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Red shift occurs when gravity is at 90*

by JerryCic
Tags: gravity, occurs, shift
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JerryCic
#1
Feb13-06, 06:14 PM
P: 6
Pardon me if i am not grasping something. but!

Tell me if this very short presentation does not give a good argument that red shift occurs between two stationary light sources where gravity is at 90 degrees.

http://creativefamily.net/science/Lecture0601.pps

This lecture is from a stanford university professor.
Professor Alexander Mayer.
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RandallB
#2
Feb13-06, 07:38 PM
P: 1,545
I have to admit it does NOT make sense to me.
Especially if it’s intended to relate some how to the GPS and Pioneer “Anomalies”.

First a “Gravitational Transverse Redshift” where two clocks sending light signals to each other at the same altitude here on earth would show a Red Shift - only our gravity is not high enough to show the effect. As the clocks are running at the same rate with respect to each other (both moving together in same frame), I highly doubt this is real. Certainly not at all like the Transverse red shift for perpendicular light between to high speed objects passing by each other.
Maybe if the gravity started to get near that of a Black Hole each clock may see the other appear to rise, but even then I’d guess no red shift.

Also, the GPS “Anomalies” look like data not adjusting for the fact the observer is not at the center of the GPS satellite orbit. Therefore, the speed in relation to the observer at both horizons will be less than the Relative speed when it is directly overhead. Considering that I’d expect a saw-toothed display they call ‘Residuals’.

Finally, I don’t see where the idea relates to the “Pioneer Anomaly” at all. In fact the wave form being complained about in the graph is not what I understand to be the Pioneer Anomaly (where it sees a small increase in sun mass after exiting the solar system).
Again this looks more like raw data not adjusted for Doppler affect of earth bound observations moving with the rotation of the earth.

Is this maybe some kind of ‘find the errors in this lecture’ scavenger hunt or quiz ??
pervect
#3
Feb13-06, 07:50 PM
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P: 7,594
I do not believe the argument holds water - I'm not familiar with Mayer, but I don't think this lecture is going to pass peer review.

A standard metric for an accelerating observer (see MTW's "Gravitation")
http://www.amazon.com/gp/product/071...Fencoding=UTF8

is (1+gz)^2 dt^2 - dx^2 - dy^2 - dz^2

It's immediately obvious from the metric that all clocks at z=constant tick at the same rate. Therefore there is no transverse red-shift due to acceleration, only due to height.

Another way to see this is to draw the same diagram the professor did but with co-moving observers, observers who instantaneously have the same velocity as the accelerating rocketship. The co-moving observer at time t=0 sees the light travel the same path as the co-moving observer at time t=1. While the inertial frame associated with the co-moving observer at t=0 is not the same inertial frame associated with the co-moving observer at t=1, all inertial frames are equivalent in that they have the same laws of physics.

[add]
While Mayer is listed on Stanford's .who

https://stanfordwho.stanford.edu/loo...&submit=Search

and his webpage is on Stanford's server, his exact qualifications are unclear. When affilition is restricted to faculty, student, staff, or other his name does not appear. (It appears when "don't know" is selected.) And he does not appear on the visiting professor page (at least not under physics visiting professors).

dicerandom
#4
Feb13-06, 08:18 PM
P: 308
Red shift occurs when gravity is at 90*

I read through his lectures a week or so back when they were posted on Slashdot.

I recall in particular that he made a big deal about his "revelation" that there is no universal "temporal direction" in a curved space, which I had thought was a given. Even in SR observers who are moving with respect to one another have time axes which point in different directions. Can anyone comment on this? Did I misunderstand what he was getting at?

His Stanford page seems to be down.
JerryCic
#5
Feb14-06, 08:01 AM
P: 6
I have added the two longer lectures that go into the math and measured effects.

http://creativefamily.net/science/

The basic thought experiment i believe shows in the simplest terms the existance of 90 degree gravitational red shift.

As the light passes an object in space, that objects zenith pull creates a blue shift when the light moves towards the body and a red shift as the light moves away. the two shifts cancel each other out and the net effect is that the light frequency is unaltered.

The dilemma is at 90 degrees (horizon). that is the perpendicular gravitational effect adds a red shift to the light path and this leaves a red shift residue that the light takes with it on to the next g-force it encounters. This residue accumulates. If so then the farthest objects in space, would have in general, the largest red shifts. And they do.

So what am i missing? why is this guy's postulates wrong? Or are they?
It is interresting to note that stanford has password protected his lectures on the web. My friend (Doug) happen to have downloaded them before they were made private.

We've all seen elaborate 'Lost in the complexity' arguments that argue this thing or that. It is amazing tome how lost some get in their work. Usually full of holes.

I just cant find the hole.

JerryC
RandallB
#6
Feb14-06, 09:38 AM
P: 1,545
Quote Quote by JerryCic
The basic thought experiment i believe shows in the simplest terms the existance of 90 degree gravitational red shift.

So what am i missing?
note that stanford has password protected his lectures

I just cant find the hole.
OK
I think I see the problem in seeing through the thought experiment A. J. Mayer presented in slide 2. Maybe he is misunderstanding his own work here.
Could be why Stanford has password protected his lectures.
OR MAYBE this is some highbrow’s idea of a little joke on students?
(The tie in to GPS & Pioneer is a little far fetched)

Anyway to understand slide 2:
Yes: clocks A & B see each other in the same Ref Frame.
Yes: As we see from observing the rocket accelerate the light moving from one clock to the other IS RED SHIFTED.
Therefore YES the clocks at A’ & B’ are still in a common Ref Frame and WILL receive Red Shifted Light.

BUT!!
A’ & B’ are NOT in the same Ref Frame as they were when they sent the light signals at positions A & B. They have accelerated! As in they are moving at some speed relative to the old A & B Ref Frame. Therefore their clocks are running a SLOWER RATE than when the light was sent at A & B.

Now adjust the “red shifted” light they are receiving; by the fact that they must observe that light with a ‘slow clock’; they will perceive the light blue shifted due to their slower rate of time. Bringing it back to exactly the same color they sent back at positions A & B.

Thus following the argument as presented to accept the idea of red shifted light is being received is OK. But just a slight of hand trick worthy of a magician, if you don’t complete the problem correctly. The trick works if you accept the clocks at A’ & B’ ( both still in a common reference frame) as being in the same reference frame as they were when at positions A & B the frame the redshift is based on.

Does this plug your hole?
pervect
#7
Feb14-06, 03:15 PM
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P: 7,594
Quote Quote by JerryCic
I have added the two longer lectures that go into the math and measured effects.

http://creativefamily.net/science/

The basic thought experiment i believe shows in the simplest terms the existance of 90 degree gravitational red shift.

As the light passes an object in space, that objects zenith pull creates a blue shift when the light moves towards the body and a red shift as the light moves away. the two shifts cancel each other out and the net effect is that the light frequency is unaltered.

The dilemma is at 90 degrees (horizon). that is the perpendicular gravitational effect adds a red shift to the light path and this leaves a red shift residue that the light takes with it on to the next g-force it encounters. This residue accumulates. If so then the farthest objects in space, would have in general, the largest red shifts. And they do.
Your argument above has done nothing, that I can see, to suggest that there should be a 90 degree redshift, but merely specualted on what might happen IF there were a 90 degree redshift.

Whatever this theory is, it's not relativity. I've also seen no evidence that the theory is peer reviewed.

I should also point out that this theory will not conserve energy. GR basically predicts that light will redshift when it goes uphill (to conserve energy), but that it will not redshift when it winds up at the same height it started at.

This theory predicts that light will constantly lose energy.

Of course GR does not conserve energy under all circumstances either, but the non-conservation of energy by this theory is considerably more gross than that of GR - GR conserves energy in asymptotically flat space-times, for instance.

This theory has not been analyzed to see how or if it can be falsified - it doesn't make any new predictions.

I suspect that several experiments already done (such as sensitive redshift experiments with centrifuges) might already have falsified the theory, but it's hard to tell, because the analysis has not been done.

By claiming that this theory is GR when it is obviously not, by not properly analyzing the experimental implications of the theory and explaining how it can be tested, by not being published first in a peer reviewed journal but by being presented directly to the lay public, this theory is poor science (if it qualifies as science at all).
JerryCic
#8
Feb14-06, 09:17 PM
P: 6
Quote Quote by RandallB
OK

BUT!!
A’ & B’ are NOT in the same Ref Frame as they were when they sent the light signals at positions A & B. They have accelerated! As in they are moving at some speed relative to the old A & B Ref Frame. Therefore their clocks are running a SLOWER RATE than when the light was sent at A & B.

Now adjust the “red shifted” light they are receiving; by the fact that they must observe that light with a ‘slow clock’; they will perceive the light blue shifted due to their slower rate of time. Bringing it back to exactly the same color they sent back at positions A & B.
I think your explaination makes sense! Thank you
RandallB
#9
Feb15-06, 09:16 AM
P: 1,545
Quote Quote by JerryCic
I think your explaination makes sense! Thank you
Your welcome

I still question if these rather long (100’s of slides?) lecture notes may not be some elaborate joke or hoax.
The author is successful either way:
1) get some to believe it
2) get others to demand the ‘theory’ produce some testable predictions.
When there is no theory here at all.

This one I think better qualifies for the concept “Not even wrong”.
It really shows when you see how poorly the GPS and Pioneer data is presented (intentional maybe?).
If Mayer is real and serious, then he’s gone to a lot of effort to fool himself.
rogerw
#10
Feb15-06, 06:01 PM
P: 1
Strangely enough, Mr Mayer is no longer listed in the Stanford who's who, and his home page there has gone 404.
clj4
#11
Feb22-06, 01:06 AM
P: 482
The whole thing has been a hoax:

http://www.bautforum.com/showthread....432#post687432
DougMar
#12
Feb25-06, 04:24 PM
P: 6
Quote Quote by RandallB
Yes: clocks A & B see each other in the same Ref Frame.
Yes: As we see from observing the rocket accelerate the light moving from one clock to the other IS RED SHIFTED.
Therefore YES the clocks at A’ & B’ are still in a common Ref Frame and WILL receive Red Shifted Light.

BUT!!
A’ & B’ are NOT in the same Ref Frame as they were when they sent the light signals at positions A & B. They have accelerated! As in they are moving at some speed relative to the old A & B Ref Frame. Therefore their clocks are running a SLOWER RATE than when the light was sent at A & B.

Now adjust the “red shifted” light they are receiving; by the fact that they must observe that light with a ‘slow clock’; they will perceive the light blue shifted due to their slower rate of time. Bringing it back to exactly the same color they sent back at positions A & B.
I have a question about this.

You say that the clocks are running at a slower rate. But travelers in the spaceship don't see that, right? If their clocks are running slower it would be in relation to some other observer, right?

If so, then would it really cancel out the red-shift that Mayer is talking about?

I don't think Mayer is talking about a red-shift due to variances in velocity like we know from SR. If that is what you meant, Randall, then I agree it is cancelled out.

I think what Mayer is suggesting is that in an accelerating ship, the path that a beam of light takes from A to B is actually curved. It is in the form of an arc, which is the same thing that happens in a gravitational field, since space is curved.

If the path of the beam is an arc, rather than a straight line, and we assume that the speed of light is constant, and the two clocks are synchronized, then wouldn't this produce a red-shift?

Doesn't the longer path of the arc, which is caused by GR produce this shift because the light is actually traveling a farther distance? Since the speed of light is constant, doesn't this cause the shift?

I don't think we need an outside observer to evaluate Mayer's theory. Both A and B are in the same reference frame and their clocks start out synchronized before acceleration. So, shouldn't we assume that their clocks stay synchronized according to SR? But if space is curved, according to GR, then doesn't the light actually travel a longer path?

Where does this fall apart? I'd really like to see the flaw better. I'm not quite grasping it.

Any suggestions?
pervect
#13
Feb25-06, 06:13 PM
Emeritus
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P: 7,594
Quote Quote by DougMar
I have a question about this.

Doesn't the longer path of the arc, which is caused by GR produce this shift because the light is actually traveling a farther distance? Since the speed of light is constant, doesn't this cause the shift?


I don't think we need an outside observer to evaluate Mayer's theory. Both A and B are in the same reference frame and their clocks start out synchronized before acceleration. So, shouldn't we assume that their clocks stay synchronized according to SR? But if space is curved, according to GR, then doesn't the light actually travel a longer path?

Where does this fall apart? I'd really like to see the flaw better. I'm not quite grasping it.

Any suggestions?
The easiest way to see that there is a flaw is to look at the energy conservation arguments for gravitational redshift.

If you allow a photon to drop in a gravitational field, you expect it to gain as much energy as allowing a material body to fall the same distance.

For a photon, E = pc = hc/lambda. Thus you expect the momentum of a photon travelling upwards in a gravitational field to decrease, i.e. it's wavelength increases as it travels upwards, i.e. it redshifts.

How much energy does a photon gain/lose when it is shot upward in a conservative gravitational field, follows an arc, and then returns to the exact same height it started from?

The answer is none, because the final height is the same as the initial height.

Therefore one does not expect a red-shift, because the photon has neither gained nor lost energy.

As far as your argument goes, there are a couple of problems:

1) You have ignored the metric when computing distances. The apparently "curved" path in the coordinate system of the accelerating observer is a path that appears "straight" in an inertial coordinate system. The reason that the straight path in an inertial coordinate system appears to be curved in the coordiante system of the accelerated observer is due to the metric coefficients that distort times and distances.

The photon really does follow the shortest path, when the metric corrections are taken into account. This can be seen by observing the path of the photon in an inertial coordinate system.

2) The coorrdinate speed of light, defined as the rate of change of the distance coordinate with respect to the time coordinate, is not constant in a gravitational field. It is only constant in an inertial coordinate system.

The local speed of light is constant relative to local clocks and rulers, but the coordinate speed of light is NOT constant. Remeber that because of the existence of the metric, local clocks keep different time from coordinate clocks.
DougMar
#14
Feb25-06, 11:36 PM
P: 6
Pervect, I follow your energy conservation argument. Mayer actually talks about this.

Mayer says that if his theory is true that all rotating bodies (because they are continously accelerating) will have a very slight loss of energy. The energy is conserved because the lost energy is radiated out from the body. As a result, he says this would mean a very slight degradation in all orbits, spinning stars, or any rotating body. He claims this meets with numerous observations that have not been explained before.I would imagine this means the same would be true for light passing perpendularly past a gravitational body.

This is clearly one place where his theory disagrees with current theories. Only real tests could prove out whether what he says is happening, especially since the amount of red-shift and radiated energy would be so incredibly small. So, I have big doubts about whether this is real.

But I would like to ask you some questions about your other points.

Let's say we put points A and B on opposite sides of a huge star. We know that light is curved by the gravity of the star. Light does take the shortest path that it can and it still looks straight to both observers, but the arc can still be measured.

So, it seems to me that if points A and B are in a space ship on Earth, the only difference is that it is a much smaller distance between them and the gravity of the Earth is much smaller than the star. In other words, the effect would have to be much smaller. But it still seems like it should be a real arc and should be measurable to both A and B.

If Einstein's equivalency of GR holds, then the same should be true in an accelerating coordinate as in a gravitational field. So, this is what it looks like from the standpoint of the accelerating coordinate system, isn't it?

You are suggesting that there is no arc from the standpoint of the inertial coordinate system. Why would this be different than when in a gravitational field? Doesn't equivalency hold here?

Yes, I also follow how the clocks of an inertial observer will not track with the clocks in the accelerating ship. But for now I would just like to look at it from the accelerating ship frame of reference, because both A and B are there and all the measurements between their clocks is done from their frame of reference. So, we should only need to look at what they measure and see, right?

So, what am I still missing?

Thanks.
pervect
#15
Feb26-06, 01:44 AM
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Quote Quote by DougMar
Pervect, I follow your energy conservation argument. Mayer actually talks about this.

Mayer says that if his theory is true that all rotating bodies (because they are continously accelerating) will have a very slight loss of energy.
I don't like Mayer's "bait and switch" tactics. If he started with different assumptions than GR, and then came to different conclusions, it would indeed be just a matter for experimental test.

The problem is that Mayer claims (at least most of the time) to start with the same assumptions as GR and come to different conclusions. This puts the "theory" on a different footing. His incorrect reasoning in deriving his theory, plus the lack of peer review, are the main problems the theory has. The theoretical predictions are a bit fuzzy, too, but perhaps I just missed that part, not having gotten past his first elementary errors. When the conclusion doesn't follow from the premises, I'm not very motivated to look at the predictions, since the logic itself is in error.

Let's say we put points A and B on opposite sides of a huge star. We know that light is curved by the gravity of the star. Light does take the shortest path that it can and it still looks straight to both observers, but the arc can still be measured.

So, it seems to me that if points A and B are in a space ship on Earth, the only difference is that it is a much smaller distance between them and the gravity of the Earth is much smaller than the star. In other words, the effect would have to be much smaller. But it still seems like it should be a real arc and should be measurable to both A and B.
Let's take an analogy. Suppose one is on the surface of the Earth, and one wants to get from New York city to Sebring Ohio, both picked because they have latitudes near 40degrees N. (Let's assume that both cities have latitudes of exactly 40 degrees N for this example, for the purposes of discussion).

The shortest path between these two points is a great circle (as is the shortest path between any two points on surface of the Earth)

When you look at the curve on a plot of latitude vs longitude though (picking a specific coordinate system in the process), the curve is not a straight line. You'll find that at the midpoint of the great circle, the "great circle" curve has a latitude that's north of the 40 degrees. So if you plot this curve on a map where lines of constant latitude are horizontal lines, this line will appear to be curved.

The same thing is happening with the accelerated observer, and with the sun.

When I say that light is taking the shortest path, what I mean unambiguously is that it is following a geodisic. This does not mean that light "really" doesn't appear to curve in some specific coordinate system. In the coordinate system of the accelerated observer, it does appear to curve. But in the coordinate system of an inertial observer, the light follows a perfectly straight path.

So the bottom line is that the English phrase "straight line" is a bit imprecise. The precise statment is that light always follows a geodesic. In the accelerated coordinate system, the light that appears to "curve" up is actually following a geodesic path, known as a null geodesic.

Light that is forced to move at a constant height coordinate in the accelerated coordinate system (perhaps with a fiber optic cable) will actually be travelling a longer path than it would if it were allowed to find its way on its own.

This finds its physical expression in Fermat's principle, the principle that light always takes the path of "least time".
JerryCic
#16
Feb26-06, 09:27 AM
P: 6
In doug's thought experiment, lets say that the source light is linearly blocked by the high gravitational source (say a dense star). Say also that the gravitational force on the light beam is enough to allow the observer to see the light source due to bending. Wouldn't this produce a ring around the dense star representing the light from the lightsource whose light is being bent equally by all paths passing the dense star on all sides?
pervect
#17
Feb26-06, 03:05 PM
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P: 7,594
Quote Quote by JerryCic
In doug's thought experiment, lets say that the source light is linearly blocked by the high gravitational source (say a dense star). Say also that the gravitational force on the light beam is enough to allow the observer to see the light source due to bending. Wouldn't this produce a ring around the dense star representing the light from the lightsource whose light is being bent equally by all paths passing the dense star on all sides?
The phenomenon is known as an Einstein ring, I believe. It requires rather special conditions to happen, but it's been observerd. The source object and observer have to be in exactly the right position.

However, the light that forms such a ring still follows the straightest possible path through space-time (a geodesic path), and obeys Fermat's p[rinciple. One can view the situation as light following "straight" (in the sense of shortest distance, i.e. geodisic) path thorough a curved space-time.

A somewhat anologous example is to suppose one was at the north pole, and traced "straight lines" (great circles) in all directions. All of the lines would converge at the south pole (assuming the Earth was a perfect sphere) - they would appear to "focus" at that point.

This requires a non-Euclidean geometry. In both cases (the Einstein ring, and the convergence of geodesics from the north pole at the south), one has an infinite number of "straight lines" connecting the same two points. This is not possible in Euclidean geometry on a plane, but it is possible in non-Euclidean geometries.
RandallB
#18
Feb26-06, 03:46 PM
P: 1,545
Quote Quote by DougMar
You say that the clocks are running at a slower rate. But travelers in the spaceship don't see that, right? If their clocks are running slower it would be in relation to some other observer, right?

If so, then would it really cancel out the red-shift that Mayer is talking about?
Your missing the point about acceleration.
When you accelerate the inertial frame you were in a few minutes ago (A & B) is NOT the inertial frame you are in now (A’ & B’). That change of frames must be accounted for. Very simple really, not compared to "some other observer" but compared to what they themselfs can no longer observe as if they were still in the frame of (A & B). If you like they could have companions with them that “let go” of the acceleration and maintain the speed & frame of the (A & B) instant.

It’s not that it cancels out the red-shift he is talking about, it IS the red shift he trying to get people to react to seriously with some success unfortunately.

How embarrassed do you think Stanford is over a hoax like this getting on their web site. Bet they wish the could get back the copies of it that have already been made.


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