The video is excellent, great information pace, no time to get board. Q. In the future, WHEN we can measure the mass of a photon, Einsteins equations will have to be modified? Is anyone publishing in this direction that I can read?
The proton mass is much larger than that of the electron, so the latter easily vibrate with the E-field of the photon and gain larger energy than the proton does
1 photon is one single sinusoide(one λ of both E and H), my deduction. I read everywhere E=hυ is energy for One Single photon. And in the right side υ is some number of [cycle/s], so E(1photon) is J.cycle. More explanatory in the equivalent formula E=hc/λ, where the wavelength(λ) is m/1cycle. Like this half of quantum energy must be in the "positive" half of the H and B sinusoidal fields, and half in the "negative" (but respecting E much bigger than H, by Free space impedance (wikipedia) E=377H). The total power of this is the Poynting vector S=EH in W/m^2 == J/sm^2.
Then, if we find out a wave of lıght length smaller than visible lıght photon vibratıon length,we strıkes or collıdes it to electron partıcle dırectly.And then what happens to material atoms ?
That depends on the material, but typically if a UV photon strikes an electron, it will impart enough energy on the electron to strip it from the atom (and thus it will ionize the atom).
Thanks your replying.My other questıon is that space time contınoum has absolute zero condıtıon,that is,0 degree Kelvin or -273 degree C,in this condition how possible does a lıght photon never freeze its vibration as material atoms does? Is it because of lıght photon has no mass and this is the cause of lıght pass through the absolute zero without any hınderence ? Thanks beforehand for yr reply.
I understand but a lambda particle has different "properties" than a photon so it seems to be counter intuitive to use the lambda symbol when talking about photons. Assuming everyone understands that a photon can act as a wavelength and a particle
Even small particles, like moving electrons have wave properties and thus a moving electron would have an associated wavelength (lambda). And thanks for the comment.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
I'm getting tired of saying how good are your lectures
We appreciate your comments.
Excellent. Excellent. Thoroughly enjoyable
Glad you enjoyed it! 🙂
The video is excellent, great information pace, no time to get board. Q. In the future, WHEN we can measure the mass of a photon, Einsteins equations will have to be modified? Is anyone publishing in this direction that I can read?
Photons do not have mass.
@@MichelvanBiezen teacher can you help me how to proof constant of physics just like planks constant and etc
The proton mass is much larger than that of the electron, so the latter easily vibrate with the E-field of the photon and gain larger energy than the proton does
1 photon is one single sinusoide(one λ of both E and H), my deduction. I read everywhere E=hυ is energy for One Single photon. And in the right side υ is some number of [cycle/s], so E(1photon) is J.cycle. More explanatory in the equivalent formula E=hc/λ, where the wavelength(λ) is m/1cycle. Like this half of quantum energy must be in the "positive" half of the H and B sinusoidal fields, and half in the "negative" (but respecting E much bigger than H, by Free space impedance (wikipedia) E=377H). The total power of this is the Poynting vector S=EH in W/m^2 == J/sm^2.
Then, if we find out a wave of lıght length smaller than visible lıght photon vibratıon length,we strıkes or collıdes it to electron partıcle dırectly.And then what happens to material atoms ?
That depends on the material, but typically if a UV photon strikes an electron, it will impart enough energy on the electron to strip it from the atom (and thus it will ionize the atom).
Thanks your replying.My other questıon is that space time contınoum has absolute zero condıtıon,that is,0 degree Kelvin or -273 degree C,in this condition how possible does a lıght photon never freeze its vibration as material atoms does? Is it because of lıght photon has no mass and this is the cause of lıght pass through the absolute zero without any hınderence ? Thanks beforehand for yr reply.
Electromagnetic radiation (photons) are not affected by temperature.
thank you sooo much
You're welcome!
Why are you using the symbol for the lambda particle when we are talking about photons? I understand we are using wavelengths but still
Lambda is not the symbol for the particle (photon), but the wavelength of the photon. Lambda is a standard symbol for the wavelength.
I understand but a lambda particle has different "properties" than a photon so it seems to be counter intuitive to use the lambda symbol when talking about photons. Assuming everyone understands that a photon can act as a wavelength and a particle
Oh, and thank you for the quick response by the way. I find everyone else dull but you however are a pleasure to watch
Even small particles, like moving electrons have wave properties and thus a moving electron would have an associated wavelength (lambda). And thanks for the comment.