# Layman's doubts about Gen Relativity

1. Jul 14, 2010

### superkan619

1) Is the space-time curvature absolute? For example is the space-time curvature caused by earth(for scientists on Earth) same when viewed from the moon?

2)Does the frequency change of a photon at a point in the gravitational field depends upon its initial position of emission?

3)When looked upon from Earth does a falling stone makes non-inertial frame of reference?

2. Jul 14, 2010

### Max™

Uh... barring consideration for reference frames in significantly different states, and avoiding going into a dynamic spacetime model in which T-symmetry is broken, yes, the gravitational pull from the Earth curving spacetime in it's vicinity is the still there when you're on the moon, simply weakened due to your distance, with the local curvature of the moon dominating your motion.

If you are asking how a photon frequency varies based on the altitude at the point of emission, then yes, two photons emitted at different altitudes would start out in slightly different parts of the gravitational gradient, but as one passed the altitude of the other, the variance beyond that point would be the same as if it had been emitted there.

When you're on the Earth, you can define it as your frame of reference, in which case a falling object could be defined as the origin of a different accelerating frame if you wished. You could also define the background stars as a frame of reference, in which case you are in motion that could provide the illusion of fictitious forces like centrifugal force, which would put you in a non-inertial frame.

If your frame of reference has a non-uniform, or accelerated motion, then the Law of Inertia will appear to be wrong, and you must be in a non-inertial frame of reference. Right now you're being pulled towards the surface of the Earth by gravity, but at rest relative to it's surface, so you feel no fictitious forces that would lead you claim you were not at rest.

3. Jul 14, 2010

### superkan619

Does this means that the time-dilation factor is fixed for every location in gravitational field regardless of the history of the photon? I thought the time dilation factor was true only for photons from infinity.

4. Jul 14, 2010

### Staff: Mentor

The curvature is represented by a tensor, and tensors are geometric quantities that are the same in all coordinate systems. Of course, the components of a tensor change in different coordinate systems, but the underlying geometric object is the same.

Yes.

No. All falling objects are inertial, meaning that they have 0 proper acceleration. This is also a tensor quantity.

5. Jul 14, 2010

### starthaus

See https://www.physicsforums.com/blog.php?b=1954 [Broken]

Last edited by a moderator: May 4, 2017
6. Jul 14, 2010

### Altabeh

Although the Riemann tensor is coordinate-dependent, as DaleSpam said, the "underlying" geometry when discussed from everybody's perspective is the same. To wit, either I'm on Earth or on a spaceship travelling to moon with a speed close to c, though I'm getting a different expression for this tensor in either case, since the curvature felt by a local observer represents the real (underlying) geometry, we say this geometry doesn't change wherever I am dealing with it. Nonetheless you see that the Riemann tensor has to change by a shift from on-Earth frame to spaceship frame.

When one says the "frequency" is red-shifted or blue-shifted it means that what has undergone a change is the "initial frequency". If not so, then with respect to what are we referring to the information on such quality of a wave at the point of reception as "altered"!?

According to the principle of equivalence (PE), all particles falling towards the surface of Earth are to be considered as "inertial" locally. They would make probabaly non-inertial frames if the distance along which you look over their motion gets really large unexpectedly as you cannot surely decide whether it's within the locality of PE though definitely inertial frames are guaranteed to exist certainly in an infinitesimal interval of their path (local flatness theorem; see, for example, Schutz B.F. A first course in general relativity. 1985 pp. 154-156 and 158-160.)

AB

7. Jul 14, 2010

### Jonathan Scott

A gravitational field does NOT change the frequency of a photon. It remains the same as it was when it was emitted, as seen by any fixed observer.

However, time runs at slightly different rates at different potentials. The fractional difference in the time rate for an object at two different positions is effectively the same as the difference in potential energy divided by the total energy.

This means that if you compare observations of the same photon from two locations at different potentials, its frequency appears to differ because of the difference in clock rates.

8. Jul 14, 2010

### Jonathan Scott

An inertial frame of reference is one in which no forces are acting on the body. There are two ways of describing gravity, either as a force acting on a body, or as an effect of the curvature of space-time. In the context of Newtonian mechanics and Special Relativity, gravity is considered as a force, so a falling stone has a force acting on it and is not an inertial frame of reference. In the context of General Relativity, gravity is normally described in the latter way, so a free falling (non-rotating) body is in an inertial frame of reference.

I was very annoyed when I got 99 out of 100 on a multiple-choice relativity exam because I said a falling stone was an inertial frame of reference, without noticing that the question was within a section on Special Relativity, so they expected it to be considered non-inertial!

9. Jul 14, 2010

### Altabeh

Hi Jonathan,

When talking of both gravitational field and frequency shifting at the same time, it behooves oneself to take into account the mechanism in which these two can be connected to each other. However I don't see something from OP's question concerning the argument that he is telling "gravitational field" generates frequency shifting! It was said that in a gravitational field such thing happens which is completely correct.

AB

10. Jul 14, 2010

### Jonathan Scott

There's more than one sort of "curvature" involved. As an analogy, consider a cone and a ball. You can make a cone with flat paper. However, you can't make flat paper match the surface of a ball except approximately over a small area.

The sort of curvature you see in a cone is loosely analogous to the way in which space-time is curved to cause a gravitational field. Locally, you can treat it as being flat, which is analogous to the fact that an observer in free fall doesn't see space-time as being curved locally (at least to first order).

The sort of curvature you see in the surface of a ball is analogous to the way in which mass-energy curves space. In general, different observers will agree about the existence of such mass and energy, although if they are moving relative to one another they may see different values.

Don't try to take this analogy too far; it's space-time that curved for gravity, not just space (otherwise objects at rest would not be affected by gravity), and the effects in 3-D space plus time are somewhat different from those that one gets with 2-D space only.

11. Jul 14, 2010

### Jonathan Scott

He mentioned "frequency change" which suggests that something about the photon changes, but this does not happen.

What typically happens is that a photon is emitted at a known frequency (for example a spectral emission line) by a process at one potential and is then observed at another. Such a photon will appear to be shifted in frequency compared with similar photons generated by the same process locally. However, the photon did not change in flight; it was emitted at the shifted frequency because of the different time rate at that location.

The apparent shift depends on the difference in gravitational potential between the location at which the photon was emitted and the location at which it was observed.

12. Jul 14, 2010

### Altabeh

The "frequency change" doesn't imply "photon" change which by itself is vague. I mean it can be inferred from your passage that you're talking about a particle change that is impossible. No one said something like a change of photon in flight and stuff like this; the OP just said something about "frequency change of a photon" in a gravitational field and that's all.

Seconded.

AB

13. Jul 14, 2010

### Staff: Mentor

I think you are reading too much into this. The standard Doppler effect could be called a "frequency change" but does not suggest anything about the photon.

14. Jul 15, 2010

### superkan619

Second question has been resolved

Does the frequency change of a photon at a point in the gravitational field depends upon its initial position of emission?

15. Jul 15, 2010

### superkan619

I never said that gravitational field changed the frequency. I was just referring to it as the vicinity of the gravitating object. Thank you Altabeh for pointing it out.

16. Jul 15, 2010

### superkan619

Please clarify the difference in opinions

As far as the lift-gravity experiment is concerned, the falling(to be assumed hereafter as non-rotating) frame feels inertial from inside. It is undoubtedly true. But, when looked from outside by some one on a perfectly-still-Earth, I "feel" Altabeh's answer is valid.

17. Jul 15, 2010

### Passionflower

A free falling (or floating) point test particle falls inertially, which means it undergoes 0 proper acceleration. This view is valid for all observers.

A free falling spatially extended test body as a whole does not fall inertially. For instance in the Schwarzschild spacetime the front and back of such a body accelerate away from the center. This acceleration increases as the body gets closer to the center of gravity.

18. Jul 15, 2010

### superkan619

This part of the sentence has enlightened my knowledge. No matter what context it is being spoken in, it encodes a fundamental thing. My laymen brain had reached a clogging point before I encountered this. Thanks Jonathan, Thanks Altabeh.

19. Jul 15, 2010

### Altabeh

As Passionflower said, this can also be felt by all observers outside the lift cabin. The point is that the system would not be felt to be inertially moving when looked upon as a whole or, to better say, when the proper acceleration is to be measured within a very large distance along which the local flatness theorem doesn't hold anymore. Sometimes it is interesting that you hit something like "Fermi normal coordinates" which produces a frame for a (timelike) geodesic wherein the proper acceleration is zero along the entire geodesic and this must be a little bit tricky because then you'd say in a free fall we can always have a vanishing proper acceleration. But note that this is only correct for all observers (who are momentarily at rest relative to the object) measuring object's proper acceleration whose frame is a Fermi-like one and in general this cannot be extended to ALL observers.

AB

20. Jul 15, 2010

### Staff: Mentor

This is not correct to the best of my understanding. It does not matter what coordinate system you use, the proper acceleration of an inertial particle is 0. You do not have to be momentarily at rest, particularly since simultaneity is completely arbitrary in GR.

$$A^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\gamma\nu}U^{\gamma}U^{\nu}$$

Which is a tensor and transforms like a tensor for all coordinate systems. Setting that gives you the geodesic equation which shows the path of an inertial object in any coordinate system.

21. Jul 15, 2010

### Altabeh

But to the best of my understanding, we both can be correct if we split the two different definitions of proper acceleration. I don't believe that we have such thing as "proper acceleration" in GR and all I hit on the internet about the kind of definition you use of this quantity in GR was the Wiki's article where the book or reference from which the material has been taken is unknown; leading me to doubt the validity of this writing. If you can prove me the "proper acceleration" in free fall in GR is the left-hand side of the geodesic equation, then I will not be unhappy using the definition of $$d^2x^a/d\tau^2$$ as the proper acceleration in GR, for which I don't have any textbook to name where it has been used/cited. If this is the case, then you clearly know that there is nothing wrong with my previous post. All I want to know now is just a valid reference where they apparently make use of the LHS of the geodesic equation to interpret the proper acceleration in GR.

AB

22. Jul 15, 2010

### DrGreg

Rindler, W (2006), Relativity: Special, General, and Cosmological, 2nd ed, Oxford University Press, ISBN 978-0-19-856732-5, p.214

defines (generalised) 4-acceleration to be the 4-vector $A^\mu$ as defined above and proper acceleration to be its magnitude

$$\sqrt{|g_{\mu\nu} A^\mu A^\nu|}$$​

(a scalar invariant).

23. Jul 15, 2010

### Staff: Mentor

Dr. Greg posted a good reference, I would also reccomend Leonard Susskind's lectures on General Relativity available on YouTube. He goes into great detail on the difference between the covariant derivative and the ordinary derivative.

Generally you will either find that form (covariant derivative of four-velocity wrt proper time) or the form that you posted (ordinary derivative of four-velocity wrt proper time). The ordinary derivative can only be used in standard Minkowski coordinates in flat spacetime, in all other cases the covariant derivative is required. If you use the ordinary derivative instead of a covariant derivatve the result is not a tensor, but the covariant derivative ensures that the result is a tensor so that all frames will agree.

24. Jul 15, 2010

### Altabeh

Yes he did a great job but I still don't believe that physics community has an agreement on such definition as Rindler is (let's say) the only one using this. I have wormed partially through Schutz, Letctures on GR by Papapetrou, D'inverno, Weinberg, the first part of Wald, MTW and recently David McMahon's Demystified Relativity but neither of them defines such thing and actually there isn't even a hint at a way one can deal with a generalized "proper" 4-acceleration in GR. I've checked almost 50 books on the internet but no result. About the YouTube reference, I've not watched any of those lectures by Susskind and am really interested to see what's going on in these lectures so I need time to check them.

There is a point behind my use of "ordinary derivative" as a proper acceleration here. If you noticed I said that such "proper acceleration" exists in SR and I do have no doubt about it but to provide a mold that this can also be used in GR I'd rather prefer to think of "local flatness" as a key to the problem and say in an infinitesimally small interval of trajectory of a freely falling object it can be proven that the "now" pronounced proper acceleration $$d^2x^a/ds^2$$ is zero for all observers and this is what EP (equivalence principle) is all about. Here we don't need to go for something like a "generalized" proper 4-acceleration to justify that such thing is always zero for particles moving along any timelike geodesic and definitely this destroys the implication of EP in GR; if the generalized proper 4-acceleration is zero everywhere along the geodesic, why do we need any of Riemann normal coordinates or Fermi normal coordinates to still have local flatness theorem hold in a variety of cases?

As Papapetrou says the Christoffel symbols describe, according to the principle of equivalence, the sum of the inertial and the gravitational accelerations and there is always a possibility to make them vanish to reduce to a vanishing proper acceleration in a free fall. I think we should treat the proper acceleration in GR this way and only deal with it when the conditions necessary for proper acceleration of SR to again play a role in GR are provided. This is my understanding of the whole issue.

AB

25. Jul 15, 2010

### Staff: Mentor

It doesn't matter for the question of this thread.
Regardless of if you call it proper acceleration or something else, a free-falling object follows a geodesic which is defined by the covariant derivative and is therefore a tensor and agreed on by all coordinate systems.