- #26

Omni

I think you all missed the bit about "Can you speak English for a lowly S/W Developer"

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- #26

Omni

I think you all missed the bit about "Can you speak English for a lowly S/W Developer"

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OK. I'm referring back to your post at the end of Page 2 of this Thread. (I think.)Originally posted by GRQC

Thank you, DW. As a follow-up, I cite a few passages from "Spacetime Physics", by Wheeler and Taylor:

The take home lesson is thatmass is invariant. No such thing as relativistic mass, anymoreso than there is "relativistic spacetime interval".

I went back again, and looked at the Physics FAQ I was referred to:

at: http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

and cut out this small section, that begins with the question:

"Is there any experimental evidence that the photon has zero rest mass?"

Below is part of the answer from that page:

"If the rest mass of the photon was non-zero, the theory of quantum electrodynamics would be "in trouble" primarily through loss of gauge invariance, which would make it non-renormalizable; also,

charge-conservation would no longer be absolutely guaranteed, as it is if photons have vanishing rest-mass. However, whatever theory says, it is still necessary to check theory against experiment.

It is almost certainly impossible to do any experiment which would establish that the photon rest mass is exactly zero. The best we can hope to do is place limits on it. A non-zero rest mass would lead to a change in the inverse square Coulomb law of electrostatic forces. There would be a small damping factor making it weaker over very large distances. The behavior of static magnetic fields is likewise modified. A limit on the photon mass can be obtained through satellite measurements of planetary magnetic fields. The Charge Composition Explorer spacecraft was used to derive a limit of 6x10-16 eV with high certainty. This was slightly improved in 1998 by Roderic Lakes in a laborartory experiment which looked for anomalous forces on a Cavendish balance. The new limit is 7x10-17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3x10-27 eV but there is some doubt about the validity of this method."

---end quote---------

It states that photons have a "rest mass". It says the "...limit of" that was measured "with high certainty." AND "This was slightly improved in 1998" to be 7x10-17 eV.

Just talk English to me.

(The equations in this thread are not impressing me, and no one seems to be able to provide the equations for what I would expect should be easy for any Physics Major. LOL!!!)

When the FAQ says that photons have a mass of so-and-so, I'm not sure where they are contradicting me original question, OR EVEN THEMSELVES.

IS "eV" an "electon Volt"?

Did they SIMPLY DO A SUBTITUTION for a MEASUREMENT OF MASS (ABOVE),

and express MASS in terms of a MEASURMENT for ENERGY?

Because >>>>THAT<<<<< IS WHAT I'm guessing is the ENTIRE source of confusion (as far as I'm concerned).

Would someone get back to English, or do you want me to start talking in C# ICL, so ain't NOBODY gonna get nuffin outta this thread! LOL!!

The other aspect of confusion was cleared up, by saying that mass is invariant, and that the variant is Energy (which I happen to recall),

but not from this thread.) I won't even bother asking what Eo means.

(I'll just assume it is when nothing is moving.)

It's just my brain rotting after reading 3 pages of stuff with equations and terminology that no one agrees on, and precious little ENHLISH!

The "take home lesson" is in what I UNDERSTAND to be the CORRECT terminology to be using. Except, lots of ppl and articles don't bother doing it, and it just makes from more questions, than answers

(on my part). However, I accept that as being the correct terminology,

despite my not understanding all of the implications of it. Except that if mass doesn't change, there should not be an increase in gravity, as I asked about. Don't bother explaining it with equations.

Before you talk equations, there needs to be quite a bit more definitons.) Just a simple few sentences not using abbreviations, as you almost did, would explain the answers in a way that EVERYONE will understand. BTW. I assume a 4 vector are the 3 Dimensions, Plus Time as the 4th Dimension. GEEEZZ. I REALLY oue some of you folks some C# ICL, and let you ponder that stuff!!! LOL!

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Careful, it does not say that photons have rest mass. It says thatOriginally posted by treat2

It states that photons have a "rest mass". It says the "...limit of" that was measured "with high certainty." AND "This was slightly improved in 1998" to be 7x10-17 eV.

By the way, yes, eV is an electron volt (unit of energy). In particle physics, mass is expressed this way because the numbers are more manageable then expressing mass in kg. When you see eV as mass, it actually means eV/c

Back to the sermon: we used to think neutrinos were massless, but there is evidence to support at least some of them have mass (on the order of a few eV). Theory originally stated that they were massless, but this was relatively easy to modify to include mass (also, there were other observed phenomena -- the solar neutrino problem -- which could be explained if neutrinos had mass (something called neutrino oscillation).

The bottom line is, as you've ascertained from that website: theory would be in trouble if the photon had mass. And, this is a well-established theory (quantum electrodynamics), which has stood up to incredible experimental verification, so the chance that something like this could kill it is pretty remote. Unlike with neutrino physics, there is nothing we observe that could be explained by a massive photon. It's more or less a pipe dream of researchers who need papers to write (and want to stir up some trouble).

Again, reporting an upper bound on the mass doesn't mean it has mass. It is just a consequence of experimental verification. We can't

Hope that was good enough.Just talk English to me.

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- #29

Nereid

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Another way to express the upper limit of the mass of a photon: it's <2x10

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1.3 x 10Originally posted by Nereid

... and 7x10^{-17}eV is how many kg?

.

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Originally posted by GRQC

Careful, it does not say that photons have rest mass. It says thatthey have (rest) mass, then it is constrained to be very, very, very small. As a contrast, the electron has a mass of 0.5 MeV (i.e. million eV).if

By the way, yes, eV is an electron volt (unit of energy). In particle physics, mass is expressed this way because the numbers are more manageable then expressing mass in kg. When you see eV as mass, it actually means eV/c^{2}.

Back to the sermon: we used to think neutrinos were massless, but there is evidence to support at least some of them have mass (on the order of a few eV). Theory originally stated that they were massless, but this was relatively easy to modify to include mass (also, there were other observed phenomena -- the solar neutrino problem -- which could be explained if neutrinos had mass (something called neutrino oscillation).

The bottom line is, as you've ascertained from that website: theory would be in trouble if the photon had mass. And, this is a well-established theory (quantum electrodynamics), which has stood up to incredible experimental verification, so the chance that something like this could kill it is pretty remote. Unlike with neutrino physics, there is nothing we observe that could be explained by a massive photon. It's more or less a pipe dream of researchers who need papers to write (and want to stir up some trouble).

Again, reporting an upper bound on the mass doesn't mean it has mass. It is just a consequence of experimental verification. We can'tmeasureexactly 0 mass, but the closer we get to 0 the more likely itis0.

Hope that was good enough.

Originally posted by GRQC

Careful, it does not say that photons have rest mass. It says thatthey have (rest) mass, then it is constrained to be very, very, very small. As a contrast, the electron has a mass of 0.5 MeV (i.e. million eV).if

By the way, yes, eV is an electron volt (unit of energy). In particle physics, mass is expressed this way because the numbers are more manageable then expressing mass in kg. When you see eV as mass, it actually means eV/c^{2}.

Back to the sermon: we used to think neutrinos were massless, but there is evidence to support at least some of them have mass (on the order of a few eV). Theory originally stated that they were massless, but this was relatively easy to modify to include mass (also, there were other observed phenomena -- the solar neutrino problem -- which could be explained if neutrinos had mass (something called neutrino oscillation).

The bottom line is, as you've ascertained from that website: theory would be in trouble if the photon had mass. And, this is a well-established theory (quantum electrodynamics), which has stood up to incredible experimental verification, so the chance that something like this could kill it is pretty remote. Unlike with neutrino physics, there is nothing we observe that could be explained by a massive photon. It's more or less a pipe dream of researchers who need papers to write (and want to stir up some trouble).

Again, reporting an upper bound on the mass doesn't mean it has mass. It is just a consequence of experimental verification. We can'tmeasureexactly 0 mass, but the closer we get to 0 the more likely itis0.

Hope that was good enough.

OK I admit it again. I'm probably even more confused.

The ususal entire equation for E=mc^2

I'm told should instead be Eo = ((Y*eV)/c2) * C^2

So I'm seening "rest energy", which I haven't gotta clue what that means, is Equal to some number Y amt times eV/C^2

and all of that is then multiplied together.

The c^2 on the right side of the equation cancel each other out, whcih tells me the equation isn't correct,

and the remaining (Y*eV) is NOW an INVALID, since it sitting over a

one of the 2 C^2, that is now ZERO,, and since you can't devide by zero..... IM TOTALLY CONVINCED i DIDN'T UNDERSTAND ANY of the flying equations. What can I say?

I can seem to get a simple equation expressed correctly.

'I still don't know what a zero rest Energy would mean,

if mass is an invariant, why not just call the thing a constant, and

multfy it by the other stuffr you want to multipfy it by.

Yep. REAL LOST. buit willing to listen to anyoen that can put this puuzel together. (Whether any does or not, I be renis if iI did't even thank them for their interest. It seems half of use leaned few new tricks.)

Long story short, if someone can put this thing into a simple equation the way it's suppossed to lock, and also iedentify the varaints) of the equation that would be meighty instructive, nad

even working through 1 example using any number wiul dbe better, cause then I cou;kd ask the the heck Ei <<<< sub 1 really means at as apposed to Eo, and W?HAT is INCREATING, AND WHAT isn't,

- #32

selfAdjoint

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And the most common factor is gamma (you mistook it for a Y), but it's [tex] \gamma = \frac{1}{\surd(1-\frac{v^2}{c^2})} [/tex].

Where v is the relative speed and c is the speed of light.

Now there are two systems of applying this factor. In one system we keep energy invariant ("conservation of energy") and apply the factor to mass. In the other we do the opposite, mass is taken as an invariant and energy gets multiplied by gamma. Each system is consistent within itself and unfortunately we have fans of both systems posting on these boards, which really sucks for the newby looking for clarification.

If the mass you measure in that something is going to vary, then you need to distinuish the mass when it's moving (relative to you) which is multiplied by gamma, from the mass when it's not moving, say relative to itself, when it doesn't have a gamma. So that latter case is called its "rest mass" in the first system I mentioned.

On the other hand if the mass is invariant, there are not two cases, just one, and mass, or "invariant mass" is always the same. Notice that numerically, invariant mass = rest mass. But as a SW developer you know how little that can mean.

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Originally posted by treat2

OK I admit it again. I'm probably even more confused.

The ususal entire equation for E=mc^2

I'm told should instead be Eo = ((Y*eV)/c2) * C^2

No, E=mc

Rest energy 'E' is the "energy equivalent" of a mass 'm'. They're the same thing.So I'm seening "rest energy", which I haven't gotta clue what that means, is Equal to some number Y amt times eV/C^2

and all of that is then multiplied together.

I hope your command of C++ structure is better than your command of English grammar.Yep. REAL LOST. buit willing to listen to anyoen that can put this puuzel together. (Whether any does or not, I be renis if iI did't even thank them for their interest. It seems half of use leaned few new tricks.) ...

Long story short, if someone can put this thing into a simple equation the way it's suppossed to lock, and also iedentify the varaints) of the equation that would be meighty instructive, nad

even working through 1 example using any number wiul dbe better, cause then I cou;kd ask the the heck Ei <<<< sub 1 really means at as apposed to Eo, and W?HAT is INCREATING, AND WHAT isn't,

If you're not going to take the time to proof-read your post, I'm not going to take the time to answer.

- #34

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Originally posted by GRQC

No, E=mc^{2}. That's it. There's your equation. The units of energy are in eV, and units of mass are in eV/c^{2}.

Rest energy 'E' is the "energy equivalent" of a mass 'm'. They're the same thing.

...

If you're not going to take the time to proof-read your post, I'm not going to take the time to answer.

Sorry about my previous post. (I can't even read what I wrote. LOL!) Given certain conditions, and a lack of allergy pills, I can become virtually blind, as you can see from my unintelligible post.Originally posted by selfAdjoint

And the most common factor is gamma (you mistook it for a Y), but it's [tex] \gamma = \frac{1}{\surd(1-\frac{v^2}{c^2})} [/tex].

Where v is the relative speed and c is the speed of light.

Now there are two systems of applying this factor. In one system we keep energy invariant ("conservation of energy") and apply the factor to mass. In the other we do the opposite, mass is taken as an invariant and energy gets multiplied by gamma. Each system is consistent within itself and unfortunately we have fans of both systems posting on these boards, which really sucks for the newby looking for clarification.

System 1

If the mass you measure in that something is going to vary, then you need to distinuish the mass when it's moving (relative to you) which is multiplied by gamma, from the mass when it's not moving, say relative to itself, when it doesn't have a gamma. So that latter case is called its "rest mass" in the first system I mentioned.

System 2

On the other hand if the mass is invariant, there are not two cases, just one, and mass, or "invariant mass" is always the same. Notice that numerically, invariant mass = rest mass. But as a SW developer you know how little that can mean.

I am must misunderstanding the use of the word "invariant", because

some posts here have said the amount of mass never varies, regardless of mass accelerating, or decelerating relative to an observer.

Other posts saying mass is an "invariant", and it varies, according to the mass accelerating or decelerating, relative to an observer outside of that "system".

If the amount of mass varies, in ANY scenario, depending upon acceleration or deceleration, and in what system the observer is in, then:

1) Why refer to mass as an invariant, in ALL systems?

2) Why did your previous posts say that the amount of mass never varies in ANY system?

3) Why refer to something as an invariant, when it does vary?

4) what about your earliest posts in which (I thought) you said that mass is an invariant (i.e. does not vary) relative to an observer in ANY system and in ANY scenario, BUT E does vary?

I'm not trying to be argumentative, but I don't understand why mass is being called an invariant, when it is being said to vary. I am equally confused by what at first appeared to be your position, as I understood it, that mass does NOT vary, relative to anyone, in ANY system, in ANY scenario, and now seems to be saying the opposite???

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- #35

selfAdjoint

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According to ONE of the two ways of formaliziing relativity, mass is invariant and energy is not. According to the OTHER formalism, mass is not invariant but energy is.

If this duality seems counterintuitive think of this. You are looking at something which is speeding by, or toward, or away from you. You observe its behavior, and set out to do equations on that. You find that the equations predict the same behavior if you plug in an invariant mass and an energy that is a certain function of speed, or if you plug in an invariant energy and have the mass be a certain function of the speed. Either way gives you the same predictions of behavior. This is actually a "duality" like the ones the string physicists talk about but SR is such a simple theory the duality has no effect, except to confuse students.

You are going to have to check each answer to your question according to who sent it. Each individual should be consistent with him/herself, but they will disagee with each other, only due to this double formality problem.

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Many thanks! At least I know that I didn't misunderstand everything, and am at least not more confused now, than after the replies began coming in. LOL!!!Originally posted by selfAdjoint

According to ONE of the two ways of formaliziing relativity, mass is invariant and energy is not. According to the OTHER formalism, mass is not invariant but energy is.

If this duality seems counterintuitive think of this. You are looking at something which is speeding by, or toward, or away from you. You observe its behavior, and set out to do equations on that. You find that the equations predict the same behavior if you plug in an invariant mass and an energy that is a certain function of speed, or if you plug in an invariant energy and have the mass be a certain function of the speed. Either way gives you the same predictions of behavior. This is actually a "duality" like the ones the string physicists talk about but SR is such a simple theory the duality has no effect, except to confuse students.

You are going to have to check each answer to your question according to who sent it. Each individual should be consistent with him/herself, but they will disagee with each other, only due to this double formality problem.

I'm reluctant to ask anything more about it, as it seems it seems that whenever I do, the reply indicates I'm varying the wrong thing. LOL! Thanks again. I'm gonna quit asking more questions, before someone says that this explanation is wrong, too! Thnaks again.

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Don't give up asking questions. I am sorry for my harsh reply from before. On these forums, it's sometimes hard to differentiate between those who are really asking for help, and those who are out to make a pest of themselves. I would encourage you to continue using this forum as a source of information. I'm sure there are many posters here who are professors (like myself) who enjoy helping those who have an interest.

I think this is a great venue for public outreach and academic discussion, and I commend the moderators for maintaining it.

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A few brief words for GRQC and everyone on appologies, then some qustions(s).Originally posted by GRQC

Don't give up asking questions. I am sorry for my harsh reply from before. On these forums, it's sometimes hard to differentiate between those who are really asking for help, and those who are out to make a pest of themselves. I would encourage you to continue using this forum as a source of information. I'm sure there are many posters here who are professors (like myself) who enjoy helping those who have an interest.

I think this is a great venue for public outreach and academic discussion, and I commend the moderators for maintaining it.

GRQC AND EVERYONE! I appreciate your responses. (I was half-kidding around, in my previous post.) There is no need to appologize, I realize that anyone who wrote a non-canned response, really took their own time to talk to me, and I appreciate your efforts, despite my earlier confusion about the 2 schools approach. I too, posted something earlier that could have been said more tactfully, and let some frustration get into the mix. I now have seen that you have been dealing with your share of Physics quacks, ego-baiters, and other ego-related/(in)sensitivities. Anyways, I'm not one of the trouble makers,

in that regard.

One thing you've noticed is that I never stop asking questions.

I also have a tendency to ask some unanswerable questions.

-----------------------------------------------------------------

9GRQC, I REALLY want to start at the very begining all over again, but for now, my objective will remain: to make it to the end of each reply w/o being confused by what was said.

---------------------------------------------------

GTQC, OR ANYONE

Please correct me entirely, when I go down the wrong path.

(Use you terminology e,g, "rest frames", or maybe that is not.

I've got a ton of new physics books I will be getting into

during the next few years. So, whatever terminialogy I will

read about, will be of interest for me to hear and become

familiar with.)

GTQC, OR SOME PPL,

you mentioned that mass remains constant while accelerating or decellerating, regardless were the obeserver is positioned. I was my understanding that you were explaining that E increases while mass remaines invariant. I'd like first know if I undersatnd that correcly, and secondly know (assuming the above is true), is E increaesing relative to an observer that is no on board, or is it increasing relative to a person that is on board?

A last question, I'm pretty sure of... when we last about E=MC^2, we are not talking about systems that being influenced by gravity. (As I see no gravitational variable in the equation, although gravity, would effect accel/decel rates in E=M*C^2) is gravity, implicitly expected to be worked into the equation, when applicatble?

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