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my other geek question for the day is how long can you hang out in the NMR room before the machine sucks the iron in your hemoglobin through your skin?

Posted
It's not yet clear whether protons decay, but if they do, it's at such a slow rate that you won't need to worry about it, Dru. tongue.gif

 

hmmm.... all the protons were created in the big bang so they are all the same age

 

what if they all hit the best before date at the same time, and get yanked?

Posted

All right dudes, I'm gonna make one serious post here, and then I'm gonna go help put an experiment together.

 

Yo Iain, you are right, speed affects the rate at which time passes for an observer, that is special relativity. In a relativistic theory of gravity, it is logically necessary that the passage of time is also affected by local spacetime geometry, which in turn is affected by the distribution of mass and energy. The obvious example of this phenomenon is gravitational redshift.

 

Wirlwind, we can measure photons because they have momentum and energy. By mass, we mean rest mass, and photons don't have any rest mass. They don't exist at rest. Now, electromagnetic fields are made of photons, and the fields carry energy, which in turn affects the geometry of spacetime, so they have all the privileges of a massive object, except for one crucial thing-there is no photon at rest. A photon at rest is a nothing, which does not have mass. You could say something cool, like 'Photons travel along null geodesics, objects with mass propagate through spacetime on time-like geodesics.' An object with mass can never go the speed of light. Light can propagate at a net speed less than the speed of light by scattering off a the atoms of a material, which is why refraction happens.

 

Just to be extra confusing and start another three pages of misinformed spray about physics, I'll mention Cerenkov radiation, which is what happens when something is moving faster than the speed of light in a given material. This is one way we detect neutrinos. The neutrinos are moving faster than the local speed of light (but still less than the speed of light in vacuum), and they radiate energy away until this is no longer true.

 

Sorry to be so serious. I'm grumpy because I kept waking up during Nuclear Astrophysics today, so I neither learned from the lecture nor got a good nap.

 

mushsmile.gifbigdrink.gifwazzup.giffruit.gif

Posted
We can measure the energy of a photon, which is just hv. And we know that m=E/c^2. Thus the "mass" of a photon is equivalent to hv/c^2, but it is rather meaningless, isn't it?

But that's just pointless. Take a 1kg rock and have it move at .5c. Now it's rest mass is greater than 1kg? Energy doesn't imply (rest) mass.

Posted

if photons don't exist at rest then of cource there would be no rest mass thats like saying the force of gravity is not a force because it does not exist when the earth stops rotating.

oh and how can nothing have momentum and energy

and isn't energy the transfer of electrons, and electrons have mass and also don't exist at rest

Posted

But that's just pointless. Take a 1kg rock and have it move at .5c. Now it's rest mass is greater than 1kg? Energy doesn't imply (rest) mass.

 

Hmmmmm young cracked u still suck.... how is moving at 0.5c at REST? the_finger.gif

Posted
We can measure the energy of a photon, which is just hv. And we know that m=E/c^2. Thus the "mass" of a photon is equivalent to hv/c^2, but it is rather meaningless, isn't it?

 

No, it's not meaningless. You're misapplying the formula. E=m*c^2 is the formula for rest mass energy. To get the rest mass of the photon, first measure its energy. Now, that was in your rest frame, in which the light was moving at c. Lorentz boost the light's energy-momentum 4-vector to the frame in which the light is at rest, and it's energy and momentum will be ZERO. The scalar magnitude of the E-p 4-vector E^2-p^2=m^2 is always zero.

 

We can reaffirm if we compare the photon's quantum mechanical energy and momentum.

p=h/lamda

E=h*f

f*lamda=c implies E=h*c/lamda

Now check the invariant Lorentz product

E^2-(p*c)^2=(h*c/lamda)^2-(h*c/lamda)^2=0

I was ignoring c's in the first paragraph above this. Forgive me, it's a very convenient habit to use c=1.

 

You were just trolling, right? fruit.gif

Posted

But that's just pointless. Take a 1kg rock and have it move at .5c. Now it's rest mass is greater than 1kg? Energy doesn't imply (rest) mass.

 

Hmmmmm young cracked u still suck.... how is moving at 0.5c at REST? the_finger.gif

It isn't...and that's the flaw in CBS's argument. Sky explained it better than me, though, but that's to be expected. He's more of a Geek_em8.gif than me, apparently.

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