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informtohagrid
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I have a doubt...energy of light is nothing but kinetic energy of light which is 1/2mc^2..but we know that E=mc^2..so where does the 1/2 gone?
The formula for kinetic energy is different in relativity than in Newtonian physics, instead of (1/2)mv^2 it's [tex](\gamma - 1)mc^2[/tex], where [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]. Also, the general formula for total energy of a particle with nonzero momentum is actually [tex]E^2 = m^2 c^4 + p^2 c^2[/tex], where m is the rest mass and p is the momentum...for an object at rest this reduces to E = mc^2, for a photon m=0 so it reduces to E = pc. Also, the general formula is equivalent to the equation [tex]E = \gamma mc^2[/tex] as long as the velocity v in [tex]\gamma[/tex] is less than c (I can show the algebra for transforming the first equation into this one if you like), so if you define the "relativistic mass" M as [tex]\gamma m[/tex] then this can be written as E = Mc^2. And note that since the general formula is [tex]E = \gamma mc^2[/tex], and kinetic energy is defined as [tex]E = (\gamma - 1)mc^2[/tex], then the total energy is just the kinetic energy plus the rest energy mc^2.informtohagrid said:I have a doubt...energy of light is nothing but kinetic energy of light which is 1/2mc^2..but we know that E=mc^2..so where does the 1/2 gone?
JesseM said:The formula for kinetic energy is different in relativity than in Newtonian physics, instead of (1/2)mv^2 it's [tex](\gamma - 1)mc^2[/tex], where [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]. Also, the general formula for total energy of a particle with nonzero momentum is actually [tex]E^2 = m^2 c^4 + p^2 c^2[/tex], where m is the rest mass and p is the momentum...for an object at rest this reduces to E = mc^2, for a photon m=0 so it reduces to E = pc. Also, the general formula is equivalent to the equation [tex]E = \gamma mc^2[/tex] as long as the velocity v in [tex]\gamma[/tex] is less than c (I can show the algebra for transforming the first equation into this one if you like), so if you define the "relativistic mass" M as [tex]\gamma m[/tex] then this can be written as E = Mc^2. And note that since the general formula is [tex]E = \gamma mc^2[/tex], and kinetic energy is defined as [tex]E = (\gamma - 1)mc^2[/tex], then the total energy is just the kinetic energy plus the rest energy mc^2.
This page says it was Einstein who derived it:informtohagrid said:Hi JesseM..i know the derivation of E=mc2 by Einstein..but how to get the derivation of E^2 = m^2 c^4 + p^2 c^2 ...and who derived it?
I don't know the exact method Einstein used to derive it or what paper it was, but you should find a derivation in any relativity textbook, and there are also two derivations on this page, look at the paragraphs above and below the equation labeled (3) about halfway down the page--that equation is equivalent to the equation for energy above (note that the author sets c=1 so it doesn't appear in the equations). Also, this paper lists 3 different approaches used in textbooks for deriving the expressions for momentum and kinetic energy in relativity, and adds a fourth one of the authors' own invention--remember that total energy is just rest energy + kinetic energy, so if you know the rest energy is mc^2 and you know the kinetic energy is [tex](\gamma - 1)mc^2[/tex], then that means the total energy must be [tex]\gamma mc^2[/tex] which a little algebra can show is equal to [tex]\sqrt{m^2 c^4 + p^2 c^2}[/tex] where [tex]p = \gamma m v[/tex].Einstein also showed that the correct relativistic expression for the energy of a particle of mass m with momentum p is E^2 = m^2 c^4 + p^2 c^2. This is a key equation for any real particle, giving the relationship between its energy (E), momentum ( p), and its rest mass (m).
Since the total energy of a particle in SR is just rest energy + kinetic energy, and a photon has zero rest energy, its kinetic energy is equal to its total energy E = pc.informtohagrid said:Thank you Neutrino...I understand that light has no rest mass but it has relativistic mass and momentum...doesn't light have kinetic energy?...do you mean since there is no rest mass it has no kinetic energy also.is this correct...
The total inertial energy, E of any particle is the sum of its kinetic energy K and its rest energy E0. Thereforeinformtohagrid said:Thank you Neutrino...I understand that light has no rest mass but it has relativistic mass and momentum...doesn't light have kinetic energy?...do you mean since there is no rest mass it has no kinetic energy also.is this correct...
The equation 1/2mc^2 is the formula for calculating the energy of a photon, which is a fundamental particle of light. This equation is important because it helps us understand the relationship between mass and energy, and how light can exhibit both wave-like and particle-like properties.
Einstein's theory of special relativity, which includes the equation E=mc^2, revolutionized our understanding of light energy. This theory explains that mass and energy are interchangeable, and that the speed of light is the maximum speed at which energy can travel.
Light exhibits properties of both a wave and a particle, a phenomenon known as wave-particle duality. This means that light can behave like a wave, with characteristics such as interference and diffraction, but it can also behave like a particle, with characteristics such as momentum and energy. This concept is essential in understanding the energy of light.
The energy of light is essential for all life on Earth. It provides us with warmth and light from the sun, allows us to see, and is used in various technologies such as solar panels and lasers. It also plays a crucial role in many scientific fields, from astronomy to quantum mechanics.
Understanding the energy of light has led to numerous technological advancements, such as solar energy and fiber optics. In the future, it may also lead to new discoveries and developments in fields such as quantum computing and communication, as well as advancements in medical imaging and treatments.