Photons Do Not Have A Mass
- Thread starter Reiku
- Start date
Are you or are you not claiming that photons have mass?
It's a simple yes or no question that an honorable gentleman should be able to easily answer, directly.
Photons have mass. Yes or no?
I don't know. The belief seems to have been, for some time, that a packet of energy will have the properties of the mass that it is equivalent to. It has weight, inertia, momentum, and gravity. I can't claim that it "has mass" except through the matter-energy equivalence which makes that point moot anyway.
Ok then.
So what we are saying is that while a photon has energy, that energy is not equal to a given mass. That energy, if converted into mass, would yield a mass of E/c^2.
Yes?
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No, it might be equal to m*c^2
There's not a lot of difference between the energy being equal to a given mass and it being equal to the energy that the mass might be converted to. I don't have a good handle on how to say what the difference is or whether there is a difference.
My mistake. See my edit above.
That's an awful lot of heming and hawing around rather than giving the correct answer: a photon has NO rest mass but DOES have relativistic mass. It's as simple as that.
MetaKron:
Let's look at that derivation closely:
They take
$$E=mc^2 \dots (1)$$
and
$$p=mv \dots (2)$$
They then combine them to get
$$p=\frac{E}{c^2}v$$
Then, to apply it to a photon they set v=c and conclude p = E/c for a photon, which happens to be correct.
The derivation, however, hides many mistakes, even though it gives the correct answer.
First, it assumes what is in question in this thread: that photons have mass m. That's the major mistake.
The subtler mistake, which is the one that makes this false derivation work, is the confusion between rest mass and relativistic mass.
When equation (1) is applied to a moving object, the equation only holds if m is the relativistic mass, and not the rest mass. In fact, the rest mass of photon is zero.
Equation (2) suffers from exactly the same problem: to hold relativistically, m must be taken in that equation to be the relativistic mass, and not the rest mass.
Since the m in all equations here is the relativistic mass, E is necessarily the total relativistic energy, and not the rest energy of the moving particle.
The final result is, of course, correct, but let's seriously try to apply one of the equations properly to a photon for a minute.
Consider p=mv. m, as we have established, is the relativistic mass. It's value is
$$m=\gamma m_0 = \frac{m_0}{\sqrt{1-(v/c)^2}}$$
where $$m_0$$ is the rest mass.
For the photon, $$m_0=0$$, but the denominator of the fraction in the last expression above is also zero. So what is m for the photon? Answer: It is undetermined by this equation.
In other words, the equation p=mv is useless for photons, as I said before.
Do you agree, MetaKron?
Let's look at that derivation closely:
They take
$$E=mc^2 \dots (1)$$
and
$$p=mv \dots (2)$$
They then combine them to get
$$p=\frac{E}{c^2}v$$
Then, to apply it to a photon they set v=c and conclude p = E/c for a photon, which happens to be correct.
The derivation, however, hides many mistakes, even though it gives the correct answer.
First, it assumes what is in question in this thread: that photons have mass m. That's the major mistake.
The subtler mistake, which is the one that makes this false derivation work, is the confusion between rest mass and relativistic mass.
When equation (1) is applied to a moving object, the equation only holds if m is the relativistic mass, and not the rest mass. In fact, the rest mass of photon is zero.
Equation (2) suffers from exactly the same problem: to hold relativistically, m must be taken in that equation to be the relativistic mass, and not the rest mass.
Since the m in all equations here is the relativistic mass, E is necessarily the total relativistic energy, and not the rest energy of the moving particle.
The final result is, of course, correct, but let's seriously try to apply one of the equations properly to a photon for a minute.
Consider p=mv. m, as we have established, is the relativistic mass. It's value is
$$m=\gamma m_0 = \frac{m_0}{\sqrt{1-(v/c)^2}}$$
where $$m_0$$ is the rest mass.
For the photon, $$m_0=0$$, but the denominator of the fraction in the last expression above is also zero. So what is m for the photon? Answer: It is undetermined by this equation.
In other words, the equation p=mv is useless for photons, as I said before.
Do you agree, MetaKron?
I didn't say anything about a rest mass, and the mass of a photon is not relativistic. It is the mass equivalent of the energy of the packet that was emitted by whatever emitted it. It is like a unit of kinetic energy that, by Planck's laws, forms a sort of particle that has a certain predictable size. Science has found out by experiment that photons are actually stretchy under common conditions, so when we talk about the wavelength of a photon we are talking about the wavelength of a photon in free or relatively free space. The "h" from the PhysicsLAB link was Planck's constant. Everyone knows that.
A photon becomes larger and larger and more diffuse the less energy that it has, and its mass equivalent and its density approach zero as its wavelength or packet size increases. It's not the same photon anyway. We're talking about the mass of a photon with energy E or momentum p. If we know its energy or its momentum we can derive the other because the velocity is always around 300,000 kilometers per second, give or take, and if we want to be precise we use whatever the speed of light is these days.
Since I was not talking about any rest mass for the photon, I am still in line with accepted 20th century science. Since you haven't seen the actual error in the derivations of the mass equivalent of the photon, I'm still a bit ahead of the curve here. AFAIK no one else has seen it either. It is actually quite obvious.
I'm talking about the "relativistic" mass if that is what you want to call it. It's the mass equivalent to the energy that is in the packet that we call the photon.
A photon becomes larger and larger and more diffuse the less energy that it has, and its mass equivalent and its density approach zero as its wavelength or packet size increases. It's not the same photon anyway. We're talking about the mass of a photon with energy E or momentum p. If we know its energy or its momentum we can derive the other because the velocity is always around 300,000 kilometers per second, give or take, and if we want to be precise we use whatever the speed of light is these days.
Since I was not talking about any rest mass for the photon, I am still in line with accepted 20th century science. Since you haven't seen the actual error in the derivations of the mass equivalent of the photon, I'm still a bit ahead of the curve here. AFAIK no one else has seen it either. It is actually quite obvious.
I'm talking about the "relativistic" mass if that is what you want to call it. It's the mass equivalent to the energy that is in the packet that we call the photon.
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The problem is partly due to the way Physics 'went'; we discovered inertia, and explained the physical properties of the world in terms of matter and heat (energy). It's still taught this way, the dynamics and kinematics are taught first; it should be the quantum stuff, or at least the nature of light, first.
Because we're addicted to mass, we get all perplexed about something that "doesn't have" any, and because we've been taught all the Newtonian stuff, so expect to see "mass" in some equation. The photon has zero for the value of this term, so its safe to ignore it, people.
But the photon does have momentum, as a "result" of what the electron does (changes its momentum). The photon is a momentum-change "message" that electrons send to each other. This can be described more succinctly with a different language (QM).
Because we're addicted to mass, we get all perplexed about something that "doesn't have" any, and because we've been taught all the Newtonian stuff, so expect to see "mass" in some equation. The photon has zero for the value of this term, so its safe to ignore it, people.
But the photon does have momentum, as a "result" of what the electron does (changes its momentum). The photon is a momentum-change "message" that electrons send to each other. This can be described more succinctly with a different language (QM).
The relationship between momentum and kinetic energy is tricky, but either way you have to have an equivalent of mass somewhere in there. It doesn't have to be "rest mass" because momentum and mass are forms of the same thing, just like a nail and a ball bearing are forms of the same thing if they have the same chemical composition. Once you have momentum you have an equivalent of mass to plug into the equations. That is, if you have the velocity. And then you also have the kinetic energy.
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MetaKron:
I have a problem with statements like "Momentum and mass are forms of the same thing".
They aren't.
If momentum and mass and energy really were all interchangeable, then physicists would use one concept and not three.
A lot of people think that the equation $$E=mc^2$$ says that mass and energy are the same thing. What it says it that there is an amount of energy associated with a mass, or that it is possible to convert a given mass to a certain amount of energy (or vice versa). It certain does not say that mass and energy are the same thing.
I think a lot of pop-science books are a bit sloppy in talking about things like "Einstein's demonstration of the equivalence of mass and energy" and similar. Mass and energy are closely related, but not "equivalent" in the sense that they are the same.
A similar argument, based on another equation that relates two variables and a constant of nature, would be the argument that energy is the same as frequency for light because E=hf for a photon. Just like in the case of energy and mass, the physical pictures of a photon's energy and its frequency are quite different.
I have a problem with statements like "Momentum and mass are forms of the same thing".
They aren't.
If momentum and mass and energy really were all interchangeable, then physicists would use one concept and not three.
A lot of people think that the equation $$E=mc^2$$ says that mass and energy are the same thing. What it says it that there is an amount of energy associated with a mass, or that it is possible to convert a given mass to a certain amount of energy (or vice versa). It certain does not say that mass and energy are the same thing.
I think a lot of pop-science books are a bit sloppy in talking about things like "Einstein's demonstration of the equivalence of mass and energy" and similar. Mass and energy are closely related, but not "equivalent" in the sense that they are the same.
A similar argument, based on another equation that relates two variables and a constant of nature, would be the argument that energy is the same as frequency for light because E=hf for a photon. Just like in the case of energy and mass, the physical pictures of a photon's energy and its frequency are quite different.
Sciencelovah
Registered Senior Member
There are reasons why physics and math are integrated in the same subforum..
In my opinion, one of the reasons is that they have to support each other.
I am having my late lunch and dont mean to post anything here but this
thread crack me up.. somebody helps me :bawl:
Sorry about that Metakron, but reaaaaaallyy..... !
In my opinion, one of the reasons is that they have to support each other.
I am having my late lunch and dont mean to post anything here but this
thread crack me up.. somebody helps me :bawl:
Sorry about that Metakron, but reaaaaaallyy..... !
Sciencelovah
Registered Senior Member
Metakron, Reiku might have produce great work, but it is of no use if nobody
understand them. To be understood, he has to use the same language (physics
and math languages with the same scientific agreement)
As far as I know (I don't have college physics background), when we discuss
exact science like physics, we have to always refers to its exact definition,
which is normally:
In non exact term, one may associate energy with heat, and heat with
temperature, just like Reiku did in this thread. That is understandable, but
scientifically is wrong. One should always look up each term to each of its
definitions, and the easiest way to check them is through their dimensions. If
energy is temperature, then what is $$25 kJ + 25^oC$$
$$50 kJ^oC$$? certainly not! :shrug:
The same applied when you equate mass with momentum. How much is 25 kg + 10 kg m/s?
35 kg?
35 m/s?
35 kg m/s?
We can't sum them up simply because they are not equal and therefore should
not be equated.
2 cats + 3 dogs are not 5 cats dogs. :shrug:
On the other hand, when one discuss about physics, one may not simply look
up at mathematical formula only. One should always check its physical context.
It is true that momentum mathematically is defined as mass of an object
times its velocity. But for which condition does it apply? For object which is at rest.
For object which is not at rest, this does not apply.
I will re-write what has many poster actually been said.
What is known as mass(m) in $$p = mv$$, is a relative mass ($$m_r_e_l$$):
$$p = m_r_e_l.v$$
When the object is at rest, its relativistic mass has a minimum value called
the "rest mass", $$m_r_e_s_t$$.
When the object has been accelerated so that it has some momentum p
and relativistic mass $$m_r_e_l$$, then its energy E turns out to be given by
$$E^2 = p^2c^2 + m_r_e_s_t^2c^4$$
You know.. when something is heavy, normally it travels slowly, right? The heavier
they are, the bigger energy is needed to transport it. And vice versa, the lighter
they are, the easier to transport. Photon is transported as fast as light velocity,
and so its mass is refered as zero. (Thats how I understand it, although I very
likely wrong).
So, given its rest mass as zero, its energy becomes:
$$E^2 = p^2c^2$$
or
$$E = pc$$
and therefore although it has no rest mass, it has momentum (p):
$$p = E/c$$
understand them. To be understood, he has to use the same language (physics
and math languages with the same scientific agreement)
As far as I know (I don't have college physics background), when we discuss
exact science like physics, we have to always refers to its exact definition,
which is normally:
- can be described as mathematical formula
- have to be dimensionally correct (either it is with dimension or dimensionless).
In non exact term, one may associate energy with heat, and heat with
temperature, just like Reiku did in this thread. That is understandable, but
scientifically is wrong. One should always look up each term to each of its
definitions, and the easiest way to check them is through their dimensions. If
energy is temperature, then what is $$25 kJ + 25^oC$$
$$50 kJ^oC$$? certainly not! :shrug: The same applied when you equate mass with momentum. How much is 25 kg + 10 kg m/s?
35 kg?
35 m/s?
35 kg m/s?
We can't sum them up simply because they are not equal and therefore should
not be equated.
2 cats + 3 dogs are not 5 cats dogs. :shrug:
On the other hand, when one discuss about physics, one may not simply look
up at mathematical formula only. One should always check its physical context.
It is true that momentum mathematically is defined as mass of an object
times its velocity. But for which condition does it apply? For object which is at rest.
For object which is not at rest, this does not apply.
I will re-write what has many poster actually been said.
What is known as mass(m) in $$p = mv$$, is a relative mass ($$m_r_e_l$$):
$$p = m_r_e_l.v$$
When the object is at rest, its relativistic mass has a minimum value called
the "rest mass", $$m_r_e_s_t$$.
When the object has been accelerated so that it has some momentum p
and relativistic mass $$m_r_e_l$$, then its energy E turns out to be given by
$$E^2 = p^2c^2 + m_r_e_s_t^2c^4$$
You know.. when something is heavy, normally it travels slowly, right? The heavier
they are, the bigger energy is needed to transport it. And vice versa, the lighter
they are, the easier to transport. Photon is transported as fast as light velocity,
and so its mass is refered as zero. (Thats how I understand it, although I very
likely wrong).
So, given its rest mass as zero, its energy becomes:
$$E^2 = p^2c^2$$
or
$$E = pc$$
and therefore although it has no rest mass, it has momentum (p):
$$p = E/c$$
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Sciencelovah
Registered Senior Member
I would also like to add, that Reiku often ignores the quote tag. When he quotes people,
he could have use the available tags, which make us (the reader) easy to separate
between his own texts and quoted texts. Sometimes just because of this simple
thing, his works is too confusing to read. :shrug:
Please take a look at this example:
Reiku's post (#11 in this thread):
http://www.sciforums.com/showpost.php?p=1656151&postcount=11
What he puts between "___" is actually a quoted text from Superluminal, 3rd paragraph:
http://www.sciforums.com/showpost.php?p=1656107&postcount=6
He must have noticed that when other people presenting quoted text, the quoted text
is appear in a grey box. Why he did not do the same? If he doesn't know, why he didn't
ask? Shall he asked, his text at least will be easier to understood.. :shrug:
he could have use the available tags, which make us (the reader) easy to separate
between his own texts and quoted texts. Sometimes just because of this simple
thing, his works is too confusing to read. :shrug:
Please take a look at this example:
Reiku's post (#11 in this thread):
http://www.sciforums.com/showpost.php?p=1656151&postcount=11
What he puts between "___" is actually a quoted text from Superluminal, 3rd paragraph:
http://www.sciforums.com/showpost.php?p=1656107&postcount=6
He must have noticed that when other people presenting quoted text, the quoted text
is appear in a grey box. Why he did not do the same? If he doesn't know, why he didn't
ask? Shall he asked, his text at least will be easier to understood.. :shrug:
Metakron, Reiku might have produce great work, but it is of no use if nobody
understand them. To be understood, he has to use the same language (physics
and math languages with the same scientific agreement)
As far as I know (I don't have college physics background), when we discuss
exact science like physics, we have to always refers to its exact definition,
which is normally:
- can be described as mathematical formula
- have to be dimensionally correct (either it is with dimension or dimensionless).
In non exact term, one may associate energy with heat, and heat with
temperature, just like Reiku did in this thread. That is understandable, but
scientifically is wrong. One should always look up each term to each of its
definitions, and the easiest way to check them is through their dimensions. If
energy is temperature, then what is $$25 kJ + 25^oC$$
The same applied when you equate mass with momentum. How much is 25 kg + 10 kg m/s?
35 kg?
35 m/s?
35 kg m/s?
We can't sum them up simply because they are not equal and therefore should
not be equated.
2 cats + 3 dogs are not 5 cats dogs. :shrug:
On the other hand, when one discuss about physics, one may not simply look
up at mathematical formula only. One should always check its physical context.
It is true that momentum mathematically is defined as mass of an object
times its velocity. But for which condition does it apply? For object which is at rest.
For object which is not at rest, this does not apply.
I will re-write what has many poster actually been said.
What is known as mass(m) in $$p = mv$$, is a relative mass ($$m_r_e_l$$):
$$p = m_r_e_l.v$$
When the object is at rest, its relativistic mass has a minimum value called
the "rest mass", $$m_r_e_s_t$$.
When the object has been accelerated so that it has some momentum p
and relativistic mass $$m_r_e_l$$, then its energy E turns out to be given by
$$E^2 = p^2c^2 + m_r_e_s_t^2c^4$$
You know.. when something is heavy, normally it travels slowly, right? The heavier
they are, the bigger energy is needed to transport it. And vice versa, the lighter
they are, the easier to transport. Photon is transported as fast as light velocity,
and so its mass is refered as zero. (Thats how I understand it, although I very
likely wrong).
So, given its rest mass as zero, its energy becomes:
$$E^2 = p^2c^2$$
or
$$E = pc$$
and therefore although it has no rest mass, it has momentum (p):
$$p = E/c$$
Thank you... You are very very wise.... including a few others round here! Even James.
Photons only have mass when either:
1. Accelerated past it's own speed = imaginary mass >
1/> or during the inflationary phase.
and 2. If a photon has a mass, it need to be stationary.