Hi!! The other day I was reading the wave theory and read about the photon. I found it very interesting, as I found out that the rest mass of photon \[m_{0}\] = 0

Also. the speed u ,of photons = speed c, of light.

Substituting u in the above realtion, we get the denominator as 0. As the rest mass is also 0. Therefore, mass of photon = \[\frac{0}{0}\] = infinite

which is impossible so either the relation is wrong or the speed or something else.

Am I correct?? or there is something i have understood wrong???

Thank You

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TopNewestThis equation only applies if \(m_0>0\), and thereby \(v<c\). Objects with mass can never travel at the speed of light, since that would require an infinite energy. – Mattias Olla · 4 years, 3 months ago

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You understood this wrong: 0/0 is not = infinite 0/0 is undefined. – Djordje Marjanovic · 4 years, 3 months ago

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– Pranjal Rs · 4 years, 3 months ago

you know according to me 0/0 can be any real number. To understand first let us take some more examples what does 6/2 = 3 mean? it means how many times 2 makes 6 i.e., 3 times two makes 6. So, 0/0 means 0 multiplied by what makes 0. Hence 0 multiplied by any real number = 0. Therefore acc. to me 0/0 = R = Complex number ( as real no.s are all complex no.s)Log in to reply

– Sreehari Vp · 4 years, 3 months ago

6/2=3 a unique solution. but 0/0 can be any value.thats why it is undefined.Log in to reply

– Vamsi Krishna Appili · 4 years, 3 months ago

Practically...In nature ,...... Nothing is left Undefined by natureLog in to reply

– Clarence Chew · 4 years, 3 months ago

Such as \(\lim_{x \rightarrow 0} \frac{x}{x} = 1\). Here , we have defined the zeros by something else, making sense to \(\frac{0}{0}\).Log in to reply

No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2. This equation often enters theoretical work in X-ray and Gamma-ray astrophysics, for example in Compton scattering where photons are treated as particles colliding with electrons. – King Vii · 4 years, 3 months ago

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No, photons do not have mass, but they do have momentum. The proper, general equation to use is E^2 = m^2c^4 + p^2c^2 So in the case of a photon, m=0 so E = p*c or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc^2. This equation often enters theoretical work in X-ray and Gamma-ray astrophysics, for example in Compton scattering where photons are treated as particles colliding with electrons.

( well , an astrophysicist says this...) – Sudha Parimala · 4 years, 3 months ago

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– Pranjal Rs · 4 years, 3 months ago

Thanx for the comment. I think I will get the anser after my graduation.Log in to reply

first of all 0/0 is not infinite............. actually it is wrong to say 0/0 is some very huge number......... well as a matter of fact there infinite real numbers between any two numbers. say between 0 and 1. so the mass of light can be a very small number too..... so 0/0 is not infinity it is some number but is not fixed. – Karthik.Ps Sharma · 4 years, 3 months ago

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who says the rest mass of photon is zero?have you ever noticed a photon at rest w.r.t you? – Sreehari Vp · 4 years, 3 months ago

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– Pranjal Rs · 4 years, 3 months ago

i think its wikipedia and even my teacher who says it is zero.Log in to reply

– Sreehari Vp · 4 years, 3 months ago

how they calculate?Log in to reply

– Pranjal Rs · 4 years, 3 months ago

i don know that , but i ll tell u after i do nuclear physics from IIT(not joking)Log in to reply

– Sreehari Vp · 4 years, 3 months ago

are you lokking for IIT. great.........Log in to reply

– Pranjal Rs · 4 years, 3 months ago

Yup!! i need to do my graduation from there. What about you?? whats ur plan??Log in to reply

Are you sure this relation is valid for photons?? I think it is true for all other particles but for a photon it is always better to equate E=hc/lambda with E=mc^2 (lambda is wavelength of the EM radiation/wave) to get the relation for mass. – Raghav Chaudhary · 4 years, 3 months ago

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– Adrian Falcone · 4 years, 3 months ago

You're correct. The equation in the original post only applies to particles with positive, non-zero rest mass.Log in to reply

According to my Field-Light theory......

When the Electron Losses a part of its mass (in emission of Light .. as electron goes from Ground state to Excited state ) ....and the Emitted mass will no longer have Charge (i.e., no Negative charge)... and thus called Photon ....

Photon = Charge Disabled Electron

None could Actually see a photon as it fastly Exists from this Universe(fixed Time rated Universe) in the Multiple collisions..

When I try to Calculate photon mass ...I got the Answer "ZERO" but I could Give dimensions of photon (Approximate)

Although Photon has some physical Dimensions...It has no mass.. (as a Fact ..we cant slow down Photon to Calculate mass....but we can Fasten Photon Speed)

I will give you the Dimensions of photon with in an hour – Vamsi Krishna Appili · 4 years, 3 months ago

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MAY BE YOU ARE CORRECT , SINCE THIS EINSTIEN THEORY OF RELATIVITY IS A BIG MYSTERY MANY SCIENTIST DO NOT ACCEPT IT YOU ARE ALSO ONE OF THIS . BUT PHOTON MAY BE HAVE INFINITE MASS SINCE WE HAVE NOT SO SOPHISTICATED EQUIPMENT TO CALCULATE MOVING MASS BUT YOU ARE WRONG ATLAST SINCE YOU THINK THAT PHOTON HAVE EQUAL SPEED AS LIGHT BUT IT IS NOT POSSIBLE ACCORDING TO EINSTIEN POSTULATE . FOR SIMPLICITY IN OTHER EQUATION WE TAKE U=C – Zahid Shekh Mohammed · 4 years, 3 months ago

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– Adrian Falcone · 4 years, 3 months ago

Not sure what you mean when you say that the theory of relativity is a big mystery. There are a lot of different experiments that verify both special and general relativity (muon lifetime, gravitational bending of starlight, minor adjustments to GPS satellites due to gravitational drag, etc). There's definitely a chance it could be inaccurate or incorrect, but I wouldn't go so far as saying "many scientist do not accept it".Log in to reply