# The mass of a photon in motion

Gold Member

## Homework Statement

Find the mass of photon in motion.

E=hf
p=hf/c

## The Attempt at a Solution

I just started studying modern physics and am aware of the above two equations. However equating E to 1/2mv^2 gives a different (wrong) answer while equating p to mv gives the correct expression when we consider v=c; is it wrong to write E=1/2 mv^2 or p=mv for a photon (something to do with relativity perhaps)?
I'd appreciate some insights, thank you

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BvU
Homework Helper
Could it be you accidentally discovered that ##E=mc^2 ## ? Or did someone provide you with the correct answer and spoiled it for you ?

By the way, there will be theoreticians who object against abusing the term 'mass' in this manner. (I'm not a theoretician )

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Gold Member
Could it be you accidentally discovered that ##E=mc^2 ## ? Or did someone provide you with the correct answer and spoiled it for you ?

By the way, there will be theoreticians who object against abusing the teerm 'mass' in this manner. (I'm not a theoretician )
Ohh! Right! I did not see it at all till you mentioned it,(I thought I wouldn't encounter this equation at all in high school). but a little fact-gathering provoked by your answer led me to the relativistic equation of E and what it implies (Although my knowledge of relativistic equations is severely limited, I know of one equation- that of dependence of mass on velocity) but it was quite illuminating.

I guess the person who wrote my textbook wasn't a theoretician either
Thank you very much for your help, entering into studying a new form of physics is wonderful :D

Gold Member
As a second thought, I'd still like to know till where KE=1/2 mv^2 is valid though

Orodruin
Staff Emeritus
Homework Helper
Gold Member
The photon is massless. That is all there is to it.

This has nothing to do with being a theoretician or not.

Krushnaraj Pandya and BvU
Orodruin
Staff Emeritus
Homework Helper
Gold Member
As a second thought, I'd still like to know till where KE=1/2 mv^2 is valid though
That depends on your experimental precision. The next term in the series expansion is proportional to mv^4/c^2. Whenever your experimental precision is sufficient to resolve that you need to take relativistic corrections into account.

Gold Member
The photon is massless. That is all there is to it.

This has nothing to do with being a theoretician or not.
What does the quantity hf/c^2 represent?

Ray Vickson
Homework Helper
Dearly Missed
Ohh! Right! I did not see it at all till you mentioned it,(I thought I wouldn't encounter this equation at all in high school). but a little fact-gathering provoked by your answer led me to the relativistic equation of E and what it implies (Although my knowledge of relativistic equations is severely limited, I know of one equation- that of dependence of mass on velocity) but it was quite illuminating.

I guess the person who wrote my textbook wasn't a theoretician either
Thank you very much for your help, entering into studying a new form of physics is wonderful :D
Nowadays we no longer speak of mass depending on velocity. That was done in the first half of the 20th century, but not anymore (at least, not by most actual physicists---but maybe still by some authors of science books for the "layman"). For particles with positive mass, we just have non-classical formulas for momentum and energy, so ##E = (1/2) m v^2## no longer applies in the relativistic world. Instead, we have ##E = m c^2 \gamma ## for the total energy of the particle, and ##K = m c^2 \gamma - m c^2 = m c^2 (\gamma - 1)## for the kinetic energy. Here,
$$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}.$$ Note that for speeds small compared to ##c## we have
$$\gamma - 1 = \frac{v^2}{2 c^2 } + \frac{3 v^4}{8 c^4} + \frac{5 v^6}{16 c^6} + \cdots \approx \frac{v^2}{2c^2}.$$
Thus, the kinetic energy is ##K \approx (1/2) m v^2## for ##|v| \ll c.##

For photons the mass is zero, and the formula ##E = m c^2 \gamma## does not hold, essentially because ##m = 0## and ##\gamma = \infty##, so the formula makes no sense at all.

BvU and Krushnaraj Pandya
Gold Member
Nowadays we no longer speak of mass depending on velocity. That was done in the first half of the 20th century, but not anymore (at least, not by most actual physicists---but maybe still by some authors of science books for the "layman"). For particles with positive mass, we just have non-classical formulas for momentum and energy, so ##E = (1/2) m v^2## no longer applies in the relativistic world. Instead, we have ##E = m c^2 \gamma ## for the total energy of the particle, and ##K = m c^2 \gamma - m c^2 = m c^2 (\gamma - 1)## for the kinetic energy. Here,
$$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}.$$ Note that for speeds small compared to ##c## we have
$$\gamma - 1 = \frac{v^2}{2 c^2 } + \frac{3 v^4}{8 c^4} + \frac{5 v^6}{16 c^6} + \cdots \approx \frac{v^2}{2c^2}.$$
Thus, the kinetic energy is ##K \approx (1/2) m v^2## for ##|v| \ll c.##

For photons the mass is zero, and the formula ##E = m c^2 \gamma## does not hold, essentially because ##m = 0## and ##\gamma = \infty##, so the formula makes no sense at all.
That clears up a lot of things, thank you very much :D

Gold Member
Nowadays we no longer speak of mass depending on velocity. That was done in the first half of the 20th century, but not anymore (at least, not by most actual physicists---but maybe still by some authors of science books for the "layman"). For particles with positive mass, we just have non-classical formulas for momentum and energy, so ##E = (1/2) m v^2## no longer applies in the relativistic world. Instead, we have ##E = m c^2 \gamma ## for the total energy of the particle, and ##K = m c^2 \gamma - m c^2 = m c^2 (\gamma - 1)## for the kinetic energy. Here,
$$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}.$$ Note that for speeds small compared to ##c## we have
$$\gamma - 1 = \frac{v^2}{2 c^2 } + \frac{3 v^4}{8 c^4} + \frac{5 v^6}{16 c^6} + \cdots \approx \frac{v^2}{2c^2}.$$
Thus, the kinetic energy is ##K \approx (1/2) m v^2## for ##|v| \ll c.##

For photons the mass is zero, and the formula ##E = m c^2 \gamma## does not hold, essentially because ##m = 0## and ##\gamma = \infty##, so the formula makes no sense at all.
What do you think the author of my textbook had in mind when he posed this problem of finding the 'mass' of a photon?

Gold Member
perhaps he meant to imply the rest mass of a photon is zero, instead of saying its mass is zero?

ZapperZ
Staff Emeritus
perhaps he meant to imply the rest mass of a photon is zero, instead of saying its mass is zero?
That makes very little sense also, because a photon is never at rest, so the "rest mass" in non-existent. It also violates Special Relativity because it implies that there exists a reference frame where the photon is at rest.

Can you type out the author, title, and publisher of your book?

Zz.

Gold Member
That makes very little sense also, because a photon is never at rest, so the "rest mass" in non-existent. It also violates Special Relativity because it implies that there exists a reference frame where the photon is at rest.

Can you type out the author, title, and publisher of your book?

Zz.
My textbook is not one with a standard aim of providing theory, it is one especially designed for an entrance called JEE Advanced. They'll often put something like this in (probably so that the reader questions it and gathers some knowledge about it?)
Its just the way competition in India is. The book is made by a private all-India coaching institute called Aakash (Interestingly, these are much deeper and better then the standard textbooks taught to us in school). It might be a deliberate question, or it may have been blissful ignorance of relativity since it is supposed to be for a high-school student perhaps

ZapperZ
Staff Emeritus
My textbook is not one with a standard aim of providing theory, it is one especially designed for an entrance called JEE Advanced. They'll often put something like this in (probably so that the reader questions it and gathers some knowledge about it?)
Its just the way competition in India is. The book is made by a private all-India coaching institute called Aakash (Interestingly, these are much deeper and better then the standard textbooks taught to us in school). It might be a deliberate question, or it may have been blissful ignorance of relativity since it is supposed to be for a high-school student perhaps
Either someone thinks that there is such a thing as "relativistic mass" for a photon, or this is a trick question in trying to get you to realize that the mass of a photon is zero.

Remember that the full relativistic equation is

E2 = (pc)2 + (mc2)2

For light, m = 0, so all you are left with is

E = pc

We have several videos and articles already written about this in this forum. One cannot use E = mc2 to argue that this is the "relativistic mass" of a photon, because then the equation above will carry more energy than there is.

I don't know if you can ask an instructor to clarify this, but someone may need to determine what is being tested here.

Zz.

Gold Member
Either someone thinks that there is such a thing as "relativistic mass" for a photon, or this is a trick question in trying to get you to realize that the mass of a photon is zero.

Remember that the full relativistic equation is

E2 = (pc)2 + (mc2)2

For light, m = 0, so all you are left with is

E = pc

We have several videos and articles already written about this in this forum. One cannot use E = mc2 to argue that this is the "relativistic mass" of a photon, because then the equation above will carry more energy than there is.

I don't know if you can ask an instructor to clarify this, but someone may need to determine what is being tested here.

Zz.
The exact question is "Find the mass of photon in motion"
It could be that my textbook is plain wrong

ZapperZ
Staff Emeritus
The exact question is "Find the mass of photon in motion"
It could be that my textbook is plain wrong
Like I said, this is someone's idea of a "relativistic mass", which is still incorrect.

A lot of people also think that since it has a "momentum", then it must have this mass, since, "p=mv". This is not valid because the definition of "momentum" in this case does not need to include a mass, because I can also write momentum as "p=hk". We use this for light and also for "crystal momentum" in solid state physics.

You should still ask an instructor if you are able to. This is something that they should not perpetuate, even with an unofficial, make-shift text that people like you are learning from. Tell your instructor that even Albert Einstein stopped using the term "relativistic mass" eventually when he realized how misleading and inappropriate that name is.

Zz.

BvU and Orodruin
Orodruin
Staff Emeritus
Homework Helper
Gold Member
What year is your textbook from?

I find that this usage is prevalent in a lot of older texts and also (unfortunately) in texts by people who have learned from those, never had to work further with the material, and then teach it as they were taught without any realisation about how the term is actually used in modern physics language (by "modern" I here mean actual modern physics, not the typical "modern physics" courses that you will get at some universities, which is about 100 years old physics).

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Tell your instructor that even Albert Einstein stopped using the term "relativistic mass" eventually when he realized how misleading and inappropriate that name is.
Also, send your instructor here to discuss the issue with us if (s)he persists.

Gold Member
Like I said, this is someone's idea of a "relativistic mass", which is still incorrect.

A lot of people also think that since it has a "momentum", then it must have this mass, since, "p=mv". This is not valid because the definition of "momentum" in this case does not need to include a mass, because I can also write momentum as "p=hk". We use this for light and also for "crystal momentum" in solid state physics.

You should still ask an instructor if you are able to. This is something that they should not perpetuate, even with an unofficial, make-shift text that people like you are learning from. Tell your instructor that even Albert Einstein stopped using the term "relativistic mass" eventually when he realized how misleading and inappropriate that name is.

Zz.
Ah! Right, its finally very clear in my head.
I don't have an instructor- I learn these things myself from books. Whenever in doubt regarding the accuracy of a sentence, I post here- and in that way I have the best instructors one could ask for :)
(while still being able to learn at my own pace, enjoying every equation in the book instead of along with a classroom teacher.)
Another thing is, this was just a trivial question at the back of the book- hardly related to the main chapter's material (just for extra knowledge I presume) and most students probably overlooked it since they didn't want to get mixed up with relativity too early but I couldn't help it.

Gold Member
What year is your textbook from?

I find that this usage is prevalent in a lot of older texts and also (unfortunately) in texts by people who have learned from those, never had to work further with the material, and then teach it as they were taught without any realisation about how the term is actually used in modern physics language (by "modern" I here mean actual modern physics, not the typical "modern physics" courses that you will get at some universities, which is about 100 years old physics).
It is certainly old enough I suppose, but the editions are revised every 6 months. I don't think they will change it in any case, and I have no way of telling the author either. And yes, we are basically studying 100 years old 'modern' physics...that's just life in high school in India

Gold Member
Also, send your instructor here to discuss the issue with us if (s)he persists.
Our syllabus is quite vast. No instructor is bothered about teaching the extra things given in a book, I just look over things that are doubtful or seem wrong in my book and ask the qualified people of PF to kindly solve my doubts :D

Gold Member
Ok,so this is what my book explicitly says- "Although a photon has no rest mass, it possesses the inertial mass m=hf/c^2, since light is deflected by gravitational field, so it is naturally assured that photons have same gravitational behavior as other particles..."

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Ok,so this is what my book explicitly says- "Although a photon has no rest mass, it possesses the inertial mass m=hf/c^2, since light is deflected by gravitational field, so it is naturally assured that photons have same gravitational behavior as other particles..."
I would throw away that book.

davenn and Krushnaraj Pandya
Gold Member
I would throw away that book.
Hahaha, alright. I'll buy a newer, more accurate book then. Thank you very much :D