Iwonderaboutthings Posted September 3, 2013 Posted September 3, 2013 What is rest mass as per QM, and even Physics. When I think of rest mass I think of it this way.. A car is moving slowly, the engine works up energy to " pull" the driver, passengers and all things in this car forward. When the car starts speeding up to a constant velocity it is thus out of its rest mass??? The mass of the car I am assuming is stationary>>? But it seems this was true when its in constant velocity>>? I think in this way because of fermions and bosons. Fermions having mass? and photons not having mass but being related to time? I have read over the internet about rest mas, but still it makes no sense whats so ever to me.. Thanks in advance
swansont Posted September 3, 2013 Posted September 3, 2013 The rest mass is the mass a particle has at rest. This is also equivalent to the energy it has at rest. I'm not sure of the history, but it seems to be a distinction between it and and another term that unfortunately cropped up, relativistic mass, which is E/c^2.
pears Posted September 3, 2013 Posted September 3, 2013 OK this is probably a really stupid question, but since things can only be at rest from a particular reference point, (inertial frame?) how does rest mass fit in with relativity of motion? Is rest mass absolute or relative? I assume it's relative because I assume rest is relative and mass is absolute? 1
ajb Posted September 3, 2013 Posted September 3, 2013 It fits in via the mass-shell constant which all physical particles obey; [math]E^{2}-p^{2}c^{2} = m^{2}c^{4}[/math]. The energy E and the 3-momentum p will be different in different inertial frames but the mass m remains the same.
Delta1212 Posted September 3, 2013 Posted September 3, 2013 OK this is probably a really stupid question, but since things can only be at rest from a particular reference point, (inertial frame?) how does rest mass fit in with relativity of motion? Is rest mass absolute or relative? I assume it's relative because I assume rest is relative and mass is absolute?Rest mass is absolute. It's called rest mass to stand in contrast to relativistic mass, but that's considered something of an out-dated concept, I believe.
pears Posted September 3, 2013 Posted September 3, 2013 Hmm - ok thanks. It will perhaps become clearer to me once I've done a bit more background reading.
ajb Posted September 3, 2013 Posted September 3, 2013 Rest mass is absolute. Right and by this we mean that it takes the same value in any inertial frame; it is a Lorentz scalar.
Iwonderaboutthings Posted September 4, 2013 Author Posted September 4, 2013 (edited) It fits in via the mass-shell constant which all physical particles obey; [math]E^{2}-p^{2}c^{2} = m^{2}c^{4}[/math]. The energy E and the 3-momentum p will be different in different inertial frames but the mass m remains the same. sorry, what is 3-momentum p stand for?? p= mv? is a product from the two from what I understand, or is this wrong? where does constant velocity come in? The rest mass is the mass a particle has at rest. This is also equivalent to the energy it has at rest. I'm not sure of the history, but it seems to be a distinction between it and and another term that unfortunately cropped up, relativistic mass, which is E/c^2. To be simple, when you say " at rest" does this mean "standing still"? Also, I assume that constant velocity is measured out of this " rest " " standing still" mass. But doesn't everything vibrate with a frequency?? Meaning that I have read that atoms, molecules, air pressure, water, thoughts, heat, x rays and etc, all have a vibration to them right? I am not really sure if what I have stated is true , but I assume this would be the case. Here is a video on this: Frequency & The Law of Vibration http://www.youtube.com/watch?v=XEX2-m8EabU Edited September 4, 2013 by Iwonderaboutthings
swansont Posted September 4, 2013 Posted September 4, 2013 To be simple, when you say " at rest" does this mean "standing still"? Also, I assume that constant velocity is measured out of this " rest " " standing still" mass. But doesn't everything vibrate with a frequency?? Meaning that I have read that atoms, molecules, air pressure, water, thoughts, heat, x rays and etc, all have a vibration to them right? I am not really sure if what I have stated is true , but I assume this would be the case. Yes, standing still. An idealized case. You subtract any relativistic effects if they are present in the experiment. However, depending on the experiment, they may be too small to actually measure. 1
ajb Posted September 4, 2013 Posted September 4, 2013 sorry, what is 3-momentum p stand for?? It is just the standard momentum in classical mechanics.
Kramer Posted September 5, 2013 Posted September 5, 2013 Sorry that i intervene in your qualified debate with my lay mans hypothesis.---The only absolute mass is the extrapolated Plank mass = M = e / (4 * pi * epsilon 0 * G) ^ 0.5 .If you see my intervention improper disregard it .
Phi for All Posted September 5, 2013 Posted September 5, 2013 Sorry that i intervene in your qualified debate with my lay mans hypothesis. ---The only absolute mass is the extrapolated Plank mass = M = e / (4 * pi * epsilon 0 * G) ^ 0.5 . If you see my intervention improper disregard it . ! Moderator Note It's always improper to hijack a scientific discussion of a mainstream science concept by introducing your own speculative ideas. You should start your own thread on your idea in Speculations.
Iwonderaboutthings Posted September 8, 2013 Author Posted September 8, 2013 (edited) Sorry that i intervene in your qualified debate with my lay mans hypothesis. ---The only absolute mass is the extrapolated Plank mass = M = e / (4 * pi * epsilon 0 * G) ^ 0.5 . If you see my intervention improper disregard it . This is just my opinion, pi ratio turns up everywhere how could this be related to mass at all?? If what I am seeing is correct then are you're saying that 4*pi ratio coupled with G is correlated with Plank Units?? Maybe this would make sense since you are referring to an extrapolated absolute rest mass, but wouldn't this be impossible using pi ratio?? Isn't pi ratio related to infinity? How do you contain infinity??? The theory of infinity and the laws of relativity between dimensions. http://hubpages.com/forum/topic/79093 If pi ratio is thus related to infinity then how can anything be extrapolated using empty space?? Are your sure the issue is not getting the limit of a physical reality of infinity such as in the case of pi itself? What pertains to this absolute rest mass?? The box of the particle? Particle in a box http://en.wikipedia.org/wiki/Particle_in_a_box I assume e to be the electron charge. From what I know e is indistinguishable?? Is this correct?? Indistinguishable Particles and Exchange http://www2.ph.ed.ac.uk/~gja/qp/qp9.pdf To add is that an exponent of .5?? Meaning that .5*2 = 1 From what I am looking at your equation appears " in-completed." The reason for this is because if at all pi ratio is being used correctly their still is the issue of precession and perhaps this explains the h-bar, give or take pi ratio turns up everywhere, how could this be related to mass at all?? But I may be wrong, what confuses me is the 4*pi ratio and the exponent of .5 Yes, standing still. An idealized case. You subtract any relativistic effects if they are present in the experiment. However, depending on the experiment, they may be too small to actually measure. So then, mass are multiples of itself, and deductions allow calculations from the positions of the multiples? Simple Example Here: In other words, 3*7 = 21/ 2 = 10.5 Where 10.5 is the length and the position of the mass in question. 3 and 7 serve as dx and dp where dp is momentum.. Now what about the uncertainty principle?? dx = unit and location, and dp = momentum, which in tern is the product of m*v but not acceleration... SO WHAT IS DOING ALL THE MULTIPLYING???????????????? Edited September 8, 2013 by Iwonderaboutthings
swansont Posted September 8, 2013 Posted September 8, 2013 So then, mass are multiples of itself, and deductions allow calculations from the positions of the multiples? Simple Example Here: In other words, 3*7 = 21/ 2 = 10.5 Where 10.5 is the length and the position of the mass in question. 3 and 7 serve as dx and dp where dp is momentum.. Now what about the uncertainty principle?? dx = unit and location, and dp = momentum, which in tern is the product of m*v but not acceleration... SO WHAT IS DOING ALL THE MULTIPLYING???????????????? I don't understand your example. One way of determining mass is with a mass spectrometer. A charged particle is given a specific amount of energy and deflected in a electric and/or magnetic field. The amount of deflection depends on the mass, the energy, and the field strength. If you know the field and the energy, measuring the deflection allows you to calculate the mass. The limits of the uncertainty principle are much, much smaller than the experimental error, so experimental error is what limits the precision of the measurement. Masses are known to about 8 significant digits. hbar is a limit out there at the 34th digit. 1
Iwonderaboutthings Posted September 15, 2013 Author Posted September 15, 2013 I don't understand your example. One way of determining mass is with a mass spectrometer. A charged particle is given a specific amount of energy and deflected in a electric and/or magnetic field. The amount of deflection depends on the mass, the energy, and the field strength. If you know the field and the energy, measuring the deflection allows you to calculate the mass. The limits of the uncertainty principle are much, much smaller than the experimental error, so experimental error is what limits the precision of the measurement. Masses are known to about 8 significant digits. hbar is a limit out there at the 34th digit. Is this correct to ask?? --->But doesn't spectroscopic-analysis rely on luminosity? If so then how can " light" be considered mass when light photons are mass-less? Yes I know about radiation and heat as energy and mass, but confused on the whole light deal. Is this something like black body radiation, or the Stephan Boltzmann method>? Stefan–Boltzmann constant http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
swansont Posted September 15, 2013 Posted September 15, 2013 Is this correct to ask?? --->But doesn't spectroscopic-analysis rely on luminosity? If so then how can " light" be considered mass when light photons are mass-less? Yes I know about radiation and heat as energy and mass, but confused on the whole light deal. A mass spectrometer is not doing spectroscopy. It's an instrument to determine the "spectrum" of mass. It works as I described. You measure the particles position when it hits a detector, not by it giving off light. http://en.wikipedia.org/wiki/File:Mass_Spectrometer_Schematic.svg http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/maspec.html 1
LaurieAG Posted September 17, 2013 Posted September 17, 2013 (edited) The rest mass is the mass a particle has at rest. This is also equivalent to the energy it has at rest. I'm not sure of the history, but it seems to be a distinction between it and and another term that unfortunately cropped up, relativistic mass, which is E/c^2. The distinction between relativistic mass and rest mass is a factor of 2 * Pi and the history is interesting. http://en.wikipedia.org/wiki/Compton_wavelength The Compton wavelength of a particle is equivalent to the wavelength of a photon whose energy is the same as the rest-mass energy of the particle. The reduced Compton wavelength is a natural representation for mass on the quantum scale. Equations that pertain to mass in the form of mass, like Klein-Gordon and Schrödinger's, use the reduced Compton wavelength. The non-reduced Compton wavelength is a natural representation for mass that has been converted into energy. Equations that pertain to the conversion of mass into energy, or to the wavelengths of photons interacting with mass, use the non-reduced Compton wavelength. A particle of rest mass m has a rest energy of E = mc^2. The non-reduced Compton wavelength for this particle is the wavelength of a photon of the same energy. Edited September 17, 2013 by LaurieAG 1
elfmotat Posted September 17, 2013 Posted September 17, 2013 It is just the standard momentum in classical mechanics. Only approximately. The form of relativistic 3-momentum is different: [math]p=\frac{mv}{\sqrt{1-v^2/c^2}}[/math]
ajb Posted September 17, 2013 Posted September 17, 2013 Only approximately. The form of relativistic 3-momentum is different: [math]p=\frac{mv}{\sqrt{1-v^2/c^2}}[/math] Right, though I did not want to say too much as people were already getting confused
Iwonderaboutthings Posted September 21, 2013 Author Posted September 21, 2013 (edited) A mass spectrometer is not doing spectroscopy. It's an instrument to determine the "spectrum" of mass. It works as I described. You measure the particles position when it hits a detector, not by it giving off light. http://en.wikipedia.org/wiki/File:Mass_Spectrometer_Schematic.svg http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/maspec.html Great Links! thanks its makes more sense with the models. Right, though I did not want to say too much as people were already getting confused I am still confused on that " 1 " in the denominator, seems like a Lorentz Transform or something.. What does this 1 in the denominator represent? Lorentz Transform? I think it means " initial start but un-like h this can be wrong considering the discrete amounts of energy from what I know. From what I gather 1 is a factor of everything right??? The distinction between relativistic mass and rest mass is a factor of 2 * Pi and the history is interesting. http://en.wikipedia.org/wiki/Compton_wavelength When you say " factor of 2 * Pi" does this mean the number 6.28?? Wouldn't that mean that pi ratio has an anti as in +1 and 1... Pi ratio is also used to measure volume from what I know, would this be the reason why for the Compton Wavelength? Along with sin, cos, tangent the speed of light, G and h, this can be very very confusing Edited September 21, 2013 by Iwonderaboutthings
elfmotat Posted September 21, 2013 Posted September 21, 2013 Great Links! thanks its makes more sense with the models. I am still confused on that " 1 " in the denominator, seems like a Lorentz Transform or something.. What does this 1 in the denominator represent? Lorentz Transform? I think it means " initial start but un-like h this can be wrong considering the discrete amounts of energy from what I know. From what I gather 1 is a factor of everything right??? What do you mean by "what does the 1 represent?" It doesn't represent anything - it's just a number. If you know that something has mass m and is moving at velocity v, then you plug m and v into the equation to get p.
Iwonderaboutthings Posted September 21, 2013 Author Posted September 21, 2013 What do you mean by "what does the 1 represent?" It doesn't represent anything - it's just a number. If you know that something has mass m and is moving at velocity v, then you plug m and v into the equation to get p. Everything has an initial start at 1 thus everything expands from there. From cells, amino acids, " time" etc. 1 has to have some connection with its " origination." Otherwise 1 as a number like you said is just a number, although used is still undefined. I have seen these 1s many many times in many equations, but I gather their must be a method to distinguish them from the origination and source per say, such as h and the Deseret Amounts of energy used in the QM... However this is just a thought, but it is logical.
LaurieAG Posted September 22, 2013 Posted September 22, 2013 When you say " factor of 2 * Pi" does this mean the number 6.28?? Wouldn't that mean that pi ratio has an anti as in +1 and 1... Pi ratio is also used to measure volume from what I know, would this be the reason why for the Compton Wavelength? The reduced Compton wavelength is the standard Compton wavelength divided by 2 * Pi. Pi equals the circumference of a circle divided by twice its radius (or its diameter). 1
Iwonderaboutthings Posted September 22, 2013 Author Posted September 22, 2013 The reduced Compton wavelength is the standard Compton wavelength divided by 2 * Pi. Pi equals the circumference of a circle divided by twice its radius (or its diameter). Ok thanks just wanted to make sure
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