# light and gravity **URL:** \ **Category:** Factual Questions\ **Created:** [August 1, 2000, 6:28am UTC](https://boards.straightdope.com/t/light-and-gravity/27674 "2000-08-01T06:28:38Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![fetaljew](https://avatars.discourse-cdn.com/v4/letter/f/90ced4/32.png) [@fetaljew](https://boards.straightdope.com/u/fetaljew)\ **Post date:** [August 1, 2000, 6:28am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/1 "2000-08-01T06:28:38Z")
when nasa sent out the voyagers i remember they talked about using jupiters strong gravitational field “like a slingshot” to give the probes an extra boost on their way towards the other planets, the idea being that the probes would speed up as they fell towards jupiter, like when rolling balls speed up when they pass a dip on a surface. so anyway… what happens when light gets bent by a gravitational field? since it is already going at the speed of light i assume that it cant go any faster, or does it? does it just gain energy and change into xrays or gamma rays or something? or does it just bend and not change in speed or energy since it virtually has no mass? ---
**Author:** ![Lazarus7](https://avatars.discourse-cdn.com/v4/letter/l/67e7ee/32.png) [@Lazarus7](https://boards.straightdope.com/u/Lazarus7)\ **Post date:** [August 1, 2000, 6:47am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/2 "2000-08-01T06:47:38Z")
Since it has no mass, one could assume that it would not change speeds at all. ---
**Author:** ![The\_Ryan](https://avatars.discourse-cdn.com/v4/letter/t/7feea3/32.png) [@The\_Ryan](https://boards.straightdope.com/u/The_Ryan)\ **Post date:** [August 1, 2000, 6:51am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/3 "2000-08-01T06:51:21Z")
First of all, I’m rather suspicious of this “slingshot” idea. It certainly is possible to use a large body to change the _direction_ of an object’s trajectory, but not the _magnitude_ (relative to the planet). Yeah, it’s going to speed as it gets closer, but it’s just going to slow down a gain when it goes away from the body. So unless the body is going the same direction that you want to go, you’re not going to get any energy out of it. Now on to light. In relativity, there is the “principle of equivalence”, which states that any gravittional field is (locally) equivalent to an accelerating frame of reference. So light does bend in a gravitational field, even though they have no (rest) mass. Now, suppose you measure the frequency of a light beam in that’s far away from a massive body. And suppose that the light beam is coming towards the object. Now, if a normal object (like yourself) were to go towards a massive object but end up at the same velocity as at the beginning, that would, by the principle of equivalence, be the same as acceelerating in empty space. So if you accelerate, you are going to measure time and length different. And so if you measure the frequency of the light beam at the surface of the massive body, you’re going to get a different answer than if you had not accelerated (because your measuring instruments are now giving different answers). It turns out that the frequency will be higher, signifying a higher energy level of the light. If you don’t accelerate (i.e, you just fall into the massive object and go splat) and measure the frequency just before hitting the ground, you’ll get the same answer as when you measured far away from the object. But someone on the ground will say that that is the “wrong” answer because you’re not “at rest”, you’re coming towards the ground at a very high speed. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [August 1, 2000, 7:09am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/4 "2000-08-01T07:09:30Z")
> [@](#): > > \*Originally posted by The Ryan \* > **First of all, I’m rather suspicious of this “slingshot” idea. It certainly is possible to use a large body to change the _direction_ of an object’s trajectory, but not the _magnitude_ (relative to the planet). Yeah, it’s going to speed as it gets closer, but it’s just going to slow down a gain when it goes away from the body. So unless the body is going the same direction that you want to go, you’re not going to get any energy out of it.** This is true if you just look at it as a two-body problem (Jupiter and spacecraft). You’re right, the spacecraft doesn’t speed up or down, it just changes direction. However, in reality, both Jupiter and the spacecraft are in orbit around the sun. If your spacecraft just grazes Jupiter on the sunlit side, what happens? Your trajectory is bent outward away from the sun, while Jupiter is pushed slightly towards the sun. So Jupiter goes into a slightly lower, and thus lower energy orbit, while your spaceship goes into a higher orbit with more energy. You just stole some energy from Jupiter’s orbital energy. This does work, and has been used on numerous NASA probes. Galileo (the Jupiter probe) made several flybys at Venus and Earth before gaining enough energy to get to Jupiter. As for light falling onto a planet, I think the momentum does increase, resulting in a wavelength shift of the incoming light. Since time runs slower on the surface, it makes sense that the observer there sees a blueshifted (faster frequency) light than an observer higher up above the planet. ---
**Author:** ![Surgoshan](https://avatars.discourse-cdn.com/v4/letter/s/e95f7d/32.png) [@Surgoshan](https://boards.straightdope.com/u/Surgoshan)\ **Post date:** [August 1, 2000, 7:13am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/5 "2000-08-01T07:13:53Z")
Just want to point out that light does have mass. A very very small mass and we don’t understand it very well yet. Otherwise it wouldn’t be affected by gravity, would it? ---
**Author:** ![Smeghead](https://avatars.discourse-cdn.com/v4/letter/s/f1d935/32.png) [@Smeghead](https://boards.straightdope.com/u/Smeghead)\ **Post date:** [August 1, 2000, 8:56am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/6 "2000-08-01T08:56:59Z")
> [@](#): > > \*Originally posted by Surgoshan \* > \*\*Just want to point out that light does have mass. A very very small mass and we don’t understand it very well yet. Otherwise it wouldn’t be affected by gravity, would it? \*\* Light does not have mass. It’s affected by gravity because gravity warps space around it. From the photon’s point of view, it’s going in a straight line, unaffected by gravity. From our point of view, it curves. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [August 1, 2000, 9:48am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/7 "2000-08-01T09:48:29Z")
Light _does too_ have mass. Light does not have what is known in relativity circles as “rest mass”, which is the mass that something has when it’s not moving. Since light can’t ever not be moving, this makes sense. Now, about the slingshot thingy. I like **scr4** ’s answer about the three-body problem, but I’ve never heard that before. Having played several space-based video games, I’d like to offer a different explanation. If you fire your rockets deep in a gravity well, you get a more efficient thrust than you do if you fire them away from any planets or things. I have no idea why this would work either, but I also read it in a Heinlein book, so it must be true. As for the OP: “or does it just bend and not change in speed or energy since it virtually has no mass?” That’s absolutely right. Light does get bent ever-so-slightly by a nearby mass. It does change in Energy for a short time, but it gets changed back when it leaves the gravity well, as **The Ryan** explained. However, the reason its path doesn’t get bent much is not due to its low mass - it’s due to its high speed. As you probably know, mass has no effect on how fast an object falls, and getting bent by a gravity well is the same deal. ---
**Author:** ![Smeghead](https://avatars.discourse-cdn.com/v4/letter/s/f1d935/32.png) [@Smeghead](https://boards.straightdope.com/u/Smeghead)\ **Post date:** [August 1, 2000, 10:02am UTC](https://boards.straightdope.com/t/light-and-gravity/27674/8 "2000-08-01T10:02:53Z")
Ok, ok. I don’t know what I was thinking. Carry on. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [August 1, 2000, 1:01pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/9 "2000-08-01T13:01:20Z")
> [@](#): > > \*Originally posted by Achernar \* > **Light _does too_ have mass.** What mass does it have? ---
**Author:** ![CandyMan](https://avatars.discourse-cdn.com/v4/letter/c/b5e925/32.png) [@CandyMan](https://boards.straightdope.com/u/CandyMan)\ **Post date:** [August 1, 2000, 1:22pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/10 "2000-08-01T13:22:28Z")
The other major thing you are forgetting in the “slingshot” move is that the trajectory away from the planet is different than the one you use to approach the planet. If you head right for the planet you can use its gravitational pull to its full effect to help you increase your velocity. Now as you pass the planet you angle yourself so that you are using some of the gravitational pull to change your direction instead of slow you down. To sum it up: As you approach the planet you attempt to line your vector up as close as possible to the planets gravitational vector, as you pass it you make a comprimise to minimize the gravitational vector to your reverse to one that redirects you in the direction you want to go. This is really just simple vector addition, but REALLY hard to do with just text. CandyMan ---
**Author:** ![PaulT](https://avatars.discourse-cdn.com/v4/letter/p/b9bd4f/32.png) [@PaulT](https://boards.straightdope.com/u/PaulT)\ **Post date:** [August 1, 2000, 1:43pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/11 "2000-08-01T13:43:39Z")
I’m with Mentock on this one – Smeg, don’t back down so easily =) Sure, light has no rest mass – but relativistic mass for objects travelling at less than c is proportional to rest mass, so the mass of a zero-rest-mass photon is still zero until you reach c. For an object going the speed of light, the relativistic mass factor is infinite – but it’s multiplied by the rest mass, which is still zero. Zero times infinity is a very indeterminate thing, which is why light carries a momentum, but I don’t think it can be reasonably argued to have a mass, since it cannot be accelerated to test what its mass might be. So to review, light has no mass to be affected by ‘gravity’, but is, as Smeghead said, simply follows straight lines in the local geometry, which is distorted by the presence of mass. Besides, Achernar, while you’re right that trajectory doesn’t depend on mass in Newtonian physics, the deflection of light that we observe is twice what Newton would have predicted (one of the best proofs of general relativity). On the _original_ topic, in addition to passing inside the orbit of a planet to steal its orbital motion, a projectile can also be sent to fly past the planet to leach some of its rotational energy (since, as pointed out above, while gravitational energy is conserved in the rotating, orbiting frame of reference of the planet, a boost can be obtained in the Sun reference frame). ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [August 1, 2000, 2:32pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/12 "2000-08-01T14:32:56Z")
> [@](#): > > I’m with Mentock on this one – Smeg, don’t back down so easily =) And I’m with PaulT on this one. Sorry Achernar but you’ve got it wrong and Smeg is actually dead on. Mass warps the fabric of space in 4-dimensions. Light ‘bends’ because it’s following a straight line in four dimensions that looks bent to us in 3-dimensions. The speed of light in a vacuum is constant. Period. If you go in the gravity well the light is in to measure it you will find this to be true. Yes, the speed changes but then time slows down or speeds up and your rulers shorten or lengthen to ensure that you get the same measurement for the speed of light. This was proven by using two atomic clocks. One was at the top of a water tower and one was at the bottom. The speed of light changes as it climbs out of earth’s gravity well. Eventually the two clocks differed on their times since one clock was running ‘faster’ than the other (or the other was running slower…take your pick…either is a correct interpretation). This time difference would mean that people on the tower and at its base will get the same measurement for the speed of light. Also, as has already been pointed out, light has no mass. ---
**Author:** ![bibliophage](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bibliophage/32/7615_2.png) [@bibliophage](https://boards.straightdope.com/u/bibliophage)\ **Post date:** [August 1, 2000, 6:40pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/13 "2000-08-01T18:40:07Z")
The controversy over whether or not a photon has mass is covered [here](http://math.ucr.edu/home/baez/physics/photon_mass.html). The photon has no **rest** mass (m[sub]0[/sub]), but it does have what some call “relativisic mass” equal to m=hf/c[sup]2[/sup]. (m is “relativistic mass,” so called; f is the frequency of the light; and h is Planck’s constant.) As for satellites and the slingshot effect: It can be thought of as an elastic collision (even though the two bodies never touch). Imagine a bowling ball is flying toward you at high speed. If you shoot it with a BB (and if the collision is elastic), the bowling ball will slow down slightly. The kinetic energy of the bowling ball declines, and the reaction is elastic, so the difference in energy must be transferred to the BB. Therefore, the BB will be moving faster after the collision than before (and in the opposite direction). On the other hand, if the bowling ball is flying away from you when you shoot it, the ball be going faster than before (and in the same direction), while the BB will be going slower than before (but in the opposite direction). Substitute “space probe” for BB and “planet” for bowling ball. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 1, 2000, 7:22pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/14 "2000-08-01T19:22:51Z")
First of all, as to light and mass: If you have a hollow container, lined with perfect mirrors (hey, this is physics, we’re allowed to use ideal materials), and you have a whole heck of a lot of photons bouncing around in there, that container will exert a stronger gravitational pull on masses around it (strictly speaking, it’ll warp space-time to a greater extent) than if there weren’t all those photons in it. Secondly, as to burning your engines deep in a gravitational well: Of _course_ Heinlein got it right (_The Rolling Stones_, right?). The explanation is actually very simple: You’re taking all that massive propellant down the well with wou, and hence getting energy out of the decrease in its kinetic energy, but then you’re leaving it down there, and not expending energy to bring it back up out of the well. This is a different effect than the slingshooting used by Voyager et al.: scr4 covered that. The probe steals some energy/momentum from the planet. ---
**Author:** ![The\_Ryan](https://avatars.discourse-cdn.com/v4/letter/t/7feea3/32.png) [@The\_Ryan](https://boards.straightdope.com/u/The_Ryan)\ **Post date:** [August 1, 2000, 7:39pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/15 "2000-08-01T19:39:54Z")
The problem that I have with the slingshot idea is that presumably these space probes are going away from the sun, not perpendicular to it. scr4 > [@](#): > > If your spacecraft just grazes Jupiter on the sunlit side, what happens? Your trajectory is bent outward away from the sun, while Jupiter is pushed slightly towards the sun. But if you’re graxing it on the sun side, then aren’t you going perpendicular to the sun, rather than away from it? > [@](#): > > If you head right for the planet you can use its gravitational pull to its full effect to help you increase your velocity. Now as you pass the planet you angle yourself so that you are using some of the gravitational pull to change your direction instead of slow you down. But the gravitational force always pulls towards the center of mass. The only way for this to not decrease your velocity is to travel perpendicularly to the mass. And if you’re traveling perpendicularly to the mass, you’ll never leave it. PaulT > [@](#): > > On the _original_ topic, in addition to passing inside the orbit of a planet to steal its orbital motion, a projectile can also be sent to fly past the planet to leach some of its rotational energy (since, as pointed out above, while gravitational energy is conserved in the rotating, orbiting frame of reference of the planet, a boost can be obtained in the Sun reference frame). But the only difference between these two frames of reference is velocity perpendicular to the sun. So yes, you can increase your perpendicular velocity, but that’s not really going to help you get away from the solar system. bibliophage > [@](#): > > Imagine a bowling ball is flying toward you at high speed. If you shoot it with a BB (and if the collision is elastic), the bowling ball will slow down slightly. The kinetic energy of the bowling ball declines, and the reaction is elastic, so the difference in energy must be transferred to the BB. Therefore, the BB will be moving faster after the collision than before (and in the opposite direction). Yes, the BB will be traveling faster in the direction that the bowling ball was traveling. But if the bowling ball is traveling perpendicularly to the sun, and you want to get awy from the sun, how does that help you? ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [August 1, 2000, 8:08pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/16 "2000-08-01T20:08:39Z")
> [@](#): > > _Bibliophage wrote:_ The controversy over whether or not a photon has mass is covered here. The photon has no rest mass (m0), but it does have what some call “relativisic mass” equal to m=hf/c2. To be fair you should have copied the quote below from the link you provided. While _technically_ correct about light having mass the page you linked to decribes the idea of photons as having mass to be a bit of a stretch using some obscure definitons of “mass”…even to the point that it is still considered correct among physicists to say light (photons) have no mass. > [@](#): > > The overwhelming consensus among physicists today is to say that photons are massless. However, it is possible to assign a “relativistic mass” to a photon which depends upon its wavelength. This is based upon an old usage of the word “mass” which, though not strictly wrong, is not used much today. \*Source: [http://math.ucr.edu/home/baez/physics/photon\_mass.html](http://math.ucr.edu/home/baez/physics/photon_mass.html) \* ---
**Author:** ![APB9999](https://avatars.discourse-cdn.com/v4/letter/a/8e8cbc/32.png) [@APB9999](https://boards.straightdope.com/u/APB9999)\ **Post date:** [August 1, 2000, 8:26pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/17 "2000-08-01T20:26:37Z")
I’ve talked about this very thing with some physicist friends of mine in the past. According to them, modern physics is moving away from the notion of “relativistic mass”. The only property of a body to which the word “mass” is properly applied is the rest mass. Relativistic mass is a concept that arose from trying to communicate relativistic ideas in classical framework, at least according to my friends, and leads to more confusion than it solves. Look at it this way. Momentum is p=mv/(1-(v/c)^2)^0.5 where m is the rest mass. The instances where relativistic mass appears are always instances where it derives from a change in momentum. There is no “relativistic momentum”, just THE momentum, as given by the above formula. Caveat: I’m not a physicist, so I can’t vouch for this being a universal take on things among physicists generally. But the two physics Ph.D.s I know have both said more or less the same thing about this at different times and in different contexts. ---
**Author:** ![PaulT](https://avatars.discourse-cdn.com/v4/letter/p/b9bd4f/32.png) [@PaulT](https://boards.straightdope.com/u/PaulT)\ **Post date:** [August 1, 2000, 8:42pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/18 "2000-08-01T20:42:46Z")
The Ryan: Of course perpendicular thrust is useful, beacuse all of our probes are in orbit around the Sun. They don’t just point their engines at the Sun and fire them continuously, trying to head straight outward. For one thing, they’re mostly launched from the Earth, and thus have an initial angular momentum around the Sun. For another, if you ever want your probe to be stable (to be able to shut off its engines) you need enough perpendicular velocity to keep it in orbit around the Sun. So even assuming that only perpendicular boosts were possible, those would still have the effect of putting the ship into a higher solar orbit. And there’s no reason the rotational boost couldn’t be done in any direction at all – it’s just a matter of which side of the planet you graze. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [August 1, 2000, 8:47pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/19 "2000-08-01T20:47:28Z")
Okay, how do you want to define mass? By the amount of gravitational pull an object exerts on other things, as **Chronos** covered? By the amount of momentum it has, as **APB9999** mentioned? (Momentum is _by definition_ mass × velocity.) The amount it weighs when you put it on a scale, divided by the acceleration due to gravity? If you choose _any_ of these ways to define mass, it’ll be identical to what’s being referred to as “relativistic mass”. For all intents and purposes, relativistic mass is mass. Having asserted myself so foolhardily, I’d like to add that I understand perfectly all the confusion. Relativity is not for those who value intuitiveness. Sorry for the extended hijack, **fetaljew** , but I think it is helpful for answering your OP. ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [August 1, 2000, 9:36pm UTC](https://boards.straightdope.com/t/light-and-gravity/27674/20 "2000-08-01T21:36:42Z")
> [@](#): > > \*Originally posted by APB9999 \* > \*\*Look at it this way. Momentum is p=mv/(1-(v/c)^2)^0.5 where m is the rest mass. The instances where relativistic mass appears are always instances where it derives from a change in momentum. There is no “relativistic momentum”, just THE momentum, as given by the above formula. > \*\* The above formula is only correct for velocities below the speed of light. When v = c, you get a 0 in the denominator in the above fraction. As far as my understanding of light, it is that light has no mass, but it does have momentum. It can be calculated by the mass-energy equivalency formula: E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup] In the above equation, E is energy, p is momentum, m is mass and c is the speed of light. Note that when p = 0, the equation degerates to the more familiar form E = mc[sup]2[/sup]. For photons, m = 0, so the equation degenerates to E = pc, or E/c = p. A photon’s energy is based upon its frequency, namely, E = hf, where h is Planck’s constant and f is the frequency of the photon. Therefore, for a photon’s momentum, p: p = hf/c In MKS units, h = 6.63 \* 10[sup]-34[/sup] J \* s, and c = 3.0 \* 10[sup]8[/sup]m/s, pretty unwieldy numbers. For this reason, many physicists like to use “God’s units” when dealing with relativistic problems. They set h = 1 and c = 1, and many of the equations get a lot easier. [Next page](https://boards.straightdope.com/t/light-and-gravity/27674.md?page=2)