# Tag Info

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Even without gravity, momentum conservation will still hold. If you elastically scatter an unknown mass $m$ with an initial (known) velocity $v$ against a known mass at rest, for instance we can take the SI standard of $1$ Kg, then from the resulting measured velocities you should be able to find $m$. Thus Yes, there will still be a mass.

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While defining "matter" we refer to "mass", but the definition of "mass" refers back to "matter". So isn't this wrong? What will be the right definitions? Circular definitions are indeed undesirable, because they talk a lot but they tell little. The solution is to break the circle by removing the requirement to define the word and refer to experience ...

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The word "matter" needs to be used with a great deal of care. In science, the word has definitely passed its use-by date. Some problems with the precise usage of the word are listed on the Wikipedia page for Matter and I quote: ...As such, there is no single universally agreed scientific meaning of the word "matter". Scientifically, the term "mass" is ...

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Assuming constant total energy, $E^2 = (m_0^2c^2)^2 + (pc)^2 = \gamma^2 m_0^2c^4$ leads to the equation $E^2 = \frac{m^2c^6}{c^2-v^2}$ and then $m = \frac{E}{c^3} \sqrt{c^2 - v^2}$, which also yields the quarter of an ellipse: $v^2 +\frac{c^6}{E^2}m^2 = c^2$, producing the following graph (ignore the number): where the vertical axis is rest mass which ...

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In my textbook the definition of matter and mass is as such: Matter: Any thing that occupies space and has mass. Mass: The amount of matter contained in a body. Why are the two definitions interlinked? Because it's not a very good textbook. Take a look at Einstein's E=mc² paper, and note this: "The mass of a body is a measure of its energy-content". It ...

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These aren't good definitions. They do help students grasp the concept, but certain students, such as you, will find it silly that one definition depends on the other. Anyway, one definition depending on the other doesn't necessarily mean that a definition is flawed. In mathematics many functions are defined over itself, yet the definition is effective in ...

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There's no one-to-one relationship. With zero rest mass, a particle must always be observed to move at $c$. A particle with nonzero rest mass, on the other hand, can move at any speed in $[0,\,c)$ (note the closed-open interval). At the risk of putting words in your mouth, I think I can recall the exact same question in my mind and it went something like ...

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Only equation I know is $$m = \frac{m_0}{\sqrt{1 - (\frac{v}{c})^2}}$$

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There are very strict limits on the mass of the photon already, so it would only affect our understanding of physics on the largest scales. The cosmologists would have some hard thinking to do, for instance. However, contrary to a comment, it would not affect relativity beyond requiring us to reconsider the usual name for $c$: not "the speed of light" but ...

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"If you double the mass of an object, its kinetic energy doubles" is simply wrong. A true statement would be: "If you double the mass of a point object, its kinetic energy doubles." Let's take for example a ball, moving with a rototraslational motion. Its kinetic energy would not be $$E_k = \frac{1}{2} mv^2$$, but it would be $$E_k = \frac{1}{2} mv^2 + ... 2 Zeldredge's answer is great, answering the question from a mathematical point of view. Since you asked this question in a physics forum, I'll just add to that by answering from a physical point of view, clarifying why the energy is linear in mass and not quadratic: Suppose you have two objects of equal mass moving parallel to each other with velocity v, ... 4 Physics has a concept of the mean free path (MFP). We can use the same principles to tackle this question. Even in the case of an infinitesimally thin pin that goes 2 mm into the body we can still calculate "roughly" how many cells are traversed. First we estimate the mean volume of a cell: 80 kg of water, 10^{14} cells, so one cell has a mass of 8\cdot ... 0 The answer is an emphatic yes! We can find the actual rest mass of things on earth. How the earth is moving with respect to the sun, galaxy,etc., is irrelevant. By saying, "on earth," the frame of reference is specified (the earth), so whatever mass a thing has on earth, (as long as it is not moving with respect to the earth), is its actual rest mass! In ... 4 Yes, doubling the mass doubles the energy, while doubling the velocity quadruples it. Your question is basically about order of operations; the exponent only applies to the variable it's immediately on. As you note, you'd have to put parentheses to make it cover the m as well. We say that energy is linear in mass, but quadratic in velocity. 3 I'm going to expand on dmckee's answer because this used to puzzle me in my younger days, and I think it's a fascinating part of modern physics. It's tempting to think of a particle like a muon as a chunk of matter whizzing around - something like a tiny billiard ball. However the physicist's description of a particle is much, much stranger. Quantum field ... 3 Mass is not conserved in that decay, but then there is no expectation that it would be. The "law" of conservation of mass is only an approximate rule that applies to low energy events and interactions. Chemists (well, the non-nuclear ones, anyway) talk about it, but physicists do not. The rule that does apply is the conservation of energy (mass being one ... -1 Am I missing something here? Well, you (among others) seem to be missing that to measure "momentum" is defined through the application of the gradient of the translation operator \nabla \hat T_{\mathbf r}[~] := \frac{d}{d \mathbf r_{\mathcal S} }[~] to what's given through observational data (e.g. concerning a particular object A under ... 1 Am I missing something here? Yes. What you're missing is "the mass of a body is a measure of its energy-content". Read Einstein's original paper, and take note of this: "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c²". Next, imagine your body is a massless photon in a gedanken mirror-box. It isn't actually at rest ... 6 They're both saying the same thing: the relativistic momentum is given by$$ \mathbf{p}=\gamma(v)\,m\mathbf{v} $$The confusion, it seems, is that you are using Feynman's m=\gamma m_0 as equivalent to the m in Resnick & Halliday's text; the actual correlation is Feynman's m_0 to Resnick's m--both of these terms are the (invariant) rest mass. ... 1 The way to define momentum in a special relativistic context is the following: Start with the trajectory of the particle parametrized by its proper time x^{\alpha} (\tau); define the four-momentum by p^\alpha = m \frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\tau}, where m is the mass of the particle (note that I'm only using one mass, not distinguishing ... 0 When the object is an elementary particle or a charged ion we can use electromagnetic interactions to measure its rest mass, given the charge in an e/m experiment. One can get the charge with Milikan's oil drop experiment. Here is a setup for the lab. 11 Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered. The relativistic mass is altered, but this is a somewhat archaic term these days, and is said to be a measure of energy. Nowadays when we say mass without qualification, we tend to mean rest mass. ... 6 The rest mass of an object, by definition is the total energy of an object as measured by an inertial observer who is at rest relative to the object. If the object is not moving uniformly, then you can measure its rest mass from a momentarily co-moving freefall frame. This rest mass is also the constant in Newton's law, i.e. the inertial mass as well. So ... 2 We do have a rough idea of the relative speed and direction of our galaxy, with respect to the other galaxies around us, the so called local group. In general relativity, which is our best theory of the universe to date, there is no such thing as absolute speed, as it depends on which frame of reference you use to measure things in. Our Earth and the ... 2 There are three types of matter/energy we consider when calculating how the universe expands: Matter - both normal matter and dark matter Radiation Dark energy We measure the expansion of the universe using a scale factor that we normally denote by a. The scale factor increases with time as the universe expands, and if we look backwards in time we see ... 0 The expansion of space is like stretching a rubber sheet. (Don't take this analogy too seriously. It works for this explanation but fails elsewhere.) The mass of the rubber sheet stays the same as it gets bigger. Space expands, but mass does not increase with it. 0 They're just applying the chain rule.$$\frac{d}{dx}f^{n}(x)=nf^{n-1}(x)\frac{df}{dx} \Rightarrow \frac{d}{dt}D^{3}(t)=3D^{2}(t)\frac{dD}{dt}$$4 Short: Given enough assumptions to make the question answerable. Feather size: Probably 300 to 500 mm for a small percentage of the feathers and closer to 50% of that for the majority Power and force: Without a complex analysis of flapping flight (with takeoff mode, soaring versus "hovering" capabilities and more) a definitive answer would be ... -1 How can you explain objects of unequal masses falling at the same rate using GR? up vote By explaining why light curves, then by using the wave nature of matter to explain why an electron falls down, and then explaining that an ensemble of electrons (and other particles) falls down at a rate that is irrespective of the number of particles in the ensemble. ... 5 If Galileo had dropped the Moon and a pebble from a very tall tower, the Moon would have fallen noticeably faster, relative to the Earth. This is true in Newtonian physics as well as GR, and it does come from the fact that the Earth falls toward the Moon too, and harder than it falls towards the pebble. The assumption that small objects do not gravitate is ... 3 The strength of gravity is given by the space-time curvature caused by all the objects in the system, in this case both the earth and the falling object. The problem is that you are ignoring the fact that the space-time curvature caused by the earth is many, many times larger than that caused by the falling object. Hence, the total curvature is to all ... -2 As often, hasn't something been overlooked. Energy & mass cannot be interchangeable with regards having the ability to generate gravitational force, because otherwise 'binding energy' would also contribute, obviously it does not. Do we have an impasse, maybe not, if photons actually possess mass after all. I have calculated algebraically that when matter ... -4 Does this equation only mean that energy has mass? No, from my understanding I can say that matter is a form of energy. What Tesla said shouldn't be treated literally. The energy have many states, it can be "heavy" or "light". The lighter is the less physical form it will have. Until it just fades from sight, but it's still there. Obviously it's not ... 29 You've taken this out of context. The context in which Tesla is writing, he's talking about kinetic energy and heat. Just a few sentences previously he states: ...according to an experimental findings and deductions of positive science, any material substance (cooled down to the absolute zero of temperature) should be devoid of an internal movement and ... 3 I think there is some confusion here. Photons are always massless. They also always move at the speed of light. Therefore every example of a photon in nature has zero mass. Perhaps you are thinking about a photon moving through a medium other than a vacuum. In this case, we can view the photon plus the interactions with the medium as a quasiparticle with a ... 26 I think that sentence shows that Tesla did not quite grasp the results from relativity. This is not unusual, as many physicists required many years to fully accept relativity, but by 1932 (the date of the writing of that text) I would expect it to be already orthodox knowledge. In any case, Tesla is known to have been a self-taught experimental genius and ... 0 Total energy has independent contributions from kinetic energy and potential energy. The kinetic energy component is determined by temperature and is not affected by particle mass (although obviously velocities will be affected). The potential energy components also do not depend on mass (e.g. Coulombic effects). It's somewhat counter-intuitive but the ... 2 Notice Since I have written the majority of this answer the question changed a lot, and now it's been put on hold. However, I won't erase parts of my answer which are answers for the original (ill posed) question, because other people might find it useful. I have extended it with parts though, which seem (to me) to answer the question the original changed ... 1 A gold brick is made of gold atoms that mutually interact. At a given temperature it weighs bit less than the weight of each piece. At a higher temperature it weighs a bit more than at a colder temperature. So the weight isn't the sum of the weights of the parts, not quite. Even in a single gold atom, it weighs a little bit less than weight of each neutron ... 6 This answer is a bit aggressive. Then again, so is the question. Euler's formula for force It's erroneous to attribute F=ma to Euler. Even though Newton never wrote F=ma in his Principia, this statement is almost universally recognized as Newton's second law. As far as I can tell, it was Jakob Hermann who first expressed Newton's words in an ... 4 You are missing something very important.$$ F(x,\dot{x},t) = m\ddot{x}(t) is a differential equation that, given that we know the forces that act, completely determines what will happen (unless the equation has multiple possible solutions, but almost all cases of application turn out to be well-behaved), i.e. the trajectory $x(t)$ from initial ...

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In the modern way of viewing things, no, (rest) mass is invariant. What happens is that the energy content of the body changes and some people still interpret this as a change in mass (which is an old point of view that, unfortunately, is fairly common). A nice discussion about this can be found here: ...

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