Particle Destruction

[Radical Edward]

when an electron and positron (or any particle/antiparticle) collide, you just end up left with energy, in essence, mass has disappeared, since both have positive mass. this is a bit confusing really, since all the other quantum numbers are conserved, such as charge, spin, lepton number etc... so why does, in the case of mass 1 + 1 = 0 ?

[Raider]

Wouldn't it be...

 

1 + 1 = lots of energy ((2c^2) + (pc))

[fafalone]

Since their matter is equal and opposite, they cancel eachother out.

[Tom Mattson]

Originally posted by Radical Edward

when an electron and positron (or any particle/antiparticle) collide, you just end up left with energy, in essence, mass has disappeared, since both have positive mass. this is a bit confusing really, since all the other quantum numbers are conserved, such as charge, spin, lepton number etc... so why does, in the case of mass 1 + 1 = 0 ?

 

It doesn't. The quandry is resolved in special relativity, in which the equivalence of mass and energy is deduced. The energy associated with the mass is entirely accounted for. For e⁺e^- annihilation, the conservation law would be written:

 

2m_ec²+KE_i=hv₁+hv₂

 

m_e=electron mass

KE_i=total initial kinetic energy

v₁=frequency of photon 1

v ₂=frequency of photon 2

c=speed of light

h=Planck's constant

 

edit: subscript bracket

 

Tom

[Tom Mattson]

double post deleted

[Radical Edward]

oh I know that the two are equivalent, and that you end up with energy left over, but my point relates mostly to effect, since mass has effects that energy doesn't, namely gravity. In effect, you have two positive gravity sources, adding together to equal no gravity source at all.

[Tom Mattson]

Originally posted by Radical Edward

oh I know that the two are equivalent, and that you end up with energy left over, but my point relates mostly to effect, since mass has effects that energy doesn't, namely gravity. In effect, you have two positive gravity sources, adding together to equal no gravity source at all.

 

No, energy does indeed gravitate. If it did not, then mass and energy would not be equivalent!

 

In GR, all mass and energy are entered into the Einstein tensor, which then determines the curvature of space time.

 

Tom

[Raider]

oh I know that the two are equivalent, and that you end up with energy left over, but my point relates mostly to effect, since mass has effects that energy doesn't, namely gravity. In effect, you have two positive gravity sources, adding together to equal no gravity source at all.

 

I do not think this is true. Kinetic energy acts as mass, making acceleration more difficult at higher speeds (Thus the closer to c, the harder it is to get closer...).

 

What evidence do you have that it does not act like mass with respect to gravity?

[Tom Mattson]

He doesn't--he is just mistaken. See my previous post. All matter and radiation contributes to the curvature of spacetime.

 

Tom

[Radical Edward]

now light having a gravitational effect is something I have never heard of.

[Tom Mattson]

Yes, the basic equations of GR are the following system of coupled, nonlinear differential equations:

 

R_uv-(1/2)Rg_uv=8:pi: GT_uv

 

These are a system of equations in the metric tensor, g_uv, which contains the geometry of spacetime. T_uv is the energy-momentum tensor, and acts as a source term. Any nonzero energy-momentum density is implied, and this includes EM radiation.

 

Explicitly, the EM energy-momentum tensor is:

 

T_uv=(1/4:pi: )(F_urF_v^r-(1/4)g_uvF_rsF^rs)

 

The important thing to note is that it is nonzero.

 

edit: Changed from Greek to Roman indices to make equation more readable. They still stand for Lorentz indices (0,1,2,3).

 

Tom

[Raider]

That is because the effect of it is minimal, edward. Kind of like you having gravity.

[Radical Edward]

I am not talking in an experimental or measured capacity, I haven't even heard of it in a theoretical capacity. Essentially, as I have heard, gravity is only present in massive bodies (bodies with mass)

 

light doesn't have mass.

[Raider]

It doesn't have rest mass. It has energy, so it does have mass.

[fafalone]

Um, no.

[Radical Edward]

I know that energy and mass are equivalent, but seriously, photons having mass? nope.

 

let us assume that they do for a second:

 

even if it wasn't rest mass, it is still mass, and hence would have to be plugged into the lorentz transforms in SR (unless you can think of a cunning reason why not). this would create a quandry, since exactly the same equation that demonstrates that only massless things can travel at c, would simultaneously demonstrate that light has mass, and therefore can't travel at c.

[fafalone]

This is another case of someone not knowing that e=mc² is really e²=(pv)² + m²c⁴ where p=hf/c (not mv).

[Radical Edward]

Originally posted by Tom

 

<quote>

 

 

you're saying photons emit gravitons?

[Tom Mattson]

Raider: "It doesn't have rest mass. It has energy, so it does have mass."

 

No, real photons are massless. If they were not, then gauge invariance could not hold, as the vector potential A would appear as a physical field in Maxwell's equations.

 

Radical Edward: "you're saying photons emit gravitons?"

 

I presented the (non-quantum) general relativity, and did not say anything about gravitions. GR says that all nonzero energy/momentum densities affect the curvature of spacetime, and thus give rise to the illusion we call "gravity" (in GR, gravity is not a force, but a geometrical effect). I don't know much about quantum gravity, but I would imagine that gravitons would have to couple to photons for the theory to be accepted. This should be clear from the way that gravity affects light (eg: light cannot escape a black hole).

 

Tom

[Radical Edward]

so, discounting diffraction, two infinitely long parallel laserbeams would bend towards one another?

[fafalone]

Is there such a thing as an anti-photon? :/

[Radical Edward]

a photon with a pi phase shift?

[Tom Mattson]

Radical Edward: "so, discounting diffraction, two infinitely long parallel laserbeams would bend towards one another?"

 

Now, you are pushing the limits of what I know (GR is not my strongest suit). On my "To Do" list is read this document:

 

http://arxiv.org/PS_cache/gr-qc/pdf/9811/9811052.pdf

 

in which they analyze the problem. I will let them speak for it.

 

fafalone: "Is there such a thing as an anti-photon? :/"

 

The photon is its own antiparticle. Antiparticle states are understood in QFT as charge conjugated states. The charge conjugation operator changes the sign of all the additive quantum numbers, but for a photon these are all zero.

 

Tom

[Tom Mattson]

If anyone wants to get started on this, let me make a quick note:

 

The energy-momentum tensor for EM:

 

T_uv=(1/4:pi: )(F_urF_v^r-(1/4)g_uvF_rsF^rs)

 

g_uv is, of course, the unknown metric for which you have to solve. But F_uv is known ahead of time. Its definition is:

 

F_uv=d_uA_v-d_vA_u, the EM field strength tensor.

 

where

 

d_u=(d/dt,-grad) is the 4-gradient

A_u=(V,A), the 4-vector potential of EM

 

For EM waves, V=0 and A should work out such that you get sinusoidal E and B waves (A is not unique, because of the gauge freedom).

 

Tom

[fafalone]

Originally posted by Tom

The charge conjugation operator changes the sign of all the additive quantum numbers, but for a photon these are all zero.

 

Isn't the spin quantum number for a photon 1?