Dynamic Mass of Photon: E=mc^2 & hv

[country boy]

I do not need sleigh of hand problems to clarify my understanding of the relation between mass and energy. There are many realistic issues that provide this clarification a lot better than the "photon in a box"

I didn't mean to imply that you need to have your understanding of anything clarified. But the photon is a good teaching tool, particularly when it comes in a box.

The photon is my favorite particle; the neutrino used to be, but then it went and got mass.

I appreciate your disdain for relativistic mass. Once one has been enlightened (or perhaps gone over to the dark side) and seen the beauty of using the four-vector approach, there is no redemption. The four velocity has major advantages, one of which is the elimination of foolish talk about relativistic mass.

However, when one first learns relativity it is easier to deal with a changing mass than with a velocity referred to proper time. Mass is a mysterious concept anyway, and we all know what velocity is, right?

So please be patient with the rest of us who still see some usefulness in the idea of relativistic mass.

 

[DonnieD]

Mass is a mysterious concept anyway, and we all know what velocity is, right?

thanks for all the answers, the discussion is very interesting. but, i wonder, isn't the concept of energy a mysterious concept also?

 

[pmb_phy]

So please be patient with the rest of us who still see some usefulness in the idea of relativistic mass.

That goes the same for some of us who are well versed in vector and tensor analysis since the general idea of mass in that context is far more complex that that of a single particle.

Pete

 

[nakurusil]

Please pause to realize that your understanding of the "photon in a box example" is incorrect. Are you suggesting that a system of particles can generate a net force on itself? Of course not. (Or at least I really hope you aren't suggesting that.)

I think you are juming to conclusions: the box on the balance is in a gravitational field (we are weighing things, right? ). Therefore, will the photon move in a straight line between the vertical walls? Or will it have a curved path? So, if it has a curved path, will its momentum have a downwards component? And if it has a momentum with a downwards component what happens when it hits the vertical wall? You can continue from here on your own.

Also, as mentioned before, the invarient mass of the box will increase when the photon is in it. So claiming the effect is just an illusion due to collision with the walls is missing the point.

Not the mass, the energy. Using a thought experiment to "prove" that the "photon has mass" is a stretch , would you agree?

But do I feel it is a particularly useful concept? No.

Excellent, then we agree.

 

[nakurusil]

Since you brought up QED I'd like to point out that no, there is more than one concept of mass. There are "intrinsic masses" for the particles, and in QED you often integrate over the momentum of particles with invarient mass not equal to their intrinsic mass. So QED doesn't predict that the photon has zero invariant mass, and in fact actually requires that it can have non-zero invarient mass to allow correct calculations.

This is a very good point, stlill hotly debated. See John Baez's comment here .
To quote him exactly:

"if you demand gauge-invariance in QED (which
is quite reasonable, since this is a symmetry of the classical
Lagrangian), you wind up getting a theory where the photon is
massless"

 

[nakurusil]

I prefer to think of physical variables as being components of tensors defined on spacetime. In this sense, $m_0$ is a scalar, and $m=\gamma m_0$ is something else. I think this approach is vindicated as relativistic mass only appears useful in a handful of problems, whereas invariant mass is a useful concept in most cases.

Totally agreed.
We can relegate "relativistic mass" to the expression $m=\gamma m_0$ that occurs in the relativistic energy / momentum:

$E=\gamma m_0c^2$
$p=\gamma m_0v$

 

[George Jones]

There are "intrinsic masses" for the particles, and in QED you often integrate over the momentum of particles with invarient mass not equal to their intrinsic mass. So QED doesn't predict that the photon has zero invariant mass, and in fact actually requires that it can have non-zero invarient mass to allow correct calculations.

This is a very good point, stlill hotly debated. See John Baez's comment here .
To quote him exactly:

"if you demand gauge-invariance in QED (which
is quite reasonable, since this is a symmetry of the classical
Lagrangian), you wind up getting a theory where the photon is
massless"

You guys are talking about the difference between internal and external lines in a Feynman diagram.

"Virtual" particles (internal lines) do not have to be on-shell, i.e., don't have to satisfy E^2 - p^2 = m^2. Integration is performed over those momenta allowed by conservation of energy/momentum at the vertices.

"Real" particles are on-shell, and gauge-invariance does demand that the (invariant-)mass of a "real" photon be zero.

 

[JustinLevy]

Please pause to realize that your understanding of the "photon in a box example" is incorrect. Are you suggesting that a system of particles can generate a net force on itself? Of course not. (Or at least I really hope you aren't suggesting that.)

I think you are juming to conclusions: the box on the balance is in a gravitational field (we are weighing things, right? ). Therefore, will the photon move in a straight line between the vertical walls? Or will it have a curved path? So, if it has a curved path, will its momentum have a downwards component? And if it has a momentum with a downwards component what happens when it hits the vertical wall? You can continue from here on your own.

So you ARE claiming a system can create a net force on itself. This is ludicrous. A system cannot create a net force on itself, as that would violate momentum conservation. You are having the photon + box system push itself down with a net force.

Let's make this even more obvious (and stray from the semi-Newtonian gravity that you are already running into problems with). Imagine a box in space with a photon in it. You measure its inertial mass by pushing it with an impulse and measuring its change in velocity. Please tell me if you believe the box would have the same, less than, or greater inertial mass compared to the box without the photon inside.

Also, as mentioned before, the invarient mass of the box will increase when the photon is in it. So claiming the effect is just an illusion due to collision with the walls is missing the point.

Not the mass, the energy. Using a thought experiment to "prove" that the "photon has mass" is a stretch , would you agree?

Yes, the rest energy AND the rest mass of the system increases.

No where did I claim this means the photon has a mass, and I do not intend to. This thought experiment is meant to show that a photon can contribute to the invarient mass of a system. That is all. I saw that you were misunderstanding the photon in a box example and am merely trying to help you rectify that.

Since you brought up QED I'd like to point out that no, there is more than one concept of mass. There are "intrinsic masses" for the particles, and in QED you often integrate over the momentum of particles with invarient mass not equal to their intrinsic mass. So QED doesn't predict that the photon has zero invariant mass, and in fact actually requires that it can have non-zero invarient mass to allow correct calculations.

This is a very good point, stlill hotly debated.

What do you consider hotly debated? There is no debate that photons with E^2 - p^2 != 0 are necessary in QED calculations.

You guys are talking about the difference between internal and external lines in a Feynman diagram.

While I know what you mean, please remember that there is no physical difference between internal and external lines in a Feynman diagram. If those particles are to be detected they must interact with more particles (and hence everything in the end is an "internal line"). How close to "on-shell" does it have to be before it is "real" verse "virtual" ... this is a continuous transformation.

But again, I know what you mean and I don't want to get side tracked on the "measurement problem" of quantum mechanics.

"Virtual" particles (internal lines) do not have to be on-shell,

Yes.

i.e., don't have to satisfy E^2 - p^2 = m^2.

Well, one comment on this...
QED is a relativistic theory, so E^2 - p^2 is still invarient. There is still an invarient mass, and E^2 - p^2 is the definition of it.

However, some textbooks do use terminology similar to what you are stating. And, if you consider "m" in that equation to mean the "instrinsic mass" instead of the invarient mass, then yes, offshell photons don't satisfy E^2 - p^2 = m^2.

Which is what I assume you meant. I don't want to quibble on semantics, I just want to make sure we're on the same page. (Which I assume we are, yes?)

 

[nakurusil]

So you ARE claiming a system can create a net force on itself. This is ludicrous. A system cannot create a net force on itself, as that would violate momentum conservation. You are having the photon + box system push itself down with a net force.

You seem to create one strawman after another and succed in beating them down. The statement of the "photon in the box problem" is that : "there is a box that is balanced pefectly on a balance. A photon is injected in the box (i.e. a photopn is added to the system . What happens to the balance?"

<rest snipped as being an obvious strawman>

What do you consider hotly debated? There is no debate that photons with E^2 - p^2 != 0 are necessary in QED calculations.

Greg Jones set you straight on this, see above.

 

[JustinLevy]

You seem to create one strawman after another and succed in beating them down. The statement of the "photon in the box problem" is that : "there is a box that is balanced pefectly on a balance. A photon is injected in the box (i.e. a photopn is added to the system . What happens to the balance?"

What happens to the balance? We agree that the box will now weigh more.

But your reason for why it weighs more is absurd. I don't understand. I know you are smart enough to know that a system cannot produce a net force on itself. So what is the problem here?

You are saying that the box + photon system still has the same mass M as before, and thus the force of gravity is still Mg, but that the photon and box system creates a net downards force because the photon is bouncing in it. Thus the total weight = Mg + "net force due to photon" > Mg.

This is absurd, and I know you are smart enough to know this is absurd, so please stop claiming it.

The weight of the system increases because the rest mass of the system DID increase.


Above you claimed that the rest energy of the system (box + photon) is greater than the box alone, but denied that the rest mass of the system is greater. Since I answered your question, please answer mine: Do you still believe the rest energy increases but not the rest mass of the system?

There is no debate that photons with E^2 - p^2 != 0 are necessary in QED calculations.

Greg Jones set you straight on this, see above.

You are the only one disagreeing here. Greg Jones and I both agree that photons with E^2 - p^2 != 0 are necessary in QED calculations.

 

[nakurusil]

What happens to the balance? We agree that the box will now weigh more.

But your reason for why it weighs more is absurd. I don't understand. I know you are smart enough to know that a system cannot produce a net force on itself. So what is the problem here?

It doesn't produce a force by itself. If you stopped trying to find fault with what I am saying and you started following the hints you could solve it by yourself.

You are saying that the box + photon system still has the same mass M as before,

Nope. The statement of the problem says that the balance was in equilibrum before the photon was inserted into the system (box) This is the whole gist of the problem, it has a very nice set of equations once you grasp the problem statement. Try listening before constructing strawmen. <rest snipped as an obvious repeated attempt to strawman construction>

 

[nakurusil]

You are the only one disagreeing here. Greg Jones and I both agree that photons with E^2 - p^2 != 0 are necessary in QED calculations.

Both John Baez and Greg Jones are telling you that "gauge-invariance does demand that the (invariant-)mass of a "real" photon be zero."

 

[cesiumfrog]

You seem to create one strawman after another and succed in beating them down. The statement of the "photon in the box problem" is that : "there is a box that is balanced pefectly on a balance. A photon is injected in the box (i.e. a photopn is added to the system

Injected? Now who's coming up with straw men?

JL, the box of photons has weight for the same reason a box of gas has weight (the particles do strike the bottom more than the top, per nakurasil's explanation). It's not an isolated system: they do so because of the Earth's gravitational field (consider perhaps how the box would look if undergoing an acceleration in flat space), and momentum is conserved by the attraction of the Earth toward the photon's gravitational field.

Totally agreed.
We can relegate "relativistic mass" to the expression $m=\gamma m_0$ that occurs in the relativistic energy / momentum:

$E=\gamma m_0c^2$
$p=\gamma m_0v$

Maybe you should print $E=\gamma m_0c^2$ on t-shirts, it seems to capure the heart of the debate. It's not a dispute of physics, just a question of which choice of variables each individual thinks make the equations prettier. Next we'll be telling others how to interpret QM.

 

[nakurusil]

Injected? Now who's coming up with straw men?

If you have difficulty in reading the problem statement go back and read it again: the whole gist of the problem is that the box is sitting on a balance and the whole thing is in equilibrum. Then a photon is added inside the box and the question is what happens to the balance. Hint: the photon does not add mass to the system. But it does add energy/momentum.

JL, the box of photons has weight for the same reason a box of gas has weight (the particles do strike the bottom more than the top, per nakurasil's explanation).

Nope, this is not what I am saying. The photon bounces between the vertical walls describing a trajectory that is curved downwards due to the presence of the gravitational field (you got to have one since you are weighting things, right?). Because each bounce is curved downwards the photon transmits a downward component of the momentum, in the form of an elementary force $$f_z=\frac {dp_z}{dt}$$ every time it hits the vertical wall. The resultant of all these elementary forces wall is what tilts the balance on the side of the platter holding the box.

It's not an isolated system: they do so because of the Earth's gravitational field

Correct, you got this one right, it is not an isolated system, the gravitational field of the Earth is key to solving the problem correctly.
While we are at it, let's dispell another myth: the box will not bounce side to side horizontally since it is highly likely that the elementary horizontal force $$f_x=\frac {dp_x}{dt}$$ is many orders of magnitude smaller than the friction between the box and the undelying balance platter.

Maybe you should print $E=\gamma m_0c^2$ on t-shirts,

Good idea!

 

[pmb_phy]

Maybe you should print $E=\gamma m_0c^2$ on t-shirts, it seems to capure the heart of the debate. It's not a dispute of physics, just a question of which choice of variables each individual thinks make the equations prettier. Next we'll be telling others how to interpret QM.

It's a bit more than that but rarely, if ever, do I see a person devle into the general cases of objects with mass including continuous media, stressed bodies, etc. Griffith and Owens have a paper in Am. J. Phys. which illustrates some of the problems one encounters under certain situations.

Who here has studied the mass of continuous media?

Pete

 

[cesiumfrog]

It's a bit more than that but rarely, if ever, do I see a person devle into the general cases of objects with mass including continuous media, stressed bodies, etc. [Griffiths and Owen]have a paper in Am. J. Phys. which illustrates some of the problems one encounters under certain situations.

Who here has studied the mass of continuous media?

I've studied the paper you're referring to (v.51 p.1120, 1983 - it's referred to in Griffith's Electrodynamics), and it is exactly what I was thinking of when I said:

alarm bells: you'd be surprised how many long standing paradoxes are actually coordinate transformation errors

Don't suppose you've studied the resolution in Phys. Rev. D 73, 104020 (2006)?

 

[pervect]

I'm surprised this thread has been going on this long, without any resolution.

I think what people need to do is to consult some textbooks on the topic. (Note that the problem doesn't really need a consideration of continuous media, though it wouldn't hurt to use one if people are up to it).

I think the following quote, from the dialog "Use and abuse of the concepts of mass" in "Spacetime physics" (Taylor & Wheeler) covers the main points.

Can a photon -- that has no mass - give mass to an absorber?

Yes. Light with energy E transfers mass m = E [ed note: in geometric units] (= E_conv/c^2 [ed note: in standard units]) to a heavy absorber. (Exercise 8.5).

Adding a photon to a hollow mirrored sphere will transfer both the energy and the momentum of the photon to the system, just as it will if the photon is absorbed by an absorber.

Adding many photons in random directions will increase the energy of the sphere, without increasing its momentum.

The only necessary formula is this. For an isolated system, the invariant mass of that system is $\sqrt{E^2 - p^2}$ in geometric units $\sqrt{\left( E^2 - \left( p c \right)^2} \right) / c^2$ in standard units.

This invariant mass will, by definition, increase if one adds energy to the system in question (by heating it up, but adding photons to it, etc) without changing the momentum.

Note that Hurkyl gave this correct answer much earlier in the thread, but seemed to be ignored.

While there is more that could be said, (especially about non-isolated systems which can be very tricky), I hope (I'm probably too optimistic) that we can get some resolution on the simple textbook problem of the invariant mass of an isolated system in special relativity.

 

[nakurusil]

I

Adding a photon to a hollow mirrored sphere will transfer both the energy and the momentum of the photon to the system, just as it will if the photon is absorbed by an absorber.

Adding many photons in random directions will increase the energy of the sphere, without increasing its momentum.

The only necessary formula is this. For an isolated system, the invariant mass of that system is $\sqrt{E^2 - p^2}$ in geometric units $\sqrt{\left( E^2 - \left( p c \right)^2} \right) / c^2$ in standard units.

The above is the well-known and a perfectly fine solution for the particular case of adding multiple photons whose random momenta cancel out. No question about it.
In unit values for c:

$$m=\sqrt{(\Sigma E)^2-<\Sigma p,\Sigma p>}$$

and since $$\Sigma p$$ is arranged to be 0 while $$\Sigma E$$ has increased due to the injection of photons in the box , it becomes obvious that the invariant mass of the system has increased.

But what about the case of the single photon? The momentum of the sole photon is non - zero , so the simple approach from Taylor and Wheeler no longer works. This is the problem we were discussing and it is a very interesting problem indeed. A different approach is needed, the one I was showing earlier, whereby the sole photon , due to the curved trajectory in the grvitational field is transfering a downward momentum to the vertical walls of the box at each collision, thus creating the effect of increased weight. The same approach can be generalized to all cases of adding photons whose momenta don't cancel out.

But wait a minute! Can't we use the fact that the added photon contributes "equally" with $$E=\sqrt{<p,p>}$$ to both "sigmas" in the formula of the invariant mass of the system? Not really, since we know that due to the gravitational field , the photon impulse changes with time , so it will be hard to draw any conclusion from

$$m=\sqrt{(\Sigma E)^2-<\Sigma p,\Sigma p>}$$

since the dot product $$<\Sigma p,\Sigma p>$$ has become variable, and, if you do all the calculations (not very difficult) one finds out that we can't determine what happened to the system's invariant mass (may have increased , may have decreased, depending on the direction of the photon momentum). So, IMHO I believe that we need to byte the bullet and use the approach that calculates the additional vertical forces.

 

[Hurkyl]

So you're also unconvinced by Pete's contrary proof?

Yep. It looks like an equivication fallacy to me: he's changed the meaning of both "momentum" and "energy", so that his argument is no longer relevant to the original question!

 

[pervect]

The above is the well-known and a perfectly fine solution for the particular case of adding multiple photons whose random momenta cancel out. Since (for c=1):

$$m=\sqrt{(\Sigma E)^2-(\Sigma p)^2}$$ and since $$\Sigma p$$ is arranged to be 0 while $$\Sigma E$$ has increased due to the injection of photons in the box , it becomes obvious that the invariant mass of the system has increased.

But what about the case of the single photon? The momentum of the sole photon is non - zero , so the simple approach from Taylor and Wheeler no longer works. This is the problem we were discussing and it is a very interesting problem indeed. A different approach is needed, the one I was showing earlier, whereby the sole photon , due to the curved trajectory in the grvitational field is transfering a downward momentum to the vertical walls of the box at each collision, thus creating the effect of increased weight. The same approach can be generalized to all cases of adding photons whose momenta don't cancel out.

The case of a heavy absorber is discussed in Taylor & Wheeler. It's not terribly hard to show that in the limit of a large mass of the absorber (m_absorber >> E/c^2, where E is the energy of the photon) that the amount of momeuntum gained by the absorber from the photon is negligible.

Consider a 400nm photon, at the upper edge of the visible spectrum. It will have an energy of about 5e-19 joules, and a momentum of 1.6e-27 kg-m/sec

If it impacts a 1 gm absorber, the absorber will move at a velocity of 1.6e-24 m/s to conserve momentum. The kinetic energy due to its motion will be negligible (about 1e-51 joules). So most of the energy of the photon goes in heating up the absorber, i.e. virtually all of the 5e-19 joules gets turned into heat.

The conservation of energy and momentum then tells us the energy, E, and momentum, p of the absorber

E = (9e13 joules + 5e-19 joules)
pc = (5e-19 joules)

We can then compute the mass sqrt(E^2 - (pc)^2)/c^2, and verify using standard series approximations that it increases by essentially 5e-19/c^2 grams due to the absorption of the photon, just as Taylor & Wheeler state.

The effect of the (pc)^2 term on the mass is negligible

I think your observation that gravity causes the photons to travel in non-straight paths is interesting, but it doesn't affect the textbook answer.

Your observation is more of an illustration of how the system weighs more in a gravitational field. One does expect a heavier object to weigh more than a lighter object, and your analysis illustrates how that happens. But we don't even need to refer to a "gravitational field" to work the problem - in fact, it is easier if we do not, if we simply stick with the standard textbook definition of invariant mass.

 

[JustinLevy]

Hint: the photon does not add mass to the system. But it does add energy/momentum.

No, that is the very point we've been trying to show you is incorrect.

A box + photon will have more invarient mass than the same box without a photon.

The above is the well-known and a perfectly fine solution for the particular case of adding multiple photons whose random momenta cancel out. No question about it.
...
But what about the case of the single photon? The momentum of the sole photon is non - zero , so the simple approach from Taylor and Wheeler no longer works.

No, the invarient mass method really is that straight-forward. In the rest frame of the empty box we have the four momentum (c=1) of (E,0,0,0) where E is the rest energy of the empty box. Now add a photon, and we have (E,0,0,0)+(e,e,0,0) where e is the energy of the photon. (E+e)^2 - e^2 is clearly GREATER than E^2. The invarient mass increases.

The point, as stated before by me and others, is that adding photons to a system can contribute to the system's invarient mass. I am not really understanding your extreme reluctance to accept this.

Everytime I bring up the invarient mass of the system you accuse me of a "strawman arguement" even though the mass increase of the system was the very question that the thought experiment brought up. Am I really misunderstanding your statements that much? From your statements quoted above it really does appear to us that you are claiming the invarient mass does not increase. But if we really are misunderstanding you somehow, please start back a little further so we can see where the problem is arising.

But we don't even need to refer to a "gravitational field" to work the problem - in fact, it is easier if we do not, if we simply stick with the standard textbook definition of invariant mass.

True. Actually I kind of feel uncomfortable with the semi-Newtonian phrasing we've been using which is why I tried to move nakurusil's argument to considering measuring the inertial mass in free space instead.

Actually, going back to the hollowed mirror sphere example you brought up, the photons add an energy density inside, but also apply pressure on the walls of the sphere which would strain the sphere. Is there someway to show how this pressure/strain would add to the inertial mass of the sphere, or would that somehow be double counting the effect of the photons?

 

[country boy]

But we don't even need to refer to a "gravitational field" to work the problem - in fact, it is easier if we do not, if we simply stick with the standard textbook definition of invariant mass.

So we are not talking about "weighing" anymore? Of course the relationship between mass and energy can be presented in terms of inertial mass alone. But I was enjoying the gravitational aspect of the discussion.

 

[pmb_phy]

Yep. It looks like an equivication fallacy to me: he's changed the meaning of both "momentum" and "energy", so that his argument is no longer relevant to the original question!

Can you fill me in and explain what an "equivication fallacy" is and how I supposedly "changed" the meaning of both momentum and energy. I go by very strict rules which I try to adhere to at all times and that rule is to make sure that at least two SR/GR textbooks back me up as to what I'm saying. I also go to the journals too and see how they define things. I also try to read several articles on the subject to make sure that one author is unique among many.

Kind regards

Pete

ps - The term "equivocal" means "subject to two or more interpretations and usually used to mislead or confuse". If you are saying that I'm intentionally try to mislead people on purpose then I resent that remark and ask that you cease on that course of reasoning where I'm thought of a person who misleads. I've nevver done anything in my posting career which was ever intended to mislead. I would consider such a post dishonorable and I never say anything which which would lead one to question my honor.

 

[nakurusil]

No, the invarient mass method really is that straight-forward. In the rest frame of the empty box we have the four momentum (c=1) of (E,0,0,0) where E is the rest energy of the empty box. Now add a photon, and we have (E,0,0,0)+(e,e,0,0) where e is the energy of the photon. (E+e)^2 - e^2 is clearly GREATER than E^2. The invarient mass increases.

Correct, it is good to see that you finally understood the problem statement.
My point was (go back and read the post) that for this particular case the approach works.
For the general case, it doesn't, this was the point I was explaining to pervect. Here is why:

Assume that the box has a number of particles of non-vanishing resultant momentum P and energy E. Then, the invariant mass of the system is :

$$m=\sqrt{ E^2-<P,P>}$$

Now add the photon of energy e and arbitrary orientation momentum p

The invariant mass becomes :

$$m'=\sqrt{(E+e)^2-<P+p,P+p>}=\sqrt{E^2-P^2+2(Ee-<P,p>}$$

The term $$2(Ee-<P,p>)=2(E\sqrt{<p,p>}-<P,p>)$$ can be positive , zero or negative, depending on the relative orientation of $$P$$ and $$p$$.

As such, $$m'>m$$, $$m'=m$$ or ...$$m'<m$$ !
Surprise, surprise, the photon doesn't always add to the invariant mass of the system!
The complete solution is further complicated by the fact that the photon momentum is not constant, as shown in my earlier post, it is time-varying. The photons will not describe straight lines, they will describe ever-descending parabolas bounded by the vertical walls. To calculate this part rigorously, you would need GR , please don't call my approach "semi -Newtonian", ok?
So, the exact contribution is a function of time:

$$2(E\sqrt{<p(t),p(t)>}-<P,p(t)>)$$

and this happens even for the very particular case you studied (P=0) :

$$2E\sqrt{<p(t),p(t)>}$$

Now, let's try another case. We will start with your simplified case (empty box) , i.e. P=0 and let's add a couple of photons with the arbitrary momenta $$p_1$$ and $$p_2$$. What happens to the invariant mass of the resulting system? I am quite sure that you can calculate it yourself after seeing the general solution.

The point, as stated before by me and others, is that adding photons to a system can contribute to the system's invarient mass.

I will make this point one last time:

1. It is a bad idea to talk about relativistic mass.
2. It is an even worse idea to talk about the relativistic mass of photons
3. It is a bad idea to use thought experiments in the style "photon in the box", which give variable results depending on initial and final conditions, depending on momentum directions in order to prove that "photons can contribute to the invariant mass of a system". Because sometimes they don't add any mass and other times they even subtract, thus making the whole issue muddled.
4. It is a good idea to say that photons add to the overall energy of the system and (vectorially) to the overall momentum of a system. We should leave it to that.

"Justin",

I understand that we are having a pedagogical dispute, you can continue teaching your way, I will continue teaching my way. I have expunged "relativistic mass" and "photon contribution the the invariant mass of a system " from my course notes and I am very happy with the results.

 

[pmb_phy]

I will make this point one last time:

1. It is a bad idea to talk about relativistic mass.
2. It is an even worse idea to talk about the relativistic mass of photons.

May I ask why? The reason it is used goes far beyond what you'll find in your intro to SR/GR classes.

Pete

 

[nakurusil]

My I ask why? The reason it is used goes far beyond what you'll find in your intro to SR/GR classes.

Pete

Because relativistic mass is not necessary in teaching relativity.
The worst thing it does, is that it brings about the notion "photons do not have rest mass but they have relativistic mass" resulting into never ending discussions that lead nowhere.

 

[nakurusil]

The case of a heavy absorber is discussed in Taylor & Wheeler. It's not terribly hard to show that in the limit of a large mass of the absorber (m_absorber >> E/c^2, where E is the energy of the photon) that the amount of momeuntum gained by the absorber from the photon is negligible.

Consider a 400nm photon, at the upper edge of the visible spectrum. It will have an energy of about 5e-19 joules, and a momentum of 1.6e-27 kg-m/sec

If it impacts a 1 gm absorber, the absorber will move at a velocity of 1.6e-24 m/s to conserve momentum. The kinetic energy due to its motion will be negligible (about 1e-51 joules). So most of the energy of the photon goes in heating up the absorber, i.e. virtually all of the 5e-19 joules gets turned into heat.

The conservation of energy and momentum then tells us the energy, E, and momentum, p of the absorber

E = (9e13 joules + 5e-19 joules)
pc = (5e-19 joules)

We can then compute the mass sqrt(E^2 - (pc)^2)/c^2, and verify using standard series approximations that it increases by essentially 5e-19/c^2 grams due to the absorption of the photon, just as Taylor & Wheeler state.

The effect of the (pc)^2 term on the mass is negligible

Yes, we are in total agreement on this.

I think your observation that gravity causes the photons to travel in non-straight paths is interesting, but it doesn't affect the textbook answer.

Your observation is more of an illustration of how the system weighs more in a gravitational field. One does expect a heavier object to weigh more than a lighter object, and your analysis illustrates how that happens. But we don't even need to refer to a "gravitational field" to work the problem - in fact, it is easier if we do not, if we simply stick with the standard textbook definition of invariant mass.

Actually, the situation is more complicated (and more interesting) , please see my answer to JustinL. I believe that we need to go thru the slightly more complicated calculations if we want an exact quantitative result.

 

[pmb_phy]

Because relativistic mass is not necessary in teaching relativity.
The worst thing it does, is that it brings about the notion "photons do not have rest mass but they have relativistic mass" resulting into never ending discussions that lead nowhere.

What tools are necessary in teaching is what may help the student in his work. Relativistic mass provides a good tool for that given all the applications it can provide and ways of looking at a problem that is difficult to look at in another way. For example; Suppose someone asked you what the center of mass of a system of particles is. You may have a hard time doing that so you might just do away with the question altogether and say that the center of mass is meaningless and has no use whereas the center of energy does have meaning. That would be a very weak arguement. But other physicists (e.g. Rindler) had no difficulty with this problem. They simply use the regular formula replacing rest mass with inertial mass. That is but one example. Now suppose that there is a particle in a static gravitational field. The particle will have constant energy. But it won't be given by E = mc² (were m is inertial mass). m would be P^0 whereas energy would be P_0. The list goes on.

Let me ask you this - How would you define the mass density of an ideal gas?

Pete

 

[nakurusil]

Let me ask you this - How would you define the mass density of an ideal gas?

Pete

Here is an http://www.owlnet.rice.edu/~jigarb/density.htm as good as any.
This is getting interesting, let's see where it leads.

 

[Hurkyl]

Can you fill me in and explain what an "equivication fallacy" is and how I supposedly "changed" the meaning of both momentum and energy.

You switched from ordinary momentum to the generalized momentum that includes a stress term, and likewise, you added a stress term to the energy that was being considered.

AFAIK, nobody has ever argued that E/c²=m if you mix in a few new energy terms and pass to some notion of generalized relativistic mass.


As an aside, it may be possible that generalized relativistic mass becomes a useful notion -- if you agree, then it would be nice to see a precise definition and an example of it actually being useful. For 3-vectors, p = mv can almost never be true for generalized momentum (and velocity might even become difficult to define, depending on the situation), so I'm highly skeptical that generalized relativistic mass is useful.

ps - The term "equivocal" means "subject to two or more interpretations and usually used to mislead or confuse". If you are saying that I'm intentionally try to mislead people

I didn't mean to imply it was intentional: I simply meant I thought it was happening.

 

[pmb_phy]

You switched from ordinary momentum to the generalized momentum that includes a stress term, and likewise, you added a stress term to the energy that was being considered.

That is not a change in definition. I was explaining what the general formula for momentum was and that reduces to what you call the "ordinary" momentum for a single particle. There was no switching going on. I was filling in where I believed tjere was a hole.

AFAIK, nobody has ever argued that E/c²=m if you mix in a few new energy terms and pass to some notion generalized relativistic mass.

Have you ever seen anyone every discuss such a situation using the most general form of mass there is at all?

As an aside, it may be possible that generalized relativistic mass becomes a useful notion -- if you agree, then it would be nice to see a precise definition and an example of it actually being useful.

What I call useful I'm sure you will have another idea of what is useful. I call "useful" that which gives the correct answer for any legitimate question in relativity. Because nobody uses it in practive today cannot be taken as any form of proof that it won't be used 300 years from now. But I did give an example of a stressed rod many times in the past. To see that derivation please take a look at the web page I created specifically for this purpose.

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

There is also an article in the Am. J. Phys. Called "The inertia of stress" which you might want to look at if you have access to this journal.

For 3-vectors, p = mv can almost never be true for generalized momentum (and velocity might even become difficult to define, depending on the situation), so I'm highly skeptical that generalized relativistic mass is useful.

Skeptical? You seemed to be saying earlier in this thread that you were 100% sure that generalized relativistic mass is useful. And there is no reason to keep referring to it as "generalized relativistic mass" since the term has a ring to it as if it was special in some sense whereas the cases like a free partilce is actually a special case.

I didn't mean to imply it was intentional: I simply meant I thought it was happening.

The term you used has that meaning. I recommend that you just say what you mean instead of using glossy language.

Best wishes

Pete

 

[rbj]

The worst thing it does, is that it brings about the notion "photons do not have rest mass but they have relativistic mass" resulting into never ending discussions that lead nowhere.

you say that, but you don't support it.

 

[pmb_phy]

you say that, but you don't support it.

I myself don;t like getting into these discussions on mass so I think it is best for me to agree to disagree with those opposed to inertial mass (aka relativistic mass) and whomever wishes to talk to me about it can PM me.

Kind regards all

Pete

 

[country boy]

I'm afraid this thread has frayed.

 

[pervect]

My $.02 - now that we've hopefully resolved the basic issues we can talk about some of the more advanced issues.

Inertial mass is actually a second rank tensor, as Hurkyl points out.

Relativistic mass is usually understood to be a scalar quantity, another name for energy (so it is not the same as inertial mass.) Pete's usage in this area is IMO non-standard, and causes a lot of confusion.

Invariant mass is also a scalar - it is the invariant length of the energy-momentum 4-vector for a point particle - or the invariant length of the energy momentum 4-vector for a system with a finite volume.

Interestingly enough, the total energy-momentum of a non-isolated system with finite volume does not transform as a 4-vector. This is behind some of what Pete is (IMO) trying to say.

If one reads Taylor & Wheeler, "Spacetime physics" closely, for instance, one will see that they are always careful to say that the mass of an isolated system is a Lorentz invariant, not that the mass of any arbitrary system is Lorentz invariant.

How does one deal with relativistic systems with a finite volume? Via the stress-energy tensor, which always transforms properly (i.e. covariantly) as a rank 2 tensor.

Given a particular frame of reference, the total energy in a given volume can be expressed as the integral of T_00, and the components of the momentum can be expressed as integrals of T_0i.

Given these volume integrals for the total energy E and mass p in some volume V in special relativity, the mass contained within a volume can be (and as far as I can tell from a close reading of the textbooks) is defined in special relativity as sqrt(E^2 - (pc)^2) / c^2, in spite of the fact that the above quantity is not always Lorentz invariant.

(I haven't seen any textbook specifically say that the mass of a non-isolated system is defined in this manner - rather, the above formula is offered as a general defintion of mass, and the comment is made additionally that the above quantity is an invariant for isolated systems. I believe it is correct to say that the quantity is still defined for non-isolated systems, but is not invariant).

In other words, in spite of the name, the "invariant mass" of a system is actually an invariant only if the system has zero volume, or if the system is isolated.

For one source for this in the literature, see http://arxiv.org/abs/physics/0505004 or the peer-reviewed

http://www.springerlink.com/content/534j31t61675w010/

by the same author.