Why doesn't a photon's mass increase to infinity?

[jbriggs444]

thinking that radios, eyes and other thing don't get more massive from absorbing more photons. That their energy is "free" of mass; is massless.

if the incoming photons' momentum net to zero on the body does that add to the mass of the body

Mass is not additive. The mass of an object that has absorbed a photon will be greater than the sum of its original mass plus the [zero] mass of the original photon.

 

[Orodruin]

In a curved spacetime, their frequency/wavelength changes as they travel.

This is an observer effect, not an effect inherent to the light signal. The wavelength also changes if the observer changes velocity but it is still the same light signal. The 4-frequency is parallel transported along the signal world line, which is as close to saying “does not change” as is possible.

What about, are the errors and uncertainties known; their causes? I'd consider the uncertainty principle to be a known unknown and kind of irrelevant with respect to experimental errors. Surely for the equipment and other parts of the experiment the source of the errors are known, mitigated and corrected for as reasonable.

I am not talking about the uncertainty principle. If you knew what the measurement errors were they would not be measurement errors. There are errors that you simply cannot get around. You might know their distribution but there is no way you can know exactly how a particular measurement was affected.

 

[PeterDonis]

The 4-frequency is parallel transported along the signal world line, which is as close to saying “does not change” as is possible.

This is a fair point; but it's worth noting that the same statement would apply to the 4-velocity of a timelike object moving on a geodesic. So this definition of "does not change" does not have the implications that @Dr Whom is thinking it does.

 

[Ibix]

This is an observer effect, not an effect inherent to the light signal. The wavelength also changes if the observer changes velocity but it is still the same light signal. The 4-frequency is parallel transported along the signal world line, which is as close to saying “does not change” as is possible.

Imagine a laser beam traveling a cosmological distance, doing a U-turn around a black hole, and returning to source. The returned beam will not have the same frequency as the enmitted beam as measured by the source (in general). I agree that all that's happened is that the photon has had its four momentum rotated, but I have difficulty not regarding that as a change.

 

[Orodruin]

Imagine a laser beam traveling a cosmological distance, doing a U-turn around a black hole, and returning to source. The returned beam will not have the same frequency as the enmitted beam as measured by the source (in general). I agree that all that's happened is that the photon has had its four momentum rotated, but I have difficulty not regarding that as a change.

Can we stop calling it “photon”? It hurts my eyes.

This is about the spacetime curvature. Of course the observer will see a different direction but there are many reasons for this. First of all, the spacetime event of emission is not the same as that of reception so you have to define what “change” means by means of parallel transport or Fermi transport of the observer frame. The spacetime curvature implies that transports will depend on the worldline, so it is not strange to find that the signal travels in a different direction according to the observer. Still, it is an observer effect. You can find any direction and frequency you like simply by changing observer.

This is a fair point; but it's worth noting that the same statement would apply to the 4-velocity of a timelike object moving on a geodesic. So this definition of "does not change" does not have the implications that @Dr Whom is thinking it does.

I agree, I just wanted to point out that many people seem to think that “frequency” is some inherent property of a light signal, but it is not.

 

[Dr Whom]

The simple answer, photons have zero rest mass. Rindler, Essential Relativity.

 

[Francis Ward]

So, to explain why I raised this question in the first place. I am not a physicist (you all fall back in amazement at that revelation - not). I read books by physicists (currently Max Tegmark - Mathmatical Universe) and I find myself questioning so much that is taken for granted. Not out some desire to question for question sake, but to try to understand. So the photon question - to which we really do not have any definitive answer, is just the microscopic end of the questions. Einsteins mass/energy equivalence, which I understand is proven experimentally, says that if an object increases in energy, it also increases in mass. Hence the rubber band analogy used by Brian Cox. So, if we take the question of a person traveling in a vehicle (his frame of reference), regardless of the speed of the vehicle, relative to his frame of reference, the person is stationary. Outside the frame, he in fact has increased kinetic energy. Therefore his mass should increase. Is this actually what happens, or does his mass remain the same? Take the scenario to it's ultimate conclusion, with the vehicle traveling close to the speed of light, the observer in the vehicle still feels stationary ... but what about his mass?

 

[Orodruin]

So, if we take the question of a person traveling in a vehicle (his frame of reference), regardless of the speed of the vehicle, relative to his frame of reference, the person is stationary. Outside the frame, he in fact has increased kinetic energy. Therefore his mass should increase. Is this actually what happens, or does his mass remain the same?

This is not about mass, it is about relativistic mass. Please see https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[Dr Whom]

Mass is relative. In the stationary frame the man's mass is his rest mass, a constant. Relative to any frame in which he is moving, his mass is greater.

 

[Dr Whom]

P.S. as for walking on water, use floating boots or a giant hamster ball. The last is great fun:)

 

[Orodruin]

Mass is relative. In the stationary frame the man's mass is his rest mass, a constant. Relative to any frame in which he is moving, his mass is greater.

In the modern usage of the word "mass", this is false. Again, the mass that is increasing is the relativistic mass. Relativistic mass is a concept that is largely not used by professional physicists.

 

[Francis Ward]

Can relativistic mass be measured?
Why is concept of relativistic mass not used?

 

[Orodruin]

Can relativistic mass be measured?

Relativistic mass is just a rescaling of an object's energy by a factor $c^2$.

Why is concept of relativistic mass not used?

Please read the PF Insight linked to in #43.

 

[Dr Whom]

Off the point. Petty semantics. As a professional physicist I have no problem with this term.

 

[Ibix]

Off the point. Petty semantics. As a professional physicist I have no problem with this term.

You regard the distinction between the modulus of a vector and one of its components as "petty semantics"?

 

[Orodruin]

Petty semantics.

It is not petty semantics. It is a question of both accuracy in communication as well as pedagogy. Having taught advanced relativity courses for years I can tell you that relativistic mass is clearly up there together with not understanding relativity of simultaneity in what people get wrong. Among laymen it leads to the misconception that an object somehow changes when it gets faster, making them draw the unfortunate conclusion that you can tell if an object has absolute motion by looking at its mass. Among professionals, it has no place because professionals discussing properties of objects generally prefer to talk about invariant properties.

I do not know what field you are in, but in fields where it matters (i.e., mainly high-energy physics) you will not see a single paper mentioning "mass" with the meaning of "relativistic mass". If you go about teaching people about relativistic mass, you are doing them a disfavour in my opinion.

 

[Dr Whom]

You believe mass is a vector?

 

[Orodruin]

You believe mass is a vector?

No, where did you get that from? Mass, as defined in SR, is the modulus of the 4-momentum of an object. Relativistic mass (or, better, just "energy") is the time-component of this 4-momentum.

 

[Dr Whom]

Which in the rest frame is the rest mass, or as you prefer rest energy, and is greater in ones in which it moves. As I stated originally.
P.S. I also checked the definition of four momentum in Rindler, Essential Relatvity. This is also mixes momentum, mass and energy in its definition. I made some margin notes years ago when I first noticed the inconsistency. There is a simple test for consistency in physics, that of the units. For a four vector to have a modulus, which it must, all components have to have the same units. The forth term in the four momentum is mc, not m. This may be a simple matter of missing universal constants out of expressions, a habit that oft times creates problems.

 

[Francis Ward]

Relativistic mass is just a rescaling of an object's energy by a factor c2
So it cannot be measured? And an object inside a vehicle traveling at speed has no increased energy over an object at rest, relative to it's frame of reference?

 

[Ibix]

Which in the rest frame is the rest mass, or as you prefer rest energy, and is greater in ones in which it moves. As I stated originally.

You are missing the point. There's a distinction between invariant mass and relativistic mass - I see you understand this. But which one you mean when you say just "mass" is not petty semantics, and you are using it in the opposite sense to the vast majority of the professional physics community.

 

[Ibix]

Relativistic mass is just a rescaling of an object's energy by a factor c2
So it cannot be measured?

It can be measured - just measure the energy.

And an object inside a vehicle traveling at speed has no increased energy over an object at rest, relative to it's frame of reference?

This is confusingly written, because you're talking about "travelling at speed" and frames of reference without making clear what's moving with respect to what. If you do make it clear, the question pretty much answers itself.

I think you are imagining one object sitting on a table and an identical object sitting in a car that drives past the table. From the point of view of the table, the object in the car is moving so has more energy than the one on the table. From the point of view of the car it's the object on the table that's moving. So in this frame each object has the same energy that the other object has in the table frame. It's symmetrical.

 

[Francis Ward]

No, I am saying that an object inside a car that is traveling at speed, has no greater energy than an object sitting in a car that is not traveling at speed relative to the objects frame of reference, which is the car

 

[jbriggs444]

No, I am saying that an object inside a car that is traveling at speed, has no greater energy than an object sitting in a car that is not traveling at speed relative to the objects frame of reference, which is the car

No one is disagreeing with that. We all agree that an object's kinetic energy in the frame of reference in which it is at rest is zero.

That does not prevent you from measuring the kinetic energy of a bullet in the frame of the target.

 

[Francis Ward]

I did not say anyone was disagreeing with it. I was clarifying the confusing way I had written my earlier post.
Where did the bullet come in?

 

[jbriggs444]

I did not say anyone was disagreeing with it. I was clarifying the confusing way I had written my earlier post.
Where did the bullet come in?

You had surmised that it was impossible to measure a photon's energy and had proceeded to reason about objects on tables or in cars. The apparent thrust of this was that we cannot measure the kinetic energy of an object on a table or of a photon.

Yet we can measure a bullet's energy.

 

[Francis Ward]

I was not aware I had surmised as such and mentioned no tables. The apparent thrust was not that either. Probably best to quit now as the discussion is not a discussion and is fruitless

 

[Mister T]

they do not [...] experience time. Which is precisely what I stated in the first place. Elementary SR.

You didn't say it precisely. To say that a massless particle doesn't experience time is not a precise way of expressing this idea.

The passage of time is not defined because any two events for which a massless particle is present are separated by an interval of zero. That is, the separation is lightlike. Proper time is the interval length, but only for events that have a timelike separation. If the separation between two events is lightlike or spacelike there is no way of assigning a proper time elapsing between the events. The very concept of proper time doesn't exist, which is quite different from saying that the elapsed proper time has a value of zero. In the latter case the implication is that the concept exists, after all, you can't assign a value (of zero or anything else) to a quantity that is undefined.

 

[PeterDonis]

I also checked the definition of four momentum in Rindler, Essential Relatvity. This is also mixes momentum, mass and energy in its definition. I made some margin notes years ago when I first noticed the inconsistency. There is a simple test for consistency in physics, that of the units. For a four vector to have a modulus, which it must, all components have to have the same units. The forth term in the four momentum is mc, not m. This may be a simple matter of missing universal constants out of expressions, a habit that oft times creates problems.

No, it's a matter of understanding what the constant $c$ actually is in relativity. It's a unit conversion factor. You can use units where $c = 1$ and then the whole issue you describe goes away; all components of 4-momentum have units of mass. Or you can use momentum units, which means you assume a value fori $c$ that is not $1$ (such as the current SI unit value), and insert a factor of $c$ as appropriate in the components of 4-momentum; or you can use energy units, which means in some places you're inserting a factor of $c^2$ (and in others you're inserting a factor of $c$). None of which changes the physics in the least. And none of which changes the point that @Ibix was making, which he stated again in post #56, and which is perfectly valid.

 

[Orodruin]

There is a simple test for consistency in physics, that of the units. For a four vector to have a modulus, which it must, all components have to have the same units. The forth term in the four momentum is mc, not m. This may be a simple matter of missing universal constants out of expressions, a habit that oft times creates problems.

This would be true if you were using a system of units where length and time dimensions were different. Rindler is not doing this and it is a very common thing to do in relativity to use the same units for length and time by using units where, by definition $c = 1$, just as it is common in quantum physics to use units where $\hbar = 1$. The typical thing to do in high-energy physics is to use units where $\hbar = c = 1$ and measure lengths and times in inverse units of energy. Understanding the system of units used is as fundamental as dimensional analysis.

In relativity, it is natural to use the same units for time and length, after all the manifold is spacetime and using the same units for length and time just assures that all your coordinates have the same physical dimension. Note that all velocities are dimensionless in those units.

If you will, this is equivalent to considering the spatial coordinates to be $x/c$ rather than $x$ (or the time coordinate to be $x^0 = ct$ rather than $t$).