morgsboi Posted March 4, 2012 Share Posted March 4, 2012 Okay, so energy is equal to mass times the speed of light squared. The problem with reaching the speed of light in a spaceship is that the more energy you gain, the more mass you will gain and you will slow down. http://zidbits.com/2011/04/why-cant-anything-go-faster-than-the-speed-of-light/ So why does the mass affect the spaceship in space if you are weightless? Link to comment Share on other sites More sharing options...

Joatmon Posted March 4, 2012 Share Posted March 4, 2012 Okay, so energy is equal to mass times the speed of light squared. The problem with reaching the speed of light in a spaceship is that the more energy you gain, the more mass you will gain and you will slow down. http://zidbits.com/2...speed-of-light/ So why does the mass affect the spaceship in space if you are weightless? Mass and weight are not the same thing. For example, if two bodies of large mass collided in space and you were between them you would surely be squashed even though they are weightless. Link to comment Share on other sites More sharing options...

morgsboi Posted March 4, 2012 Author Share Posted March 4, 2012 Mass and weight are not the same thing. For example, if two bodies of large mass collided in space and you were between them you would surely be squashed even though they are weightless. Ah okay. Yeah I know they aren't the same thing. So would it be possible to make it expand instead of contract? Mass and weight are not the same thing. For example, if two bodies of large mass collided in space and you were between them you would surely be squashed even though they are weightless. Actually, it still doesn't make sense to me how even if the mass is great, what would cause the drag? Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted March 4, 2012 Share Posted March 4, 2012 Actually, it still doesn't make sense to me how even if the mass is great, what would cause the drag? It's not drag or friction that's the problem. More massive objects take more energy to accelerate, even when they're in space. As objects move faster and faster, it takes more and more energy to speed them up. When an object is moving near the speed of light, you can give it huge amounts of energy and it accelerates very little. 1 Link to comment Share on other sites More sharing options...

morgsboi Posted March 4, 2012 Author Share Posted March 4, 2012 It's not drag or friction that's the problem. More massive objects take more energy to accelerate, even when they're in space. As objects move faster and faster, it takes more and more energy to speed them up. When an object is moving near the speed of light, you can give it huge amounts of energy and it accelerates very little. Ah, that really helps, thanks. Link to comment Share on other sites More sharing options...

D H Posted March 4, 2012 Share Posted March 4, 2012 (edited) When you write [math]E=mc^2[/math] the mass to which you are referring is called relativistic mass. Relativistic mass is a concept that most physicists don't think is necessary. All that dragging in the concept of relativistic mass does is confuse the student. What [math]E=mc^2[/math] is saying is that relativistic mass is a synonym for energy -- so why not just use energy? Better is to use rest mass, or intrinsic mass -- the mass an object appears to have to an observer at rest with respect to the object. In terms of intrinsic mass, the mass-energy equivalence formula becomes [math]E^2 = (m_0c^2)^2 + (pc)^2[/math] where [math]p=m_0v/\sqrt{1-(v/c)^2}[/math] is relativistic momentum. Now this is saying something: Energy is a function of rest mass and relativistic momentum. Another way to write this latter expression for energy is [math]E^2 = \left(m_0c^2\right)^2\left(1 + \frac{(v/c)^2}{1-(v/c)^2}\right) = \left(m_0c^2\right)^2 \frac{1}{1-(v/c)^2}[/math] Now look what happens to the rightmost term as v approaches c. It gets ever larger and larger because the denominator [math]1-(v/c)^2\to0\ \text{as}\ v\to c[/math]. The relativistic momentum grows without bounds, and thus so does the energy. Another way to look at it is that it takes an unbounded amount of energy to make something with non-zero mass approach the speed of light. Edit Fixed the numerator and simplified. Edited March 4, 2012 by D H 2 Link to comment Share on other sites More sharing options...

elfmotat Posted March 4, 2012 Share Posted March 4, 2012 (edited) Even more simply: By the work-energy theorem, the energy required to bring a mass to a certain velocity is equal to the change in its kinetic energy. Kinetic energy in Special Relativity is given by: [math]E_k=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2[/math] (m refers to the rest mass.) As you can see, as v grows larger the energy required to increase v grows without bound. If you wanted to bring something to the speed of light, the change in kinetic energy becomes infinite. Edited March 4, 2012 by elfmotat 1 Link to comment Share on other sites More sharing options...

granpa Posted March 5, 2012 Share Posted March 5, 2012 (edited) it is only longitudinal mass that increases. transverse mass is not affected. http://en.wikipedia.org/wiki/Transverse_mass http://en.wikipedia.org/wiki/Mass_in_special_relativity#Transverse_and_longitudinal_mass KE ∝ m(gamma - 1) Edited March 5, 2012 by granpa -1 Link to comment Share on other sites More sharing options...

D H Posted March 5, 2012 Share Posted March 5, 2012 Longitudinal and transverse mass are old, defunct concepts. There is no reason to use them whatsoever. Link to comment Share on other sites More sharing options...

granpa Posted March 5, 2012 Share Posted March 5, 2012 so is relativistic mass Link to comment Share on other sites More sharing options...

swansont Posted March 5, 2012 Share Posted March 5, 2012 Okay, so energy is equal to mass times the speed of light squared. The problem with reaching the speed of light in a spaceship is that the more energy you gain, the more mass you will gain and you will slow down. Relativity doesn't claim that you will slow down. As objects move faster and faster, it takes more and more energy to speed them up. When an object is moving near the speed of light, you can give it huge amounts of energy and it accelerates very little. That's not even a relativistic concept; it holds even in the classical regime where KE=1/2 mv^2. Each incremental amount of added KE gives an increasingly smaller increase in v. In relativity, though, the added speed is even smaller. Link to comment Share on other sites More sharing options...

IM Egdall Posted March 5, 2012 Share Posted March 5, 2012 (edited) Okay, so energy is equal to mass times the speed of light squared. The problem with reaching the speed of light in a spaceship is that the more energy you gain, the more mass you will gain and you will slow down. http://zidbits.com/2...speed-of-light/ So why does the mass affect the spaceship in space if you are weightless? I found this example helped me understand how mass increases with speed per special relativity. Hope it helps you. Astronauts in space are effectively weightless, but they still have mass. They use a contraption called an Inertial Balance Spring Scale to determine if their mass is changing during their time in space. This is a seat set on springs. An astronaut sits on the seat, grabs onto the rails, and pulls him or herself down --compressing the springs. Then the astronaut lets go. The spring scale chair then oscillates up and down as the springs release and compress. The more massive he/she is, the slower the spring scale oscillates. Now imagine a rocket in outer space with a spring scale and astronaut inside. Say the rocket flies by you at 87% the speed of light. Per special relativity, from your point-of-view time inside the rocket goes at half the rate as your time (time dilation). So you see the astronaut on the spring scale oscillating at half the rate than if the rocket were at rest. Since the spring scale oscillates slower by a factor of two, you conclude the astronaut has twice the mass than at rest. This is a relative effect. From the astronaut's point-of-view (reference frame), the uniformly moving rocket is at rest. So his/her mass is unchanged. But to you, the rocket is moving. So you see time on the rocket running at a slower rate and the astronaut's mass as greater. And the greater the rocket's velocity with respect to you, the more you see its time running slower, and the greater the astronaut's mass. The effect is not linear. It goes as the square root (1 - v ^2) where v is the velocity as a percentage of the speed of light. As the rocket's velocity, v approaches the speed of light, its mass approaches infinity. Edited March 5, 2012 by IM Egdall Link to comment Share on other sites More sharing options...

swansont Posted March 5, 2012 Share Posted March 5, 2012 I found this example helped me understand how mass increases with speed per special relativity. Hope it helps you. Astronauts in space are effectively weightless, but they still have mass. They use a contraption called an Inertial Balance Spring Scale to determine if their mass is changing during their time in space. This is a seat set on springs. An astronaut sits on the seat, grabs onto the rails, and pulls him or herself down --compressing the springs. Then the astronaut lets go. The spring scale chair then oscillates up and down as the springs release and compress. The more massive he/she is, the slower the spring scale oscillates. Now imagine a rocket in outer space with a spring scale and astronaut inside. Say the rocket flies by you at 87% the speed of light. Per special relativity, from your point-of-view time inside the rocket goes at half the rate as your time (time dilation). So you see the astronaut on the spring scale oscillating at half the rate than if the rocket were at rest. Since the spring scale oscillates slower by a factor of two, you conclude the astronaut has twice the mass than at rest. This is a relative effect. From the astronaut's point-of-view (reference frame), the uniformly moving rocket is at rest. So his/her mass is unchanged. But to you, the rocket is moving. So you see time on the rocket running at a slower rate and the astronaut's mass as greater. And the greater the rocket's velocity with respect to you, the more you see its time running slower, and the greater the astronaut's mass. The effect is not linear. It goes as the square root (1 - v ^2) where v is the velocity as a percentage of the speed of light. As the rocket's velocity, v approaches the speed of light, its mass approaches infinity. You can't conclude that mass increases and time slows from this factor of two. Time dilation completely explains the slower oscillations in this example and the frequency varies as [math]\omega = \sqrt{\frac{k}{m}}[/math], so it wouldn't change by a factor of two anyway, if the mass changed it would slow by the square root. As D H already mentioned, the "mass increases" concept comes from not separately accounting for translational KE. It's from using m=E/c^2, where E is the total energy and m is the relativistic mass, rather than [math]E^2=m^2c^4 + p^2c^2[/math], where m is now the rest mass, which is invariant. Or in terms of momentum, you have [math]p=\gamma mv[/math] and you are redefining mass as [math]\gamma m[/math] , making it velocity-dependent. In the oscillating astronaut-on-a-spring example, if you use relativistic mass you have the curious circumstance of mass varying with time, and being dependent on the orientation of the oscillator. Link to comment Share on other sites More sharing options...

DrRocket Posted March 5, 2012 Share Posted March 5, 2012 When you write [math]E=mc^2[/math] the mass to which you are referring is called relativistic mass. Relativistic mass is a concept that most physicists don't think is necessary. All that dragging in the concept of relativistic mass does is confuse the student. What [math]E=mc^2[/math] is saying is that relativistic mass is a synonym for energy -- so why not just use energy? That depends on the physicist and the problem at hand. Mass is not a cut-and-dried concept and the various different definitions of "mass" each have a useful place. There are three common uses of "mass" in physics. 1) Rest mass. This notion of mass, [math]m_0[/maath] or [math]m[/math] when taken in a fixed context, is what is most often seen in the context of quantum field theory where the rest mass of an elementary particle is particularly important. It is also commonly seen in elementary texts on special relativity. It has the disadvantage that rarely is a frame available in which all particles are at rest, and therefore is problematic when dealing with systems of particles and in particular with macroscopic bodies composed of elementary particles in which case the sum of the (rest) masses of the particles does not equal the measured mass of the body (at the nuclear level the rest masses of the constiuent quarks are much less than the mass of the nucleons which they comprise). When applied to macroscopic bodies the notion of rest mass has the disacvantage that the rest mass of the composite system is no the rest mass of the particles that comprise it. 2) Relativistic mass. This notion of mass [math]\gamma m_0[/math] or [math]m[/math] when taken in an understood context, was advocated by Tolman as the concept most deserving of the term "mass", probably because the equation [math] F=\frac {dmv}{dt}[/math] holds in special relativity when [math]m[/math] is the relativistic mass. It makes the simple and memorable formula [math]F=mc^2[/math] valid. A significant disadvantage is the difficulty in addressing the energy of particles of zero rest mass, such as photons. It has the advantage that, as it is essentially total energy, the relativistic mass of a system of particles is the sum of the masses of it parts. 3) Invariant mass. This notion of mass is used for systems of particles, quite commonly in accelerator experiments, and is the relativisitic mass of the system when measured in coordinates in which the net system momentum is zero. Invariant mass of a macroscopic object is what is measured on a laboratory scale (and yes a object when heated gains mass in this context). As invariant mass corresponds to total system energy the formula [math]E=mc^2[/math] is also valid when mass is interpreted as invariant mass. Better is to use rest mass, or intrinsic mass -- the mass an object appears to have to an observer at rest with respect to the object. In terms of intrinsic mass, the mass-energy equivalence formula becomes [math]E^2 = (m_0c^2)^2 + (pc)^2[/math] While that formula is true, it is simply untrue that the mass that a macroscopic object has in a frame at rest with respect to that object is meaningfully related to the the rest mass of its constituents, but rather is the invariant mass of the system of particles that comprise it. The calculation of invariant mass requires knowing the relativistic mass of those particles. This is conveniently ignored in introductory texts on special relativity. The plain fact is that one ought not be dogmatic and demand that one and only one definition of "mass" is correct. All three of the above definitions are used, and used correctly and effectively, in the proper context with the proper understanding. No one concept is superior to the others in all situations. http://en.wikipedia.org/wiki/Mass_in_special_relativity Link to comment Share on other sites More sharing options...

D H Posted March 5, 2012 Share Posted March 5, 2012 Invariant mass of a macroscopic object is what is measured on a laboratory scale (and yes a object when heated gains mass in this context). You are correct. Given an ideal laboratory scale, an object will register an increase in mass as its temperature is raised. Note well: An ideal scale is needed. This increase in mass is immeasurably small in any terrestrial application. However, you need to keep the target audience in mind. One does not need to be a graduate student in physics to understand special relativity; some aspects of special relativity are now taught at the high school level. The original poster is someone who is confused by the mass m in E=mc^{2}, so it is best to assume a high school or a freshman level understanding of physics. Relativistic mass is an unneeded complication at this level of understanding, as is the general relativistic concept of invariant mass. Link to comment Share on other sites More sharing options...

DrRocket Posted March 5, 2012 Share Posted March 5, 2012 (edited) The original poster is someone who is confused by the mass m in E=mc^{2}, so it is best to assume a high school or a freshman level understanding of physics. Relativistic mass is an unneeded complication at this level of understanding, as is the general relativistic concept of invariant mass. 1) Relativistic mass has been taught at all levels for decades. The recent fashion to treat "mass" as rest mass is indeed fairly recent. In fact the formulas using relativistic mass are more easily remembered than those that use a combination of rest mass and momentum. [math]m= \gamma m_0[/math] and [math]E=mc^2[/math] vs [math] m=m_0[/math] and [math] E^2 =( mc^2)^2 + (pc^2)[/math] Which of course are equivalent as one sees by substituting [math] p = \gamma m_0 v [/math] You also get the formula for force that is famililar from Newtonian mechanics [math]F= \frac {d(mv)}{dt} = \frac {dp}{dt}[/math] 2) Invariant mass is not a concept from general relativity. It has nothing to do with general relativity. In fact in general relativity it can be rather difficult to define the term "mass" at all. Invariant mass is a concept from special relativity. It the concept of mass that is used, usually implicitly and without remark, to describe macroscopic objects, since that is what labratory scales actually measure. http://en.wikipedia....eral_relativity http://en.wikipedia....cial_relativity I have found that one of the surest ways to confuse those just learning a subject is to make positive, but wrong, statements that must be corrected later. Better to tell them that there are several useful notioins of "mass", but that for such and such a particular class we will just use this one (insert whatever you intend to use). That alerts them that there are alternatives, but does not require them to grapple with all of them simultaneously. Mass in special relativity is a particular problem because of the fame of the equation [math] E=mc^2[/math] which is stated in terms of relativistic mass. Edited March 5, 2012 by DrRocket Link to comment Share on other sites More sharing options...

IM Egdall Posted March 11, 2012 Share Posted March 11, 2012 You can't conclude that mass increases and time slows from this factor of two. Time dilation completely explains the slower oscillations in this example and the frequency varies as [math]\omega = \sqrt{\frac{k}{m}}[/math], so it wouldn't change by a factor of two anyway, if the mass changed it would slow by the square root. As D H already mentioned, the "mass increases" concept comes from not separately accounting for translational KE. It's from using m=E/c^2, where E is the total energy and m is the relativistic mass, rather than [math]E^2=m^2c^4 + p^2c^2[/math], where m is now the rest mass, which is invariant. Or in terms of momentum, you have [math]p=\gamma mv[/math] and you are redefining mass as [math]\gamma m[/math] , making it velocity-dependent. In the oscillating astronaut-on-a-spring example, if you use relativistic mass you have the curious circumstance of mass varying with time, and being dependent on the orientation of the oscillator. Thank you for the enlightenment. Link to comment Share on other sites More sharing options...

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