# Mass of Light

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$\text{Since light has energy, there is a small amount of mass equal to that energy.}$

$Given: f=1.0\times10^{15} \text{ Hz}, h=6.626\times10^{-34} J\cdot s, c=299792458 \text{ m/s}$

$Find: \text{mass of photon } m_p$

$E=hf, E=mc^2$

$\text{Equate the 2 values of E:}$

$hf=mc^2$

$m_p=\frac{hf}{c^2}$

$=\frac{\left(6.626\times10^{-34}\right)\left(1.0\times10^{15}\right)}{\left(299792458\right)^2}$

$=7.37242\times10^{-36} kg$

$\text{Compare with mass of electron }m_e=9.10939\times10^{-31} \text{ kg}$

$m_e/m_p=123,560$

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So who is supposed to care? And why?

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First of all, that's a terribly arbitrary value for the photon's frequency. Second, what you've derived is the relativistic mass of a photon, which as far as most people are concerned is basically useless. Your answer is actually frame-dependent because the frequency changes when you transform to another reference frame. The rest mass of a photon (which is what matters) is zero.

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Since light has energy, there is a small amount of mass equal to that energy.

Not by what is meant by mass. As elfmotat says, that's relativistic mass. If you don't add the qualifier you're confusing two different definitions.

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A cold body absorbing photons will gain mass,and visa versa a hot body emitting photons will lose mass.

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Why you used so high frequency?

Shouldn't better use the smallest frequency such as 1? So E=h*1
Or reverse - find the largest wave length.

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Why you used so high frequency?

Shouldn't better use the smallest frequency such as 1? So E=h*1

Or reverse - find the largest wave length.

That's not consistent with the (flawed) framing of the problem. If you use relativistic mass, the value will be frequency dependent. (besides, the smallest value is zero)

A cold body absorbing photons will gain mass,and visa versa a hot body emitting photons will lose mass.

True, but that does not mean that photons have mass. Mass is not a conserved quantity.

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I am not implying that photons have mass,but if I drop a cannon ball it falls to earth,and if I fired a photon horizontal to the earth's surface,in the initial instance they would both accelerate towards the earth at the same rate.

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who's talking to who?

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I am not implying that photons have mass,but if I drop a cannon ball it falls to earth,and if I fired a photon horizontal to the earth's surface,in the initial instance they would both accelerate towards the earth at the same rate.

Yes, absolutely. However, the OP is speaking of photons having mass. Without further context, it could be interpreted that you are supporting the claim, because there's nothing inherent in them that debunks it.

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I thought photons do not have mass? and that photons can never be at rest....

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I thought photons do not have mass? and that photons can never be at rest....

They don't.

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It's an oxymoron: photons cannot be at rest point.

Discussing the rest mass of a photon is like discussing the color of a unicorn.

And stating that the rest mass of a photon is null is like saying the unicorns have no color (because there are no unicorns).

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It's an oxymoron: photons cannot be at rest point.

Discussing the rest mass of a photon is like discussing the color of a unicorn.

And stating that the rest mass of a photon is null is like saying the unicorns have no color (because there are no unicorns).

No. We have an equation $E^2 = p^2c^2 + m^2c^4$ and we know that for a photon, p=E/c. What can you conclude about m?

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No. We have an equation $E^2 = p^2c^2 + m^2c^4$ and we know that for a photon, p=E/c. What can you conclude about m?

You are putting the equation above all.

Can a photon be at rest?

if Yes you can apply an equation, if No, no equation can stand.

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You are putting the equation above all.

Science.

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Can a photon be at rest?

No.

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All massless particles travel at c and all particles which travel at c are massless. This follows from the math, which is where all the physics is.

A semantic argument doesn't change any of that.

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Yes or No please.

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if Yes you can apply an equation, if No, no equation can stand.

The equation swansont gave applies in all inertial frames, you do not need a rest frame for that equation to make sense.

Yes or No please.

In Minkowski space-time, there are no inertial coordinates such that a photon can be considered at rest.
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The equation swansont gave applies in all inertial frames,

E=p*c

p=E/c

Honestly this way everything can be explained..

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The equation swansont gave applies in all inertial frames, you do not need a rest frame for that equation to make sense.

In Minkowski space-time, there are no inertial coordinates such that a photon can be considered at rest.

Yes or No, please.

Besides, if not in Minkowski space-time what is the difference?

besides#2, "all inertial frames" ; does that include the inertial frame in which a photon is at rest?

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(edit) forget the besides #1 & #2. A Yes or No will do. I really don't understand you problem. ACG52 has no problem answering a so simple question.

Can a photon be at rest?

Edited by michel123456
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If you don't get another answer it's because it was answered once, but also because you're asking the wrong question*. The name you give the term is one of semantics and convenience, but it does not take precedence over the equation. The equation is the physics. Based on what we know, the value for mass which you assign to a photon is zero. For convenience, we call this rest mass, because for massive particles we want to use E=mc^2 properly, which requires that p = 0. Photons, then, are a special case, but anyone familiar with physics already knows this. For photons, there is no frame in which p = 0. For massive particles, this is the rest frame.

*The truth of the matter is that, for this particular discussion, it's not a yes or no question, even though you have disguised it as one. The real answer, in the context of the discussion, is "it doesn't effing matter". (The repetition of the question here remind me of the word games one hears on school playgrounds, the purpose of which is so that the protagonists can laugh at someone and feel better about themselves. I thought we were here to discuss physics, so that's disappointing.)

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You are correct.

It is disappointing.

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So you are arguing that I asked a wrong question.

Swansont wrote:

For photons, there is no frame in which p = 0.

IOW there is no frame in which a photon has zero mass.

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