Inertial mass, invariant mass and the photon?

pivoxa15

In some cases, inertial mass does not equal invariant mass? What is the relation between the two?

So the photon can have non zero inertial mass but always 0 invariant mass?

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rbj

pivoxa15 said:
In some cases, inertial mass does not equal invariant mass? What is the relation between the two?
$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where $m_0$ is the "rest mass" or "invariant mass", $m$ is the "relativistic mass" or "inertial mass" relative to some observer watching it fly by, $v$ is the speed it flies by, and $c$ is the speed of light in a vacuum.

some here are uncomfortable with relativistic mass (they refer only to invariant mass when the term "mass" is brought up) and thus with any reference to the mass of a photon (which you can get from its energy). they prefer to say that "although the photon is massless, it has momentum of:

$$p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$"

i prefer to say that the momentum of anything is

$$p = m v$$

and keep in mind that, by "mass", we mean relativistic or inertial mass.

So the photon can have non zero inertial mass but always 0 invariant mass?
$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$

even if $m>0$, when $v=c$ then $m_0$ is always zero. we assume that photons travel at the same speed of light. because their rest mass or invarinat mass is zero, some like to call photons "massless particles" (i don't)

Parlyne

The relativistic momentum formula that you cited simply does not work for photons.

$$p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ works just fine when you have a massive object; but, photons have $$m_0 = 0$$ and $$v = c$$. So, this form for the momentum gives you $$\frac{0}{0}$$, which is simply indeterminate.

A more straightforward way to address energy and momentum is to use the relativistic dispersion relation:

$$E^2 = p^2 c^2 + m_0^2 c^4$$

Here, there are no ambiguities; and, we can begin to see why rest mass is a far more powerful concept than relativistic mass. In most applications of relativity (or, at least ones where more than just time dilation is at work), problems can be solved using energy and momentum and only translating back to more familiar things like velocity after the fact. And, we can see that the dispersion relation would only become more complicated if we insisted on using relativistic mass rather than inertial.

rbj

Parlyne said:
The relativistic momentum formula that you cited simply does not work for photons.

$$p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ works just fine when you have a massive object; but, photons have $$m_0 = 0$$ and $$v = c$$.
yes, and the first follows from the second. it's better expressed as

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$

which shows that $m_0 = 0$ when $v = c$

So, this form for the momentum gives you $$\frac{0}{0}$$, which is simply indeterminate.

A more straightforward way to address energy and momentum is to use the relativistic dispersion relation:

$$E^2 = p^2 c^2 + m_0^2 c^4$$
wanna how that gets derived from

$$E = m c^2$$
and
$$p = m v$$
and

Here, there are no ambiguities;
except you have momentum and energy without mass. that can be a little confusing when

$$E = m c^2$$
and
$$p = m v$$
and
$$m = m_0 \gamma = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$?

it's easier for me to remember that than to rememeber

$$E^2 = p^2 c^2 + m_0^2 c^4$$ .

and, we can begin to see why rest mass is a far more powerful concept than relativistic mass. In most applications of relativity (or, at least ones where more than just time dilation is at work), problems can be solved using energy and momentum and only translating back to more familiar things like velocity after the fact. And, we can see that the dispersion relation would only become more complicated if we insisted on using relativistic mass rather than inertial.
relativistic mass is the inertial mass.

what dispersion relation do you mean (that gets more complicated)?

$$E^2 = p^2 c^2 + m_0^2 c^4$$

is more complicated than

$$E = m c^2$$.

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pivoxa15

I wasn't clear about what inertial mass was. You say it is relativitic mass so that makes more sense. Would you say gravitiational mass like inertial mass is relativitic mass? So there are only two types of masses in the universe, relativitic mass and invariant mass?

Just to make sure, the v in the equations is the speed of the object relative to the observer (who 'thinks' he is stationary)?

rbj said:
$$we assume that photons travel at the same speed of light. because their rest mass or invarinat mass is zero, some like to call photons "massless particles" (i don't) Which one is it 1. photons travel at speed of light therefore 0 mass (from relativity). Or 2. Phtons have 0 mass therefore travel at speed of light (from relativity). I say photons were defined to represent light so by definition they travel at speed of light. At this stage we don't know whether they have 0 invariant mass or not. We then apply relativity and find that they have 0 invariant mass no matter the inertial mass they may have. So 1. Although you seem to suggest 2. How do we know the photon have 0 invariant mass in the first place, it isn't evident from its definition is it? Last edited: Parlyne We don't know for sure that the photon is truly massless. However, experimental and observational results lead us to conclude that the photon mass is constrained by [tex]m_\gamma < 6 \times 10^{-17} \mathrm{eV} (= 1.1 \times 10^{-52} \mathrm{kg})$$.

The relativistic energy formula ($$E = \frac{m_0 c^2}{\sqrt{1-(\frac{v}{c})^2}}$$) makes it pretty clear that a massless particle can only have non-zero energy if it travels at c. So, in fact, it is most proper to say that photons travel at c because they are massless.

As for the relativistic dispersion relation, you can show that $$E^2 - p^2 c^2 = C$$, where C is a Lorentz invarient constant, just by knowing that energy and momentum together transform as a 4-vector. I will concede that you need the formulas including velocity to show that $$C = m_0^2 c^4$$. However, in doing that, there's no ambiguity in the case of massless particles, because the term $$1 - (\frac{v}{c})^2$$ appears in both the numerator and the denominator; so, it can be eliminated before chosing the mass.

The overall point I was trying to make is that the relativistic mass of a photon is not well defined in terms of rest mass and velocity; so, you have to force it, by hand, to obey whatever relations you want and, even then, you have to go backwards and define it in terms of the energy or momentum. But, if you have to do that, you lose its utility anyway.

rbj

Parlyne said:
We don't know for sure that the photon is truly massless. However, experimental and observational results lead us to conclude that the photon mass is constrained by $$m_\gamma < 6 \times 10^{-17} \mathrm{eV} (= 1.1 \times 10^{-52} \mathrm{kg})$$.
what you're writing about here is invariant mass a.k.a. rest mass. when physicists say "photons are massless" they normally mean that the rest mass is zero, but not in the case here that you mention.

there is a semantic dispute about the term "mass" always meaning the rest mass or invariant mass. i can point to college level textbooks (about 20 years old) that freely discuss the mass of photons:

$$m = \frac{E}{c^2} = \frac{h \nu}{c^2}$$

this mass does behave as passive graviational mass (photons are diverted, from a Euclidian perspective, by another mass in proximity) and active graviational mass (photons count in the energy density that curve spacetime in GR and they do so equivalently to a particle of mass shown above).

The relativistic energy formula ($$E = \frac{m_0 c^2}{\sqrt{1-(\frac{v}{c})^2}}$$) makes it pretty clear that a massless particle can only have non-zero energy if it travels at c.
now you're the one doing 0/0 division. i was not. the reason that photons have no rest mass (assuming that they fly about at the same speed in a vacuum as the propagation of E&M waves in a vacuum) is

$$m_0 = m \sqrt{1-\frac{v^2}{c^2}} = \frac{h \nu}{c^2} \sqrt{1-\frac{v^2}{c^2}} = \frac{h \nu}{c^2} \sqrt{1-\frac{c^2}{c^2}} = 0$$

pretty simple.

So, in fact, it is most proper to say that photons travel at c because they are massless.
"most proper" is your opinion. i think 0/0 division is less proper than multiplying a non-zero quantity by zero resulting in zero. the reason that photons have no rest mass is because they (ostensibly) move at the speed of $c$.

As for the relativistic dispersion relation, you can show that $$E^2 - p^2 c^2 = C$$, where C is a Lorentz invarient constant, just by knowing that energy and momentum together transform as a 4-vector.
and you can show it much more simply without 4-vectors.

I will concede that you need the formulas including velocity to show that $$C = m_0^2 c^4$$.
you only need:

$$E = m c^2$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

and

$$p = m v$$

However, in doing that, there's no ambiguity in the case of massless particles,
you mean rest mass. photons have inertial mass which is why they have momentum and energy.

because the term $$1 - (\frac{v}{c})^2$$ appears in both the numerator and the denominator; so, it can be eliminated before chosing the mass.

The overall point I was trying to make is that the relativistic mass of a photon is not well defined in terms of rest mass and velocity;
of course not, because it has no rest mass (unless that experiment above that you cite eventually shows that they travel at a speed less than $c$.

so, you have to force it, by hand, to obey whatever relations you want and, even then, you have to go backwards and define it in terms of the energy or momentum. But, if you have to do that, you lose its utility anyway.
dunno what utility i lost, but deriving

$$E^2 = (m c^2)^2 = p^2 c^2 + m_0^2 c^4$$

from

$$E = m c^2$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

and

$$p = m v$$

is pretty straightforward and valid even if it seems to be not in vogue of late.

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rbj

pivoxa15 said:
I wasn't clear about what inertial mass was. You say it is relativitic mass so that makes more sense. Would you say gravitiational mass like inertial mass is relativitic mass? So there are only two types of masses in the universe, relativitic mass and invariant mass?
i dunno how many types of mass there is. some papers discuss various different properties of mass that don't, by definition, have to be equivalent. but i think that the Equivalence Principle of GR says that they are equivalent.

Just to make sure, the v in the equations is the speed of the object relative to the observer (who 'thinks' he is stationary)?
yup.

Which one is it

1. photons travel at speed of light therefore 0 mass (from relativity).

Or

2. Phtons have 0 mass therefore travel at speed of light (from relativity).
they're equivalent. i prefer (1.) because the explanation is much more straightforward. Parlyne prefers (2.). Parlyne believes that

$$E^2 = p^2 c^2 + m_0^2 c^4$$

is more straightforward than

$$E = m c^2$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$p = m v$$

i don't think the top equation is straightforward to remember but it is straightforward to derive from the bottom three.

I say photons were defined to represent light so by definition they travel at speed of light.
we believe that they do, but there is a possibility that the photons travel ever so slightly than the speed of E&M propagation (as in "light"). but i doubt that is the case. if they do travel at ever so slightly less than $c$, then there is no correct meaning to the saying that "photons are massless".

At this stage we don't know whether they have 0 invariant mass or not. We then apply relativity and find that they have 0 invariant mass no matter the inertial mass they may have. So 1. Although you seem to suggest 2.
no, i do not. Parlyne does, but i think (1.) is the more fundamental way to look at it.

How do we know the photon have 0 invariant mass in the first place, it isn't evident from its definition is it?
i dunno how they measure the speed of photons (relative to $c$) nor what experimental result shows that they might have non-zero invariant mass (i think it only shows a top limit to the invariant mass which is so small that the difference in speeds between the particle and the wave might never be measurable).

Parlyne

rbj said:
what you're writing about here is invariant mass a.k.a. rest mass. when physicists say "photons are massless" they normally mean that the rest mass is zero, but not in the case here that you mention.
I'm not taking any sort of cases. I'm stating the experimental upper bound on the rest mass of the photon. If the photon had non-zero mass, but smaller than I cited, we wouldn't be able - at our current capabilities - to tell the difference between that and a massless photon. And if, in fact, it did have non-zero mass, it wouldn't actually travel at the "speed of light" (at least, not as the term is used in relativity).

there is a semantic dispute about the term "mass" always meaning the rest mass or invariant mass. i can point to college level textbooks (about 20 years old) that freely discuss the mass of photons:

$$m = \frac{E}{c^2} = \frac{h \nu}{c^2}$$

this mass does behave as passive graviational mass (photons are diverted, from a Euclidian perspective, by another mass in proximity) and active graviational mass (photons count in the energy density that curve spacetime in GR and they do so equivalently to a particle of mass shown above).
The deflection of photons by gravity requires no consideration of "inertial mass." We can understand this simply as a consequence of non-Euclidean geometry. Almost any path that is locally a "straigh line" (geodesic) will curve globally. I'll also point out that once GR is involved energy and momentum are only well defined locally and depend on your choice of coordinates; but, invarient quantities - like rest mass - are always well defined independent of coordinates.

now you're the one doing 0/0 division. i was not. the reason that photons have no rest mass (assuming that they fly about at the same speed in a vacuum as the propagation of E&M waves in a vacuum) is

$$m_0 = m \sqrt{1-\frac{v^2}{c^2}} = \frac{h \nu}{c^2} \sqrt{1-\frac{v^2}{c^2}} = \frac{h \nu}{c^2} \sqrt{1-\frac{c^2}{c^2}} = 0$$

pretty simple.
Actually, what I was pointing out is that 0/X = 0 for all non-zero X. All I said was that a massless particle can't have any energy unless it moves at c, in which case, the energy formula tells us nothing about its energy. It could, in fact still be 0. It could also be infinite, or anywhere in between.

My problem with your argument, is that you're confusing the relativistic "speed of light" with the speed photons travel at. Certainly, if photon actually have no rest mass, these are the same speed; but, conceptually they're quite different. Your argument seems to come down to "light move at the 'speed of light,' so $$\frac{1}{\gamma} = 0$$." But, there is nothing that a priori forces the relativistic "speed of light" to be the same as photon velocity. We could do special relativity with a much higher c and we'd still get the same fundamental results without needing to reference light. That we see the two speed to be the same is evidence that photons are massless, not the reason they are. We don't have that kind of freedom to define mass.

"most proper" is your opinion. i think 0/0 division is less proper than multiplying a non-zero quantity by zero resulting in zero. the reason that photons have no rest mass is because they (ostensibly) move at the speed of $c$.
Rest mass is an inherent property of an object. The speed it moves at can't cause its mass to take on a certain value.

and you can show it much more simply without 4-vectors.

you only need:

$$E = m c^2$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

and

$$p = m v$$
Your definition of m is still not valid for particles with 0 rest mass. The 4-vector derivation is valid for any mass, since it depends on nothing other than the geometry of space-time.

you mean rest mass. photons have inertial mass which is why they have momentum and energy.
Unless you intend to attribute mass to non-propagating electric and magnetic fields, you have a little problem here; because they can have energy and momentum too.

of course not, because it has no rest mass (unless that experiment above that you cite eventually shows that they travel at a speed less than $c$.
If it's not well defined, I just don't see how you can claim that all one needs to know are the formulas you keep citing.

rbj

Parlyne said:
I'm not taking any sort of cases. I'm stating the experimental upper bound on the rest mass of the photon.
that's the case of "rest mass". not the case of "relativistic mass".

If the photon had non-zero mass, but smaller than I cited, we wouldn't be able - at our current capabilities - to tell the difference between that and a massless photon.
you mean "rest mass". photons have mass, but (assuming they move at $c$ for any observer) no rest mass.

And if, in fact, it did have non-zero mass, it wouldn't actually travel at the "speed of light" (at least, not as the term is used in relativity).
if, by "mass", you mean rest mass, then i agree.

Actually, what I was pointing out is that 0/X = 0 for all non-zero X. All I said was that a massless particle can't have any energy unless it moves at c, in which case, the energy formula tells us nothing about its energy. It could, in fact still be 0. It could also be infinite, or anywhere in between.
the equation you depicted, yes. it has a 0/0 problem. that's what i pointed out. but we know that these photons have energy

$$E = h \nu = \hbar \omega$$

which means they have (relativistic) mass

$$m = \frac{h \nu}{c^2}$$

which is not zero.

My problem with your argument, is that you're confusing the
relativistic "speed of light" with the speed photons travel at.
no, i also separated the two concepts. can you read english? would you like me to point out what i said? however, photons travel very nearly to $c$ and if they do not travel at $c$ (which i think is unlikely, but who knows), then they have non-zero rest mass.

Certainly, if photon actually have no rest mass, these are the same speed;
or you could say the converse: if photons move at speed $c$, even if they have inertial mass, they have no rest mass.

but, conceptually they're quite different. Your argument seems to come down to "light move at the 'speed of light,' so $$\frac{1}{\gamma} = 0$$." But, there is nothing that a priori forces the relativistic "speed of light" to be the same as photon velocity.
nothing "forces" $c$ to be anything. it is a primary quantity that ultimately is only a reflection of the units we use to measure it. the experimental question is "do photons move at speed $c$?" if they do, they cannot have non-zero rest mass. nothing forces photons to move at speed $c$, but they appear to do that to such a degree experimenters cannot tell the difference in speeds.

We could do special relativity with a much higher c and we'd still get the same fundamental results without needing to reference light.
$c$ being a dimensionful quantity, there is no meaning to that statement. there is no special relativity at any speed other than $c$. we can take $c$, $G$, $\hbar$, and $4 \pi \epsilon_0$ out of all equations of physical law (we'd be measuring everything in terms of Planck Units). there is no $c$ to change to a much higher value.

That we see the two speed to be the same is evidence that photons are massless, not the reason they are.
you mean rest mass. they're not massless. and you're also incorrect that, it's not just evidence of not having rest mass, it is the reason and i pointed out why before.

Rest mass is an inherent property of an object.
i agree with that.

The speed it moves at can't cause its mass to take on a certain value.
you keep starting your points with "rest mass" and then, by slight of hand, finish them with just "mass". The speed it moves at can't cause its rest mass or invariant mass to take on a different value. that's why they call it "invariant".

Your definition of m is still not valid for particles with 0 rest mass.
sure it is. but it's not about rest mass.

The 4-vector derivation is valid for any mass, since it depends on nothing other than the geometry of space-time.
feel free to derive it here.

I just don't see how you can claim that all one needs to know are the formulas you keep citing.
i never said that. i said

$$E^2 = p^2 c^2 + m_0^2 c^4$$

can be derived from:

$$E = m c^2$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$p = m v$$

and here is how:

$$m^2 = \frac{m_0^2}{1-\frac{v^2}{c^2}}$$

$$m^2 \left( 1 - \frac{v^2}{c^2} \right) = m_0^2$$

$$m^2 - \frac{(mv)^2}{c^2} = m^2 - \frac{p^2}{c^2} = m_0^2$$

$$m^2 = \frac{p^2}{c^2} + m_0^2$$

$$E^2 = (m c^2)^2 = m^2 c^4 = \left( \frac{p^2}{c^2} + m_0^2 \right) c^4$$

$$E^2 = p^2 c^2 + m_0^2 c^4$$

that's not so hard now, is it?

now Parlyne, i have a question for you:

there is a serious proposal (Mohr & Taylor) to change the definition of the kilogram so that Planck's constant would be defined, not measured (similarly to how the meter was redefined to fix the speed of light to 299792458 m/s):

The kilogram is the mass of a body at rest whose equivalent energy equals the energy of a collection of photons whose frequencies sum to 13563926667 x 1040 hertz.

now, suppose i was able to put that collection of photons in a perfectly mirrored (internally) and closed box of negigible mass, and then put the kilogram prototype in another box of negligible mass and put those two boxes on a balance scale. which way would the scale tip? toward the kg prototype box?

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Parlyne

rbj said:
that's the case of "rest mass". not the case of "relativistic mass".
It's not a "case." It's a fundamental property. And when I say "mass" I mean (just like every other physicist I know) what you insist on calling "rest mass." That's what I mean every time. If I want to talk about relativistic mass, that's what I'll say.

the equation you depicted, yes. it has a 0/0 problem. that's what i pointed out. but we know that these photons have energy

$$E = h \nu = \hbar \omega$$
My whole point was that 0/0 is indeterminate, while 0/(anything else) is 0, which is why defining relativistic mass as $$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$ is not valid for particles with $$m_0 = 0$$. And, of course photons have energy; but that's not something you can derive from relativity alone. Or, to put it a different way, from the standpoint of relativity photon energy is a totally free parameter. The fact that we can use quantum mechanics to relate energy to frequency doesn't change this.

which means they have (relativistic) mass

$$m = \frac{h \nu}{c^2}$$

which is not zero.
You seem unconcerned that you previously defined energy in terms of relativistic mass; but, now you're flipping your definition to account for the special case of $$m_0 = 0$$.

no, i also separated the two concepts. can you read english? would you like me to point out what i said? however, photons travel very nearly to $c$ and if they do not travel at $c$ (which i think is unlikely, but who knows), then they have non-zero rest mass.
If your argument is that photons have no rest mass because they travel at c, then you have failed to separate the concept of "speed at which light travels" from "speed characteristic of special relativity."

And, not only can I read English, but I can speak it fluently, which is more than I can say for most people who have it as their first language.

or you could say the converse: if photons move at speed $c$, even if they have inertial mass, they have no rest mass.

nothing "forces" $c$ to be anything. it is a primary quantity that ultimately is only a reflection of the units we use to measure it. the experimental question is "do photons move at speed $c$?" if they do, they cannot have non-zero rest mass. nothing forces photons to move at speed $c$, but they appear to do that to such a degree experimenters cannot tell the difference in speeds.

$c$ being a dimensionful quantity, there is no meaning to that statement. there is no special relativity at any speed other than $c$. we can take $c$, $G$, $\hbar$, and $4 \pi \epsilon_0$ out of all equations of physical law (we'd be measuring everything in terms of Planck Units). there is no $c$ to change to a much higher value.
You're ignoring my point. But, since you're going to be pedantic about units I'll make the same argument in unitless quantities. Let the relativistic "speed of light" be $$c$$ and the photon propagation speed be $$c_\gamma$$. If we assume that $$\frac{c}{c_\gamma} > 1$$ we will still get the same kinematic theory of special relativity. Nothing about relativity forces $$c$$ and $$c_\gamma$$ to be the same. Yet, your argument for photon rest mass depending on photons travelling at $$c$$ assumed this. All I've stated is that if a particle (such as a photon) has $$m_0 = 0$$, it must propagate at speed $$c$$.

you mean rest mass. they're not massless. and you're also incorrect that, it's not just evidence of not having rest mass, it is the reason and i pointed out why before.
It is a totally standard terminology to refer to particles of 0 rest mass as "massless." If I call something "massless," that is what I mean. Always.

And I've already given you the reasons, multiple times, why travelling at $$c$$ is evidence for, not the cause of, masslessness.

i agree with that.

you keep starting your points with "rest mass" and then, by slight of hand, finish them with just "mass". The speed it moves at can't cause its rest mass or invariant mass to take on a different value. that's why they call it "invariant".
But, that's the whole point I'm making. You can't say that its rest mass is 0 because it moves at the speed of light. That implies that rest mass depends on speed, which it doesn't. The only reasonable causal statement to make is that "it moves at the speed of light because its rest mass is 0."

sure it is. but it's not about rest mass.

feel free to derive it here.
See below.

i never said that. i said

$$E^2 = p^2 c^2 + m_0^2 c^4$$

can be derived from:

$$E = m c^2$$

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$p = m v$$

and here is how:

$$m^2 = \frac{m_0^2}{1-\frac{v^2}{c^2}}$$

$$m^2 \left( 1 - \frac{v^2}{c^2} \right) = m_0^2$$

$$m^2 - \frac{(mv)^2}{c^2} = m^2 - \frac{p^2}{c^2} = m_0^2$$

$$m^2 = \frac{p^2}{c^2} + m_0^2$$

$$E^2 = (m c^2)^2 = m^2 c^4 = \left( \frac{p^2}{c^2} + m_0^2 \right) c^4$$

$$E^2 = p^2 c^2 + m_0^2 c^4$$

that's not so hard now, is it?
That derivation is fine when $$m_0 \neq 0$$. But, since neither E nor p is well defined under your definitions when $$m_0 = 0$$, the derivation is not valid for that case. The way to derive it rigorously, so that it applies in all cases is as follows.

We know that an object's energy and momentum are components of the 4-vector:

$$p^\mu = \left (\frac{E}{c}, \vec{p} \right )$$

The geometry of space-time lets us find the magnitude of $$p^\mu$$ by:

$$p^\mu \eta_{\mu \nu} p^\nu = \frac{E}{c} (1) \frac{E}{c} + (-1) \vec{p} \cdot \vec{p}$$

Or, more concisely:

$$p^\mu p_\mu = \frac{E^2}{c^2} - |\vec{p}|^2$$

Since this must a Lorentz invarient, we can evaluate it in the object's rest frame, where $$\vec{p} = 0$$. Then:

$$p^\mu p_\mu = \frac{E_{rest}^2}{c^2}$$

But, we know from Einstein that $$E_{rest} = m_0 c^2$$. Thus,

$$p^\mu p_mu = m_0^2 c^2$$

This, of course, gives (with a little rearranging) $$E^2 = p^2 c^2 + m_0^2 c^4$$.

now Parlyne, i have a question for you:

there is a serious proposal (Mohr & Taylor) to change the definition of the kilogram so that Planck's constant would be defined, not measured (similarly to how the meter was redefined to fix the speed of light to 299792458 m/s):

The kilogram is the mass of a body at rest whose equivalent energy equals the energy of a collection of photons whose frequencies sum to 13563926667 x 1040 hertz.

now, suppose i was able to put that collection of photons in a perfectly mirrored (internally) and closed box of negigible mass, and then put the kilogram prototype in another box of negligible mass and put those two boxes on a balance scale. which way would the scale tip? toward the kg prototype box?
It wouldn't tip. But, this isn't a question of relativistic mass. Rather it's about how the rest mass of a composite object depends on the energies of its components. After all, the dispersion relation holds for any system. The way we look at it here would be something like:

$$m_0^2 c^4 = (\sum_i E_i)^2 - |\sum_i \vec{p_i}|^2 c^2$$

Then, since the composite object is at rest, we can impose the constraint that all the momenta added together cancel - $$\sum_i \vec{p_i} = 0$$ - leaving the statement that:

$$m_0 = \frac{\sum_i E_i}{c^2}$$

This really is no different from the idea of considering mass a form of internal energy.

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rbj

Parlyne said:
It's not a "case." It's a fundamental property.
it's semantics. semantics of yours that you are insisting on everyone else taking. the OP said "inertial mass" and in that context, it had to be relativistic mass. when you referred to "photon mass constrained by $m_\gamma < 6 \times 10^{-17} \mathrm{eV} (= 1.1 \times 10^{-52} \mathrm{kg})$" you were referring to rest mass and i was clarifying that. the "mass" the OP was referring to is not constrained by 10-52 kg.

And when I say "mass" I mean (just like every other physicist I know) what you insist on calling "rest mass." That's what I mean every time. If I want to talk about relativistic mass, that's what I'll say.
but you need to deal with other people (like the OP) that may mean something else when they said mass.

and when you say: "Rest mass is an inherent property of an object. The speed it moves at can't cause its mass to take on a certain value." that's fine and good except when the "object" is a photon believed to travel at a speed of $c$. and you can't persuasively say that my use of

$$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$

was invalid.

My whole point was that 0/0 is indeterminate, while 0/(anything else) is 0, which is why defining relativistic mass as $$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$ is not valid for particles with $$m_0 = 0$$.
sure it is (and you were incorrect about this from the beginning). but you turn it around and say

$$m_0 = m \sqrt{1-\frac{v^2}{c^2}}$$

instead. there is, to start with, no assumption about the rest mass, just about the speed of the photons. if for whatever reason, assumption or empirical evidence, you decide that the speed of the particles (photons) are the same as the speed of the wave, that is $c$ (for any observer), from that and the equation above you can legitimately conclude that the rest mass is zero.

And, of course photons have energy; but that's not something you can derive from relativity alone.
i never derived photon energy from relativity at all. it's $E = h \nu$ and that comes from emprical evidence (the photoelectic effect) that they have codified into a physical law since it seems to be true all of the time. what i did derive, from special relativity, is that photons have (relativistic) mass that is $m = E/c^2$ and such did not originate from me. then i also pointed out (from SR) that the rest mass is zero, of course with the reasonable assumption that the speed of these kind of particles (photons) move at the speed of $c$.

Or, to put it a different way, from the standpoint of relativity photon energy is a totally free parameter.
only if they move at a speed of $c$.

The fact that we can use quantum mechanics to relate energy to frequency doesn't change this.
of course not, it affirms it.

You seem unconcerned that you previously defined energy in terms of relativistic mass; but, now you're flipping your definition to account for the special case of $$m_0 = 0$$.
no i haven't. i haven't done that once. (and the impetus is on you to quote where i have. just to be clear, the energy i referred to was not just the rest energy.)

If your argument is that photons have no rest mass because they travel at c, then you have failed to separate the concept of "speed at which light travels" from "speed characteristic of special relativity."
i haven't done that either. in fact i accept the possibility that photons might travel at a speed ever so slightly less than $c$ and, if they do, they are not "massless" in any sense of the term since they would have some (but extremely small, that we cannot presently measure it) rest mass.

And, not only can I read English, but I can speak it fluently, which is more than I can say for most people who have it as their first language.
well, you missed and misread or, at least, misrepresented several things i have written.

You're ignoring my point.
no, i'm disputing it.

But, since you're going to be pedantic about units I'll make the same argument in unitless quantities. Let the relativistic "speed of light" be $$c$$ and the photon propagation speed be $$c_\gamma$$. If we assume that $$\frac{c}{c_\gamma} > 1$$ we will still get the same kinematic theory of special relativity.
sure, fine. but then photons are not "massless" by any definition.

Nothing about relativity forces $$c$$ and $$c_\gamma$$ to be the same.
but i never said that relativity forced them to be the same. i'm only saying that the rest mass of photons being zero is a consequence of them being the same, assuming they are the same. or, at least, that the zero rest mass and $c_\gamma = c$ are equivalent.

Yet, your argument for photon rest mass depending on photons travelling at $$c$$ assumed this.
yes. and if photons travel more slowly than $c$ (for any real observer), then they are not "massless" in any sense of the word.

All I've stated is that if a particle (such as a photon) has $$m_0 = 0$$, it must propagate at speed $$c$$.
i never disagreed with that either. but you said more. you insist that the causality must be:

"massless" $\Longrightarrow$ "travels at speed of $c$"

when the fact is it is just as reasonable to say

"massless" $\Longleftarrow$ "travels at speed of $c$"

(and i shown why) and probably the best thing to say is

"massless" $\iff$ "travels at speed of $c$".

It is a totally standard terminology to refer to particles of 0 rest mass as "massless." If I call something "massless," that is what I mean. Always.
so then you get people (like the OP) asking: "In some cases, inertial mass does not equal invariant mass? ... So the photon can have non zero inertial mass but always 0 invariant mass?"

you, and other physicists need to qualify or clarify what you mean when you call something (that has mass in some sense of the word) "massless".

And I've already given you the reasons, multiple times, why travelling at $$c$$ is evidence for, not the cause of, masslessness.
repeating a failed argument doesn't strengthen it. essentially, as i see it, you keep insisting that it has to be:

$$v = c \sqrt{1 - \frac{(m_0 c)^2}{h \nu}}$$

and i keep responding that

$$m_0 = \frac{h \nu}{c^2} \sqrt{1 - \frac{v^2}{c^2}}$$

is just as valid. simply rearranging an equation so the desired quantity becomes the "lvalue" does not prove causality. both arrangements are equally valid and i've shown why the second is also just as valid.

But, that's the whole point I'm making. You can't say that its rest mass is 0 because it moves at the speed of light.
i can and i did and it is correct.

That implies that rest mass depends on speed, which it doesn't.
no, i never said nor implied that. but a particle's relativistic mass (as observed by the so-called "stationary" observer) depends on speed (relative to that observer).

edit: ooops. in fact i am saying that, in the special case where the particle with finite energy travels at the speed of c, the rest mass must be zero because of the oft repeated relativistic mass equation above.

The only reasonable causal statement to make is that "it moves at the speed of light because its rest mass is 0."
wrong. an equally reasonable (in fact more reasonable) is that its rest mass is zero because it moves at a speed of $c$ (or is believed to).

That derivation is fine when $$m_0 \neq 0$$.
no, it's fine even when $m_0 = 0$

But, since neither E nor p is well defined under your definitions when $$m_0 = 0$$, the derivation is not valid for that case.
sure it was. E was defined externally ($E = h \nu$). p was a result.

It wouldn't tip.
ah, so that collection of "massless" photons weigh as much as a kilogram? sounds like saying "massless" without qualification when refering to photons is immune to confusion. especially if we find out that they may have rest mass around 10-52 kg and that they have inertial mass of $m = (h \nu)/c^2$ in any case. saying "massless" without qualification is so consistently clear and correct.

But, this isn't a question of relativistic mass.
it was for the OP. he (or she) posted:

pivoxa15 said:
In some cases, inertial mass does not equal invariant mass? What is the relation between the two?

So the photon can have non zero inertial mass but always 0 invariant mass?
i assert that when the OP said "inertial mass" in the context of that post, that said "inertial mass" is relativistic mass.

This really is no different from the idea of considering mass a form of internal energy.
no disagreement there.

but i'm done with the thread. besides a disagreement about fact: my "improper" use of the relativistic mass equation and your insistence that the (rest) masslessness of photons must come first (and then the consequence of that is that they move at the speed of the E&M propagation), there were dozens of times that you literally misrepresented what i said, and you haven't proven a single statement wrong. there were several things you tried to imply that i said that would be wrong, but i didn't say them. i only said what i said.

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TDS

My Compliments!

@rbj and Parlyne,

First, allow me to say that I have enjoyed reading your "discussion" that has been going on here. You both seem to be well versed in your chosen discipline. You both have been using formulas to support your individual points. Formulas that are between 50 and 100 years old! Start thinking outside the box. Just because someone writes a formula that describes a function and says that this is the way it is does not make it so. Remember, the Wright Brother were told that they would not be able to fly.

Second, allow me to say that I did not attend any Formal Schools with the exception of the School of Hard Knocks with a double major in Logic and Common Sense.

Having said this, I know that the photon has mass, both inertial and relativistic. If it did not have mass, it could not be affected by a gravitational force.

One thing I do not understand is this. How can it be said that an object that travels faster than light gains mass infinitely when we cannot test this theory. We do not have anything that can travel that fast. We have no idea what type of energy field it is that surrounds the photon. Does the photon start out at the speed of light or does it slow down to the speed of light when we detect it. Is the photon a photon prior to being released from the source or does it become a photon after being released from the source?

I am not wanting to take away from pivoxa15 original question, so please continue trying to answer it.

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