Bending of Light due to Gravity

[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.