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Calling pmb_phy

by Aer
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Aer
#73
Aug5-05, 01:05 AM
P: 214
Quote Quote by JesseM
are you or are you not disagreeing with the assertion that the theory of relativity says the inertia of a compound object is proportional to its total energy?
I provided you a link that stated otherwise - that is, the assertion you are making above is incorrect.

Quote Quote by JesseM
So you agree the binding energy contributes to the inertial mass of the reactants, that their inertial mass is not solely the sum of the rest masses of all the particles involved?
Where do you think binding energy comes from? It comes from the particles rest masses, not any type of kinetic or potential energy! That is fundamental.

After the objects bind, this binding energy is forever lost and thus the inertial mass of the new object is less than the combined inertial masses of the two objects before binding. Again - nothing to do with kinetic and potential energy increasing an objects inertia.

I never thought I'd have to remember concepts learned in chemistry class on a relativity forum.
Aer
#74
Aug5-05, 01:13 AM
P: 214
Quote Quote by εllipse
Actually, there is. Inertial mass is caused by a body's resistance to acceleration, so if increase in speed = increase in inertia = increase in inertial mass = increase in gravitational mass = increase in weight is correct, which it seems to be to me from the reasoning I outlined earlier, then relativistic mass would seem to atleast be related to inertial mass, if not equivalent.
As an object is accelerating within Earth's gravitational field, if it approaches .9c, it will still accelerate all the same with the same amount of force in its own frame. (i.e. relativistic mass does not equal gravitational mass) and inertial mass = rest mass
JesseM
#75
Aug5-05, 01:23 AM
Sci Advisor
P: 8,470
Quote Quote by Aer
I provided you a link that stated otherwise - that is, the assertion you are making above is incorrect.
The link you provided was ambiguous, and I provided four links to back up what I'm saying. And as I said, I emailed the author of the page you referred to, if he ends up supporting my position will that change your mind in any way?
Quote Quote by Aer
Where do you think binding energy comes from? It comes from the particles rest masses, not any type of kinetic or potential energy!
Huh? The binding energy is the energy it takes to pull the atoms apart, and the atoms are held together by electromagnetic forces. So, the binding energy is just the difference between the electromagnetic potential when the atoms are in the bound state vs. the electromagnetic potential when they are moved arbitrarily far apart (the system naturally stays bound because the potential energy is lower, although in some cases there may be a 'hump' in the potential where the potential becomes higher when you start to move them apart but then goes lower when they're even farther apart, so some molecules can release energy when broken apart, as described on this page).
Aer
#76
Aug5-05, 01:28 AM
P: 214
Quote Quote by JesseM
The link you provided was ambiguous, and I provided four links to back up what I'm saying. And as I said, I emailed the author of the page you referred to, if he ends up supporting my position will that change your mind in any way?
I did not find it ambiguous, what part of, -mass is frame independent- is unclear. Or did you just choose to ignore that?

Quote Quote by JesseM
Huh? The binding energy is the energy it takes to pull the atoms apart,
Ahh yes, you can tell I am not a Chem person. This is true - I stated it backwards. Anyway, the binding energy still comes from the rest mass of the object that is being split. My argument was still correct in substance.
JesseM
#77
Aug5-05, 01:33 AM
Sci Advisor
P: 8,470
Quote Quote by Aer
I did not find it ambiguous, what part of, -mass is frame independent- is unclear. Or did you just choose to ignore that?
Like I said, there was also the issue of whether he was talking about a compound object where all the parts were at rest with regard to each other. Again, if he ends up supporting my position will this change your mind at all? I want a commitment in advance on this!
Quote Quote by Aer
Ahh yes, you can tell I am not a Chem person. This is true - I stated it backwards. Anyway, the binding energy still comes from the rest mass of the object that is being split.
No, as I said it comes from the difference in potential energy between the bound state and the unbound state. The sum of the rest masses of the particles doesn't change when you split them apart.
Aer
#78
Aug5-05, 01:37 AM
P: 214
Quote Quote by JesseM
Like I said, there was also the issue of whether he was talking about a compound object where all the parts were at rest with regard to each other. Again, if he ends up supporting my position will this change your mind at all? I want a commitment in advance on this!
Why would I make up an opinion based only on some random person's opinion? Only experimental proof will change my mind - find that.


Quote Quote by JesseM
No, as I said it comes from the difference in potential energy between the bound state and the unbound state. The sum of the rest masses of the particles doesn't change when you split them apart.
That is part of the rest mass, is it not?
Aer
#79
Aug5-05, 01:41 AM
P: 214
From wikipedia:

Some books follow this up by stating that "mass and energy are equivalent", but this is somewhat misleading. The mass of an object, as we have defined it, is a quantity intrinsic to the object, and independent of our current frame of reference. The energy E, on the other hand, varies with the frame of reference; if the frame is moving at a high velocity relative to the object, E will be very large, simply because the object has a lot of kinetic energy in that frame. Thus, E = mc2 is not a "good" relativistic statement; it is true only in the rest frame of the object.
So doesn't this mean either the above is incorrect or "mass of a compound object is a measure of its total energy" is incorrect.
Aer
#80
Aug5-05, 01:43 AM
P: 214
Quote Quote by JesseM
The sum of the rest masses of the particles doesn't change when you split them apart.
From wikipedia:
Because a bound system is at a lower energy level, its mass must be less than its unbound constituents. Nuclear binding energy can be computed from the difference in mass of a nucleus, and the sum of the mass of the neutrons and protons that make up the nucleus. Once this mass difference (also called the mass defect) is known, Einstein's formula (E = mc˛) can then be used to compute the binding energy of any nucleus.
JesseM
#81
Aug5-05, 01:50 AM
Sci Advisor
P: 8,470
Quote Quote by Aer
Why would I make up an opinion based only on some random person's opinion? Only experimental proof will change my mind - find that.
Shifting the goalposts again. As I've said over and over, I am only asking about what the theory of relativity predicts. Surely the opinion of professional physicists should have some influence on whether you accept my claim that the theory predicts inertia is proportional to total energy, no? If all the physicists in the world claimed that the theory does predict this, would you still somehow argue that they are all wrong about what the theory "really" predicts?
Quote Quote by Aer
That is part of the rest mass, is it not?
Uh, no, potential energy is not part of the rest mass, since I just told you the sum of rest masses doesn't change when you change the potential.
Quote Quote by Aer (quoting wikipedia)
Some books follow this up by stating that "mass and energy are equivalent", but this is somewhat misleading. The mass of an object, as we have defined it, is a quantity intrinsic to the object, and independent of our current frame of reference. The energy E, on the other hand, varies with the frame of reference; if the frame is moving at a high velocity relative to the object, E will be very large, simply because the object has a lot of kinetic energy in that frame. Thus, E = mc2 is not a "good" relativistic statement; it is true only in the rest frame of the object.
Of course, I agree 100% with this. What's your point?
Quote Quote by Aer (quoting wikipedia)
Because a bound system is at a lower energy level, its mass must be less than its unbound constituents. Nuclear binding energy can be computed from the difference in mass of a nucleus, and the sum of the mass of the neutrons and protons that make up the nucleus. Once this mass difference (also called the mass defect) is known, Einstein's formula (E = mc^2) can then be used to compute the binding energy of any nucleus.
Here I think they are talking about inertial mass (or gravitational mass, which would be the same), not rest mass (or you could say they are talking about the rest mass of the nucleus, but with the understanding that the 'rest mass' of a compound system is defined as its total energy divided by c^2).
Aer
#82
Aug5-05, 01:52 AM
P: 214
Quote Quote by JesseM
Here I think they are talking about inertial mass (or gravitational mass, which would be the same), not rest mass.
Inertial mass IS rest mass

Even when you are shown to be wrong, you still claim you are right!

What other mass would they be refering to? Mass only has one definition in the equation E = mc^2 and that is inertial mass.

Here, maybe numbers will help you:

A deuteron is the nucleus of a deuterium atom, and consists of one proton and one neutron. The masses of the constituents are:

mproton = 1.007276 u (u is Atomic mass unit)
mneutron= 1.008665 u
mproton + mneutron = 1.007276 + 1.008665 = 2.015941 u

The mass of the deuteron is:

Atomic mass 2H = 2.013553 u
And if rest mass is not inertial mass, then which of the following is it?

Strictly speaking, there are three different quantities called mass:

* Inertial mass is a measure of an object's inertia: its resistance to changing its state of motion when a force is applied. An object with small inertial mass changes its motion more readily, and an object with large inertial mass does so less readily.
* Passive gravitational mass is a measure of the strength of an object's interaction with the gravitational field. Within the same gravitational field, an object with a smaller passive gravitational mass experiences a smaller force than an object with a larger passive gravitational mass. (This force is called the weight of the object. In informal usage, the word "weight" is often used synonymously with "mass", because the strength of the gravitational field is roughly constant everywhere on the surface of the Earth. In physics, the two terms are distinct: an object will have a larger weight if it is placed in a stronger gravitational field, but its passive gravitational mass remains unchanged.)
* Active gravitational mass is a measure of the strength of the gravitational field due to a particular object. For example, the gravitational field that one experiences on the Moon is weaker than that of the Earth because the Moon has less active gravitational mass.
JesseM
#83
Aug5-05, 02:00 AM
Sci Advisor
P: 8,470
Quote Quote by Aer
Inertial mass IS rest mass
For a compound system, this is only true if you define its total "rest mass" as its total energy divided by c^2. This is how physicists define things, but if you choose to make up your own idiosyncratic definition where the compound system's rest mass is just the sum of the rest mass of its parts, then inertial mass and rest mass will not be the same. And of course, if you do this, you'll be hard-pressed to explain why the reactants in a chemical reaction have a different inertial mass than the products, even though all the constituent particles are the same (as far as I know massive particles like protons, electrons and neutrons are not created or destroyed in chemical reactions--photons may be, but they have zero rest mass).
Quote Quote by Aer
Even when you are shown to be wrong, you still claim you are right!
That source doesn't "show" that inertial mass IS rest mass, it doesn't address the issue at all--you are just making an assumption.
learningphysics
#84
Aug5-05, 02:03 AM
HW Helper
P: 4,124
Quote Quote by Aer
From wikipedia:

Because a bound system is at a lower energy level, its mass must be less than its unbound constituents. Nuclear binding energy can be computed from the difference in mass of a nucleus, and the sum of the mass of the neutrons and protons that make up the nucleus. Once this mass difference (also called the mass defect) is known, Einstein's formula (E = mc^2) can then be used to compute the binding energy of any nucleus.
Yes... this precisely shows that the rest mass of the nucleus is not simply the sum of the rest masses of the constituent particles. You need to take into account the energy content.

Nuclear binding energy is just one form of energy that leads to mass... it is by no means the only one.

A hydrogen atom weighs slightly less than the sum of the masses of a proton and electron. The difference in mass is due to kinetic energy, and electrostatic potential energy between the electron and proton.
JesseM
#85
Aug5-05, 02:12 AM
Sci Advisor
P: 8,470
Quote Quote by Aer
What other mass would they be refering to? Mass only has one definition in the equation E = mc^2 and that is inertial mass.
No, the m in that equation is rest mass.
Quote Quote by Aer
Here, maybe numbers will help you:
A deuteron is the nucleus of a deuterium atom, and consists of one proton and one neutron. The masses of the constituents are:

mproton = 1.007276 u (u is Atomic mass unit)
mneutron= 1.008665 u
mproton + mneutron = 1.007276 + 1.008665 = 2.015941 u

The mass of the deuteron is:

Atomic mass 2H = 2.013553 u
What's your point? I am sure they are defining the mass of the deuteron as its total energy divided by c^2. If they were just defining its mass as the sum of the rest masses of all the particles, then why do you think its mass is not 2.015941 u, assuming you acknowledge that "binding energy" does not involve any extra particles with nonzero rest mass?
Quote Quote by Aer
And if rest mass is not inertial mass, then which of the following is it?
Strictly speaking, there are three different quantities called mass:

* Inertial mass is a measure of an object's inertia: its resistance to changing its state of motion when a force is applied. An object with small inertial mass changes its motion more readily, and an object with large inertial mass does so less readily.
* Passive gravitational mass is a measure of the strength of an object's interaction with the gravitational field. Within the same gravitational field, an object with a smaller passive gravitational mass experiences a smaller force than an object with a larger passive gravitational mass. (This force is called the weight of the object. In informal usage, the word "weight" is often used synonymously with "mass", because the strength of the gravitational field is roughly constant everywhere on the surface of the Earth. In physics, the two terms are distinct: an object will have a larger weight if it is placed in a stronger gravitational field, but its passive gravitational mass remains unchanged.)
* Active gravitational mass is a measure of the strength of the gravitational field due to a particular object. For example, the gravitational field that one experiences on the Moon is weaker than that of the Earth because the Moon has less active gravitational mass.
This is from the wikipedia entry on mass, and they say in the section on relativity that "the quantity m has a simple physical meaning: it is the inertial mass of the object as measured in its rest frame, the frame of reference in which its velocity is zero." This definition implies that the inertial mass of the object when measured in a frame other than its rest frame will not be equal to the "m" in E=mc^2 (ie the rest mass)--and of course this is true, objects with large velocities are harder to accelerate than objects that have the same rest mass but smaller velocities, which by definition means they have different inertial masses.
εllipse
#86
Aug5-05, 02:15 AM
P: 195
Quote Quote by Aer
What other mass would they be refering to? Mass only has one definition in the equation E = mc^2 and that is inertial mass.
Actually, isn't the m in [tex]E=mc^2[/tex] relativistic mass (if applied to other reference frames)? I thought the invariant mass version of the equation was [tex]E^2=p^2c^2+m^2c^4[/tex].
Aer
#87
Aug5-05, 02:15 AM
P: 214
Quote Quote by JesseM
And of course, if you do this, you'll be hard-pressed to explain why the reactants in a chemical reaction have a different inertial mass than the products, even though all the constituent particles are the same (as far as I know massive particles like protons, electrons and neutrons are not created or destroyed in chemical reactions--photons may be, but they have zero rest mass).
It takes energy to bind the proton and neutron together, no?
Aer
#88
Aug5-05, 02:17 AM
P: 214
Quote Quote by JesseM
No, the m in that equation is rest mass.
From this page
The rest mass (m) of a particle is the mass defined by the energy of the isolated (free) particle at rest, divided by c 2 . When particle physicists use the word ``mass,'' they always mean the ``rest mass'' (m) of the object in question
JesseM
#89
Aug5-05, 02:18 AM
Sci Advisor
P: 8,470
Quote Quote by Aer
It takes energy to bind the proton and neutron together, no?
Yup, and of course this fits with my claim that the inertial mass of the deuteron is equal to its total energy divided by c^2, but it doesn't fit too well with your claim that its inertial mass is dependent only on the sum of the rest masses of its parts, and not on any other forms of energy that may be in the deuteron.
Aer
#90
Aug5-05, 02:20 AM
P: 214
Quote Quote by JesseM
What's your point? I am sure they are defining the mass of the deuteron as its total energy divided by c^2. If they were just defining its mass as the sum of the rest masses of all the particles, then why do you think its mass is not 2.015941 u, assuming you acknowledge that "binding energy" does not involve any extra particles with nonzero rest mass?
From this page
the energy that holds a nucleus together; the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus.

So is the energy not a part of the combined system?


Sorry that I have to keep refering to authority - Chemistry is not my strong point.


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