Bending of Light due to Gravity

[avito009]

Why does light bend due to gravity? I thought hard, read a lot and I found 3 reasons I can give as the answer. But first let me tell you what Newton said.

According to Newton light (Photons) is massless so light can't bend due to gravity because only things with mass can be affected by gravity. But as per General Relativity Einstein says light bends due to gravity. Below are the reasons as per my understanding as to why light bends.

Reasoning 1

In this scenario we assume Newton was right. So we say that massless particles aren't affected by gravity. But light has mass which is called "Effective Mass". In solid state physics, a particle's effective mass is the mass that it seems to have when responding to forces. This effective mass can be acted upon by gravity which only cares how much mass a particle has; alternately, gravity only cares about how much mass or EQUIVALENT ENERGY a particle has given by E = m c^2. Also, if you prefer the particle description of physics over the wave description, you can approximate all photons as 'bullets' each carrying a mass of m = hf/c^2 and traveling at the speed of light.

EFFECTIVE MASS, is the same as the KINETIC ENERGY of photons divided by c2 (C is the speed of light).

Thus, even though light has no REST MASS (because it can never be at rest!), it does have an effective mass which (it turns out) has all the properties one expects from MASS - in particular, it has weight in a gravitational field [photons can "fall", which happens in a black hole] and exerts a gravitational attraction of its own on other masses.

Reasoning 2

Here we say that even though photons (Light Particles) have no mass they still have momentum. Photons cannot have mass because they travel at the speed of light, but they must have momentum for events like photoionization to occur. Photoionization means the removal of one or more electrons from an atom or molecule by absorption of a photon of visible or ultraviolet light. Also known as atomic photoelectric effect. (Remember Einstein's photoelectric equation).

It is generally agreed that mass is not a requirement for having momentum. Particles made up entirely of energy can also have momentum according to modern physics. So as you know gravity affects anything that has energy and Light has energy as well as momentum.

So you will say that momentum= mass x velocity. But remember Momentum is also p=E/c where p is the momentum and E is the energy also C is the speed of light. Photons have energy, and c is a constant.

Reasoning 3

Light gets bent because it travels in space time that is warped around massive objects.

Light sometimes passes through space (or space-time) that is warped or bent because of a nearby object having very strong gravity. The light passes through this space in what (from the light's point of view) is a straight line. To other observers the light may appear to have followed a bent path. So gravity warps space-time, and light appears to bend as it travels through this warped space-time. The light isn't doing anything except following what is a completely natural path through space.

What general relativity says is that any massive object warps the spacetime around it. You can think of this with a simple analogy. Imagine a stretched rubber sheet that is completely flat. This represents the spacetime when there is no mass. Now, if you put a heavy ball in the rubber sheet, it will cause a distortion in the sheet. This is exactly what happens in space, except that it is in 3 dimensions instead of two.

Further, a photon always travels by the shortest distance between two points. As spacetime is warped, the light appears to bend around a massive object. In reality, it is not that the object is attracting light, but it is just that the photons are traveling by the shortest distance in a curved spacetime.

Photons of light are not technically affected by large gravitational fields; instead space and time become distorted around incredibly massive objects and the light simply follows this distorted curvature of space.


Please tell me which of the above reasons is correct.

 

[phinds]

#3 is definitely correct but I'm not clear that the others are totally Incorrect

 

[ShayanJ]

Reasoning 3 is the right one and it seems you have a clear understanding of it.
In GR, any kind of energy(including mass, which after SR, is just another kind of energy) distorts Space-Time and so for distorting Space-Time you need to have some kind of energy but for being affected by that distortion, there is no requirement beyond being inside Space-Time!
As phinds said, you can't say the other two are wrong. You may build theories on top of GR that tell you those, but they don't seem needed and so we don't consider them.

 

[A.T.]

But first let me tell you what Newton said.

According to Newton light (Photons) is massless so light can't bend due to gravity because only things with mass can be affected by gravity.

Where did Newton say this? I read the opposite:

http://www.colorado.edu/philosophy/vstenger/Cosmo/NewPho.pdf

In a note at the end of Optiks, published in 1704, Newton suggests that the
particles in his corpuscular theory of light should be affected by gravity just as
ordinary matter. In 1801, German astronomer and physicist Johann Georg von
Soldner (1776 – 1833) estimated that the deflection of a beam of corpuscles
skimming along the surface of the sun would be 0.9 arc-seconds.

Note that Newtonian gravitational acceleration is independent of the accelerated test mass, so it doesn't matter that the test mass is zero.

Please tell me which of the above reasons is correct.

As others said #3 is how GR states it: by geometry.

 

[ChrisVer]

Well Newton happened to be right at first, but the correct answer is taken from GR...
Using Newtonian calculations you get different bending's (I think with a factor of ~2), so they are not correct.. the correct result is taken from GR (geometry!)

 

[edguy99]

I like number 1:

This effective mass can be acted upon by gravity which only cares how much mass a particle has; alternately, gravity only cares about how much mass or EQUIVALENT ENERGY a particle has given by E = m c^2. Also, if you prefer the particle description of physics over the wave description, you can approximate all photons as 'bullets' each carrying a mass of m = hf/c^2 and traveling at the speed of light.

But this is absolutely correct:

Well Newton happened to be right at first, but the correct answer is taken from GR...
Using Newtonian calculations you get different bending's (I think with a factor of ~2), so they are not correct.. the correct result is taken from GR (geometry!)

If you try to view the photon as a bullet, it has to be a pretty fancy bullet (one with a number of internal variables or properties) that allow the photon to be out by a factor of 2 in its gravitational pull, plus it has some other properties like wave/particle duality, where energy is stored as if the photon was a harmonic oscillator.

 

[Jonathan Scott]

It is important to be aware that photons passing a gravitational source are effectively being bent by two effects.

Firstly, Newtonian gravity applies to everything including photons, giving an acceleration of $g$. This can be considered to be due to the curvature of space with respect to time, in that if you plot the radial distance against time, a free fall line curves towards the source.

Secondly, space itself is curved by gravity, which has the effect that something moving through it is accelerated proportionally to $v^2/r$ where $r$ is the radius of curvature. According to GR, the radius of curvature for a weak Newtonian gravitational field $g$ is $c^2/g$ so the resulting acceleration component due to velocity is $v^2/((c^2)/g) = (v^2/c^2) g$. This means that a photon is effectively accelerated by twice as much as a slow-moving object.

Photons are not special in this way. A particle moving at very nearly the speed of light will follow a similar path to a photon, and a photon trapped between the walls of a reflective box whose sides appear to be parallel using local rulers will fall with the same local acceleration as a brick (noting that the rulers measuring horizontal distance will appear curved to an external observer).

 

[jmnew51]

Photons and gravity

If photons have no mass and are not subject to the dilation of their mass, when they travel at the speed of light, then why are they affected by gravity. I thought for something to be affected by gravity it had to have mass.

Thanx

Jim

 

[jmnew51]

I think reason #3 is the most correct one, but I also agree with phinds in that I'm not sure that #1 and #2 are totally incorrect.

 

[BruceW]

I think it is even more subtle than #3 implies. light itself bends gravity too. so, even if there was only light in the universe, light would bend gravity and influence the path of other light, due to the effect of light on the gravitational field.

 

[Dale]

According to Newton light (Photons) is massless so light can't bend due to gravity because only things with mass can be affected by gravity.

This is a common misunderstanding. The acceleration due to gravity is independent of the mass in Newtonian gravity, so a massless particle is accelerated by gravity just as much as a massive one.

 

[avito009]

Reasoning 4

Reasoning 4

Mass isn't the only thing affected by gravity, energy is too. Einstein's famous equation E=mc^2 tells us the equivalence of energy and mass.

In actual fact, mass and energy are the same things, much like ice and water are the same things, just slightly different forms of the same thing. In fact, mass is just a very static version of energy. All energy travels at the speed of light, but mass travels at less than the speed of light. Energy would also create a gravitational field, except that it's moving so fast that it doesn't have a chance to sink into space and produce a static source of gravity, like mass does. Energy is moving through space like a lizard which can run across a surface of water, without sinking in, because it moves so fast.

 

[TumblingDice]

Einstein introduced the Equivalence Principle in 1907, when he observed that the acceleration of bodies towards the center of the Earth at a rate of 1g is equivalent to the acceleration of an inertially moving body - such as observed in a hypothetical elevator in free space being accelerated at a rate of 1g.

He then combined the EP with special relativity to predict that light rays bend in a gravitational field, even before he developed the concept of curved spacetime.

Imagine a hole on one side of the accelerating elevator just large enough to allow a beam of light to enter. As the beam crosses to the opposite side of the elevator, the acceleration will cause the elevator to 'travel' faster than it was when the light beam entered. The beam will end up striking the opposite wall of the elevator lower than the height it entered on the other side.

I think that light will "fall" in a gravitational field just as quickly as a bowling ball. It's just that light moves so darn fast that it's difficult to see/measure. I'm looking towards our expert members to please check me on this!

 

[A.T.]

I think that light will "fall" in a gravitational field just as quickly as a bowling ball. It's just that light moves so darn fast that it's difficult to see/measure. I'm looking towards our expert members to please check me on this!

Yes, as Jonathan said, light is not special.

 

[mal4mac]

#1 and #2 might be reasonable arguments from the Newtonian model, but the Newtonian model breaks down when it comes down to considering how light actually bends. The famous Eddington solar eclipse expedition had predictions for bending of light due to the Newtonian model, and bending of light due to the Einsteinian model. The Einsteinian model predicted about twice the bending of the Newtonian model, and that's the bending that was observed. So #3 is the actual model that fits reality best (so far...)

 

[jartsa]

In empty space Joe shoots with a laser gun and with a revolver on a target. Joe and target are inertial and co-moving.

Bob, who is accelerating and observing the shooting, says that the bullet and the laser pulse and the target are all accelerating the same way.

Right?

And then Bob must conclude that shooting happened in non-curved space?


_____________________



Let's put Joe in a container that is free falling in the gravity field of the earth. The same shooting experiment works the same way. So an observer standing on the surface of the Earth must conclude that the space is not curved near the earth?

 

[mal4mac]

If photons have no mass and are not subject to the dilation of their mass, when they travel at the speed of light, then why are they affected by gravity. I thought for something to be affected by gravity it had to have mass.

Photons have no rest mass. But they have energy - which is why you shouldn't stand in front of an industrial laser. By mass-energy equivalence (e=mc²) photons have a mass equivalent to their energy.

 

[A.T.]

Let's put Joe in a container that is free falling in the gravity field of the earth. The same shooting experiment works the same way.

Only if the box is small. The equivalence principle only holds locally.

So an observer standing on the surface of the Earth must conclude that the space is not curved near the earth?

Locally the tidal effects of intrinsic curvature are negligible, so one treat small areas as flat.

 

[CKH]

It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.

Newtonian gravity is a force that exists between masses ($F=G m_1 m_2/r^2$). In his equations there would be zero gravitational force acting on zero mass. Since $F=ma$, $F/m=a$, but in this case that is $0/0=a$ which is undefined.

 

[Dale]

It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.

Newtonian gravity is a force that exists between masses ($F=G m_1 m_2/r^2$). In his equations there would be zero gravitational force acting on zero mass. Since $F=ma$, $F/m=a$, but in this case that is $0/0=a$ which is undefined.

$$F=ma$$
$$\frac{G M m}{r^2} = m a$$
$$g m = m a$$
$$g = a$$

The acceleration is independent of mass, and is well defined.

 

[avito009]

Acceleration due to gravity.

You are right mentor. I think I can explain that so that others can understand.

On further reading I realize that I was wrong about Newtons theory where I mentioned massless particles aren't effected by gravity. But actually bending of light could be calculated by Newton's theory but bending of light under general relativity would be twice that of Newtons theory.

In Newtonian mechanics, the acceleration of object A toward object B is not dependent on the mass of object A but on the mass of object B and the distance between objects A and B. Because the mass of object A does not affect its acceleration due to the gravitational influence of object B, does Newtonian mechanics predict that a massless particle (e.g. a photon) would be gravitationally affected by an object with mass?

Newton obviously knew that the mass of an object falling under the influence of Earth's gravity has no effect on its acceleration, i.e., all objects accelerate toward Earth at 32 ft/sec/sec regardless of their mass ("weight"). Therefore it follows that an object with no mass, such as a photon, would follow the same rule: it accelerates to Earth at 32 ft/sec/sec. (The reason we don't notice this is that photons spend so little time between the object we see and our eyes due to their extreme speed.)
Newton obviously knew this, and logically concluded that photons from distant stars grazing the Sun's limb (edge) would "fall" just a bit towards the Sun as they passed by, resulting in a slightly curved trajectory.

Now I explain why the acceleration is independent of mass of the object falling down.

By Newton's Universal Law of Gravitation, the force between two object due to gravity

F=GMm/r2

where M and m are the masses of the two bodies attracting each other, r the radius of the Earth (or distance between the center of the Earth and you, standing on its surface), and G is the gravitational constant.. Let's say M is the mass of the Earth and m of the object we're dropping.

Using F=ma=GMm/r2

We can rearrange to find a=GM/r2 (m gets canceled out), independent of the mass of the object.

Thus all object accelerate at the same rate under gravity alone.

 

[CKH]

After posting my message it occurred to me that this is true as $m>0$ (as $m$ approaches 0, not sure of correct notation). My argument leads to an undefined result whereas your algebra (which I admit is correct) does not.

 

[A.T.]

$$g m = m a$$
$$g = a$$

Can you make that step, if m = 0?
$$2 * 0 = 0 * 1$$
$$2 = 1$$

 

[Jonathan Scott]

Can you make that step, if m = 0?
$$2 * 0 = 0 * 1$$
$$2 = 1$$

Even if the maths is a little dodgy, the physics holds. In general relativity, the rate of change of momentum is proportional to the total energy, not the rest mass, but the momentum also involves the total energy, not the rest mass. So in that case both sides are still non-zero and equal.

 

[Dale]

You can make that step if m is arbitrary to show that a does not depend on m. If it does not depend on an arbitrary m then it does not depend on a zero m.

Your above equation should be:

$2\;m=m\;1$
$(2-1)\;m = 0$
$m=0$

So your equation above is not a proof that 2=1, but that m=0. In other words, that equation is not independent of m.

 

[Jonathan Scott]

The formula for the coordinate acceleration $d\mathbf{v}/dt$ in GR is messy because the coordinate speed of light decreases as an object falls to a lower potential. However, for the simple weak-field Newtonian approximation using isotropic coordinates (where the spatial scale factor relative to local measurements is the same in all directions), there is a simple equation of motion which does not involve the mass or energy of the test object, where all of the variables are relative to the coordinate system (including $c$, which here represents the coordinate speed of light which varies slightly with potential):
$$\frac{d}{dt} \left ( \frac{\mathbf{v}}{c^2} \right ) = \frac{1}{c^2} \, \mathbf{g} \, \left ( 1 + \frac{v^2}{c^2} \right ) $$
That is, if you use $\mathbf{v}/c^2$ instead of $\mathbf{v}$, it looks very similar to ordinary Newtonian gravity except for the extra term $v^2/c^2$ for space curvature, and the rate of change is exactly in the direction of the field, regardless of the direction of motion.

 

[CKH]

You can make that step if m is arbitrary to show that a does not depend on m. If it does not depend on an arbitrary m then it does not depend on a zero m.

Your above equation should be:

$2\;m=m\;1$
$(2-1)\;m = 0$
$m=0$

So your equation above is not a proof that 2=1, but that m=0. In other words, that equation is not independent of m.

This is tangential to the tread, but your algebraic argument is more subtle than it appears. The rule that you can cancel equal multipliers on two sides of equations requires a special argument for the case of 0. That argument might be that as the multiplier approaches zero, the equation continues to validate cancellation. Since this is true for arbitrary small multipliers, we conclude that at the limit of zero it is true (treating 0 as an infinitesimal number).

Reminds me of Bishop Berkeley's arguments with Newton concerning calculus. From a wiki article:

Historically, one of the earliest recorded references to the mathematical impossibility of assigning a value to a/0 is contained in George Berkeley's criticism of infinitesimal calculus in The Analyst ("ghosts of departed quantities").

 

[Dale]

It is not that subtle: a=g independent of m.

In fact, a=g independent of m is the defining characteristic of gravity. To go from that defining characteristic of gravity to a force law we multiply both sides by m to get (for any m):

$a=g$
$ma=mg$
$F=mg$

You can do the order of operations either way, but the independence from m is clear from the outset. The force law is constructed in order to achieve that independence, not the other way around.

 

[CKH]

What I'm pointing out concerns the justification for the algebra that you are using.

What can you conclude about the relationship between $x$ and $y$ in $0 x = 0 y$?

I'm not saying you're wrong, just that the algebra of real numbers requires a justification in the case that $m=0$ which considers of what happens as $m ->0$. I had never noticed that before.

 

[Dale]

You cannot conclude anything about x and y in that case, but that is not the case we have here.

The case that we actually have is $mx=my$ for any m. Since m is arbitrary you can indeed conclude that the relationship between x and y is $x=y$.

 

[A.T.]

The case that we actually have is $mx=my$ for any m. Since m is arbitrary you can indeed conclude that the relationship between x and y is $x=y$.

This assumes that there one and the same relationship between x and y, for all m, including m = 0. How do you justify that there is "the relationship", and not different relationships?

 

[Dale]

I am not sure what you are saying there. Could you express that mathematically? Are you considering something like $mx=my \; f(m)$?

 

[Dale]

Or maybe you are more considering something like a 3D space where the set of points satisfying $mx=my$ is the union of the x=y plane and the m=0 plane?

 

[A.T.]

I am not sure what you are saying there.

You say that from

mx = my

we can conclude that the relationship between x and y is

x=y for all m

But how do we know the relationship isn't actually:

x = y for m != 0
x = 3y for m = 0

This would also satisfy mx = my.

 

[ChrisVer]

Well for Newtonian mechanics, the main thing is that there is nothing more fundamental than the other..
For a rough example, take the 2nd Newton's law and the Hamilton Least Action Principle... they are the same thing (they give the same equations of motion) through different formalisms... nothing is more fundamental than the other when you consider classical mechanics. It's just that the one or the other is more useful.
Now then, one can take as fundamental in Newtonian gravity the force, thus he would immediately find F=0 (from the global gravitational force), and think "ah then, a massless particle won't feel any force". But nothing stops you from taking as fundamental the gravitational acceleration, which is mass-independent.
Nevertheless, the Newton just HAPPENED to have guessed the bending of light. By happened I mean it was an accidental prediction, which was right but started from the wrong place.
For example GR would allow the gravitational interaction between two beams of light, that wouldn't be true in any way through Newtonian gravity (even the gravitational acceleration depends on the mass of the source, send it to zero and you won't find anything). So Newton's starting point was wrong, but predicted a physical procedure that was correct - that's amazingly lucky for the fellow guy ...

As for the discussion saying mx=my and stuff, that's totally wrong way to put m=0, because you indeed get no result... what you actually do is:
x= \frac{my}{m}
and take the limit m \rightarrow 0
Then you will get your correct result... The first way (m=0) only a mathematician would have guessed... A physicist works with limits...
x= lim_{m \rightarrow ∞} \frac{(1/m) y}{(1/m)} = lim_{m \rightarrow ∞} \frac{m y}{m}=y

 

[BruceW]

Newton obviously knew this, and logically concluded that photons from distant stars grazing the Sun's limb (edge) would "fall" just a bit towards the Sun as they passed by, resulting in a slightly curved trajectory.

Really? It seems much more likely that Newton did not even consider how gravity might affect light. Having said that, I haven't actually read Principia, so I don't know for sure.

Using F=ma=GMm/r2

We can rearrange to find a=GM/r2 (m gets canceled out), independent of the mass of the object.

Thus all object accelerate at the same rate under gravity alone.

the m doesn't get canceled out if m=0. But, in the limit of m being very small, you do get a=GM/r2. Really, I think Newtonian mechanics doesn't tell us anything about the motion of bodies with zero mass. It can tell us what happens to bodies with very small mass, but as soon as you start saying things like m=0, I would not call that Newtonian mechanics.

 

[ChrisVer]

and what is actually the difference for physics to say m=0 or m \rightarrow 0 or \frac{1}{m} \rightarrow ∞?

 

[Dale]

You say that from

mx = my

we can conclude that the relationship between x and y is

x=y for all m

That is not quite what I am saying. I am saying that given:

$mx=my$ for all $m$

we can conclude that the relationship between $x$ and $y$ is $x=y$.

I agree that the "for all m" does not automatically follow if it is not given, but if it is given then I am pretty sure that the conclusion $x=y$ does in fact follow.

 

[CKH]

and what is actually the difference for physics to say m=0 or m \rightarrow 0 or \frac{1}{m} \rightarrow ∞?

The issue (which is tangential) is how mathematics allows us to cancel the m's in the very algebra that physics employs to predict things even though in $0a=0g$ you can make no claim about the relationship between $a$ and $g$.

If you dismiss the algebra that you use as not physics, how can you claim to compute correct physics using algebra? Shouldn't the two agree on the rules?

This issue (of what happens as $m->0$) was the center of a lively debate in Newton's time. The same argument that Newton used to found the calculus of infinitesimals is needed to support the cancellation of $m$ in the equations in the case of $0$.

Mathematicians do use the exact same reasoning and spend plenty of time studying the use of limits. They do this for fun and to help assure that the formalisms we use in physics are valid.

What you can do when a variable is 0 is not always well defined, even in physics.

 

[ChrisVer]

In physics you exact some formula... to see how things behave at singular or not well defined points, you take limits...you don't go and and put your parameters the singular values.
And I was talking about Newton and about the Newtonian calculations. I find them basically wrong, since the correct answer is given by GR itself.
As for the algebra as you give 0a=0g it doesn't work...what works is having:
ma=mg --> a= g (mass-independent)
who stops you from starting immediately by a=g?

 

[WhatIsGravity]

Kinda off key with this thread, but it's cool to realize that if light (energy) is bent by a gravitational field... then that light also creates a gravitational field.

I think a=g, in this context, is the equivalency principle?

 

[jartsa]

It is important to be aware that photons passing a gravitational source are effectively being bent by two effects.

Firstly, Newtonian gravity applies to everything including photons, giving an acceleration of $g$. This can be considered to be due to the curvature of space with respect to time, in that if you plot the radial distance against time, a free fall line curves towards the source.

Secondly, space itself is curved by gravity, which has the effect that something moving through it is accelerated proportionally to $v^2/r$ where $r$ is the radius of curvature. According to GR, the radius of curvature for a weak Newtonian gravitational field $g$ is $c^2/g$ so the resulting acceleration component due to velocity is $v^2/((c^2)/g) = (v^2/c^2) g$. This means that a photon is effectively accelerated by twice as much as a slow-moving object.

Photons are not special in this way. A particle moving at very nearly the speed of light will follow a similar path to a photon, and a photon trapped between the walls of a reflective box whose sides appear to be parallel using local rulers will fall with the same local acceleration as a brick (noting that the rulers measuring horizontal distance will appear curved to an external observer).

I think this is wrong. Because I don't see anything about tidal effects there. See posts #16 and #18.

I mean, there should be Δg in the equations. Space curves -> g changes -> tidal effect -> extra bending of light

 

[A.T.]

I think this is wrong. Because I don't see anything about tidal effects there.

Tidal effects occur when comparing two initially parallel light beams, at slightly different r, and therefore different g's.

Space curves -> g changes -> tidal effect -> extra bending of light

No, g is not related to spatial geometry, but to the time geometry (gravitational time dilation and its derivatives).

g affects all objects equally, regardless of their speed through space, while spatial geometry affects only objects already moving through space. The extra bending is due to spatial geometry, which only affects objects that move through space.

See this for more info:
http://mathpages.com/rr/s8-09/8-09.htm

 

[A.T.]

I am saying that given:

$mx=my$ for all $m$

we can conclude that the relationship between $x$ and $y$ is $x=y$.

We could just as well assume the relationship between x and y is:

x = y for m != 0
x = 3y for m = 0

Both relationships are consistent with mx=my for all m. So why should your conclusion be the only right one?

 

[A.T.]

My summary of the Newtonian light bending issue:

1) It's OK to formulate Newtons gravity based on universal acceleration, without the force. This automatically applies to light. So you simply postulate that light is affected too.

2) You can start with the force equations and work with the limit m -> 0. Then make a physicists argument, that nature doesn't like discontinuities, so it's save to extrapolate the limiting value ti m = 0.

3) What I don't like, is starting with the force equations and then using division by m, to make claims about the situation when m = 0.

 

[Dale]

We could just as well assume the relationship between x and y is:

x = y for m != 0
x = 3y for m = 0

Both relationships are consistent with mx=my for all m. So why should your conclusion be the only right one?

Your second relationship does not hold for all m, and you are given that the relationship, whatever it is, holds for all m. Remember, you are given "$mx=my$ for all m". If x=1 and y=3 that holds for m=0, but not for m=1. Therefore x=1 and y=3 is not consistent with what is given.

Even if some mathematician would object (which I doubt), it is clear that in physics canceling variables on both sides of an equation is a common and well-established technique. It is part of nearly every non-trivial proof that I have ever seen. So even if a mathematician would object to $mx=my$ for all m implying $x=y$ I don't think that any physicist would object to $ma=mg$ implying $a=g$.

 

[Dale]

1) It's OK to formulate Newtons gravity based on universal acceleration, without the force. This automatically applies to light. So you simply postulate that light is affected too.

This actually is my preferred approach, and I think the one that is the most historically and theoretically accurate. The observation of universal acceleration came first, and the force law was developed to be consistent with that observation.

The quick derivation that you dislike simply arises because most people who have this confusion are starting with the force law and assuming that f=0 implies a=0, which is not true for a massless particle.

 

[avito009]

Air Resistance

Ok people now I know the question you will ask. You will say that if we drop a feather from some height and if we drop a brick from the same height (at the same time) then the feather will come down later than the brick.

So I have mentioned that all objects accelerate toward Earth at 32 ft/sec/sec regardless of their mass. So why the time lag between the feather touching the ground and the brick coming down?

So here is my answer. The air resistance (Drag) for a feather is going to be greater because it has a larger surface area then a brick in proportion to its weight. Air resistance will slow its decent even with gravity.

As the brick and the feather begin to gain speed while falling down, they encounter the upward force of air resistance. Air resistance is the result of an object plowing through a layer of air and colliding with air molecules. The more air molecules which an object collides with, the greater the air resistance force. Subsequently, the amount of air resistance is dependent upon the speed of the falling object and the surface area of the falling object. Based on surface area alone, it is safe to assume that (for the same speed) the brick would encounter more air resistance than the feather.

But why then does the brick, which encounters more air resistance than the feather, fall faster? After all doesn't air resistance act to slow an object down? Wouldn't the object with greater air resistance fall slower?

You need an understanding of Newton's first and second law and the concept of terminal velocity. According to Newton's laws, an object will accelerate if the forces acting upon it are unbalanced; and further, the amount of acceleration is directly proportional to the amount of net force (unbalanced force) acting upon it. Falling objects initially accelerate (gain speed) because there is no force big enough to balance the downward force of gravity. Yet as an object gains speed, it encounters an increasing amount of upward air resistance force. In fact, objects will continue to accelerate (gain speed) until the air resistance force increases to a large enough value to balance the downward force of gravity. Since the brick has more mass, it weighs more and experiences a greater downward force of gravity. The elephant will have to accelerate (gain speed) for a longer period of time before there is sufficient upward air resistance to balance the large downward force of gravity.

Once the upward force of air resistance upon an object is large enough to balance the downward force of gravity, the object is said to have reached a terminal velocity. The terminal velocity is the final velocity of the object; the object will continue to fall to the ground with this terminal velocity. When the air resistance force equals the weight of the object, the object stops accelerating and falls at a constant speed called the terminal velocity. In the case of the brick and the feather, the brick has a much greater terminal velocity than the feather. As mentioned above, the brick would have to accelerate for a longer period of time. The brick requires a greater speed to accumulate sufficient upward air resistance force to balance the downward force of gravity. In fact, the brick never does reach a terminal velocity; there is still an acceleration on the brick the moment before striking the ground.

The feather quickly reaches a balance of forces and thus a zero acceleration (i.e., terminal velocity). On the other hand, the brick never does reach a terminal velocity during its fall; the forces never do become completely balanced and so there is still an acceleration. If given enough time, perhaps the brick would finally accelerate to high enough speeds to encounter a large enough upward air resistance force in order to achieve a terminal velocity. If it did reach a terminal velocity, then that velocity would be extremely large - much larger than the terminal velocity of the feather.

So in conclusion, the brick falls faster than the feather because it never reaches a terminal velocity; it continues to accelerate as it falls (accumulating more and more air resistance), approaching a terminal velocity yet never reaching it. On the other hand, the feather quickly reaches a terminal velocity. Not requiring much air resistance before it ceases its acceleration, the feather obtains the state of terminal velocity in an early stage of its fall. The small terminal velocity of the feather means that the remainder of its fall will occur with a small terminal velocity.

 

[CKH]

In physics you exact some formula... to see how things behave at singular or not well defined points, you take limits...you don't go and and put your parameters the singular values.

Then we agree that you must take limits to justify cancellation of $m$ in your equations. Formal mathematics does the same thing as physics so you can rely on it to calculate physical results.

The only reason I brought it up is that, in my naive original conclusion, I failed to take into account what happens as the mass approaches zero, but instead asked "what can you conclude at exactly 0"?

Then I realized that algebra itself requires the same arguments (understanding 0 as an infinitesimal value) for the rule that multiplier variables can be canceled even if they can assume the value 0.

AT states:

[2) You can start with the force equations and work with the limit m -> 0. Then make a physicists argument, that nature doesn't like discontinuities, so it's save to extrapolate the limiting value ti m = 0.

And formal mathematics of real numbers, apart from physics, makes a similar argument. Mathematicians apparently don't like needless discontinuities either.

I think we all agree.

 

[BruceW]

Even if some mathematician would object (which I doubt), it is clear that in physics canceling variables on both sides of an equation is a common and well-established technique. It is part of nearly every non-trivial proof that I have ever seen. So even if a mathematician would object to $mx=my$ implying $x=y$ I don't think that any physicist would object to $ma=mg$ implying $a=g$.

I have a good counter-example. Suppose we're doing some quantum mechanics. Since the Hamiltonian operator is Hermitian, we have:
$H|\psi_i > = E_i |\psi_i >$ and $<\psi_k|H=E_k<\psi_k|$
Therefore, $<\psi_k|H|\psi_i>=E_i<\psi_k|\psi_i>$ and $<\psi_k|H|\psi_i>=E_k<\psi_k|\psi_i>$
Therefore, $E_i<\psi_k|\psi_i>=E_k<\psi_k|\psi_i>$
Therefore if $<\psi_k|\psi_i>$ is non-zero, we have $E_i=E_k$. But when $<\psi_k|\psi_i>$ is zero, we will generally have $E_i$ and $E_k$ not equal.

This is useful because if we can establish that the energy eigenstates are non-degenerate, this means that the energy eigenstates are orthogonal to each other. So anyway, this is a good example where taking $ma=mg$ to imply $a=g$ would ruin the physics.