Scientists stop light

[KrazyHorse]

Originally posted by Japher
This helped me

Though it still doesn't make a whole lot of sense in that all equations for energy require a mass, and that "rest masses" have been calculated for a photon, yet they still say it is zero.

E^2 = m^2*c^4 + p^2*c^2

So yes, the rest energy of a photon is zero. It's a good thing that it travels at the speed of light, because no sane reference frame can catch up to it...

[KrazyHorse]

Originally posted by Kramerman
ok, new question



i was taught that mass is what warps space time, adn that light has no gravity itself, but does get 'bent' when traveling thru warps made by a mass, like a star or somethn. is this energy deal somethn relatively new or was my physics teacher a moron? it could very possiblly be the second

The energy contained in an EM field (i.e. a photon) most certainly does contribute to bending spacetime. As a matter of fact, the Maxwell stress-energy tensor (the basic thing you use to figure out what your metric is) was first used as a convenience in classical electromagnetic field theory.

[KrazyHorse]

Originally posted by Big Crunch


I always figured that it had to be energy that causes gravity, if it isn't I can't comprehend how the recipricol force/momentum transfer is going to take place when objects gravitationally pull on massless particles.

I also wonder what would happen to the gravitational fields generated by matter and anti-matter after they annihilate. Surely they still exist even when all that is left is photons. Else you could have perpetual motion machines?

The basic thing is energy density. Mass is part of it...

[Rogan Josh]

Originally posted by KrazyHorse
It's a good thing that it travels at the speed of light, because no sane reference frame can catch up to it...

Why do you think it is a good thing? I am not sure what the physical consequences of the speed of light being slightly less than c would be, but I don't think they would be severe (if say for example the mass of the photon was 10^-8 eV). [Although our physics theories would all fall apart - the Standard Model insists that the photon is exactly massless because it is a gauge boson of a local U(1) symmetry.]

I think one of the problems students have in their understanding of c as a barrier, is that they don't understand that it really has nothing to do with light. The c barrier is there with or without the existence of light. It just so happens that light has no mass, and so it travels at c, but then so does a gluon and we don't call it 'gluon speed'. In principle, we could imagine a universe where all particles had mass and nothing travelled at c.

[KrazyHorse]

Uhh...i was pointing out that photons wouldn't exist if they had no mass and travelled at less than c, because then they'd have 0 energy

Photons having a mass leads to some odd things, but I know so little about high energy symmetries that I'm out of my league talking about it...

I know c has nothing to do with light...

[Rogan Josh]

Originally posted by KrazyHorse
I know c has nothing to do with light...

Yeah - I know. I wan't meaning you.

[Az]

I do have this subconscious connection.

What sort of a coinsidence is it? photons travelling at precisely the maximum speed in the universe? wtf?



Oh, and btw, I wanted to ask the physicists a question.

Say, we have a very long object. 10 million mile long rod. If we'll give it a little push on one side, will it simultaneously move at the other side of the rod?

[Rogan Josh]

Originally posted by Azazel
What sort of a coinsidence is it? photons travelling at precisely the maximum speed in the universe? wtf?

It depends what physics theories you believe. It is no coincidence that massless particles travel at c. That is a given, so the question really is: is it a coincidence that the photon is massless.

I would say no, becuase it is a gauge field. How much quantum mechanisc to you know? Are you aware that probabilities of QM transitions are given by the absolute value of "matrix elements" squared? In other words,

P = |M|^2 where P is a probability and M is something the theory tells us.

Now if you multiply M by a phase, exp(i f) where 'f' is any real number, P stays the same. Following me still?

So there is a symmetry in the theory - we can multiply the M by the phase and get the same answer. In this case it is called a U(1) symmetry (the U stands for 'unitary', meaning that the absolute value of the multiplication factor is 1, while the 1 is because it is a 1x1 matrix, just a number) [Another symmetry group for example is SU(2) - 2x2 matrix, which is unitarty and 'special' (has determinant 1)]

OK, so we have a U(1) symmetry, but relativity tells us that we on Earth don't know what people on the other side of the galaxy are doing right now, because it takes time to pass a light signal to them. In other words, information can only travel at speeds < c). So we can't make this phase shift the same everywhere. f must be a function od space and time. f=f(x). If the symmetry still holds, then we call it a 'local' symmetry because it lets us change the phase only locally.

But then the symmetry is no longer true UNLESS you introduce a new particle (called a gauge particle) whith special properties. It turns out that these special properties are exactly the properties of light. So light is a direct consequence of requiring the laws of physics to be locally U(1) symmetric.

In this sense, it is no coincidence that light is massless, since this is one of the requirements.

Say, we have a very long object. 10 million mile long rod. If we'll give it a little push on one side, will it simultaneously move at the other side of the rod?

No it won't. The 'push' will travel uo the object like a wave.

[Az]

So there is a symmetry in the theory - we can multiply the M by the phase and get the same answer. In this case it is called a U(1) symmetry (the U stands for 'unitary', meaning that the absolute value of the multiplication factor is 1, while the 1 is because it is a 1x1 matrix, just a number) [Another symmetry group for example is SU(2) - 2x2 matrix, which is unitarty and 'special' (has 0 determinant)]

lost you here.

No it won't. The 'push' will travel uo the object like a wave.

I figured as much, but I wasn't sure.

I am not certain why, though.

[Rogan Josh]

Originally posted by Azazel
lost you here.

OK. Say we have a set of rules, which we apply to find out some physical quantity (like the probability of a particle decay). Now if we have a different set of rules which gives the same answer and gives the same answer for every possible thing we can calculate, then the second set of rules is as valid as the first.

Now, changing from one set of rules to the other is a symmetry of the theory. Just like turning a square by 90 degrees makes it look exactly the same, changing from one set of rules to the other makes the universe look the same.

Agree?

Good. So, now imagine everything (I am simplifying the true picture here) we can calculate with our rules only affects the universe by its absolute value squared. That is if I have a complex number M = a + i b, I take its complex conjugate M* = a - i b, and physical things only depend on M*M = (a-ib)(a+ib) = a^2+b^2.

Now, if I multiply the M by exp(i f), then M = exp(i f) (a+ib) and M*=exp(- i f) (a-ib), so M*M = exp(-if) (a-ib) exp(if) (a+ib) = a^2+b^2

So the physical thing M*M doesn't change. The 'alternate theory' where everything is multiplied by exp(i f), called a 'phase', is just as valid because it gives the same answers. And the act of multiplying everything by the phase, is the symmetry transformation (like rotating the square by 90 degrees).

Notice that I didn't need to specify a value for f - it could be anything.

Are you with me so far?

[Az]

yep ( reminded me of that molecular symmetry course I took).

[Japher]

What the **** are you on?

The same thing Einstein was on?

Is what these people say right?

[Rogan Josh]

OK, but it turns out that the laws of physics aren't quite like that because we can't just multiply by exp(i f) everywhere at once. Since information can only travel at c or slower, we are not allowed to multiply at point A by the smae phase as at point B (or rather they needn't be the same, since if we are at A, we don't know the value of f at B).

In other words f must depend on where you are - it is a function of position x: f=f(x).

Now, lets just imagine a quantum mechanics system with only electrons in it. If we had only electrons, and made this transformation, we would find that the physical results do change. It is not a symmetry. But if we include photons in the theory, the transformation on the photons exactly cancels out the transformation on the electrons and we have a symmetry again.

In other words, by insisting that the symmetry transformation be 'local' (ie position dependent), we insist on the existence of light, and in fact the electromagnetic force just drops out too.

In fact, all the forces come this way. Electromagnetism comes from insisting on a U(1) symmetry, the weak nuclear force comes from a local SU(2) symmetry (the symmetry group of a sphere), and the strong nuclear force is local SU(3). It is thought (but not yet proven) that gravity is actually local supersymmetry (supersymmetry is a symmetry between particles with different spins).

So God didn't say 'Let there be light'. He said 'Let there be a local U(1) symmetry'.

[Rogan Josh]

Originally posted by Japher
Is what these people say right?

The first guy, Mr Moustris, is an engineer. Never trust engineers talking about physics. He uses his last equation as a definition of mass, and then says that photons have mass (which they do by his definition). But all physicists worth their salt would call 'mass' what he refers to as 'rest mass'. I suppose it is just a matter of semantics.....

The second guy is right.

[Japher]

Well that explains it, I'm an engineer...

I can understand 'rest mass' being zero... now.

thanks

I leave the physic-ing to physicists...