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Rasalhague
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Some naive questions about the meaning of this expression.
In what follows, I'll use the word mass for "rest mass", the magnitude of the energy-momentum 4-vector. (Answers in terms of "relativistic mass" are fine, just let me know what definitions you're using. Taylor & Wheeler in Spacetime Physics sometimes use the term rest mass, as in the quote below, to avoid ambiguity, even though they depricate the term relativistic mass.)
In what sense is this equivalence to be understood? Presumably not in the obvious sense that any [itex]E[/itex] in an equation can be replaced with [itex]m[/itex] or [itex]mc^2[/itex]. Not the former because it leads to this not-generally true conclusion:
[tex]E = \gamma m[/tex]
[tex]E = mc^2[/tex]
[tex]\therefore \gamma = c^2[/tex]
Or if
then treating [itex]E[/itex] as equivalent to [itex]m[/itex] would lead to the conclusion that [itex]2E[/itex] and [itex]2m[/itex] are conserved while [itex]E[/itex] and [itex]m[/itex] are not, and that therefore 2 is not conserved!
And a photon is said to have no mass, and therefore its energy, [itex]E[/itex], is equal to its 3-momentum, [itex]p[/itex] because
[tex]\left(m_{phot}\right)^2 = \left(E_{phot}\right)^2 - \left(p_{phot}\right)^2[/tex]
[tex]0 = \left(E_{phot}\right)^2 - \left(p_{phot}\right)^2,[/tex]
which argument wouldn't work if [itex]m[/itex] or [itex]mc^2[/itex] were notationally equivalent to [itex]E[/itex].
So is "energy and mass are equivalent" merely a statement of the fact that, once the appropriate factors of [itex]c[/itex] are included, energy and mass have the same units, in which case could we just as well say that mass and energy and 3-momentum are "equivalent" to each other, in this sense, although not, of course, equivalent in the obvious sense that a statement about mass would be true if and only if it's true also of energy and 3-momentum.
Or is the equivalence something to do with a change from mass to energy, or vice versa, in a physical interaction? In Spacetime Physics, in chapter 2, section 13 "Equivalence of energy and rest mass" (which a previous owner of my copy has corrected to "Equivalence of rest energy and mass"), Taylor and Wheeler give the example of an inelastic collision between two lumps of putty which merge as a result of the collision. Before the collision, the first lump is moving, the second at rest. They use T for kinetic energy, subscript 1 for properties of the first lump, subscript 2 for properties of the second, and subscripts "initial" and "final" for the total mass, energy and 3-momentum of the system before and after the collision. From the conservation of energy and 3-momentum, and from the fact that the energy of a body at rest is equal to its mass, they reason:
[tex]E_{initial} = E_{final} = E_1 + E_2 = E_1 + m_2[/tex]
[tex]p_{initial} = p_{final} = p_1[/tex]
[tex]\left(m_{final}\right)^2 = \left(E_{final}\right)^2-\left(p_{final}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \left(E_1 + m_2\right)^2-\left(p_1\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \mathbf{\left(E_1\right)^2 - \left(p_1\right)^2} + \left(m_2\right)^2 + 2E_1m_2 = \left(m_{initial}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \mathbf{\left(m_1\right)^2} + \left(m_2\right)^2 + 2\left(m_1 + T_1\right)m_2 = \left(m_{initial}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \left(m_1\right)^2 + \left(m_2\right)^2 + 2m_1m_2 + 2T_1m_2 = \left(m_{initial}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \left(m_1+m_2\right)^2+2T_1m_2 = \left(m_{initial}\right)^2[/tex]
(I've filled in some of the steps there.) Conclusion:
In connection with this, they say on p. 134 that "rest mass often changes in an inelastic encounter", contrasting the property of (frame-)invariance with conservation.
My question here: isn't the rest mass of the whole system always conserved? How can it not be, since it depends only on the energy of the whole system (which is conserved) and the momentum of the whole system (which is conserved)?
Perhaps I'm missing something, but the idea that rest mass "changes" in an inelastic collision, seems to rely on adopting an inconsistent convention whereby, before the collision, one uses the word "mass" for the sum of the individual masses of the separate bodies, whereas after the collision one switches the referent of "mass" to the mass of the whole system.
The following quote deals with a similar, putty-based example:
I'm not sure what "effective mass" means in Taylor & Wheeler's terms. At first I thought it might be energy, give or take the relevant conversion factor, that some people call relativistic mass, the time component of the energy-momentum 4-vector, since this changes from frame to frame as [itex]\gamma[/itex] changes. But from the situation being described, this "effective mass" seems rather to play the role of Taylor & Wheeler's rest mass of the system. For them, the famous equation means, in natural units, "rest energy equals mass", i.e. in a frame where the total momentum of a system is zero, its rest mass is equal to the sum of the energies of its individual particles.
Or does the statement that mass and energy are equivalent relate to the fact that mass is the source of gravity in Newtonian mechanics, whereas in GR, the source of gravity is energy-stress, whose components include energy density and energy flux density, among other things, and therefore energy, in GR, is partly responsible for what mass is wholly responsible for in Newtonian mechanics, and therefore total relativistic energy is very roughly "equivalent" to Newtonian mass in some simplified special case?
In what follows, I'll use the word mass for "rest mass", the magnitude of the energy-momentum 4-vector. (Answers in terms of "relativistic mass" are fine, just let me know what definitions you're using. Taylor & Wheeler in Spacetime Physics sometimes use the term rest mass, as in the quote below, to avoid ambiguity, even though they depricate the term relativistic mass.)
bcrowell said:Let's say you have a photon of energy E at a distance r from a material particle of mass M. Then the force between them can be found by using mass-energy equivalence, [itex]E=mc^2[/itex], giving [itex]F=GME/c^2r^2[/itex].
Mass and energy are equivalent (B. Crowell: Simple Nature 7.3.5).
In what sense is this equivalence to be understood? Presumably not in the obvious sense that any [itex]E[/itex] in an equation can be replaced with [itex]m[/itex] or [itex]mc^2[/itex]. Not the former because it leads to this not-generally true conclusion:
[tex]E = \gamma m[/tex]
[tex]E = mc^2[/tex]
[tex]\therefore \gamma = c^2[/tex]
Or if
[...] in relativity there are no separate laws of conservation of energy and conservation of mass. There is only a law of conservation of mass plus energy (referred to as mass-energy). In natural units, [itex]E+m[/itex] is conserved, while in ordinary units the conserved quantity is [itex]E+mc^2[/itex]" (Simple Nature 7.3.5)
then treating [itex]E[/itex] as equivalent to [itex]m[/itex] would lead to the conclusion that [itex]2E[/itex] and [itex]2m[/itex] are conserved while [itex]E[/itex] and [itex]m[/itex] are not, and that therefore 2 is not conserved!
And a photon is said to have no mass, and therefore its energy, [itex]E[/itex], is equal to its 3-momentum, [itex]p[/itex] because
[tex]\left(m_{phot}\right)^2 = \left(E_{phot}\right)^2 - \left(p_{phot}\right)^2[/tex]
[tex]0 = \left(E_{phot}\right)^2 - \left(p_{phot}\right)^2,[/tex]
which argument wouldn't work if [itex]m[/itex] or [itex]mc^2[/itex] were notationally equivalent to [itex]E[/itex].
So is "energy and mass are equivalent" merely a statement of the fact that, once the appropriate factors of [itex]c[/itex] are included, energy and mass have the same units, in which case could we just as well say that mass and energy and 3-momentum are "equivalent" to each other, in this sense, although not, of course, equivalent in the obvious sense that a statement about mass would be true if and only if it's true also of energy and 3-momentum.
Or is the equivalence something to do with a change from mass to energy, or vice versa, in a physical interaction? In Spacetime Physics, in chapter 2, section 13 "Equivalence of energy and rest mass" (which a previous owner of my copy has corrected to "Equivalence of rest energy and mass"), Taylor and Wheeler give the example of an inelastic collision between two lumps of putty which merge as a result of the collision. Before the collision, the first lump is moving, the second at rest. They use T for kinetic energy, subscript 1 for properties of the first lump, subscript 2 for properties of the second, and subscripts "initial" and "final" for the total mass, energy and 3-momentum of the system before and after the collision. From the conservation of energy and 3-momentum, and from the fact that the energy of a body at rest is equal to its mass, they reason:
[tex]E_{initial} = E_{final} = E_1 + E_2 = E_1 + m_2[/tex]
[tex]p_{initial} = p_{final} = p_1[/tex]
[tex]\left(m_{final}\right)^2 = \left(E_{final}\right)^2-\left(p_{final}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \left(E_1 + m_2\right)^2-\left(p_1\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \mathbf{\left(E_1\right)^2 - \left(p_1\right)^2} + \left(m_2\right)^2 + 2E_1m_2 = \left(m_{initial}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \mathbf{\left(m_1\right)^2} + \left(m_2\right)^2 + 2\left(m_1 + T_1\right)m_2 = \left(m_{initial}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \left(m_1\right)^2 + \left(m_2\right)^2 + 2m_1m_2 + 2T_1m_2 = \left(m_{initial}\right)^2[/tex]
[tex]\left(m_{final}\right)^2 = \left(m_1+m_2\right)^2+2T_1m_2 = \left(m_{initial}\right)^2[/tex]
(I've filled in some of the steps there.) Conclusion:
The rest mass of the final system is greater than the sum of the rest masses of the original colliding objects. [...] The increase in rest mass measures precisely the energy which has gone into heat and whirling and any other form of internal excitation of the final system" (Taylor & Wheeler:Spacetime Physics, p. 121.)
In connection with this, they say on p. 134 that "rest mass often changes in an inelastic encounter", contrasting the property of (frame-)invariance with conservation.
My question here: isn't the rest mass of the whole system always conserved? How can it not be, since it depends only on the energy of the whole system (which is conserved) and the momentum of the whole system (which is conserved)?
Perhaps I'm missing something, but the idea that rest mass "changes" in an inelastic collision, seems to rely on adopting an inconsistent convention whereby, before the collision, one uses the word "mass" for the sum of the individual masses of the separate bodies, whereas after the collision one switches the referent of "mass" to the mass of the whole system.
The following quote deals with a similar, putty-based example:
Now we know that mass is invariant, and no molecules were created or destroyed, so the masses of all the molecules must be the same as they always were. The change is due to the change in [itex]\gamma[/itex] with heating, not to a change in [itex]m[/itex]. But how much does the mass appear to change? On page 858 we prove that the perceived change in mass exactly equals the change in heat energy between two temperatures, i.e., changing the heat energy by an amount [itex]E[/itex] changes the effective mass of an object by [itex]E[/itex] as well. This looks a bit odd because the natural units of energy and mass are the same. Converting back to ordinary units by our usual shortcut of introducing factors of [itex]c[/itex], we find that changing the heat energy by an amount E causes the apparent mass to change by [itex]m=E/c^2[/itex]. Rearranging, we have the famous [itex]E=mc^2[/itex]" (B. Crowell: Simple Nature 7.3.5).
I'm not sure what "effective mass" means in Taylor & Wheeler's terms. At first I thought it might be energy, give or take the relevant conversion factor, that some people call relativistic mass, the time component of the energy-momentum 4-vector, since this changes from frame to frame as [itex]\gamma[/itex] changes. But from the situation being described, this "effective mass" seems rather to play the role of Taylor & Wheeler's rest mass of the system. For them, the famous equation means, in natural units, "rest energy equals mass", i.e. in a frame where the total momentum of a system is zero, its rest mass is equal to the sum of the energies of its individual particles.
Or does the statement that mass and energy are equivalent relate to the fact that mass is the source of gravity in Newtonian mechanics, whereas in GR, the source of gravity is energy-stress, whose components include energy density and energy flux density, among other things, and therefore energy, in GR, is partly responsible for what mass is wholly responsible for in Newtonian mechanics, and therefore total relativistic energy is very roughly "equivalent" to Newtonian mass in some simplified special case?
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