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Difference Between General & Special Theory Of Relativity


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#1 thewonderer

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Posted 25 January 2010 - 07:14 PM

Can someone explain me the difference between general and special relativity?
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:doh:

#2 insane_alien

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Posted 25 January 2010 - 07:24 PM

general relativity takes into account a non-flat local space-time manifold.

basically, general relativity should be used when in a gravitational field. if there is no gravitational field then general relativity reduces to special relativity.

special relativity is called special simply because it is a special case of general relativity.

general relativity is called such because it can be applied generally.
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#3 thewonderer

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Posted 25 January 2010 - 07:42 PM

can some1 explain it mathematically?
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#4 ajb

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Posted 25 January 2010 - 09:52 PM

can some1 explain it mathematically?


Basically, in special relativity the space-time is taken to be a manifold equipped with the Minkowski metric. That is it formally resembles Euclidean space. In a similar way to Euclidean space there is a perfected collection of coordinates called the "inertial coordinates". The metric in such coordinates is (up to signs for the components) essentially the identity matrix. The admissible coordinates between inertial coordinates preserve the metric. These are the Lorentz transformations, which are linear.

Now, in general relativity we lose the preferred collection of coordinates and have to consider non-inertial ones. (Think of inertial coordinates are rectangular coordinates and non-inertial ones as more general coordinates). The transformations we need to consider are now more general diffeomorphisms. Furthermore, the metric is now something more general. The space-time is not necessarily flat, that is it does not really resemble Euclidean space outside the neighbourhood of a point.


Physically we think of the curvature of space-time as gravity.
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#5 michel123456

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Posted 26 January 2010 - 01:10 PM

Here correct me if I am wrong:
Special Relativity works in absence of gravitational field.
Special Relativity works in Inertial frame Of Reference.
Free fall is a situation of Inertial frame Of Reference.

But free fall is a situation where you are accelerated under the action of gravity, inside a gravitational field.
Isn't that confusing?
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Michel what have you done?


#6 ajb

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Posted 26 January 2010 - 02:26 PM

Special Relativity works in absence of gravitational field.


I believe you can do something like special relativity in the presence of an external Newtonian potential. Physically the trouble will be that instantaneous action does not sit well with spacial relativity. Mathematically, I am sure you run into problems.

Special Relativity works in Inertial frame Of Reference.


Special relativity states that the inertial coordinates provide a preferred collection of coordinates. Such coordinates are related via Lorentz transformations. However, one is free to consider any coordinates one likes.

Free fall is a situation of Inertial frame Of Reference.


Yes, ok.

But free fall is a situation where you are accelerated under the action of gravity, inside a gravitational field.
Isn't that confusing?


What is assumed is that the gravitational field does not change. So for example, Einstein's lift under free fall provides an inertial frame provided the gravitational field is constant in the lift.

In reality this is not true. It is approximately true for "small lifts" in "weak fields". It can be made mathematically restated as a local theorem. In essence, this states that space-time is a Riemannian manifold, i.e. there exists coordinates in the neighbourhood of a point such that at that point the metric is the Minkowski metric and in that coordinate system the Christoffel symbols vanish. The physics part states that all non-gravitational physics reduces to that of special relativity. This places restrictions on how one can couple things to the gravitational field.
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#7 ranjan_adarsh

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Posted 5 February 2010 - 06:23 AM

Simply Special Relativity deals with object travelling with constant velocity comparable to light speed
And General Relativity deals with accelerations in space-time
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#8 swansont

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Posted 5 February 2010 - 09:22 AM

Simply Special Relativity deals with object travelling with constant velocity comparable to light speed
And General Relativity deals with accelerations in space-time


It's actually a little more subtle than that. You can deal with accelerations in SR.
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#9 ajb

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Posted 5 February 2010 - 10:33 AM

Simply Special Relativity deals with object travelling with constant velocity comparable to light speed
And General Relativity deals with accelerations in space-time


As swansont says, there is the notion of proper acceleration.

In special relativity you are free to consider frames that are not inertial. In doing so the formalism takes you most of the way to general relativity, but not quite all the way. The metric is still the Minkowski metric.
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#10 Bob_for_short

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Posted 5 February 2010 - 12:15 PM

GR treats the gravity as a curvature of space-time. It is not a filed but a Riemann geometry effect. The curvature of such a space-time is not zero: R > 0.

A theory of gravity in a flat (Minkowski) space-time was developed by A.A. Logunov, Russian academician: http://arxiv.org/abs/gr-qc/0210005. No troubles on this way but advantages.

Edited by Bob_for_short, 5 February 2010 - 12:55 PM.

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#11 Obelix

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Posted 14 March 2010 - 05:23 AM

can some1 explain it mathematically?


It's all about geometry. The fundamental feature of the latter, according to Riemann, is the metric tensor g_{ab} (a, b assuming the values 0,1,2,3, i.e. g_{00}, g_{01},...g_{11}, g_{12},... etc.) which, appart form specifing the way distances, angles, etc. are to be measured, it is the building block of all other quantities: Connection (the quantity that acts on vectors transforming them "parallel to themselves" in a general curved background) Curvature (appart from measuring what its name indicates, it is also a measure of the intensity of the gravitational field) etc.

Nowthen: In Special Relativity (SR) the metric tensor is the Minkowski Tensor \eta_{ab}, with \eta_{00}=-1, \eta_{11}=\eta_{22}=\eta_{33}=1 and al otehr components equal to zero. That is, the metric of SR has the form (in Cartesian coordinates t,x,y,z: ds^2=-dt^2+dx^2+dy^2+dz^2.

In General Relativity (GR) the components of the metric tensor can be more complicated functions of the coordinates (whatever the system used). Besides, components like g_{01}, g_{23} etc. can be different to zero, i.e. the metric may include cross terms (dxdy, dtdx, etc). The metric tensor of the Schwarzschild Spacetime, describing an isolated spherical distribution of mass in an otherwise empty space (appropriate, therefore to describe the spacetime around the Sun in excellent approximation) has the form: ds^2=-(1-\frac{2Gm}{c^2r})dt^2+\frac{1}{(1-\frac{2Gm}{c^2r})}dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2), i.e., here we have: g_{00}=-(1-\frac{2Gm}{c^2r}), g_{11}=\frac{1}{(1-\frac{2Gm}{c^2r})}, g_{22}=r^2, g_{33}=r^2\sin^2\theta, in spherical coordinates which are the best for a problem with spherical symmetry.

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#12 Simpleton

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Posted 19 March 2010 - 06:35 AM

Would there be a difference in mass between a large, active molten core body and the exact cold duplicate out in interstellar space giving of only black body radiation.
Please and thank you
Sorry it was supposed to be a new question or post.
To much in a rush.
Sorry again

Merged post follows:

Consecutive posts merged
I would very much appriciate if some one could please move this where it belongs. I am sorry but I can not fix it.

Edited by Simpleton, 18 March 2010 - 10:18 PM.
rong place

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