Original post by student438
p=mv
a photon is massless so m=0, therefore p=0?
a photon is massless so m=0, therefore p=0?
Energy mass equivalency.
Original post by j0rd4nn
well, you're truely stupid.
if p = m, then x = s which equates to the compression of t, which renders l. Now, when l = t + (s - x) / p = m.
geesh.. read a book once in a while?
if p = m, then x = s which equates to the compression of t, which renders l. Now, when l = t + (s - x) / p = m.
geesh.. read a book once in a while?
wow... yeah, i'm stupid... because i'm clearly saying p=m... rather than p=mv and p=0 because (0)v =0....
It's to do with the fact that essential everything comes down to energy.
If you takes planks relationship
E=hf and the fact that c=fl (where l is lambda = wavelength)
Rearranging gives f=c/l
Substituting into the first equation gives
E = hc/l => E/c = h/l
The units of energy/velocity reduce to kg.m/s which is momentum.
Thus for a photon of wavelength l
p = h/l
If you takes planks relationship
E=hf and the fact that c=fl (where l is lambda = wavelength)
Rearranging gives f=c/l
Substituting into the first equation gives
E = hc/l => E/c = h/l
The units of energy/velocity reduce to kg.m/s which is momentum.
Thus for a photon of wavelength l
p = h/l
momentum p= mV
photon energy is E = hf
when photon is travelling its energy is equal to kinetic energy
E= 0.5mV^2 rearrange to obtain photons mass
use mass and velocity to find photons momentum
but photon is travelling too fast so momentum formula is not very accurate
photon energy is E = hf
when photon is travelling its energy is equal to kinetic energy
E= 0.5mV^2 rearrange to obtain photons mass
use mass and velocity to find photons momentum
but photon is travelling too fast so momentum formula is not very accurate
(edited 13 years ago)
Original post by j0rd4nn
well, you're truely stupid.
if p = m, then x = s which equates to the compression of t, which renders l. Now, when l = t + (s - x) / p = m.
geesh.. read a book once in a while?
if p = m, then x = s which equates to the compression of t, which renders l. Now, when l = t + (s - x) / p = m.
geesh.. read a book once in a while?
Trololololol
OP, remember that energy and mass are interchangeable.
Original post by j0rd4nn
well, you're truely stupid.
if p = m, then x = s which equates to the compression of t, which renders l. Now, when l = t + (s - x) / p = m.
geesh.. read a book once in a while?
if p = m, then x = s which equates to the compression of t, which renders l. Now, when l = t + (s - x) / p = m.
geesh.. read a book once in a while?
What rubbish are you on about? At least troll properly. Dear me.
Original post by Shmeiks
That is correct for classical physics, but at speeds close to the speed of light it becomes obvious that the relation p=mv is just an approximation. At these aforementioned speeds it becomes apparent that something needs only energy/ velocity to have momentum.
relativistic momentum.
OK, some people have argued that p=mv is only valid for low velocities. This not true, P=mv is valid for all velocities and will give the correct value provided that the relativistic mass is used, which becomes more apparent at higher velocities.
Now, the reason that a photon has momentum is because it has energy
Energy of photon E=HC/L (let L stand for lambda, the wavelength, c the speed of light and H plank's constant),
then by using Velocity = f.L , f = frequency, you get that:
P = H/L , essentially this is because of the equivalency between energy and mass.
Now, the reason that a photon has momentum is because it has energy
Energy of photon E=HC/L (let L stand for lambda, the wavelength, c the speed of light and H plank's constant),
then by using Velocity = f.L , f = frequency, you get that:
P = H/L , essentially this is because of the equivalency between energy and mass.
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