004

### Sammywu

This debate is really interesting. So I started with Hubble redshift.

Many people might already know this. I evaluate it by taking the energy-momentum tensor of a photon. -E^2+Pc^2=0 when taking the flat-space metric to calculate its scalar product and because its rest mass is believed to be zero. Assuming a photon from a remote star to us is a energy-momentum tensor, from the far remote star's coordinates, E=h/alpha, alpha is the wave length. By lorentz transformation this tensor to the viewpoint of us assuming the star is moving from as in v and beta=v/c. The E' from our viewpoint will be gamma*(1-beta)E. By taking E'=h(alpha+diff(alpha)), I can derive the diff(alpha)/alpha=sqrt((1+beta)/(1-beta))-1, matching the published Hubble redshift fomula.

That answered my question to how Hubble redshift shall be explained as Energy change or not. The Energy-Momentum tensor shows a photon can be viewed as different wave length and energy by a moving observer in flat space.

13. Feb 17, 2004

### Sammywu

In L. B. Okun's paper that PMB_PHY showed above, the writer simply said Energy-Momentum vector is covariant and (t,x,y,z) is contravariant. It's no doubt that (t,x,y,z) is contravariant. I took what Okun said is true that Energy-momentum shall be treated as covariant. Later, this bothered me. I can't show that Energy-momentum shall be a covariant tensor or rank(0,1).

Does any one have ideas why L. B. Okun said that?

14. Feb 17, 2004

### Peterdevis

The energy momentum vector is covariant because the four velocity vector is. The four velocity vector is the tangent vector of the path that is parameterizased with the proper time.
The energy momentum vector can defined as the four velocity vector x massa (for a particle)

I dont know what you mean be (t,x,y,z) is contravariant? Where does it stand for?

15. Feb 17, 2004

### Sammywu

(t,x,y,z) is coordinate or I intended it for the vector dX. The vector dX for a infinitesmal distance between two event points shall be contravariant.

I also tried to think in this way. Notice in classic theory, F*D=W or dE and F*t=dP. So, It seems that change of energy-momentum tensor, now is (dE, dP) , would equal to somethings like @(i)T(lower-j)*dX(upper-J), the product of derivatives of a certain tensor and dX. Of course, you can see the tensor T as rank(0,1) is the energy-momentum tensor of another object. It appeared to be describing the energy-momentum transfer of two objects.

This might not be orthodox, but it did appear that energy-momentums need to have a lower index.

16. Feb 18, 2004

### Peterdevis

The infinitesimal distance is not a vector(tensor (0,1))Vectors are always related to on point (in differential geometry)and they point not to anything.
the definition of the infinitesimal interval =

$$ds^2 = \eta_{\mu\upsilon}dx^\mu dx^\upsilon$$

where $$\eta_{\mu\upsilon}$$ is the metric

17. Feb 18, 2004

### Sammywu

Peter,

I am sorry. that was not what I meant even though that's what I wrote. It's rather the infinitesmal vector pointing from the current event point to another event point. Actually the tangent vector of the world-line curve may be the better answer to what I meant. Does that cosntitute a tensor?

My thought about Energy-momentum tensor was wrong too. The relationship between work as F*D and momentum as F*T are purely from the negative signature between time and space.

The question remained is why the 4-velocity shall be the formula of dX(upper-i)/d(tau). The proper time as the denominator is the one making it covariant. Isn't it?

If we write a velocity as dX(upper-i)/dt or @(j)X(upper-i)*dX(upper-j)/dt, doe sthise look just like a vector instead of a 1-form. A vector is contravariant, isn't it?

18. Feb 18, 2004

### Sammywu

Actually my thought of force as related to a tensor of rank(0,2) was not completely wrong. The EM force tensor F(low-uv) has Electric field in the time-space elements and magnetic field in the spatial elements. The electric field E(1) is related to the partial derivative to t of magnetic field flux over the supersurface dy*dz. Likewise , B(3) could be related to partial derivative to x of electric field flux over dt*dy.