[alancalverd]

Nothing to do with belief. I'm an experimental physicst, and when my photon beams in air or tissue exceed 1.022 MeV we get pair production, signalled by the appearance of two 511 keV photons, exactly as predicted by E = mc² and the known mass of the electron (another experimental value).

Whether you believe it or not, it happens every damn time, which is how we diagnose tumors on the one hand, and calibrate small accelerators on the other.

Dear Alan,

What does that have to do with relativistic mass?

relativistic mass = γm = m√(1-v²/c²)

so we start with a positronium pair with rest mass 2m_e and end up with two photons of zero mass, thus γ = 0 and v = c.

OK, it's cheating to multiply by zero, but when we accelerate massive particles to relativistic speeds, the incremental energy required to do so turns out to be proportional to γm, not m. You can do this experiment most easily (!) in a cyclotron but relativistic effects are just measurable with megavolt electrons, which you can generate with a simple Van de Graaff. Which, come to think of it, is also cheating as I suspect few of our other correspondents have one in the backyard, so we'll have to look for an astronomical example, or you can take my word for it.

[lightarrow]

Nothing to do with belief. I'm an experimental physicst, and when my photon beams in air or tissue exceed 1.022 MeV we get pair production, signalled by the appearance of two 511 keV photons, exactly as predicted by E = mc² and the known mass of the electron (another experimental value).

Whether you believe it or not, it happens every damn time, which is how we diagnose tumors on the one hand, and calibrate small accelerators on the other.

Dear Alan,

What does that have to do with relativistic mass?

relativistic mass = γm = m√(1-v²/c²)

so we start with a positronium pair with rest mass 2m_e and end up with two photons of zero mass, thus γ = 0 and v = c.

OK, it's cheating to multiply by zero, but when we accelerate massive particles to relativistic speeds, the incremental energy required to do so turns out to be proportional to γm, not m. You can do this experiment most easily (!) in a cyclotron but relativistic effects are just measurable with megavolt electrons, which you can generate with a simple Van de Graaff. Which, come to think of it, is also cheating as I suspect few of our other correspondents have one in the backyard, so we'll have to look for an astronomical example, or you can take my word for it.

That equation is not correct for photons. You say "cheating" where I'd say "writing incorrect equations".
The one which is always true in SR is the one I've already written:

E² = (mc²)² + (cp)²     (1)

where "m" is the (invariant, aka proper, aka rest) system's mass. I prefer "invariant" because, for particles moving at c, the terms "proper" and "rest" are meaningless.

Now let's apply it to a system of a couple e⁺ e^- which disintegrates into a couple of γ photons in the couple's centre of mass frame of reference:

a. Before disintegration.
System's mass m is 511 keV/c² + 511 keV/c² = 1022 keV/c².
System's momentum p is zero because we are in the couple's centre of mass. Using equation (1) we have:
→ E = 1022 keV

b. After disintegration in a couple of γ photons.
The system's energy E is conserved and so E is the same as before, 1022 keV (when I say "system's energy" I'm obviously referring to the total energy of the system of the two photons taken together).
Let's see the system's mass m.
From (1):

m = (1/c²)sqrt[E² - (cp)²]

Because of symmetry, the two photons have the same energy and opposite direction of propagation, so their total momentum p sum up to zero. So:

m = (1/c²)sqrt[E² - 0] = 1022 keV/c².

So, invariant mass is conserved, in this case.
I didn't have any need of relativistic mass.
It was just *an example* of the fact that people usually (I'm not referring to you) have the incorrect idea that "energy is transformed in mass" or the other way round, because the concept of "mass" is not as straightforward as we usually think.

--
lightarrow

[lightarrow]

...and when my photon beams in air or tissue exceed 1.022 MeV we get pair production..

When you wrote this you made it precisely clear that something else was needed and that made it clear that the photons were not alone. In that case lightarrow was wrong in attempting to "correct" you. We all know that air and tissue contain nuclei.

Ok, but the fact a photon is travelling in air or in a biological tissue does not ensure that it collides with a nucleus, unless you specify that it has done it. Indeed, how could they make radiographies to you, if the absorption coefficient of tissues/bones for X-rays photons were 100%?

And gamma photons are even more penetrating...

--
lightarrow

[lightarrow]

I'm an experimental physicst, and when my photon beams in air or tissue exceed 1.022 MeV we get pair production, signalled by the appearance of two 511 keV photons, exactly as predicted by E = mc² and the known mass of the electron (another experimental value).

I imagine you have already made that computation even in the frame of reference of a fast moving nucleus on which the photon collides, as high energy particle physicists do every day. Did you use E = mc² in that case?
In case, please ask to a particle or nuclear physicist, which is the equation he uses...

--
lightarrow

[alancalverd]

I lioke my patients to be stationary. Sometimes we anaesthetise them, and from time to time they are even dead. The dead ones don't produce good functional data but there's no doubt about their mass  - the padre sees to that.

[PmbPhy (OP)]

Ok, but the fact a photon is travelling in air or in a biological tissue does not ensure that it collides with a nucleus, unless,

Why did you think he posted that part about air and tissue? It's because he wanted to make it clear that there'd be nuclei in the way for a chance for a collision to take place. I knew that. The question in my mind is why you didn't.

[PmbPhy (OP)]

relativistic mass = γm = m√(1-v²/c²)

That's an equality for relativistic mass and proper mass for a tardyon. The definition for relativistic mass is m= p/v which holds for photons.

If you were to look in SR texts which use relativistic mass then you'd see them use this definition of the relativistic mass to find the rel-mass of a photon. E.g. see

http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm

[lightarrow]

I lioke my patients to be stationary. Sometimes we anaesthetise them, and from time to time they are even dead. The dead ones don't produce good functional data but there's no doubt about their mass  - the padre sees to that.

And all this means what? That you can use E = m*c² ? Of course you can use an equation which is valid only in a special case, if you use it only in that special case. For example I can say that Earth gravity doesn't depend on height if height variations are small and the distance from the Earth centre is large; or I can say that pendulum oscillations are isochronous if the oscillation angles are <<1, or I can say that a body's total energy doesn't vary with its speed, if its kinetic energy is much smaller than its mass multiplied by c². If you want to restrict physics to such special cases, why talking about relativistic effects at all and so, for example, talking about E = m*c² ?

Edited the format. 07/12/2014

--
lightarrow

[lightarrow]

Ok, but the fact a photon is travelling in air or in a biological tissue does not ensure that it collides with a nucleus, unless,

Why did you think he posted that part about air and tissue? It's because he wanted to make it clear that there'd be nuclei in the way for a chance for a collision to take place. I knew that. The question in my mind is why you didn't.

I know, you know and he knows. Do everyone who read it understand that it's necessary a nucleus and that an electron is not enough, for example?

--
lightarrow

[yor_on]

Lightarrow, you're one of the brightest lights here, not always comfortable though.
Whoever told me that science is comfortable?

So do your thing, I will do mine, and in the end we might get a good laugh of it.

[PmbPhy (OP)]

Ok, but the fact a photon is travelling in air or in a biological tissue does not ensure that it collides with a nucleus, unless,

Why did you think he posted that part about air and tissue? It's because he wanted to make it clear that there'd be nuclei in the way for a chance for a collision to take place. I knew that. The question in my mind is why you didn't.

I know, you know and he knows. Do everyone who read it understand that it's necessary a nucleus and that an electron is not enough, for example?

--
lightarrow

If that's the case, i.e. you were worried about the forum not realizing that then you could have simply pointed  that out to the forum instead of Alan. Saying that to Alan makes it appear as if our friend Alan doesn't know what he's talking about and at least I know better than that.

[lightarrow]

If that's the case, i.e. you were worried about the forum not realizing that then you could have simply pointed  that out to the forum instead of Alan. Saying that to Alan makes it appear as if our friend Alan doesn't know what he's talking about and at least I know better than that.

Excuse me, Sir, but how can I know if he knows it or not? I do not have gifts of divination  []
How many times we correct people and we are corrected from others (me more than you, certainly) simply because of what we/others have written? Or should we always have to write "in order to inform people reading this post, I have to correct this phrase/equation/concept/reasoning even if I know (or presume to know) that the poster knows very well the subject"?
 []

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lightarrow

[lightarrow]

Lightarrow, you're one of the brightest lights here,

because of the nickname?  []

not always comfortable though.
Whoever told me that science is comfortable?
So do your thing, I will do mine, and in the end we might get a good laugh of it.

Ok.

--
lightarrow

[PmbPhy (OP)]

Excuse me, Sir, but how can I know if he knows it or not?

Because in post #5 he wrote I'm an experimental physicist,.. and as such he'd know it.

[alancalverd]

Trust me. I'm an experimental physicist.

If pair production occured in vacuo the cosmos would be full of 511 keV gamma rays. I'm sure that most of the people who follow this forum know it isn't.

[lightarrow]

Excuse me, Sir, but how can I know if he knows it or not?

Because in post #5 he wrote I'm an experimental physicist,.. and as such he'd know it.

Ah, ok. So if I say that I'm a theoretical physicist you stop talking about relativistic mass?
 []

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lightarrow.

[Bill S]

Pete, when you have contributed so generously to my “poll”, it feels a bit mean not to have reciprocated by joining yours.  However, I’m sure you will understand that I don’t want to get too far out of my depth. 

Before I read this thread I thought relativistic mass was effectively equivalent to inertia. 

Now I’ve read the thread, I thing “I don’t know nuffin' ”.   []

[PmbPhy (OP)]

I was speaking to a friend of mine, Dr. Wolfgang Rindler, who's an authority in in the field of relativity and asked him

When I start writing my paper I need to get a good feeling of how many people either use relativistic mass or who find it useful etc. I can't figure out what gives all these physicists who claim that it's not used anymore the idea that such is the case. So how do I go about finding out what percentage of relativists use it? Any ideas? Thanks.

He gave me permission to quote his response as follows

Most relativists use relativistic mass, whereas particle physicists use rest mass, and they are the main consumers of SR!  Best,  W

[lightarrow]

"On the Abuse and Use of Relativistic Mass".
Gary Oas.

http://arxiv.org/abs/physics/0504110

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lightarrow

[alancalverd]

Excuse me, Sir, but how can I know if he knows it or not?

Because in post #5 he wrote I'm an experimental physicist,.. and as such he'd know it.

Ah, ok. So if I say that I'm a theoretical physicist you stop talking about relativistic mass?
 []

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lightarrow.

Alas, theoretical physics provides plenty of insights but no actual knowledge.