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Isn't light as an electromagnetic radiation massless, thus not affected by gravitational pull?

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- Thread starter wavingerwin
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Isn't light as an electromagnetic radiation massless, thus not affected by gravitational pull?

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- #2

sylas

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Even in classical physics, there is an expected divergence of light with gravity. I've never fully understood this myself, but as a start, note that the acceleration of a very light body in a gravitational field is independent of its mass.

Isn't light as an electromagnetic radiation massless, thus not affected by gravitational pull?

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If you drop a one kilogram weight, it falls at the same speed as a two kilogram weight.

This is because there is an equivalence of "inertial mass" (which says how much force is required to make something diverge a bit) and "gravitational mass" (which says how much force is applied in a gravitational field).

Light may be massless... but that's both inertial mass and gravitational mass, in classical physics. I think. You can apply a limit as both internal and gravitational mass tend to zero, and that (I think) is the classical expectation for light... assuming the equivalence of inertial and gravitational mass. (Someone correct me if I have this wrong).

Consider a particle of mass m being pushed around by a large mass M. The force involved is GmM/r^2. The amount of gravitational acceleration is F/m, or GM/r^2.

Note that the "m" term is now gone.

Although you actually need relativity to analyze a blackhole properly, you can calculate the boundary beyond which light cannot escape using classical physics.

Potential energy of a lightweight particle is -GMm/r. Kinetic energy is 0.5mv^2.

A particle is gravitational bound when GMm/r > 0.5mv^2, so the escape velocity at distance r is v=sqrt(2GM/r). The quantity "m" does not appear. If you use "c" as the escape velocity and solve for r, you actually get the Schwartzchild radius of a blackhole: 2GM/c^2

Classical physics is an approximation. For a full treatment, you need some quantum physics in which light does have inertia in the sense of having momentum; and some relativity to deal properly with gravity in terms of geometry rather than a force in Euclidean space. But it turns out you still get the same result for the limit at which light is trapped at a blackhole.

Cheers -- sylas

- #3

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Another way of answering the question is that, in relativity, "mass" and "energy" are really the same thing (remember E = mc^2 ?), so it is perfectly logical that light should be influenced by gravity the same way as matter is.

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rcgldr

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vanesch

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x(t) = vx t + x0

y(t) = vy t + y0

z(t) = vz t + z0

which is nothing else but "straight line motion" (uniform motion) in flat space, as you've learned it in high school.

But when spacetime is warped, the curvature of spacetime tells you what are now the geodesics. Light follows special geodesics, which are "null geodesics" (which correspond, in flat spacetime, to uniform straight motion such that vx^2 + vy^2 + vz^2 = c^2).

And it turns out that if you start out in certain regions in the spacetime of a black hole, you cannot leave a certain region - in fact all null geodesics end up in the singularity, no matter "in what direction" they start out. (at least for a Schwarzschild black hole ; for rotating black holes, it is way more involved).

Matter points follow "timelike geodesics" which is the equivalent in flat space of having vx^2 + vy^2 + vz^2 < c^2. Also all timelike geodesics starting out in an event in the neighbourhood of the black hole end up in the singularity, and are not able to leave a certain region.

I'm not 100% of this, but I think that there exist geodesics that can leave a point near a black hole, only they are not "timelike" nor "null geodesics" but are spacelike. But I could be wrong on this.

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sylas

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There is a distance at which light that is tangential to the hole will follow a circular path curving around the black hole. For a non-rotating black hole, this "photon sphere" is at 1.5 Schwartzchild radii.

It's a bit like a conventional circular orbit, in which the speed remains the same but the velocity vector is endlessly pulled sideways, orthogonal to the direction of motion.

This shows that the classical ideas I used above have broken down -- because such photon is "gravitationally bound" in a sense, and yet outside the point where classical physics says c is the escape velocity.

As everyone has noted, the proper description of gravity in these extremes is geometric, using relativity. Different co-ordinate systems do strange things near the Schwatzchild radius. If you try and set up co-ordinates in which a photon moving radially is always moving at "c", you end up with something that looks like space itself getting sucked into the hole. Very weird; it makes my head hurt.

Cheers -- sylas

- #7

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And it's energy. Light climbing up out of a gravity well shifts red. Whimsically put, if you shine a flashlight straight up, the beam gets tired.From what I've read, the speed of light isn't affect by gravity, just it's direction

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Regardless of whether it actually exists or not.

- #9

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one is:

it's got to do with the inertial and gravitational equivalence principle.

According to it, i suppose, there is no physical experiment that can determine if a reference frame is an accelerated frame or is in a gravitational field having the same value of g as that of accl frame

so if you imagine a straight light beam in a lift accelerating upwards, you should see it bends slightly, like in projectile motion as if you throw a ball

Einstein said that the equivalence principle demanded that the same is true if gravity is present because you cannot distinguish between the two inside the frame.

that's one of the reasons to believe gravity bends light, since velocity of light is so large, you need a lot of acceleration or equivalently a strong gravitational field to actually notice the slight bend.

a black hole is just the extreme case, it bends light so forcibly that it never actually escapes, as in you consider escape velocity of a particle, same here.

another way:

light has to spend energy while traveling in a gravitational field. if you shine yellow light at the moon, the beam slowly loses energy on the way, and at some distance, might appear red. if you consider a black hole , its the extreme case.

Although these are what is called as the 'layman's way' of understanding gravity and black holes. If you know deep physics, then you'll find a mathematical expression too.

As pointed out, black holes were invented by mathematicians (at least they gave the idea)

- #10

DaveC426913

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No. They exist. And they do capture light. The reason light cannot escape is because strong gravity bends space-time. Light follows curved space-time. Inside the extremely curved space-time of a black hole, it is so curved that all paths bend back on themseves to the singularity. Light, which follows these paths through space-time, bends back on itself and ends up at the singularity.

Regardless of whether it actually exists or not.

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You say that all path are bent so that light end up back in the black hole. So, if you imagine a line/path goint radial out from the center of a black hole, in what direction is it bent? up? down? left? right?No. They exist. And they do capture light. The reason light cannot escape is because strong gravity bends space-time. Light follows curved space-time. Inside the extremely curved space-time of a black hole, it is so curved that all paths bend back on themseves to the singularity. Light, which follows these paths through space-time, bends back on itself and ends up at the singularity.

I know that there is no absolute up and such, but what i mean is in what direction is the path bent from our perspective?

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The biggest reason for which light cannot escape a black hole is the fact that the black hole bends space. Light always travels in a straight line in relation to space, so if the space is curved, then light will curve with it. Light is simply following the curved space into the black hole. Light will never come back out again because space is constantly warping even more at the speed of light. Its like running on a treadmill that always goes as fast as you do: you will never move in relation to it.

Oh and to answer your question, space would be squashed.

Oh and to answer your question, space would be squashed.

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DaveC426913

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In that case, it is infinitely red-shifted.You say that all path are bent so that light end up back in the black hole. So, if you imagine a line/path goint radial out from the center of a black hole, in what direction is it bent? up? down? left? right?

- #14

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I see, so you are saying that if light wasn't redshifted, then it could escape. Isn't it just a matter of time before the light find the radial path then?In that case, it is infinitely red-shifted.

if yes, doesn't that mean that without redshift, it would be shining just as bright as a normal star?

The reason I ask is that I have not hear that the only reason that light can't escape a black hole is because of redshift. Is redshift the only reason that light can't escape?

What happens with the light when it's red shifted? Is it destroyed?

- #15

DaveC426913

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No it couldn't.I see, so you are saying that if light wasn't redshifted, then it could escape.

But I see your point. I can't really describe what would happen to infinitely red-shifted light.

Isn't it just a matter of time before the light find the radial path then?

if yes, doesn't that mean that without redshift, it would be shining just as bright as a normal star?

The reason I ask is that I have not hear that the only reason that light can't escape a black hole is because of redshift. Is redshift the only reason that light can't escape?

What happens with the light when it's red shifted? Is it destroyed?

- #16

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Okay. Thanks for trying anyway.No it couldn't.

But I see your point. I can't really describe what would happen to infinitely red-shifted light.

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