Does flat spacetime actually exist around us?

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Special relativity should be a special case of general relativity, for flat spacetime manifolds. For locally flat manifolds, special relativity should however give approximate results.

But even Earth is a non-inertial frame. So that would mean that special relativity can only be observed for small distances on Earth, when observed along the surface? Muon experiment should not be affected by Earth rotation, for example (IMHO).

I am asking this because it seems that Michelson-Morley experiment (MMX) should have returned speeds different from c, similar to a Sagnac experiment when observed from the rotating, non-inertial frame.

In other words, did MMX work only because distances happened to be short enough?
 

Ich

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I am asking this because it seems that Michelson-Morley experiment (MMX) should have returned speeds different from c, similar to a Sagnac experiment when observed from the rotating, non-inertial frame.
The Sagnac effect is proportional to the area enclosed by the light path.
 

bcrowell

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There are two different issues here: flat versus curved spacetime, and accelerated versus unaccelerated observers.

Special relativity is not limited to unaccelerated observers. The rotation of the earth doesn't mean that you can't analyze things using SR.
 
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I am asking this because it seems that Michelson-Morley experiment (MMX) should have returned speeds different from c, similar to a Sagnac experiment when observed from the rotating, non-inertial frame.
No, it can be shown that SR predicts a null result for MMX in rotating frames.
On the other hand, if you insist in treating MMX as a Sagnac effect, bear in mind that the area closed by the light polygon in MMX is...zero!
 
There are two different issues here: flat versus curved spacetime, and accelerated versus unaccelerated observers. Special relativity is not limited to unaccelerated observers. The rotation of the earth doesn't mean that you can't analyze things using SR.
It can be shown that SR predicts a null result for MMX in rotating frames. On the other hand, if you insist in treating MMX as a Sagnac effect, bear in mind that the area closed by the light polygon in MMX is...zero!
The Sagnac effect is proportional to the area enclosed by the light path.
I will quote Wikipedia:

Special relativity:
Special relativity is the physical theory of measurement in inertial frames of reference

Sagnac effect:
For ω != 0 this frame of reference is non-inertial, which is why the speed of light at positions distant from the observer (at r = 0) can vary from c.

In other words, speed of light can vary from c in non-inertial frames. So null result for MMX should not imply that speed of light is constant in a rotational frame (because it isn't, and it rarely is when spacetime is not flat).

Starthaus and Ich, I am not sure what does area have to do with this. It shouldn't change anything if you added a couple of additional mirrors to MMX, because you are not rotating the setup during measurement. On the other hand, area closed by the light polygon in Sagnac is important because this area is perpendicular to the axis of rotation.

Performing MMX on Earth (since it is rotating) is like placing the whole setup at the edge of the Sagnac disk, miniaturized, with source at a very close distance to a detector, and then rotating the disc at constant speed.

So, if speed of light can vary from c in rotational frames, it seems that it should also vary along circles of latitude at the surface of Earth, shouldn't it?
 

Ich

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Starthaus and Ich, I am not sure what does area have to do with this.
It means that the classical MMX experiment is insensitive to rotation. This answers your question.
So, if speed of light can vary from c in rotational frames, it seems that it should also vary along circles of latitude at the surface of Earth, shouldn't it?
I think I see your misunderstanding: The (measured, "real") speed of light is locally always c. If you read something like
Wikipedia said:
For ω != 0 this frame of reference is non-inertial, which is why the speed of light at positions distant from the observer (at r = 0) can vary from c.
it means that you can choose coordinates in which "the speed of light" is not what it used to be. "speed" meaning dx/dt here, where x and t can be quite arbitrary functions of space and time. However, if you translate from these coordinates to standard coordinates, everything is fine again. The measurable speed of light in any (small!) region is still c, no matter what fancy coordinates you use.
It's just if you measure in some extended region that the result may differ from c. That's why area is important in the Sagnac effect.
So, if speed of light can vary from c in rotational frames, it seems that it should also vary along circles of latitude at the surface of Earth, shouldn't it?
As an example for this coordinate issue: say the latitudinal speed of light is 2700 degrees per second at the equator. It is then some 1800 degrees per second where I live. Quite different, and totally irrelevant.
 
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The impact of any effect on a measurement depands on two factors:
1) The size of the effect at the device
2) The sensitivity of the device to the effect

For the MMX spacetime is not very curved, the rotation is slow, and it is not very sensitive to either curvature or rotation. It can be analyzed with SR just fine.
 
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Performing MMX on Earth (since it is rotating) is like placing the whole setup at the edge of the Sagnac disk, miniaturized, with source at a very close distance to a detector, and then rotating the disc at constant speed.

So, if speed of light can vary from c in rotational frames, it seems that it should also vary along circles of latitude at the surface of Earth, shouldn't it?
Ich already gave you some excellent answers. I will address only the above.
-MMX and Sagnac are two very different experiments, the only thing that they have in common is that they use a light closed loop.
-In the Sagnac experiment, the area enclosed by the loop is non-zero, resulting into a movement of the interference fringes.
-In MMX the area enclosed by the light loop is zero, resulting into a null effect
-In order to find out why the above is true you need to study SR, the mathematical formalism is not very difficult
-Finally, I can prove to you in about half a page that MMX, though executed in the rotating frame of the Earth, produces, as predicted by SR, a null result. You won't find this in any textbook since textbooks deal only with MMX in translational frames. Before I do so, I will ask you to learn SR, otherwise my post would be a waste of time.
 
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Hi.
Special relativity should be a special case of general relativity, for flat spacetime manifolds. For locally flat manifolds, special relativity should however give approximate results.

But even Earth is a non-inertial frame. So that would mean that special relativity can only be observed for small distances on Earth, when observed along the surface? Muon experiment should not be affected by Earth rotation, for example (IMHO).
Everywhere in the universe, frame of inertia is local, not universal. You are right that special relativity can only be observed for small distances compared to astronomic distances that Hubble's law is applicable.
Regards.
 
I still don't understand why area should be important in MMX. MMX setup is not being constantly rotated, so I don't see why you should mention area as the reason for the null result.

If Sagnac wheel was not rotated, it would also have null result.

Does the following setup have zero area (taken from Wikipedia), for example:
[URL]http://upload.wikimedia.org/wikipedia/commons/0/06/Michelson-Morley_experiment_%28en%29.svg[/URL]

Ich said:
It means that the classical MMX experiment is insensitive to rotation.
Insensitive to rotation around the axis perpendicular to the device, yes. But I was referring to Earth's rotation.

But now that I think of it, after the initial beam splits into two parts, beams travel the same path in the direction and contrary to the direction of Earth rotation. Since it gives a null result anyway even in emission theory, it's obvious that the effect of travelling in two directions must cancel at the end.
 
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I still don't understand why area should be important in MMX. MMX setup is not being constantly rotated, so I don't see why you should mention area as the reason for the null result.
The area (0) is the reason why MMX is insensitive to rotation. It does not matter if MMX were rotated or not because it is insensitive to rotation. One of the hallmarks of a good measuring device is that it is designed to be insensitive to extraneous factors whenever possible.
 
For the MMX spacetime is not very curved, the rotation is slow, and it is not very sensitive to either curvature or rotation.
With my current calculations, this is the only answer I believe is correct: that this difference can be ignored for small speeds. I still hope I did a mistake somewhere.

The area (0) is the reason why MMX is insensitive to rotation. It does not matter if MMX were rotated or not because it is insensitive to rotation. One of the hallmarks of a good measuring device is that it is designed to be insensitive to extraneous factors whenever possible.
But now you are talking about area as everyone else, and I have the feeling you still don't understood what I am asking.

I am not talking about rotating the MMX device. I am talking about Earth rotating while the experiment is performed.

Consider how the MMX experiment is performed. Device is placed at Earth surface. Earth is rotating, so it is not an inertial frame.

If you consider the distance between splitter and forward mirror, it is moving in the direction of Earth's rotation, and it is not in an inertial frame.

If distance between beam splitter and upper mirror is D, then from the moment the blue beam leaves the splitter, t=D/c time will pass until it hits the upper mirror. Additional t=D/c time will pass until it gets back to beam splitter, making its total travel time 2D/c.

On the other hand, if distance between beam splitter and forward mirror is also equal to D, from the moment the red beam leaves the beam splitter, t=D/(c+Rw) time will pass before it hits the forward mirror, because it's "moving away" from the beam splitter. (applying same logic from the http://en.wikipedia.org/wiki/Sagnac_effect#Theory" article for an observer looking at Earth). On its way back, it will take shorter time, t=D/(c-Rw).

The problem is that D/(c-Rw) + D/(c+Rw) does not equal 2D/c.

So, obviously for w<<c it can be approximated to 2D/c, but it is not equal, so it appears like (if speed light is constant) there should exist very small interference (due to small w).

Similar to the logic in Sagnac effect article, SR doesn't even need to be applied if you pretend you are observing Earth from a "stationary" spaceship. In that case everything is moving, light speed is constant and you don't need to worry. Or is Wikipedia Sagnac effect calculation wrong?

I also suspect that Sagnac effect calculations don't apply completely, since path is linear between beam splitter and forward mirror, so I think that the actual distance which light passes in both cases is r=2R*sin[D/(2R)], if D is the curved length along the surface, but that complicates the calculation a bit and I still don't get a null result.

starthaus said:
I can prove to you in about half a page that MMX, though executed in the rotating frame of the Earth, produces, as predicted by SR, a null result. Before I do so, I will ask you to learn SR, otherwise my post would be a waste of time.
I believe I know SR good enough already, I am familiar with LT and so far haven't found any problem with the experiments. I certainly don't know GR, but if you can explain this using SR, that would be great.
 
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I have the feeling you still don't understood what I am asking.

I am not talking about rotating the MMX device. I am talking about Earth rotating while the experiment is performed.
Yes, I understood exactly that. I cannot speak for everyone else, but it was sufficiently clear from the very beginning that you were talking about exactly this scenario that I would guess that everyone here understood.

On the other hand, if distance between beam splitter and forward mirror is also equal to D, from the moment the red beam leaves the beam splitter, t=D/(c+Rw) time will pass before it hits the forward mirror
But the distance is not D. You have forgotten length contraction.
 

Ich

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You have forgotten length contraction.
Not only length contraction, also the skewed up/down light path. Further, even if the times were different: you would normally not even have equal path lengths in an MMX experiment. That's not how it works.
psmitty said:
I believe I know SR good enough already
Yes, that's a problem. You have to overcome this belief to make progress.

Here's a list of the things you got wrong with your description:
  • What you're calculating is inertial motion. Rotation is irrelevant.
  • You failed to calculate the up/down light travel time correctly. Look up the MMX math, there are many good explanations.
  • You failed to take Lorentz contraction into account.
  • You didn't understand how MMX works. It not about the presence or absence of interference fringes. Which would have nothing to do with different path lengths anyway. It's about shifting fringes.
Try to get these points straight. If you get stuck, ask here.
 
Not only length contraction, also the skewed up/down light path. Further, even if the times were different: you would normally not even have equal path lengths in an MMX experiment. That's not how it works.

Yes, that's a problem. You have to overcome this belief to make progress.

Here's a list of the things you got wrong with your description:
  • What you're calculating is inertial motion. Rotation is irrelevant.
Why is it irrelevant? Because it is slow compared to other values? What if I want to make a precise calculation?

  • You failed to calculate the up/down light travel time correctly. Look up the MMX math, there are many good explanations.
  • You failed to take Lorentz contraction into account.
What about the Sagnac explanation in Wikipedia? It doesn't take L. contraction into account when it describes that detector is "moving away" from the source.

  • You didn't understand how MMX works. It not about the presence or absence of interference fringes. Which would have nothing to do with different path lengths anyway. It's about shifting fringes.
Could you explain the difference?

Try to get these points straight. If you get stuck, ask here.
I did some math and I did ask, but I hoped to get a more concrete answer. The only answer I got here was "area is zero". Does area have anything to do with my question?

Could you at least point me to some article with a correct explanation? Googling it finds all sorts of weird ideas.
 
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Not only length contraction, also the skewed up/down light path.
Oops, you are right, I missed that.

What about the Sagnac explanation in Wikipedia? It doesn't take L. contraction into account when it describes that detector is "moving away" from the source.
The derivation for the Sagnac effect in Wikipedia is non-relativistic. It explicitly states that it is assuming v<<c in the theory section.

The only answer I got here was "area is zero". Does area have anything to do with my question?
Yes, the area is zero therefore the apparatus is insensitive to rotation.
 
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Ich

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What if I want to make a precise calculation?
Then try to get the O(v^2) effects right instead of fiddling about the O(v^6) (or whatever) contributions. Really. You have to understand MMX first.
Could you explain the difference?
You will always see some arbitrary fringe pattern. The pattern contains exactly zero information.
What you're looking for is a shift in the pattern when you rotate the experiment. If there were a classical ether, you'd see different patterns depending on your orientation relative to the velocity vector of the ether.
I did some math and I did ask, but I hoped to get a more concrete answer.
You showed your math after asking. Not until then it became clear that your problem is not about rotation, but about understanding the classical linear MMX.
The only answer I got here was "area is zero". Does area have anything to do with my question?
It has all to do with what you asked, but nothing to do with what you meant.
Could you at least point me to some article with a correct explanation? Googling it finds all sorts of weird ideas.
I see, there are crackpot sites dominating.
I found http://www.upscale.utoronto.ca/PVB/Harrison/SpecRel/SpecRel.html#MMExpt", try to follow their derivation with the swimmers.
 
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JesseM

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I see, there are crackpot sites dominating.
I found http://www.upscale.utoronto.ca/PVB/Harrison/SpecRel/SpecRel.html#MMExpt", try to follow their derivation with the swimmers.
This site also has two helpful animations, showing why aether theories implied that if the device was at rest to the aether, two split peaks of a light wave would recombine perfectly aligned after going through the arms of the interferometer, but if the device was in motion relative to the aether, there would be a shift in the peaks when the split beams traveling on the two arms were recombined, which would mean a shift in the fringe pattern.

psmitty, if you can see how this works, then you should also be able to see why, if an aether theory was correct and the interferometer was in motion relative to the aether at some constant velocity, then rotating the arms to different angles should change the amount of shift in the fringe pattern; this is basically what the Michelson-Morley experiment tried to do (with the added twist that they tried it during different times of year, so that even if the device happened to be nearly at rest relative to the aether at one time it should have a non-negligible speed at different point's in the Earth's orbit)
 
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