Bending of Light due to Gravity

[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.

 

[Dale]

We have already covered that case, it is not what I am claiming. Again, the case under consideration is when you are given $mx=my$ for all m.

In your example, the equivalent statement would be where you are given:
$E_i <\psi_k|\psi_i>=E_k <\psi_k|\psi_i>$ for all $<\psi_k|\psi_i>$.

In which case you do, in fact, have $E_i = E_k$.

 

[CKH]

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.

Thanks a lot. Now I'm hopelessly lost and again wondering what we really can conclude about a truly massless particle from Newton's laws.

 

[CKH]

Why is it OK to conclude that Newton's gravity can act on a massless particle while a charged body cannot act on an uncharged particle?

$F = G mM/r^2$
$F = K qQ/r^2$ (where $K = 1/4 ∏ ε_0$)

Should we conclude that a charged object can affect the path of an uncharged particle such as a photon?

I certainly believe that (this form of) Newton's equations can tell us the affect on any particle that has the property mass (non-zero mass), no matter how small that mass is so long as there is some mass. But a particle that has no mass experiences no force. It seems magical to predict exactly how it will be affected by these laws.

Unfortunately we cannot test this because Newton's law is wrong.

Perhaps it's more logical to say that since $F=ma$, $a = F/m$. When $m$ vanishes, $a$ is undefined regardless of whether F is 0?

Newton could have proposed that a = GM/r2 and stated that this equation applies to all bodies. Then we can immediately conclude that photons are affected as claimed.

I don't know how he originally formulated his law of gravitation.

 

[Dale]

Why is it OK to conclude that Newton's gravity can act on a massless particle while a charged body cannot act on an uncharged particle?

$F = G mM/r^2$
$F = K qQ/r^2$ (where $K = 1/4 ∏ ε_0$)

The answer is in the equations you posted. For gravity:

$F = G mM/r^2$
$ma = G mM/r^2$
$a = GM/r^2$
So a is not a function of m.

For Coulomb's law:
$F = K qQ/r^2$
$ma = K qQ/r^2$
$a = KqQ/mr^2$
So a is a function of q.

Despite the superficial similarity of the force laws the physical observations that led to the development of the force laws are completely different. For gravity it was observed that acceleration was independent of mass. For Coulomb's law it was observed that acceleration was proportional to charge.

 

[dextercioby]

Why is it OK to conclude that Newton's gravity can act on a massless particle [..]

Who says that ?

 

[Dale]

Who says that ?

I do and Élie Cartan did, and I believe that Newton did also.

http://en.wikipedia.org/wiki/Newton–Cartan_theory

 

[CKH]

OK DaleSpam, but as far as I can tell Newton never made any claim about the behavior of massless particles. This came much later:

http://mathpages.com/rr/s6-03/6-03.htm

Newton’s theory of universal gravitation already predicted that the path of any material particle [I read this as matter which always has mass] (regardless of its composition) moving at a finite speed is affected by the pull of gravity. However, the finite speed of light was not well established in Newton’s time, as discussed in Section 3.3, and it was far from clear that light consists of material particles. These uncertainties precluded Newton from making any definite prediction about whether and how light is affected by gravity. By the late 18th century the finite speed of light was well established so, although the constitution of light was still unknown, it was possible to apply Newton’s law to compute the deflection of light by gravity – under the assumption that a pulse of light responds to gravitational attraction as does a particle of matter moving at the same speed.

According to this author Newton's equations do not predict the effect of gravity on massless particles. Additional assumptions are needed, in particular the assumption that a gravitation field accelerates particles of light in the same way as particles of matter.

That suggests that the rote use of algebra is not always correct. In particular cancelling multipliers is based on dividing both sides of the equation by the multiplier. When the multiplier is 0, division is undefined so the cancellation rule cannot be applied to 0.

I must retract my agreement with the limit argument. The algebra you used is wrong for m=0.

This has been an illuminating discussion, at least for me.

 

[CKH]

Dale (is it OK to call you Dale?), more specifically:

$mx=my$ => $x=y$ only when $m$ is not 0.

 

[Dale]

We have been over this ad nauseum.
$mx=my$ for all m does imply $x=y$ and for gravity it is a physical observation that $a=g$ is independent of mass from which the force law is derived.

 

[CKH]

What can I say? Your algebra is correct except for the case of m=0.

See http://math.stackexchange.com/questions/67994/why-should-you-never-divide-both-sides-by-a-variable-when-solving-an-equation

 

[Vanadium 50]

The experimental observation that led to Newtonian Gravity is a = g. That implies ma = mg. To then say that m = 0 somehow invalidates a = g makes no sense whatever.

 

[WhatIsGravity]

Doesn't everything have a gravitational field? Even pure energy like a photon, still creates a gravitational field? m=0 for anything observable, doesn't really make sense. e=mc^2.

 

[ChrisVer]

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.

And also, you don't need to have $E_{i}=E_{j}$ but $E_{i} \in [E_{j}-2 \Gamma , E_{j}+2 \Gamma]$
to say that E_{i}=E_{j}. But well, bringing in discrete objects [existing in quantum mechanics] makes the thing more difficult...
Also there you can have something like: $<x|x'> = \delta(x'-x)$...
well the delta function can be defined as 0 everywhere and infinity at the point where the argument vanishes.
It can also be defined as a gaussian where you send the standard deviation to zero.

 

[BruceW]

And also, you don't need to have $E_{i}=E_{j}$ but $E_{i} \in [E_{j}-2 \Gamma , E_{j}+2 \Gamma]$
to say that E_{i}=E_{j}. But well, bringing in discrete objects [existing in quantum mechanics] makes the thing more difficult...

you've lost me here. I was just using discrete states, for this simple example. Also, are those gamma functions? I don't understand what they are for.

 

[CKH]

The experimental observation that led to Newtonian Gravity is a = g. That implies ma = mg. To then say that m = 0 somehow invalidates a = g makes no sense whatever.

I'm sorry, but you are completely misinterpreting my posts. I never said any such thing. Please, if you want to tell me I'm making no sense you should at least read my posts carefully first.

What I said was that ma=mg does not imply a=g for the single case of m=0. It is most certainly valid for all cases in which m is not zero.

More generally, what I am claiming is that you cannot prove from Newton's laws (as stated by Newton) that particles with no mass are affected by gravity.

Why do you think you can, if you do? Note that post-Newtonian theories of gravity are not what I am discussing here.

Is that clear now?

 

[WhatIsGravity]

Darn. Do you guys really need to delve into equations to answer this question?

 

[ChrisVer]

you've lost me here. I was just using discrete states, for this simple example. Also, are those gamma functions? I don't understand what they are for.

They are the linewidths...in QM in order to say A=B it's enough for them to be almost equal up to their intrinsic uncertainties.
So for me, if $i \ne j$ so $<\psi_i | \psi_j> =0$ you can still have $E_i = E_j$ as long as the spectrum becomes continuous, that's in fact the reason it does...

^EDITED

 

[BruceW]

We have already covered that case, it is not what I am claiming. Again, the case under consideration is when you are given $mx=my$ for all m.

In your example, the equivalent statement would be where you are given:
$E_i <\psi_k|\psi_i>=E_k <\psi_k|\psi_i>$ for all $<\psi_k|\psi_i>$.

In which case you do, in fact, have $E_i = E_k$.

hmm... But I do have $E_i <\psi_k|\psi_i>=E_k <\psi_k|\psi_i>$ for all $<\psi_k|\psi_i>$ and it is not always true that $E_i=E_k$.

But I think I understand what you're aiming at. I'm guessing you mean that if we assume the same relationship between all $E_i$ and all other $E_k$ then it is true that every $E_i$ has the same value, independent of $i$. (because we have the trivial $E_i=E_i$ and then can extend from there to $E_i=E_k$ for every k).

 

[TumblingDice]

m=0 for anything observable, doesn't really make sense. e=mc^2.

E=mc^2 isn't the complete equation. It only includes the portion of energy related to the mass, but not the contribution of momentum. It's O.K. for m to be zero for light, which is always measured locally as traveling at c.

Ben created a FAQ in this forum that explains this relationship well:
How can light have momentum if it has zero mass?

 

[BruceW]

They are the linewidths...in QM in order to say A=B it's enough for them to be almost equal up to their intrinsic uncertainties.
So for me, if $i \ne j$ so $<\psi_i | \psi_j> =0$ you can still have $E_i = E_j$...

ah right, that's slightly more complicated than I was thinking. I like to stick with the easier examples first :) hehe.