OT:Tempted to get this book

[narf poit chez BOOM]

If light has mass and goes at lightspeed, how come we're not all squished?
Best answer I can think of is that light ignores that rule and simply exerts force according to it's mass. What that might mean, I don't know.

[Jack Simth]

narf poit chez BOOM said:
If light has mass and goes at lightspeed, how come we're not all squished?
Best answer I can think of is that light ignores that rule and simply exerts force according to it's mass. What that might mean, I don't know.

It's been a while since I've taken physics but ....
Light has 0 rest mass, and thus 0 rest energy. However, it travels at the speed of light. The equation for the energy (mass, energy; essentially the same thing in modern physics) of an object is something like E = sqrt(1/(1-(v^2/c^2)))mc^2. For light, m = 0, but v = c. The equation then translates to
E = sqrt(1/(1-(c^2/c^2)))(0)c^2
= sqrt(1/(1-1))(0)
= sqrt(1/0)(0)
= (1/0)(0)
= (0/0)
Which is mathamatically undefined. However, the universe comes up with an answer! It's a particular constant over the wavelength of the light, for any given photon (there's something similar for the inertia of the photon). Fortunately, it's a VERY SMALL constant, so it takes a LOT of photons to have any measureable impact when they hit you.

[narf poit chez BOOM]

If light has 0 rest mass and 0 mass from lightspeed, how can it have any mass at all?

[Jack Simth]

The basic energy equation:
E = sqrt(1/(1-(v^2/c^2)))mc^2.
Forget the mc^2 part for a moment, and focus on the stuff inside the sqrt
1/(1-(v^2/c^2))
Now, for light, v = c. So, we get, in the innermost set of parentheses,
c^2/c^2
which is, of course, 1.
So, we replace the innermost parenthesis with the result, and get
1/(1-1)
inside the sqrt. 1-1 = 0, so when we make that replacement, we get
1/0
which, other than being an odd comic, is mathamatically undefined. However, we do know that in the equation
y = 1/x, as x approaches 0, y approaches infinity. So, if we replace 1/0 with infinity, we get infinity inside the square root.
sqrt(infinity)
as the sqrt of infinity is still infinity, we can drop the sqrt function entierly, and the energy equation then becomes
E = infinity*mc^2
As m represents the rest mass, it's 0. If we make that replacement:
E = infinity*0*c^2.
In math, infinity * 0 isn't defined, and the c^2 isn't germain to this conversation (if you like, we can say the 0 eats it up and doesn't notice). The universe, however, makes it's own definitions, and comes up with a result for light contigent on the wavelength of the photon in question. However, that answer only applies when the photon is traveling at the speed of light. Plug any lower real velocity in there (well, any real velocity between c and -c, not inclusive of the endpoints), and the sqrt function comes up with a real answer, and the m=0 eats up everything, leaving a big fat 0 for E.

[narf poit chez BOOM]

I think I sorta got it.