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How does general relativity describe gravity as the curvature of spacetime?

The equivalence principle, general relativity as the curvature of spacetime by mass and energy, the bending of light, black holes and the Schwarzschild radius.

An SQA Advanced Higher Physics answer on general relativity, covering the equivalence principle, gravity as the curvature of spacetime, the bending of light by gravity, black holes, and calculating the Schwarzschild radius.

Generated by Claude Opus 4.813 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. The equivalence principle
  3. Gravity as curved spacetime
  4. Black holes and the Schwarzschild radius
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to state and use the equivalence principle, describe general relativity's picture of gravity as the curvature of spacetime caused by mass and energy, know that light is affected by gravity, understand that very massive, dense objects form black holes, and calculate the Schwarzschild radius of a non-rotating black hole. This is a largely conceptual key area with one calculation.

The equivalence principle

Einstein's "happiest thought" was that a person in free fall feels no gravity, and a person in an accelerating rocket feels an apparent gravity. No local experiment can tell a steady gravitational field from a steady acceleration. This single principle is the seed of general relativity: it immediately predicts that light bends in gravity, because a light beam crossing an accelerating lift appears to curve, so it must curve in an equivalent gravitational field too.

Gravity as curved spacetime

The Newtonian idea of a gravitational force is an excellent approximation in weak fields, but general relativity is more fundamental and is needed where gravity is strong or where high precision matters. Because energy as well as mass curves spacetime, even light, which has energy, is deflected. This curvature is not a metaphor: it produces measurable effects such as the bending of starlight near the Sun and the slowing of clocks deep in a gravitational field.

Black holes and the Schwarzschild radius

You can motivate the Schwarzschild radius from the Newtonian escape-velocity formula by setting the escape speed equal to cc: 2GMr=c\sqrt{\frac{2GM}{r}} = c rearranges to r=2GMc2r = \frac{2GM}{c^2}, which (remarkably) matches the full relativistic result. Anything that crosses inside rsr_s cannot return. Black holes form when the most massive stars collapse at the end of their lives, the link back to stellar physics.

Examples in context

GPS satellites must correct their clocks for general-relativistic time dilation, or positions would drift by kilometres a day. Gravitational lensing, where a massive galaxy bends light from a more distant one into arcs, is a direct image of curved spacetime. The Event Horizon Telescope photographed the shadow of the supermassive black hole in galaxy M87, at the scale of its Schwarzschild radius. Gravitational waves from merging black holes, detected by LIGO, are ripples in spacetime itself.

Try this

Q1. State what general relativity identifies as the cause of gravity. [1 mark]

  • Cue. The curvature of spacetime by mass and energy.

Q2. State what happens to light that crosses the event horizon of a black hole. [1 mark]

  • Cue. It cannot escape; nothing, including light, can leave from inside.

Q3. Write the relationship for the Schwarzschild radius of a non-rotating black hole. [1 mark]

  • Cue. rs=2GMc2r_s = \frac{2GM}{c^2}.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style4 marksCalculate the Schwarzschild radius of a black hole of mass 8.0×1030 kg8.0 \times 10^{30}\ \text{kg}. Take G=6.67×10−11 N m2kg−2G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{kg}^{-2} and c=3.0×108 m s−1c = 3.0 \times 10^{8}\ \text{m s}^{-1}.
Show worked answer →

The Schwarzschild radius is rs=2GMc2r_s = \dfrac{2GM}{c^2}.

Substitute: rs=2×6.67×10−11×8.0×1030(3.0×108)2r_s = \dfrac{2 \times 6.67 \times 10^{-11} \times 8.0 \times 10^{30}}{(3.0 \times 10^{8})^2}.

Numerator: 2×6.67×10−11×8.0×1030=1.07×10212 \times 6.67 \times 10^{-11} \times 8.0 \times 10^{30} = 1.07 \times 10^{21}.

Denominator: (3.0×108)2=9.0×1016(3.0 \times 10^{8})^2 = 9.0 \times 10^{16}.

So rs=1.07×10219.0×1016=1.2×104 mr_s = \dfrac{1.07 \times 10^{21}}{9.0 \times 10^{16}} = 1.2 \times 10^{4}\ \text{m}.

Markers reward the correct relationship, squaring cc in the denominator, and the value with unit.

SQA AH style3 marksState the equivalence principle and give one observable consequence of it.
Show worked answer →

The equivalence principle states that the effects of a uniform gravitational field are indistinguishable, locally, from those of a constant acceleration of the reference frame.

An observer in a sealed, accelerating spaceship cannot tell, by any local experiment, whether they are being accelerated or sitting in a gravitational field.

One consequence is that light is bent by gravity, because a light ray crossing an accelerating frame follows a curved path, so it must also curve in a gravitational field.

Markers reward a correct statement of the principle and a valid consequence such as the bending of light or gravitational time dilation.

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