<p>Light travels in straight lines in spacetime, but not necessarily straight lines through space, and the same is true for free-falling orbits of test particles like satellites (when thrust-less). However, in relativity what constitutes 'space' means taking some 'slice of <em>now</em>' through spacetime, which is of course depends on the reference frame one chooses to use.</p>

<p>For weak, slowly changing gravitational fields appropriate to the Sun, a conventional choice is the static weak-field metric
$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2\right)}_{\mathrm{d}S^2}\text{,}$$
where $\Phi$ is the Newtonian gravitational potential. Thus, we can compare to a hypothetical flat spacetime, with a flat Euclidean space given by $\mathrm{d}S^2$. The trajectory of light follows null geodesics ($\mathrm{d}s = 0$), so the light travels <em>as if</em> traveling through a medium of <a href="https://en.wikipedia.org/wiki/Refractive_index" rel="nofollow">refractive index</a>
$$n = c\frac{\mathrm{d}t}{\mathrm{d}S} = c\sqrt{\frac{1-2\Phi/c^2}{1+2\Phi/c^2}}\approx 1 - 2\frac{\Phi}{c^2}\text{.}$$
A changing refractive index does make light bend; this is exemplified in the simplest case by <a href="https://en.wikipedia.org/wiki/Snell%27s_law" rel="nofollow">Snell's law</a>, although the continuously varying case found here needs the more general <a href="https://en.wikipedia.org/wiki/Fermat%27s_principle" rel="nofollow">Fermat's principle</a>.</p>

<p>You shouldn't take the analogy of a refractive medium too literally. Some differences between an actual medium is that there is no dispersion as the refractive index is independent of frequency, and that there is an additional effect of <em>gravitational redshift</em> not present in refractive media. But the latter is separate issue from the trajectories of light rays, so we can ignore it for our immediate purposes.</p>

<p>Light deflection is quantified by an <em><a href="https://en.wikipedia.org/wiki/Impact_parameter" rel="nofollow">impact parameter</a></em> $b$, which can be thought of as the perpendicular distance between the Sun and a hypothetical flat-spacetime trajectory, and a <em>scattering angle</em> $\theta$, the angle between that and its actual path in the limit of infinite distance. (Note that the illustration in wikipedia assumes a repulsive rather than attractive potential, but otherwise conceptually it's the same situation.)</p>

<p>Modeling the Sun as having a spherically symmetrical gravitational potential $\Phi = -GM/r$, the leading-order light deflection is,
$$\begin{eqnarray*}
\theta = 4\frac{GM}{c^2b} &amp;\quad&amp;\left(\text{if}\;\frac{GM}{c^2b}\ll 1\right)\text{,}\end{eqnarray*}$$
which is twice what one would get for a particle at light speed under Newtonian mechanics. For light ray just grazing the outer edge of the Sun (or being emitted from there), this angle is about $1.7$ seconds of arc. Completely radial light rays would not be bent in this spherically-symmetric case.</p>

CC BY-SA 3.0
2015-04-16T22:57:15.320Z
2015-04-16T22:57:15.320Z
Stan Liou
<p>When we travel through space, we use a curved path following an orbit to reach our destinations.
<img src="https://i.sstatic.net/Pq8fP.gif" alt="Like So" /></p>
<p>As far as I'm aware, we could take a more straight line approach but it would require much more deltaV (both in reaching the speed and in slowing down when reaching the target to be captured.)</p>
<p>However the path would still be slightly curved because of the curvature of space and because of orbital tracjectory when leaving Earth.</p>
<p>Does the light escaping the Sun move fast enough to follow a straight line path until it hits Earth, or is it curved in those 8 minutes?</p>
<p>What about to somewhere further like Pluto?</p>

<orbit><light><speed>
Does the Sun's light travel fast enough to have a straight path to Earth?
Ezra Bailey
# Does the Sun's light travel fast enough to have a straight path to Earth?

(Asked by: username_0 on 2015-04-16)

When we travel through space, we use a curved path following an orbit to reach our destinations.
![Like So](https://i.sstatic.net/Pq8fP.gif)

As far as I'm aware, we could take a more straight line approach but it would require much more deltaV (both in reaching the speed and in slowing down when reaching the target to be captured.)

However the path would still be slightly curved because of the curvature of space and because of orbital tracjectory when leaving Earth.

Does the light escaping the Sun move fast enough to follow a straight line path until it hits Earth, or is it curved in those 8 minutes?

What about to somewhere further like Pluto?

**Tags:** orbit, light, speed


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## 1 answers

### Answer by username_1 on 2015-04-16. Score: 4

Light travels in straight lines in spacetime, but not necessarily straight lines through space, and the same is true for free-falling orbits of test particles like satellites (when thrust-less). However, in relativity what constitutes 'space' means taking some 'slice of *now*' through spacetime, which is of course depends on the reference frame one chooses to use.

For weak, slowly changing gravitational fields appropriate to the Sun, a conventional choice is the static weak-field metric
$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2\right)}\_{\mathrm{d}S^2}\text{,}$$
where $\Phi$ is the Newtonian gravitational potential. Thus, we can compare to a hypothetical flat spacetime, with a flat Euclidean space given by $\mathrm{d}S^2$. The trajectory of light follows null geodesics ($\mathrm{d}s = 0$), so the light travels *as if* traveling through a medium of [refractive index](https://en.wikipedia.org/wiki/Refractive_index)
$$n = c\frac{\mathrm{d}t}{\mathrm{d}S} = c\sqrt{\frac{1-2\Phi/c^2}{1+2\Phi/c^2}}\approx 1 - 2\frac{\Phi}{c^2}\text{.}$$
A changing refractive index does make light bend; this is exemplified in the simplest case by [Snell's law](https://en.wikipedia.org/wiki/Snell%27s_law), although the continuously varying case found here needs the more general [Fermat's principle](https://en.wikipedia.org/wiki/Fermat%27s_principle).

You shouldn't take the analogy of a refractive medium too literally. Some differences between an actual medium is that there is no dispersion as the refractive index is independent of frequency, and that there is an additional effect of *gravitational redshift* not present in refractive media. But the latter is separate issue from the trajectories of light rays, so we can ignore it for our immediate purposes.

Light deflection is quantified by an *[impact parameter](https://en.wikipedia.org/wiki/Impact_parameter)* $b$, which can be thought of as the perpendicular distance between the Sun and a hypothetical flat-spacetime trajectory, and a *scattering angle* $\theta$, the angle between that and its actual path in the limit of infinite distance. (Note that the illustration in wikipedia assumes a repulsive rather than attractive potential, but otherwise conceptually it's the same situation.)

Modeling the Sun as having a spherically symmetrical gravitational potential $\Phi = -GM/r$, the leading-order light deflection is,
$$\begin{eqnarray\*}
\theta = 4\frac{GM}{c^2b} &\quad&\left(\text{if}\;\frac{GM}{c^2b}\ll 1\right)\text{,}\end{eqnarray\*}$$
which is twice what one would get for a particle at light speed under Newtonian mechanics. For light ray just grazing the outer edge of the Sun (or being emitted from there), this angle is about $1.7$ seconds of arc. Completely radial light rays would not be bent in this spherically-symmetric case.


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