0
$\begingroup$

I always thought special relativity is only there because it is the building block for "The General theory of Relativity", until recently I encountered a text from my course book given below.

A more accurate formula for Doppler effect which is valid even when the speeds are close to light, requires the use of Einstein's special theory of relativity.

Is special relativity mentioned here only because this problem doesn't need to deal with the space-time curvature(I am quite new to relativity, please look it from a beginner point of view)? Or is there any other case in special relativity, which is not a part of the general relativity?

Improve this question
$\endgroup$
10
  • 2
    $\begingroup$ Why is linear algebra still in use, if calculus is more broad? $\endgroup$
    – WillO
    Commented Nov 10, 2015 at 16:56
  • 1
    $\begingroup$ When the stress-energy tensor is not too big (that is, usually, when gravitational fields are "weak" and your interactions and processes are not super high energy), and your frame of reference is inertial, special relativity is a really fine model of reality. For instance, quantum electrodynamics uses special relativity, and its predictions are incredibly accurate. You don't need general relativity to know that one apple and another apple add up to two apples; sometimes a simpler model gives the same results as a more complex one within the desired precision. So you choose the simple one. $\endgroup$
    – nabla
    Commented Nov 10, 2015 at 16:59
  • 1
    $\begingroup$ same as Newton's laws, they are easier to use and still accurate for some applications $\endgroup$
    – user83548
    Commented Nov 10, 2015 at 17:00
  • 2
    $\begingroup$ @WillO While I completely agree with the sentiment, calculus and linear algebra deal with two quite different things. Therefor unlike the example here where special relativity is obtained as a special limit of general relativity, linear algebra and calculus are actually different areas. One does not imply the other. $\endgroup$
    – Prahar
    Commented Nov 10, 2015 at 17:01
  • 2
    $\begingroup$ @prakharlondhe yes, it's a similar idea. Of course you can use Maxwell's equations to obtain more precise results, or even quantum optics, but usually ray optics are just fine for our purposes and, of course, it is much easier to develop an intuition about a simpler model $\endgroup$
    – nabla
    Commented Nov 11, 2015 at 12:36

3 Answers 3

Reset to default
4
$\begingroup$

General relativity is locally just special relativity. Recall general relativity includes gravity, via curved spacetime, but special relativity excludes gravity and assumes flat spacetime. But as Einstein realised, a person freely falling in a uniform gravitational field wouldn't feel their weight (see the equivalence principle); to them, spacetime is locally flat.

Now consider the relativistic Doppler effect in special relativity. This is a "redshift" of light etc caused by relative motion of two observers -- emitter and receiver. But you can apply the exact same formula to two observers in general relativity. We just require they be at the same event (that is, same place and same time), otherwise it is not clear the effect is due to relative motion alone, and not a gravitational redshift for instance. (My Master's thesis showed the interpretation is somewhat flexible.) The reference frame of each observer is described by a set of orthonormal vectors. Then, with caveats I haven't explained, you can apply results from special relativity.

There are various reasons special relativity is still important. Firstly it is simpler, so if you had some difficult research topic you might want to solve it in this arena before attempting curved spacetime. Another is that quantum physics is unified with special relativity but not general relativity, so the physics is not known. In particle accelerators gravity is not important over the short duration of particle collisions.

Improve this answer
$\endgroup$
1
$\begingroup$

Expanding on WillO's comment, that special relativity holds locally is a required assumption in general relativity, much like the theory of how straight lines (and other linear transformations) behave is a required assumption in calculus.

It is technically correct to say that general relativity on flat spacetime is special relativity, but this cannot be a definition of special relativity -- if it were, it would be circular.

Improve this answer
$\endgroup$
0
$\begingroup$

Only when you consider gravity beyond Newton's theory, physics makes use of GRT. In all other cases SRT is used. GRT is complex and incompatible with field theory, specifically with quantum field theory.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.