Conversation started Dec 8, 2020 at 12:04.
Dec 8, 2020 12:04
I have seen the phrase "but it doesnt explain gravity" many times
just what is the conundrum with gravity and modern theories?
Do you mean the conflict between general relativity and quantum field theory?
yeah
GR is a field theory. The field is the metric tensor, which is a function of position in spacetime just like any other field. So just like any other field in principle it could be quantised to produce a quantum theory of gravity.
The trouble is that quantising a field is a mathematically tricky thing to do.
what do you mean by "quantizing"
for instance, what does "quantizing the EF" mean
i know quantizing means restricting to discrete values, but what does it mean in this context?
nb EF means electric field
Do you know what a Fourier transform is and what it does?
Dec 8, 2020 12:10
decompose any F into sines and cosines?
Yes. Suppose we have some function that is a function of time - any function of time.
This function can be written as a sum of sines and cosines of different frequencies.
corrrect
So suppose we have an EM field that is a function of position in spacetime, then it can be written as a sum of sines and cosines by Fourier transforming it. This is a 3D transform, and the "sines and cosines" are plane waves that can have any direction in space, but the basic principle is the same.
OK so far?
Dec 8, 2020 12:14
yes
Now, when you solved the Schrodinger equation for a free particle you discovered that the solution was a plane wave. Yes?
ill take your word for it
Well, let's do it. The SE for a free particle (in 1D for simplicity) is $$ \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = i\hbar\frac{d\psi}{dt} $$
Suppose we make a guess that the solution is a plane wave $\psi(x,t) = e^{i(kx-\omega t)}$
Then $\frac{d^2\psi}{dx^2} = -k^2\psi$. Yes?
And likewise $\frac{d\psi}{dt} = -i\omega\psi$
Dec 8, 2020 12:21
yes
And we can substitute these derivatives back into the SE to get $$\frac{-\hbar^2}{2m}k^2\psi = -\hbar\omega\psi $$
So we have a solution as long as $$ \frac{\hbar^2}{2m}k^2 = \hbar\omega $$
right
So QM tells us that the wavefunction for a free particle is a plane wave. The wavelength and frequency are related to the momentum and energy of the particle.
Dec 8, 2020 12:24
right.
The momentum and energy can be any value, but for any one particle its momentum and wavelength have specific values so it has a specific wavefunction.
Anyhow ...
2 questions, although these are better suited for etomorrow: 1) what is the wave function, 2) what is a free particle
If the EM field can be written as an infinite sum of plane waves then is it possible that the EM field can be written as an infinite sum of free particles i.e. photons?
And the answer is that yes indeed the EM field can be written as an infinite sum of free particles i.e. photons.
So we quantise the EM field by writing it as a sum of the wavefunctions we get for free particles from the Schrodinger equation.
Sounds easy doesn't it :-)
hmm now i understand "quantised" in this context
Yes
But ...
Oh deary me ..
Photons aren't free particles. Photons interact with anything that is charged, e.g. electrons, and that means they aren't free. You have to include these interactions in your wavefunction, and that produces equations too hard to solve. In fact there is still some argument as to whether the mathematics of quantum field theory actually makes sense.
Dec 8, 2020 12:30
........
@JohnRennie oh
@JohnRennie what are "gravitons"?
So we can only do calculations in QFT by using an approximation. Basically we start with the free field and introduce the interactions as perturbations of the free field.
You've probably heard of Feynmann diagrams. Yes?
Well each Feynman diagram is a representation of one term in the perturbations that we have to sum to actually do a calculation.
You have to calculate and add up an infinite number of Feynman diagrams to do the calculation, but luckily the infinite series converges pretty fast so in practice these calculations are fairly straightforward.
Does this all make sense so far?
Dec 8, 2020 12:34
yeah, kinda
But there's a problem. If you sum up the Feynman diagrams in the obvious way they don't converge. The answer is always infinity.
But it turns out there is a fix for this, and this is a process called renormalisation.
Basically this is taking the infinite sum you get from the first calculation and subtracting infinity from it to give a finite result.
Now this sounds pretty dodgy maths, and for a long time people were pretty suspicious of the renormalisation process.
However it has now been put on a sound mathematical footing, and it works. Using renormalisation we find we can do calculations in QFT and get answers that agree with experiment.
damn
In fact QFT now works so well that no experiment anywhere has ever disagreed with calculations using QFT. Not at the LHC - not anywhere. It is a spectacularly successful theory.
8 mins ago, by satan 29
@JohnRennie what are "gravitons"?
@satan29 We took the EM field and figured out that it could be written as a sum of photons. Yes?
Dec 8, 2020 12:39
yes
So the question is whether we can do the same for the gravitational field? Can we write the quantised gravitational field as a sum of gravitons?
If we could then we'd have a quantum theory of gravity.
But ...
Do you remember that I said above we need to use renormalisation to actually do calculations with QFT?
Well renormalisation doesn't work with the gravitational field. We still get infinities.
The mathematical reasons why it doesn't work are well understood, but what we don't know is what does this mean physically.
ohhh
Is it that we just aren't doing the calculation in the right way - remember that the maths behind QFT is still considered a bit iffy.
 
Conversation ended Dec 8, 2020 at 12:42.