somecallmegenius Posted November 8, 2012 Share Posted November 8, 2012 I have always wondered why the energy of a photon in vacuum is equal to E = pc (where p is the momentum of the photon, and c is the speed of light in vacuum) and not E = 1/2 pv (where for a photon v = c) as is the case for the kinetic energy of any moving mass. Of course, I understand that photons are massless, but can anyone clearly explain how E = mc^2 and not E = 1/2 mc^2 and prove that in a theoretical non-empirical way?? Link to comment Share on other sites More sharing options...

ydoaPs Posted November 8, 2012 Share Posted November 8, 2012 (edited) E^{2}=(mc^{2})^{2}+(pc)^{2} Photons are massless. Bam, E=pc edit: typo Edited November 8, 2012 by ydoaPs 1 Link to comment Share on other sites More sharing options...

somecallmegenius Posted November 8, 2012 Author Share Posted November 8, 2012 E^{2}=(mc^{2})+(pc)^{2} Photons are massless. Bam, E=pc I already knew that part. Now, as a start it would be helpful if you told me where Einstein derived that equation from and what it logically means, and no I will not accept the standard pythagorean theorem explanation, because if you are gonna go that route, you'll have to tell me why E, mc^{2}, and pc form a right-angle triangle. -1 Link to comment Share on other sites More sharing options...

imatfaal Posted November 8, 2012 Share Posted November 8, 2012 Before we move on let's correct the typo to avoid confusion [math] E^2 = (mc^2)^2 + (pc)^2 [/math] 1. [math] E^2=m^2c^4 [/math] 2. [math] p = mv [/math] [math] m^2c^4 = \frac{m_0^2c^4}{1-\frac{v^2}{c^2}} [/math] [math] m^2c^4\left(1-\frac{v^2}{c^2}\right)= m_0^2c^4 [/math] [math] m^2c^4 - \frac{m^2v^2c^4}{c^2} = m_0^2c^4 [/math] [math] m^2c^4 - (mv)^2c^2 = m_0^2c^4 [/math] sub in the equations 1 and 2 from above [math] E^2 - p^2c^2 = m_0^2c^4 [/math] [math] E^2= m_0^2c^4 + p^2c^2 [/math] 2 Link to comment Share on other sites More sharing options...

swansont Posted November 8, 2012 Share Posted November 8, 2012 I have always wondered why the energy of a photon in vacuum is equal to E = pc (where p is the momentum of the photon, and c is the speed of light in vacuum) and not E = 1/2 pv (where for a photon v = c) as is the case for the kinetic energy of any moving mass. Of course, I understand that photons are massless, but can anyone clearly explain how E = mc^2 and not E = 1/2 mc^2 and prove that in a theoretical non-empirical way?? Why should the factor of 1/2 appear in either equation? In the classical equation it appears from the integration of (mv dv), from the definition of work. 2 Link to comment Share on other sites More sharing options...

derek w Posted November 8, 2012 Share Posted November 8, 2012 as a start it would be helpful if you told me where Einstein derived that equation from and what it logically means Herman Minkowski was Einstein's tutor. Link to comment Share on other sites More sharing options...

somecallmegenius Posted November 9, 2012 Author Share Posted November 9, 2012 Why should the factor of 1/2 appear in either equation? In the classical equation it appears from the integration of (mv dv), from the definition of work. Ok, thank you, that is a good reminder of a fact which I forgot I knew. It also allows me to properly rephrase my question. What I mean is, does the definition of momentum change when going from classical to relativistic physics? I mean does the equation E = 1/2 pv, change as v approaches the speed of light (relativistically relevant speeds..)? Does the kinetic energy of a moving object slowly shift from E = 1/2 pv towards E = pv as v increases? Link to comment Share on other sites More sharing options...

swansont Posted November 9, 2012 Share Posted November 9, 2012 Ok, thank you, that is a good reminder of a fact which I forgot I knew. It also allows me to properly rephrase my question. What I mean is, does the definition of momentum change when going from classical to relativistic physics? I mean does the equation E = 1/2 pv, change as v approaches the speed of light (relativistically relevant speeds..)? Does the kinetic energy of a moving object slowly shift from E = 1/2 pv towards E = pv as v increases? The classical momentum and kinetic energy equations are the first-order approximations of the relativistic equations. If you expand them in orders of v/c, you ignore the terms where the powers of (v/c)<<1 So the definitions never change, but the approximation that gives you the simple form of the equation is no longer valid. 2 Link to comment Share on other sites More sharing options...

somecallmegenius Posted November 9, 2012 Author Share Posted November 9, 2012 The classical momentum and kinetic energy equations are the first-order approximations of the relativistic equations. If you expand them in orders of v/c, you ignore the terms where the powers of (v/c)<<1 So the definitions never change, but the approximation that gives you the simple form of the equation is no longer valid. Perfect. Thank you. The answer I was looking for. So, basically, the approximations become more and more inaccurate as the value of v/c increases. So, swansont is the real deal after all... Link to comment Share on other sites More sharing options...

swansont Posted November 9, 2012 Share Posted November 9, 2012 Perfect. Thank you. The answer I was looking for. So, basically, the approximations become more and more inaccurate as the value of v/c increases. Generally speaking, around v/c > 0.1 is where the classical approximations start to noticeably fail. The expansions I can think of are in even powers of v/c, so that's a 1% value for the (v/c)^2 term. Link to comment Share on other sites More sharing options...

juanrga Posted November 9, 2012 Share Posted November 9, 2012 I have always wondered why the energy of a photon in vacuum is equal to E = pc (where p is the momentum of the photon, and c is the speed of light in vacuum) and not E = 1/2 pv (where for a photon v = c) as is the case for the kinetic energy of any moving mass. Of course, I understand that photons are massless, but can anyone clearly explain how E = mc^2 and not E = 1/2 mc^2 and prove that in a theoretical non-empirical way?? Starting from [math]E = \sqrt{m^2c^4 + p^2c^2 }[/math] for a massless particle such as the photon set [math]m = 0[/math] and you obtain [math]E = |p|c[/math] for a non-relativistic particle expand the square root in a power series and ignore the higher order terms because [math]v \ll c[/math] [math]E = \sqrt{m^2c^4 + p^2c^2 } = mc^2 + \frac{p^2}{2m} + \cdots [/math] you obtain the 1/2 factor characteristic of the non-relativistic theory. Link to comment Share on other sites More sharing options...

elfmotat Posted November 10, 2012 Share Posted November 10, 2012 The formulas for relativistic momentum and kinetic energy are different from the approximations used in Newtonian mechanics. The difference becomes more and more drastic as v approaches c. Here's a simple way to derive E=pc starting with only the formula for relativistic momentum and a few assumptions (i.e. the Work-Energy Theorem holds valid in relativity, our final equation holds for massless particles). The formula for relativistic momentum can be found by analyzing collisions with the Lorentz transformation, and what you end up with is [math]p=\gamma mv[/math] where [math]\gamma=(1-v^2/c^2)^{-1/2}[/math]. From there you can find the formula for relativistic kinetic energy by using the Work-Energy Theorem, calculating the work done in bringing a mass at rest to a velocity v: [math]E_k=\int Fdx=\int \frac{dpdx}{dt}=\int vdp=pv-\int pdv=\gamma mv^2-m\int \frac{vdv}{\sqrt{1-v^2/c^2}}[/math] Evaluating that integral gives: [math]E_k=\gamma mc^2+\varphi [/math], where [math]\varphi[/math] is some constant of integration. It's easy to solve for [math]\varphi[/math]; all you have to do is take the fact that when v=0 (i.e. the object is at rest) its kinetic energy is taken to be zero, and [math]\gamma=1[/math]. So what you get is: [math]0=mc^2+\varphi~~~ \Rightarrow ~~~\varphi =-mc^2~~~\Rightarrow~~~ E_k=\gamma mc^2-mc^2[/math]. The term "mc^{2}" looks like some intrinsic energy associated to a mass (and is appropriately called "rest energy"). Adding this term to both sides of the kinetic energy formula gives the total energy of a body: rest energy + kinetic energy: [math]E=E_k+mc^2=\gamma mc^2[/math] Now that we have the formulas for the total relativistic energy and the relativistic momentum of a moving body, we can be a bit tricky to find some relationships between them. We can start by solving for [math]\gamma[/math] in both equations: [math]\gamma =\frac{E}{mc^2}=\frac{p}{mv}~~~\Rightarrow ~~~ Ev=pc^2[/math] So there's a neat little formula relating energy and momentum. We can also use this result to get to the equation you're looking for. Start by squaring the energy equation and rearranging: [math]E^2=\frac{(mc^2)^2}{1-v^2/c^2}~~~\Rightarrow ~~~E^2-\frac{(Ev)^2}{c^2}=(mc^2)^2[/math] Now substitute pc^{2} in for Ev: [math]E^2-\frac{(pc^2)^2}{c^2}=(mc^2)^2~~~\Rightarrow ~~~E^2-(pc)^2=(mc^2)^2[/math] As you're aware, this equation holds for all particles, including massless ones. Setting m=0 gives E=pc. Link to comment Share on other sites More sharing options...

somecallmegenius Posted November 11, 2012 Author Share Posted November 11, 2012 The formulas for relativistic momentum and kinetic energy are different from the approximations used in Newtonian mechanics. The difference becomes more and more drastic as v approaches c. Here's a simple way to derive E=pc starting with only the formula for relativistic momentum and a few assumptions (i.e. the Work-Energy Theorem holds valid in relativity, our final equation holds for massless particles). The formula for relativistic momentum can be found by analyzing collisions with the Lorentz transformation, and what you end up with is [math]p=\gamma mv[/math] where [math]\gamma=(1-v^2/c^2)^{-1/2}[/math]. From there you can find the formula for relativistic kinetic energy by using the Work-Energy Theorem, calculating the work done in bringing a mass at rest to a velocity v: [math]E_k=\int Fdx=\int \frac{dpdx}{dt}=\int vdp=pv-\int pdv=\gamma mv^2-m\int \frac{vdv}{\sqrt{1-v^2/c^2}}[/math] Evaluating that integral gives: [math]E_k=\gamma mc^2+\varphi [/math], where [math]\varphi[/math] is some constant of integration. It's easy to solve for [math]\varphi[/math]; all you have to do is take the fact that when v=0 (i.e. the object is at rest) its kinetic energy is taken to be zero, and [math]\gamma=1[/math]. So what you get is: [math]0=mc^2+\varphi~~~ \Rightarrow ~~~\varphi =-mc^2~~~\Rightarrow~~~ E_k=\gamma mc^2-mc^2[/math]. The term "mc^{2}" looks like some intrinsic energy associated to a mass (and is appropriately called "rest energy"). Adding this term to both sides of the kinetic energy formula gives the total energy of a body: rest energy + kinetic energy: [math]E=E_k+mc^2=\gamma mc^2[/math] Now that we have the formulas for the total relativistic energy and the relativistic momentum of a moving body, we can be a bit tricky to find some relationships between them. We can start by solving for [math]\gamma[/math] in both equations: [math]\gamma =\frac{E}{mc^2}=\frac{p}{mv}~~~\Rightarrow ~~~ Ev=pc^2[/math] So there's a neat little formula relating energy and momentum. We can also use this result to get to the equation you're looking for. Start by squaring the energy equation and rearranging: [math]E^2=\frac{(mc^2)^2}{1-v^2/c^2}~~~\Rightarrow ~~~E^2-\frac{(Ev)^2}{c^2}=(mc^2)^2[/math] Now substitute pc^{2} in for Ev: [math]E^2-\frac{(pc^2)^2}{c^2}=(mc^2)^2~~~\Rightarrow ~~~E^2-(pc)^2=(mc^2)^2[/math] As you're aware, this equation holds for all particles, including massless ones. Setting m=0 gives E=pc. Got it. Thank you elfmotat for the clear derivation and everyone else for your contributions. So, from the above derivation, E = mc^{2} is the energy resulting from unleashing the rest energy of a certain mass. However, that brings a question to mind, since c is the speed of light in vacuum, do we have to divide the rest energy by the index of refraction of the medium in which the conversion happens to obtain the actual resulting energy? Link to comment Share on other sites More sharing options...

swansont Posted November 11, 2012 Share Posted November 11, 2012 However, that brings a question to mind, since c is the speed of light in vacuum, do we have to divide the rest energy by the index of refraction of the medium in which the conversion happens to obtain the actual resulting energy? No Link to comment Share on other sites More sharing options...

apixy Posted May 22, 2016 Share Posted May 22, 2016 (edited) I came up with the equation, for the test in 2 days, and to verify it I searched it up and found this. I came up using this... [math] E = pc [/math] We know 2 equations that are true, and they are: [math] E=\frac{hc}{\lambda} [/math] [math]... \lambda=\frac{h}{p} [/math] Substitute and you get your equation really... [math] E=\frac{hc}{\frac{h}{p}} [/math] Which is equivalent to [math] E=\frac{hpc}{h} [/math] The 2 'h' cancel out to leave you with [math] E=pc [/math] Edited May 22, 2016 by apixy Link to comment Share on other sites More sharing options...

Markus Hanke Posted May 22, 2016 Share Posted May 22, 2016 (edited) I will not accept the standard pythagorean theorem explanation, because if you are gonna go that route, you'll have to tell me why E, mc^{2}, and pc form a right-angle triangle. Perhaps it would help to start at the most basic entity for a system in motion, being its Lagrangian. The action for a relativistically moving particle is of the form [latex]\displaystyle{S=\int \left ( -mc^2\sqrt{1-\frac{v^2}{c^2}} \right )ds}[/latex] From this, you can then derive a quantity called the 4-momentum by taking [latex]\displaystyle{p_{\mu}=-\frac{\partial S}{\partial x^{\mu}}=\left ( \frac{E}{c},-\mathbf{p} \right )}[/latex] The usual energy-momentum relation is then quite simply the norm of this 4-vector : [latex]\displaystyle{p^{\mu}p_{\mu}=\left | \mathbf{p} \right |^2-\frac{E^2}{c^2}=-m^2c^2}[/latex] So, the basic idea is that energy-momentum is a 4-vector, and the norm of that 4-vector is defined to be the proper mass of the particle. Because we are in flat Minkowski spacetime, calculating the norm of a 4-vector is equivalent to applying the Pythagorean theorem to the three elements of the equation ( proper mass, energy = time component of vector, and momentum = magnitude of spatial component of vector ). The first step ( Lagrangian to 4-momentum ) can also be made mathematically precise via the calculus of variations. Edited May 22, 2016 by Markus Hanke Link to comment Share on other sites More sharing options...

David Cartano Posted June 12, 2016 Share Posted June 12, 2016 For a photon, E = pc, where p is momentum, and momentum is mass x velocity. But since a photon has no mass, why doesn't this equation reduce down to E = 0. Does anyone know what radiation momentum is for a photon? Link to comment Share on other sites More sharing options...

swansont Posted June 12, 2016 Share Posted June 12, 2016 For a photon, E = pc, where p is momentum, and momentum is mass x velocity. But since a photon has no mass, why doesn't this equation reduce down to E = 0. Does anyone know what radiation momentum is for a photon? It's E/c. It can be deduced from E&M equations mv only applies to massive particles; it's a nonrelativistic formula Link to comment Share on other sites More sharing options...

Peter Manchester Posted January 22, 2018 Share Posted January 22, 2018 The orginal question confuses the rest energy (mc^2) of a particle with mass, with kinetic energy (1/2 mv^2). A particle with mass m at rest, and no kinetic energy, still has a rest energy derived from its mass. The wikipedia article https://en.wikipedia.org/wiki/Energy–momentum_relation is a very general discussion of the topic, covering both massive and massless particles. Link to comment Share on other sites More sharing options...

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