What exactly is Gravity?

Arjun Artro · November 30, 2012

If gravity is a force acting between anything that has mass, why does gravity from black holes pull in light (photons) which have no mass at all? And even if we consider light as a particle and not a wave, any particle with a finite mass needs infinite energy to accelerate to light speed. And one more thing. If gravity can pull in light, what happens to a beam of light coming towards a black hole? Would it accelerate due to gravity? And would light break its own top speed?

juanrga · November 30, 2012

If gravity is a force acting between anything that has mass, why does gravity from black holes pull in light (photons) which have no mass at all? And even if we consider light as a particle and not a wave, any particle with a finite mass needs infinite energy to accelerate to light speed. And one more thing. If gravity can pull in light, what happens to a beam of light coming towards a black hole? Would it accelerate due to gravity? And would light break its own top speed?

The usual Newtonian expression [math]F = -\frac{GMm}{r^2}[/math] is a non-relativistic expression valid for massive particles, but not for massless particles as photons. In reality gravity acts between anything with a nonzero energy-momentum. Light (i.e., photons) is affected by gravity because has a nonzero energy-momentum. No, light does not break "its own top speed".

Arjun Artro · December 1, 2012

The usual Newtonian expression [math]F = -\frac{GMm}{r^2}[/math] is a non-relativistic expression valid for massive particles, but not for massless particles as photons. In reality gravity acts between anything with a nonzero energy-momentum. Light (i.e., photons) is affected by gravity because has a nonzero energy-momentum. No, light does not break "its own top speed".

Thank you . But i have this doubt . Anything that is pulled by gravity gets accelerated. And for different masses, there are different constant values of acceleration by which all objects, regardless of their mass gets accelerated. So wouldn't light too? I know the answer might be no coz . .there's no proof to prove otherwise. But i want to know why it doesn't.

timo · December 1, 2012

In reality gravity acts between anything with a nonzero energy-momentum.

I assume with "nonzero energy-momentum" you mean "anything with either non-zero momentum or non-zero energy". While you run into technical issues with the p->0 limit for light (which apart from a possible potential energy implies E->0), the effect of gravity on a light pulse is independent on its momentum (in the sense that the equation of motion can be sensibly written down as the 4-acceleration being a function of the 4-position and the 4-velocity and not including the 4-momentum). In that respect, I tend to disagree with your statement.

I do agree with the statement that "gravity is a force between objects with mass" is only the everyday-world simplification of a conceptually greater principle, though.

@Arjun: I haven't gone through the calculation, so I am slightly hypothesizing here: Certain aspects of gravity that you learn in non-relativistic gravity are simply not true in relativistic gravity. You mentioned that all objects get accelerated equally, irrespective of their mass. The statement implies an even more astonishing assumption that is so fundamental to much of non-relativistic physics that people tend to not even recognize it as a strong assumption: The assumption that the acceleration due to gravity is independent of the current velocity [*]. Now, if you actually look at the equation of motion in relativistic gravity (you should readily find it as "geodesic equation" on Google or Wikipedia) you will find that the acceleration of an object does depend on its velocity [**]. As the velocity of a massless object strikingly differs from that of a massive one, so can its acceleration.

[*] In some sense it is rather remarkable that people take this as granted, given that friction is a well-know everyday-phenomenon no satisfying this assumption.

[**] There is the possibility that due to extra constraints the velocity as you know it may actually cancel out. This is exactly the part where I am hypothesizing in assuming it not to always cancel.

derek w · December 1, 2012

If light travels at the speed of light across 3 dimensional space,and does not travel forward in time.The event horizon of the black hole is the point where light has stopped travelling across 3 dimensional space and is travelling forward in time with the mass of the black hole.

Arjun Artro · December 1, 2012

If light travels at the speed of light across 3 dimensional space,and does not travel forward in time.The event horizon of the black hole is the point where light has stopped travelling across 3 dimensional space and is travelling forward in time with the mass of the black hole.

But for light, and inside a black hole, there exist no time. Isn't that right?

derek w · December 1, 2012

If the speed of light is constant,and light cannot escape from a black hole,then inside the black hole t=c?

Arjun Artro · December 1, 2012

If the speed of light is constant,and light cannot escape from a black hole,then inside the black hole t=c?

t=c???

can you explain?

If the speed of light is constant,and light cannot escape from a black hole,then inside the black hole t=c?

t=c???

can you explain?

derek w · December 1, 2012

If space-time is 4 dimensional,(x,y,z,t),and if the speed of light is constant,if light travels across 3 dimensional space at C then it cannot travel along the 4th dimension T=0.

However if it has zero velocity in 3 dimensional space,and the speed of light is constant then it must be travelling along the T axis at the speed of light T=C.

Arjun Artro · December 1, 2012

If space-time is 4 dimensional,(x,y,z,t),and if the speed of light is constant,if light travels across 3 dimensional space at C then it cannot travel along the 4th dimension T=0.

However if it has zero velocity in 3 dimensional space,and the speed of light is constant then it must be travelling along the T axis at the speed of light T=C.

From my knowledge, light cannot exist with zero velocity. But yes, it has to come down to zero velocity inside a black hole. Or what will happen to all the beams of light that get sucked inside a black hole? So which one would be true?

Inside a black hole, if there's no time, then light can travel only in 3 dimensions.That means, movement with no time running. Is that possible?

juanrga · December 1, 2012

Thank you . But i have this doubt . Anything that is pulled by gravity gets accelerated. And for different masses, there are different constant values of acceleration by which all objects, regardless of their mass gets accelerated. So wouldn't light too? I know the answer might be no coz . .there's no proof to prove otherwise. But i want to know why it doesn't.

Yes the acceleration of a test body depends on the mass of the source body. This is also true for light. Light bending around the Sun depends on the mass of the Sun: more mass more bending.

I assume with "nonzero energy-momentum" you mean "anything with either non-zero momentum or non-zero energy". While you run into technical issues with the p->0 limit for light (which apart from a possible potential energy implies E->0), the effect of gravity on a light pulse is independent on its momentum (in the sense that the equation of motion can be sensibly written down as the 4-acceleration being a function of the 4-position and the 4-velocity and not including the 4-momentum). In that respect, I tend to disagree with your statement.

I do agree with the statement that "gravity is a force between objects with mass" is only the everyday-world simplification of a conceptually greater principle, though.

The gravitational interaction is [math]T^{\mu\nu}h_{\mu\nu}[/math], where [math]T^{\mu\nu}[/math] is the energy-momentum tensor. Therefore anything with a nonzero energy-momentum tensor will feel gravity. What I am saying is not exactly identical to what you did mean "anything with either non-zero momentum or non-zero energy", because I am also including systems for which both momentum and energy are nonzero.

The effect of gravity on a light pulse depends on its momentum. The variation of momentum of a light signal moving near a gravitational field (source mass M at rest) is

[math]\frac{4GMp}{c^2r}[/math]

and this gives the measured deflection angle for a ligth signal bending around the Sun

[math]\alpha = \frac{\Delta p}{p} = \frac{4GM}{c^2r}[/math]

Edited December 1, 2012 by juanrga

michel123456 · December 1, 2012

In a BH light is not sucked, it is spacetime that is so much curved that a ray of light following the geodesic will enter again the BH.

Arjun Artro · December 1, 2012

In a BH light is not sucked, it is spacetime that is so much curved that a ray of light following the geodesic will enter again the BH.

You mean... the ray of light will make a loop?

timo · December 1, 2012

The gravitational interaction is [math]T^{\mu\nu}h_{\mu\nu}[/math], where [math]T^{\mu\nu}[/math] is the energy-momentum tensor.

I don't know what "the gravitational interaction is <some number>" is supposed to mean. And I don't know what "h" is, either. I certainly didn't doubt that you can write down expressions involving the energy-momentum tensor.

What I am saying is not exactly identical to what you did mean "anything with either non-zero momentum or non-zero energy", because I am also including systems for which both momentum and energy are nonzero.

In physics, math and computer science (and presumably other sciences, too) "or" usually means an inclusive or, not an exclusive or (xor). That's the sense I meant here, not least since momentum usually implies kinetic energy, anyways.

The effect of gravity on a light pulse depends on its momentum. [...] this gives the measured deflection angle for a ligth signal bending around the Sun
[math]\alpha = \dots = \frac{4GM}{c^2r}[/math]

It's kind of interesting that I had chosen the very same example to claim that the effect of gravity does not depend on energy and momentum of the light pulse.

Edited December 2, 2012 by timo

zapatos · December 2, 2012

I would have thought there was no light in a black hole. That is, light travels at c until the photon is absorbed by some matter in the black hole, then it ceases to exist as light. No different than a photon being absorbed by the earth.

Edited December 2, 2012 by zapatos

derek w · December 2, 2012

I would have thought there was no light in a black hole. That is, light travels at c until the photon is absorbed by some matter in the black hole, then it ceases to exist as light. No different than a photon being absorbed by the earth.

Yes,but it would add to the mass of the black hole,where as the earth would radiate it back off.

STeve555 · December 2, 2012

gravity is the only force that can communicate with other universa, if out there at all.

There shall be a gravity radio soon enough. And this will be our soul apparatus to talk with parallel universa.

The brane we currently live on is not semipermeable, it is totally in reach by the multitude of messengers from other universa.

People've deemed me a numbskull for 20 or more years, but gravity radio will be the future. It will be like seance, but this time it is real.

michel123456 · December 2, 2012

You mean... the ray of light will make a loop?

Sort of. Have a look at the 3rd diagram on the link below. Beware to understand the first ones also.

http://www.phy.syr.edu/courses/modules/LIGHTCONE/schwarzschild.html

Arjun Artro · December 2, 2012

So... does light move inwards even after crossing the event horizon? I think it shouldn't. If there's any movement inside the black hole, that means, there's an existence of dimension , leading to the existence of time. But inside a black hole, time is non existent.

???

derek w · December 2, 2012

Sort of. Have a look at the 3rd diagram on the link below. Beware to understand the first ones also.

http://www.phy.syr.e...warzschild.html

In the first diagram the arrow of time is pointing in the direction in which the event horizon is travelling,and light has zero velocity in 3 dimensional space at the event horizon,so T=C?

The event horizon is travelling along the T-axis at the speed of light?

juanrga · December 2, 2012

I don't know what "the gravitational interaction is <some number>" is supposed to mean. And I don't know what "h" is, either.

What I wrote is not a number but a physical quantity (with units of energy). "h" is the potential of the gravitational field.

In physics, math and computer science (and presumably other sciences, too) "or" usually means an inclusive or, not an exclusive or (xor).

Agree but you wrote "anything with either non-zero momentum or non-zero energy" and In English, the construct "either ... or" is usually used to indicate exclusive or.

It's kind of interesting that I had chosen the very same example to claim that the effect of gravity does not depend on energy and momentum of the light pulse.

The angle does not depend on the momentum p in a first approximation, when you add higher order corrections to the bending formula the angle depends on the momentum as well. In any case, the variation of the momentum of the light signal depends on its initial momentum p at any order, as shown above.

Edited December 2, 2012 by juanrga

timo · December 2, 2012

What I wrote is not a number but a physical quantity (with units of energy).

"The gravitational interaction is given by <some unspecified physical quantity with units of energy>" isn't exactly more helpful to me than "the gravitational interaction is given by <some number>".

The angle does not depend on the momentum p in a first approximation, when you add higher order corrections to the bending formula the angle depends on the momentum as well.

I doubt it.

In any case, the variation of the momentum of the light signal depends on its initial momentum p at any order, as shown above.

Sure. In the very same sense, the effect of classical gravity on a free-falling object depends on that object's mass (except that in non-relativistic physics the comparison of momenta at different locations is properly defined).

juanrga · December 4, 2012

"The gravitational interaction is given by <some unspecified physical quantity with units of energy>" isn't exactly more helpful to me than "the gravitational interaction is given by <some number>".

The term [math]T^{\mu\nu}h_{\mu\nu}[/math] is not "unspecified" but a well-known term with precise properties. Next picture is from Feynman textbook on gravitation

Note that Feynman adds an extra lambda coupling constant, because this helps him in a perturbation expansion in orders of lambda, but usually lambda is absorbed into h and the expansion made in terms of orders of h.

dripto biswas · December 4, 2012

you see MR . Arjun Artro, as far as i can imagine, you can use general relativity and its geometrical picture of our universe to find a solution to your problem.

if you do not know why according to general relativity no light can escape from a black hole, then i'll try to explain it to you. if you take a stretched rubber sheet and put objects of different masses onto its surface , you will see that the heavier the object the more is the curvature of the rubber sheet. now, here we take that the rubber sheet represents the very nature of the space-time in our universe. we all know that a black hole is a mass of infinite density and therefore it will act like an object of infinite 'weight' and thus will form an infinite curve in the rubber membrane(space-time). so when light passes near the curvature of the black hole, it starts to fall within the curve of the black hole. but since it is an infinite curve, then theoretically, we should need infinite energy(impossible) to over come the curve

and leave through the other side. now, think of an object moving with uniform velocity in circular orbit. its direction changes every instant but its velocity does not.

same is the case here. the light travels with constant velocity but when it 'changes its direction towards the infinite depth of the black hole's 'gravity well', the velocity does not change.