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Difference between General and Special relativity

I know that general relativity applies in non-inertial reference frames whereas special relativity does not but what difference does this make? does the math of special relativity work in the general relativity. if not what is the difference?

also does the theory hold in a reference frame undergoing NON-UNIFORM acceleration?
(edited 11 years ago)
Reply 1
Special relativity is a special case of general relativity where there is no gravity. So in special relativity we have flat spacetime and the metric is ds2=c2dt2dx2dy2dz2ds^2 = c^2dt^2 - dx^2 -dy^2 -dz^2 whereas in general relativity you can have f(x,t) in front of c2dt2,dx2,dy2,dz2c^2dt^2, dx^2, dy^2, dz^2 depending on the gravitational object you're dealing with.

It is actually a misconception to think that special relativity can't be used to describe non-inertial reference frames - it can be and is used to describe accelerating particles (and no doubt non-uniformly accelerating particles too). However, the link between gravity and spacetime curvature is not made in special relativity, and can only be found in the more general theory.


The reason special relativity works so well is described by the equivalence principle. This says that in any point in spacetime a frame can be found in which the laws of physics obey special relativity. Why this is true can be seen by imagining a particle above the earth, in its gravitational field. The particle will accelerate due to gravity but in a frame accelerating at that same rate, the particle will be stationary. And so we get special relativity.

However, whilst at any point in spacetime a frame can be found reducing the laws of physics to those of special relativity, the frame of reference needed for each point will be different. This change in reference frame is manifest in the curvature of spacetime and the resulting gravity.
(edited 11 years ago)
Reply 2
Original post by 3nTr0pY
whereas in general relativity you can have f(x,t) in front of c2dt2,dx2,dy2,dz2c^2dt^2, dx^2, dy^2, dz^2 depending on the gravitational object you're dealing with.


Just saying, people tend to use the form c2dτ2=gμνdxμdxνc^2 d\tau^2 = g_{\mu\nu}dx^\mu dx^\nu, where gμνg_{\mu\nu} denotes the metric describing the space-time, because if you use c2dτ2=Adt2Bdx2Cdy2Ddz2c^2 d\tau^2=A dt^2 - B dx^2 - C dy^2 - D dz^2, you imply the metric tensor is diagonal, which is not really the case for most metric, for instance, Kerr metric.
(edited 11 years ago)
Reply 3
Original post by jsmith6131
I know that general relativity applies in non-inertial reference frames whereas special relativity does not but what difference does this make? does the math of special relativity work in the general relativity. if not what is the difference?

also does the theory hold in a reference frame undergoing NON-UNIFORM acceleration?


General Relativity works for everything. If there is no acceleration, the it simplifies to Special Relativity :smile:.

The maths of Gen Rel will work in Spec Rel (it'll just simplify to Spec Rel maths), but the opposite is not true...
Reply 5
Original post by Qwertish
General Relativity works for everything.


It works for everything about gravity.

General relativity describes gravity via geometry of space-time.

As a supplementary note, general relativity becomes special relativity at the limit of weak field in which one can do the following approximation:

gμν=ημν+hμνg_{\mu\nu}=\eta_{\mu\nu}+h_{\mu \nu} and that hμν1||h_{\mu\nu}||\ll1

In words:

A metric gμνg_{\mu\nu} describes space-time and the metric in a flat space-time aka Minkowski metric is the metric in special relativity which is

Unparseable latex formula:

\eta=\begin{bmatrix}[br] 1 & 0 & 0 & 0 \\[br] 0 & -1 & 0 & 0 \\[br] 0 & 0 & -1 & 0 \\[br] 0 & 0 & 0 & -1[br] \end{bmatrix}



The really awesome feature of this metric is that it is constant, so maths will be really simple, and that's also the reason why the maths in SR is much simpler than in GR.

On the other hand, hμνh_{\mu \nu} is the perturbation term, which is used to alter the original Minkowski metric. hμν1||h_{\mu \nu}||\ll 1 means the "magnitude" of the metric is really small so that it can still be approximated as a linear field.
Reply 6
Original post by agostino981
Just saying, people tend to use the form c2dτ2=gμνdxμdxνc^2 d\tau^2 = g_{\mu\nu}dx^\mu dx^\nu, where gμνg_{\mu\nu} denotes the metric describing the space-time, because if you use c2dτ2=Adt2Bdx2Cdy2Ddz2c^2 d\tau^2=A dt^2 - B dx^2 - C dy^2 - D dz^2, you imply the metric tensor is diagonal, which is not really the case for most metric, for instance, Kerr metric.

Lol, yeah, but didn't want to scare him off with gμνg_{\mu\nu}. And writing down spherical polars in latex is a pain in the arse! :tongue:
Reply 7
Original post by Qwertish
General Relativity works for everything. If there is no acceleration, the it simplifies to Special Relativity :smile:.

The maths of Gen Rel will work in Spec Rel (it'll just simplify to Spec Rel maths), but the opposite is not true...

Actually, it should be:

"If there is no gravity, then it simplifies to Special Relativity"


Special Relativity can handle pure electromagnetic accelerations perfectly well. It's only when you have gravitational accelerations that the curvature of spacetime needs to be accounted for.
Reply 8
Original post by agostino981
It works for everything about gravity.

General relativity describes gravity via geometry of space-time.

As a supplementary note, general relativity becomes special relativity at the limit of weak field in which one can do the following approximation:

gμν=ημν+hμνg_{\mu\nu}=\eta_{\mu\nu}+h_{\mu \nu} and that hμν1||h_{\mu\nu}||\ll1

In words:

A metric gμνg_{\mu\nu} describes space-time and the metric in a flat space-time aka Minkowski metric is the metric in special relativity which is

Unparseable latex formula:

\eta=\begin{bmatrix}[br] 1 & 0 & 0 & 0 \\[br] 0 & -1 & 0 & 0 \\[br] 0 & 0 & -1 & 0 \\[br] 0 & 0 & 0 & -1[br] \end{bmatrix}



The really awesome feature of this metric is that it is constant, so maths will be really simple, and that's also the reason why the maths in SR is much simpler than in GR.

On the other hand, hμνh_{\mu \nu} is the perturbation term, which is used to alter the original Minkowski metric. hμν1||h_{\mu \nu}||\ll 1 means the "magnitude" of the metric is really small so that it can still be approximated as a linear field.


Original post by 3nTr0pY
Actually, it should be:

"If there is no gravity, then it simplifies to Special Relativity"


Special Relativity can handle pure electromagnetic accelerations perfectly well. It's only when you have gravitational accelerations that the curvature of spacetime needs to be accounted for.


Ah, okay :smile:. Thanks.
Reply 9
Original post by Qwertish
Ah, okay :smile:. Thanks.

No problem. The only caveat is that the electromagnetic force is caused by photons and photons have energy and energy curves spacetime. So having a pure electromagnetic force is technically impossible. There will always be (at least) a tiny contribution from gravity.
Original post by 3nTr0pY
No problem. The only caveat is that the electromagnetic force is caused by photons and photons have energy and energy curves spacetime. So having a pure electromagnetic force is technically impossible. There will always be (at least) a tiny contribution from gravity.


Literally all interactions mediated by virtual particles curve space-time, but it will get really complicated as general relativity does not work well with quantum field theories which in turns implies that you need a quantized version of gravity such as string theory or loop quantum gravity. We are now talking beyond the standard model and GR already.

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