# Special Relativity vs. General Relativity

1. Jun 6, 2005

### IndustriaL

What is the difference between Special Relativity and General Relativity?

2. Jun 6, 2005

### robphy

The modern view [used by working relativists] is that General Relativity works with a spacetime consisting of an arbitrary 4-manifold $$M$$ with a lorentzian metric tensor field $$g_{ab}$$, whereas Special Relativity is the special case where the spacetime is the Minkowski spacetime consisting of $$R^4$$ and the [flat] Minkowski metric $$\eta_{ab}$$.

Last edited: Jun 6, 2005
3. Jun 6, 2005

### dextercioby

Take a look the the axioms for each of them.You'll then see that the geometry of space-time depends on the mathematical interpretation of one postulate in each theory.

You may draw a comparison with QM and its first postulate.

Daniel.

4. Jun 6, 2005

### mathman

From a layman's point of view, special relativity is concerned only with inertial systems (no acceleration), while general relativity does not have that restriction. In particular G.R. is a theory of gravity.

5. Jun 6, 2005

### dextercioby

Nonetheless,special relativity deals very well with the problem of a pointlike particle moving through space-time with a constant acceleration "a".

Daniel.

6. Jun 6, 2005

### robphy

Yes, indeed!... even with non-constant acceleration! The standard twin paradox situation is a non-constant acceleration problem.

As used here, "special relativity" really refers to "application of the Lorentz Transformations".

Yes, G.R. is a theory of gravity. As such, the spacetime of GR is often said to be a dynamical one... the equations of motion being the Einstein Field Equations.

7. Jun 6, 2005

### pmb_phy

Special relativity applies only to coordinate systems which correspond to inertial frames while general relativity applies to all coordinate systems.

The special cases of the vacuum domain wall, straight cosmic string and uniform g-field make you really think about these ideas and why they became to be defined as such.

Pete

8. Jun 6, 2005

### εllipse

To put it in even simpler, very generic terms. Special relativity is the theory of what happens at very fast speeds, while general relativity is the theory of what happens with very dense masses. To go further, special relativity solves the problems Newtonian physics has with high speeds; general relativity solves the problems Newtonian physics has with very large gravitational fields. Of course, this is a generalization. There are many more applications.

9. Jun 6, 2005

### robphy

εllipse,
Your statement reminds me of something called the "Bronstein Cube"
which this first link attributes to Penrose
http://www.physik.fu-berlin.de/~lenzk/PICS/bronsteincube.gif

http://physics.syr.edu/research/hetheory/minnowbrook/stachel.html [Broken]
(click on the thumbnails with cubes)

Essentially, this diagram suggests relationships between the different theoretical regimes using certain limits of fundamental constants.

Last edited by a moderator: May 2, 2017
10. Jun 7, 2005

### pmb_phy

No. That is a misconception. Special relativity was created in part to explain things happing at low speeds. Even at low speeds Lorentz contractions play a role in the electric field of a slowly moving wire. To determine the seperation of events in spacetime as measured in a moving frame you can't neglect relativity if the events have a large spatial seperation.
GR addresses all sorts of motion. In fact, according to Einstein, you can have a flat spacetime - change frames of reference and you "produce" a gravitational field. In this case the only mass working here is the mass of the "distance stars."

Pete

11. Jun 7, 2005

### pmb_phy

Thanks for the link Rob. Please note that Stachel does not adhere to the "The modern view [used by working relativists] ..." comment you made. He adhere's to Einstein's views on GR and not to the view found in, say, Wald. Remind me in the future to e-mail his article on this point to you.

Pete

Last edited by a moderator: May 2, 2017
12. Jun 7, 2005

### εllipse

Do you think someone asking what the difference in SR and GR is will know what Lorentz contractions are? I posted an answer in words anyone could understand.

13. Jun 7, 2005

### pmb_phy

Do you think the same person would understand Rob's post?

The main fact I'm pointing out is that SR is not just for high speed motion. That was pretty clear in my post

14. Jun 7, 2005

### robphy

While it's been argued that this part is in accurate:
I think we can agree that this part of the εllipse's description is fine:

Without much context on what the original poster already knows, it is my preference to first give a precise though-possibly-advanced answer (which can be simplified with clarifications as needed) rather than give an imprecise and possibly-misleading answer that has to be cleaned-up or thrown out later.

[I must also admit that one reason for that first answer I gave was to try to defeat misconceptions, particularly those stemming from the historical development of the subject, and to advocate the modern terminology and interpretations used in practice.]

15. Jun 7, 2005

### pmb_phy

Please don't get me wrong Rob. I see nothing with your post. I think its nice to have different people here posting different views at different levels. A discussion works best that way here. I was unable to determine the level of sophistication of the poster but he seems to read alot about physics and relativity from his profile so it seems he'd at least have heard of the Lorentz transformation.

I think different definitions give different answers. E.g. I think some would say that no Riemann -> No g-field while others would say no $\Gamma$ -> no g-field. The first comes from MTW while the second comes from MTW and Wald. This last part is, of course, confusing. To see the latter part see MTW page 467.

Btw - what do you find wrong with historical matter if it works better for some people? In his text "Concepts of space," Max Jammer has an foreword by Einstein in which he discusses the importance of history in science. It is well worth your read. I can scan and e-mail if you wish as always.

Pete

16. Jun 7, 2005

### robphy

Here is part of a quote I like from J.L. Synge:
The history of SR/GR is interesting... there's some good stories in there and there is stuff to learn from it.

However, today [in practice], a lot of the ideas have been formulated neatly with some precise definitions... let's use them! In teaching others, I feel we (as a whole) go further in understanding and advancing the subject by teaching the modern formulation (appropriately simplified for the audience) and building upon it rather than stumbling over the same mistakes made in the past. (Certainly, it may necessary to take folks through a few mistakes to get them to appreciate things... but I think we need to streamline the path somewhat.)

my \$0.02

17. Jun 7, 2005

### pmb_phy

Rob - That is not what I meant. Speaking of the practicing scientist referring to the historian, Einstein said
A great example is to be found in the American Journal of Physics. With so many people saying "Gravity is a curvature in spacetime" some people take that to mean that a uniform gravitational field will have spacetime curvature. There is an article in AJP by someone who assumes the Riemann tensor must be zero for such a field and when he gets a non-zero value he is pleased with himself. The problem was that "uniform gravitational field" means "zero Riemann tensor." His lack of knowledge and relying on such ideas as "Gravity is a curvature in spacetime" and ignoring what a gravitational field really is (as Einstein knew all too well) led him to make this serious error in his article. The author, the editor and the referees all got it wrong since it was published.

The article I refer to is
Nonequivalence of a uniformly accelerating reference frame and a frame at rest in a uniform gravitational field, Edward A. Desloge, Am. J. Phys., Vol. 57, No. 12, Dec 1989, page 1121-1125

Einstein would roll over in his grave if he read that article!

Of course other authors assume a vanishing Riemann tensor such as
Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173

In my own experience it took me a very long time to learn that E does not always equal mc^2 (recall that stress contributes to momentum and thus to inertial mass aka "relativistic mass" = m = p/v). I should have read Rindler's text first. It would have saved me a LOT of time.

I went back to Einstein's original papers and there it all was in his 1907 paper (or was it 1906?). What a genius Einstein was! But in all the dicussions over the last 7 years I've had on the concept of mass nobody ever mentioned these basic concepts. Probably because the stress-energy-momentum tensor is never used in SR texts as applied to simple bodies such as a capacitor. But leave it to Rindler to do so!

See

Another example is my current efforts to write a text (I'll be finished in about 40 years :rofl:). I want to start off by presenting a (an operational??) definition of "space," "time" and "spacetime." Not an easy task. One article I keep putting off is that regarding the Hopi Indians. In there language it is said that there is no concept of time. All good stuff. Hans Riechenbach has a great book called "The Philosophy of Space and Time" which I will be reading someday. I pick it up once and a while when I need to switch gears for a moment.

Pete

Last edited: Jun 7, 2005
18. Jun 10, 2005

### Zanket

Agreed. Such quotes are an example of how many sources on GR make the subject way more complicated than it is. I use the following definitions, which are simplified from a cross section of books:

curved spacetime nonuniform gravitational field (see)
flat spacetime uniform gravitational field (see)
inertial frame A local frame in free fall
local In a vicinity or having a volume throughout which the tidal force is negligible
nonlocal In a vicinity or having a volume within which the tidal force is significant
nonuniform gravitational field (curved spacetime) The gravitational field of a nonlocal frame
tidal force The relative gravitational acceleration of two test particles in free fall
uniform gravitational field (flat spacetime) The gravitational field of a local frame

A region of flat spacetime as large as our observable universe, say, could be moving at relativistic velocity and accelerating toward and relative to some mass outside of it. The region can be deemed an inertial frame so long as the tidal force imparted by the mass is negligible throughout the region (i.e., the mass does not curve the region). The reason that inertial frames must be infinitesimal in GR is a technicality to guarantee that the tidal force within them is negligible; practically speaking there is no size limit. Once that is understood, it follows that flat spacetime is synonymous with a uniform gravitational field, and curved spacetime is synonymous with a nonuniform gravitational field.

19. Jun 10, 2005

### pmb_phy

I hope you understand how happy you've made me by this comment.

"curved spacetime nonuniform gravitational field" - I agree.

"flat spacetime uniform gravitational field" - I agree.

"inertial frame A local frame in free fall" - If you add that the frame need not be local if the spacetime is flat then I agree.

"local In a vicinity or having a volume throughout which the tidal force is negligible" - I agree. But its better if you add that the volume is a 4D volume of spacetime and not of simply space.

"nonlocal In a vicinity or having a volume within which the tidal force is significant" - I agree.

"nonuniform gravitational field (curved spacetime) The gravitational field of a nonlocal frame" - I disagree. Nonlocal can include flat spacetime. The correct definition is a gravitational field with non-zero tidal forces/gradients.

"tidal force The relative gravitational acceleration of two test particles in free fall" - I agree.

A wee bit-o-math for yee!

http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm

"uniform gravitational field (flat spacetime) The gravitational field of a local frame" - I disagree. A uniform gravitational field is a gravitational field where the tidal forces are zero.

"Once that is understood, it follows that flat spacetime is synonymous with a uniform gravitational field," - I disagree. Whether there is a uniform gravitational field is present will depend on the coordinates you choose. Thus you can "produce" a uniform gravitational field by changing coordinates to that associated with a uniformly accelerating frame of reference.

"and curved spacetime is synonymous with a nonuniform gravitational field." - I agree.

Curved Pete :rofl:

20. Jun 11, 2005

### Zanket

Welcome back!

I think some of your disagreement stems from my definitions of local vs. nonlocal. Let me know what you think:

In the definitions I gave, a local frame comprises only flat spacetime, whereas a nonlocal frame contains some curved spacetime, disqualifying it as an inertial frame.

“Local” is a potentially confusing term, for in English it means both “nearby” (e.g. a local store) and “not widespread” (e.g. a local anesthetic) and both meanings are used when discussing local frames in my books. Something “not widespread” may or may not be nearby. We are used to the dual usage of “local” in general English, and I’ve gotten used to it in GR. The definition I gave for “local” encompasses both situations. I expanded the definition beyond that of a typical GR book, to say that what is deemed local or nonlocal is based on the whether or not the tidal force is negligible. The typical GR book is wishy-washy, using qualifiers like “very small” or “small enough” when they’re really getting at the tidal force as the differentiator.

Good catch. I’ve been studying GR for years now and it still seems that my books use “space” and “spacetime” interchangeably. Perhaps they are just being sloppy.

I think I cover “Nonlocal can include flat spacetime” in the definition I gave for nonlocal. When I say “within which the tidal force is significant,” I mean that somewhere within, the tidal force is significant—it’s not negligible throughout, as it is when local. It could be negligible (flat spacetime) in all but a small fraction of it.

Maybe a wee bit for you, but not for me!

I’m using the practical definition rather than the technical definition. I’m saying that a uniform gravitational field is a gravitational field where the tidal forces are negligible. I think the practical definition, which allows a local frame to be of any size, is better since, technically, flat spacetime—a uniform gravitational field—exists nowhere in the universe except infinitesimally. The practical definition is what is used in experiments of special relativity, which is what we compare the theory to.

I don’t get your comment here. Can you elaborate as to when flat spacetime is not synonymous with a uniform gravitational field? IMO when a noninertially uniformly accelerating rocket (in deep space--flat spacetime of indefinite extent) "produces" a uniform gravitational field, the spacetime within the rocket is just as flat as when its engines are off. When its engines are off, the local gravitational acceleration is zero; the rocket accelerates noninertially and uniformly at a = 0.

Last edited: Jun 11, 2005
21. Jun 11, 2005

### pmb_phy

Thanks! I got the right drugs

Yes. I believe the difference in our notions are due to differences in the way we choose to define "local." I typical go by the book on most occations. I will specify when texts different with respect to a definition. But the term "local" has a well defined definition which is uniform throughout the literature.

No. This is definitely wrong. In a small region of spacetime about a point P in spacetime is refered to as a local spacetime coordinate system if the gravitational force vanishes at P and the gravitational force at small distances from P may be ignored. Local pertains only to free-fall motion. So local can contain both flat and curved spacetime. In fact One can say that at a point P in spacetime one can choose coordinates which can be locally flat at P but also curved at P. I'm standing on the Earth which we will consider to be a perfect sphere for the sake of arguement. Think of yourself as standing on this sphere with rulers to measure distances and the ruler has a certain degree of precision. With your ruler you constuct a "small" equilateral triangle. But ignoring small variations you determine that the sum of the interior angles is 180 degrees. Then you say "However if you beef up the precision of your measurements you will determine that it really is a sphere. You can say that your coordinate is restricted to the "small" region and it is therefore is is locally flat coordinate system. But you know that the surface really isn't flat so you also say that the survace is curved where the triangle is located.

There is controversy regarding local coordinate systems in the GR literarture. The 100% correct definition of "local" means that at the origin of the coordinate system the gravitational force vanishes. If the spacetime is curved then points away from the origin have a non-gravitational force in general.

As defined in a differential geometry text most, if not all, define local as follows
In spacetime if you're in a globally inertial frame then the entire spacetime is flat. If you're in a frame in which an enormous region is flat but outside that region there is curvature then ther is no term defined for this example.
No! Because I messed up. Thanks for catching that.

Pete

Last edited: Jun 12, 2005
22. Jun 12, 2005

### Zanket

Agreed, except for the last sentence, because my books disagree with you that “local” pertains only to free-fall motion. Granted I have a small library of physics books, and they are laymen’s books, but they are authoritative. In two of them, The Riddle of Gravitation and Black Holes and Time Warps, a local frame is a frame having a negligible tidal force throughout, but it need not be free-falling. In the former book a free-falling local frame is called a “free-falling frame” and in the latter it is called a “local inertial frame” (distinct in the book from the “inertial frame” of special relativity, where free fall is not mentioned). The book Exploring Black Holes calls this a “free-float frame” or an “inertial frame”. Four terms for the same thing, one of which overrides the “inertial frame” of special relativity. Maddening! I think it’s best to do away with the superfluous “free-float frame” and keep only the “inertial frame” that mentions free fall, since the inertial frame of special relativity does not exist in reality. Then it is clearer that a free-falling local frame is the inertial frame for practical applications of special relativity.

Where my books disagree on terms, I feel free to define my own, ideally from a melding of them where the term can be simply defined. I like that a local frame need not be free-falling, because then I have a term for a frame throughout which the tidal force is negligible, and which is being noninertially accelerated, such as the frame of the room I’m in now, about which I can say that, for practical purposes, the local gravitational acceleration is 1g throughout. What is your term for such a frame?

And note that if “local frame” implies free-falling, then “local gravitational acceleration” can confuse.

I’m confused. Doesn’t a tidal force (what I assume you mean by “gravitational force”) vanish at any point? If so, why the conditional “if”? And if a “local spacetime coordinate system” is anchored by a fixed point P, then how can “local” pertain only to free-fall motion? Is P free-falling (i.e., is the coordinate system in free fall)?

This goes to a distinction between “negligible”, “infinitesimal” and “zero”. My books tend to be wishy-washy on these. To me, “negligible” is the only one useful for practical purposes, so I reject the others. Then I would say that the surface is flat locally, where the triangle is located, where “local” is defined as a region having negligible curvature throughout.

When “local” is defined as negligible tidal force throughout, there will not be confusing (to me) statements like “local can contain both flat and curved spacetime”. After reading your Earth example I interpret this to mean that “local contains both flat and curved spacetime, as one alternates between deeming the tidal force throughout to be negligible or not”. I think a definition serves better when it is unambiguous.

Not sure I get that. I interpret it to mean that the 100% correct definition of “local” is an “infinitesimally small region”, and spacetime is curved in any larger region. That's what a lot of my books imply.

There is a term for that in the first two books I mention above. It’s a “local frame”, which is finite in extent, and throughout which the tidal force is negligible (well, The Riddle of Gravitation talks only about the "immediate vicinity of some point"), not necessarily infinitesimal or zero. Exploring Black Holes also has this concept, except therein it’s defined only in the form of the “inertial frame”, a local frame in free fall. Wherever I see definitions using words like "very small" or "immediate vicinity", I know that the definition can be improved since those are ambiguous words.

Isn’t there some international standards committee that hashes this stuff out so everyone can be on the same page? It seems that physics definitions in general are all over the map. For that reason I prefer books that have a decent glossary. For my paper I put my definitions in a "conventions herein" section, to emphasize that they are my definitions, not necessarily those of others.

Last edited: Jun 12, 2005
23. Jun 12, 2005

### pmb_phy

I'll have to get back later since I have to wait until I'ma more awake

Pete

24. Jun 12, 2005

### pmb_phy

Can you provide the exact quote. E.g. some authors use the term "space" to refer to "spacetime".

Where my books disagree on terms, I feel free to define my own, ideally from a melding of them where the term can be simply defined. I like that a local frame need not be free-falling, because then I have a term for a frame throughout which the tidal force is negligible, and which is being noninertially accelerated, such as the frame of the room I’m in now, about which I can say that, for practical purposes, the local gravitational acceleration is 1g throughout. What is your term for such a frame?

I'm not sure what you're trying to find out. Regarless of the strength of the tidal forces the region of spacetime mus be kept small. That means that you can't run an experiment for too long or you will measure tidal gradients
Keep in mind that they are not simply refered to as "local" amd that makes quite difference. No matter ho small the spatial portion of space is,given enough time the experiment will violate the Principle of
Equivalence.
If referres to whether the region of spacetime you're interesting in is small enough so that not enough time has passed so that the curvature that you speak of can't be measured.
The reference frame of interested is a frame of reference attached to the origin which the coordinate system is in free-fall along with the point P.
I agree that it may confuse people. In this case if the experiment runs for a long enough time then the tidal forces cannot be neglected

I don't that believes we can be helped - However you might call your local representative.

Let me add that tidal force is different to gravitational force. The object is under investigation. When tidal forces are present then you can't say whether something is local of not. I've seen someone show me this and seeing can of help.

Pete

25. Jun 12, 2005

### pmb_phy

Sure it does. Where did you great that idea??

I do agree that different texts make any distinguishment ment.