Bending of Light due to Gravity

[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: &lt;x|x&#039;&gt; = \delta(x&#039;-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 &lt;\psi_i | \psi_j&gt; =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 &lt;\psi_i | \psi_j&gt; =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.

 

[ChrisVer]

I editted it ( just to make sure you read what I meant).

 

[BruceW]

...
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.
...

very nice. good reasoning. But did the brick transform into an elephant while it was falling?! lol

 

[BruceW]

I editted it ( just to make sure you read what I meant).

right, so the states you are talking about are 'almost discrete', but not quite, so they are not fully energy eigenstates, but have some spread around the most likely value. (Is that right?)

edit: or I guess we could have discrete energy eigenstates, but the states you are talking about might be made up of some very tight spread of energy eigenstates, all which have very similar energy values, so you have an effective blurring.

 

[Dale]

Wow, the tangents have sprouted tangents. Thread closed.