# Special relativity for accelerated trajectories

'There is no truth to the rumor that SR is unable to deal with accelerated trajectories and general relativity must be invoked.' Can some one explain it. I used to think SR is for flat spacetime and GR is for curved spacetime.

• Rindler Coordinates Jul 21, 2022 at 17:58
• A plane is flat. It still contains curves that are not lines. Jul 22, 2022 at 5:06
• Who are you quoting?
– hft
Jul 22, 2022 at 15:19

Yes, special relativity deals with the flat spacetime $$(\Bbb{R}^4,\eta)$$, and GR deals generically with curved spacetimes $$(M,g)$$. So, the distinction between SR and GR comes in the form of curvature of (the Levi-Civita connection of) the metric, and it is the curvature which models gravity, i.e SR deals with zero (or 'weak') gravity while GR deals with non-zero gravity (even on the astronomical scale).

Now, we can talk about acceleration just as well in either case. The concept of acceleration of a curve $$\gamma$$ can be defined in every Lorentzian manifold- flat or not- via the simple expression $$\nabla_{\dot{\gamma}}\dot{\gamma}$$, where $$\nabla$$ is the Levi-Civita connection of the metric. Let me emphasize the following point though: the notion of acceleration depends on the metric tensor (because the metric tensor determines the Levi-Civita connection $$\nabla$$). So you can consider different spacetimes $$(M,g_1),(M,g_2)$$ with the same underlying manifold but different metric tensor. If you consider the same curve $$\gamma:\Bbb{R}\to M$$, then you may find that $$\gamma$$ has zero acceleration with respect to $$g_1$$, but non-zero acceleration with respect to $$g_2$$.

If a curve $$\gamma$$ satisfies $$\nabla_{\dot{\gamma}}\dot{\gamma}=0$$, then we say the curve is not accelerating; the terminology in differential geometry is that $$\gamma$$ is a geodesic. In SR, the spacetime and underlying geometry is so simple that all the mathematical machinery reduces to the naive things we expect: geodesics in $$(\Bbb{R}^4,\eta)$$ are precisely the straight lines in $$\Bbb{R}^4$$, and any curve which is not a straight line thus has non-zero acceleration.

So, to reiterate, there is no contradiction between the statements "SR deals with flat spacetimes $$(\Bbb{R}^4,\eta)$$" and "we can talk about accelerating observers in SR". Things may sound confusing if you start excessively working exclusively with coordinates in the beginning because the introduction of coordinates obscures the geometry (and hence the physics).

There is no real acceleration in a flat spacetime as a whole; Since in spacetime our velocity is always equal to the speed of light, we can't speed up or slow down inside spacetime as a whole...

Sure we can go faster in space, but we will also go slower in time... Meaning that we can accelerate in space but by doing so, we will also desselerate in time... At the end, our net speed will continue to be equal to the speed of light no matter what we do!

That is called a hyperbolic rotation, and its the only way to move in spacetime.

• "Since in spacetime our velocity is always equal to the speed of light" what? and what do you mean there is no real acceleration in a flat spacetime? Jul 26, 2022 at 10:50
• If you combine your speed in space, with your speed in time, the outcome will always equal to the speed of light. That's your speed in spacetime, and its always equal to C=1ls/s. youtu.be/au0QJYISe4c?t=301
– Nuke
Jul 26, 2022 at 10:54
• well, that is some non-standard play on words. What the video is talking about with the quarter circle is the factor which pops up in time-dilation (namely $\sqrt{1-\left(\frac{v}{c}\right)^2}$) Jul 26, 2022 at 11:25
• Well, standard words, don't seem to get the meaning of Einstein's relativity across very well though to new people. That's why so many new people get confused, frustrated and missinformed by using standard words. When in reality, Einstein's relativity is so simple... Everything is moving in spacetime with one speed, the speed of 1ls/s , also known as "C". And that speed is constant (so it can't be changed) and invariant (so it doesn't depend on the frame of reference). No matter who's observing that speed will be 1ls/s for everyone. Something that ultimately preserves causality in the theory.
– Nuke
Jul 26, 2022 at 23:59