Post by p4wn3r

Posted: 2/27/2018 1:55 PM

Quote

p4wn3r

Player (42)

Joined: 12/27/2008
Posts: 873
Location: Germany

marzojr wrote:

That is exactly backwards: any theory can be written in diffeomorphically invariant form: there is even a formulation of Newtonian mechanics (plus Newtonian gravity) written in such a form, and this one is valid (i.e., had the same form) on all reference frames. As a bonus, gravity is even related to the curvature of space-time in this theory as well (FYI, this version of Newtonian gravity is a lot more complex than I am letting on).

Please provide a reference for this statement, and forward me the specific theory you have in mind. It is generally acknowledged that the local gauge symmetry completely defines the physically measurable content of a theory, where GR's "gauge group" is the diffeomorphism group. In any case, I should let you know that this sounds like typical crackpottery from gravity people that no one should waste their time on, but I'll give it a chance nonetheless.

Posted: 2/27/2018 9:01 PM

Quote

marzojr

He/Him

Experienced player (760)

Joined: 9/29/2008
Posts: 964
Location: 🇫🇷 France

p4wn3r wrote:

marzojr wrote:
That is exactly backwards: any theory can be written in diffeomorphically invariant form: there is even a formulation of Newtonian mechanics (plus Newtonian gravity) written in such a form, and this one is valid (i.e., had the same form) on all reference frames. As a bonus, gravity is even related to the curvature of space-time in this theory as well (FYI, this version of Newtonian gravity is a lot more complex than I am letting on).
Please provide a reference for this statement, and forward me the specific theory you have in mind.

When you ask for a reference, do you mean the part about it being possible to write any theory being in diffeomorphically invariant form, or the part about the formulation of Newtonian mechanics + gravity in diffeomorfism invariant form? Assuming the latter: I did have a theory in mind, it is called 'Newton-Cartan theory'. It was first developed by Élie Cartan, a french mathematician, and published in the early 1920s; it has the distinction of being the first non-relativistic diffeomorphism-invariant theory of gravity. In this theory, spacetime is non-metric (you have a spatial metric, and you have universal time, but it is not possible to define a spacetime metric), a non-metric connection which yields a nonzero Riemann tensor, and this connection is compatible with the spatial metric. There is a good summary in Misner, Thorne, Wheeler (Gravitation), chapter 12.

Marzo Junior

Posted: 2/27/2018 9:57 PM

Quote

p4wn3r

Player (42)

Joined: 12/27/2008
Posts: 873
Location: Germany

From what I see, it is just a conceptual confusion. Newton-Cartan theory is simply the statement that the equations of motion for Newtonian gravity can be cast in the form of a geodesic equation for a given metric. That's not really surprising. However, this has nothing to do with the diffeomorphism invariance of the Einstein Field Equations, the EFE specifiy conditions that the metric tensor must satisfy. In Newton-Cartan theory, you get the gravitational field, which should be computed in an inertial frame, and from it you derive the metric that corresponds to such field. However, if you were to repeat the same procedure in an arbitrary frame, the equations to generate the correct field would be different, and the metric you would obtain from the equations of motion would definitely not be equivalent to the one in an inertial frame. So, the metric you find in Newton-Cartan theory is different for non-inertial frames (it can have different curvature, etc.), and the way you determine it is not diffeomorphically invariant at all. EDIT: See eq. (3.20) in this paper. The LHS is diffeomorphically invariant, because it is a curvature tensor, but the RHS is not. The equation is covariant with respect to Lorentz transformations, but not with respect to diffeomorphisms. So, the metric equation for NC gravity is not diffeomorphically invariant like the EFE are.

Posted: 2/27/2018 11:23 PM

Quote

marzojr

He/Him

Experienced player (760)

Joined: 9/29/2008
Posts: 964
Location: 🇫🇷 France

To quote a movie: amazing, everything you just said is wrong. Except maybe for the first line, but you are applying it to the wrong person; so I will give it partial credit. By what you said above, it is evident to me that you lack anything being a basic grasp of differential geometry. This i assessment comes from the last two paragraphs you wrote, in particular, but the second half of the third paragraph is also telling. You also failed to appreciate that Newton-Cartan theory does not have a metric; it has a connection (Christoffel symbols/covariant derivative), but no space-time metric. While writing my other reply, there was a much longer post that ended up being scraped because it was a bit of a rant and I thought it might not have been necessary. I now see I was wrong. Diffeomorphism invariance is not a property only of the EFE. The entire theory, and all equations that come from it, have this property. It was called "principle of general covariance" by Einstein; and it basically boils down to a requirement that the laws of physics must be written in terms of geometric objects (scalars, vectors, tensors) so that they (the equations) will have the same form on all coordinate systems (and yes, I am aware of the distinction between active diffeomorphisms and passive diffeomorphisms; they don't matter here). This principle had come under fire since 1917 (at least) because many authors felt it had no real content - because any physical theory can be written in such a way. You can read about this on any graduate-level textbook on GR, or on lecture notes for such a course. For example, check Sean Carroll's (https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll5.html). Now comes the part were it is a bit of a rant: usual formulations of Newtonian mechanics and SR are not written correctly: they are written assuming a certain class of coordinates, inertial frames, and lack things that would make them lose diffeomorphism invariance. Specifically, they do not make use of covariant derivatives, and they do not use a general metric. This additionally includes not making distinctions between vectors and 1-forms, and etc, in Newtonian mechanics. When you correct these deficiencies, you get versions of Newtonian mechanics and SR whose laws are the same on all coordinate systems. Newton-Cartan theory is one of these. The laws for corrected SR match the versions used by GR, by the way, except that the metric is not a solution of the EFE, but is ultimately obtained by doing coordinate transformations from a Minkowski metric. The usual rules for "lifting" laws from SR to GR (partial derivatives to covariant derivatives, etc) are not needed with a proper formulation of SR because they should have been baked into SR instead.

Marzo Junior

Posted: 2/27/2018 11:33 PM

Quote

p4wn3r

Player (42)

Joined: 12/27/2008
Posts: 873
Location: Germany

marzojr wrote:

To quote a movie: amazing, everything you just said is wrong. Except maybe for the first line, but you are applying it to the wrong person; so I will give it partial credit.

Dear Marzo, All the garbage you wrote after this line is just meaningless. I don't care about your trolling remarks. The equations are clearly not diffeomorphic invariant, end of story. It's irrelevant if you gravity people somehow convinced yourselves that by just writing something with metrics, giving funny names to things or just stating with no reason that formulations that you don't like are wrong, changes this basic fact. You apply a coordinate change the equations are different. This is the basic statement of gauge symmetry which is violated by the equations once you write them down. I don't feel I will do anything constructive by addressing your posts, so I consider this matter closed.

Posted: 2/28/2018 6:10 AM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

p4wn3r wrote:

All the garbage you wrote after this line is just meaningless. I don't care about your trolling remarks.

I have no idea what either of you is saying, as it goes well above my head. However, I do know that if someone dismisses and completely skips reading somebody's arguments based solely on an off-hand remark, they tend to be avoiding responding to the actual argument being presented. In other words, it's a kind of a form of argumentum ad hominem (completely dismissing, and not responding to, someone's arguments because they said something you perceive as a personal attack, which smells like dodging the question, or the argument in this case).

Posted: 2/28/2018 4:03 PM

Quote

p4wn3r

Player (42)

Joined: 12/27/2008
Posts: 873
Location: Germany

It definitely feels like that, I will elaborate if you want. The property of an equation being diffeomorphically invariant is a clear mathematical statement, which is either true or false, and can be verified easily without any knowledge of differential geometry. It is clear that the equations of GR are diffeomorphic invariant and the ones in the theory he provided are not. In fact, I have not seen the claim that those equations are diffeomorphic invariant anywhere in the references I have read. What is actually said is that Newton-Cartan theory is a geometrical theory of Newtonian gravity, which is not the same thing as diffeomorphically invariant. Besides, I do not perceive as personal attacks when someone points out that what I am saying is incomplete or even wrong. That has happened recently, I uploaded a paper to the arXiv and three days later a physicist contacted me saying that they were already working on the same thing, and that the model I proposed had been put forward in a different context some years earlier. His email came full of references, was not in a "you're wrong" tone and actually helped me learn more. It even helped me generalize the model and I even managed to get another paper written, which is now under review in J Math Phys, where I explicitly acknowledge this physicist for his help. A very different thing is just claiming that someone is wrong without providing any meaningful references and, quite frankly, his arguments rest mostly on the fact that derivations of GR presented in the most used textbooks are in grave error, this is a statement which would get you laughed at at any serious physics department. In short, while I would definitely acknowledge the discussion on how someone would arrive at the position that Marzo has, the habit of not providing meaningful references when you contest someone's position and replying to clear mathematical statements with outright trolling and strings of buzzwords just indicates sloppy research on his part and casts doubt on the constructive nature of his criticisms in general. Since I am not his instructor and this forum is not a place to correct his academic conduct, I don't feel it is worth to keep the discussion going.

Posted: 2/28/2018 7:49 PM

Quote

marzojr

He/Him

Experienced player (760)

Joined: 9/29/2008
Posts: 964
Location: 🇫🇷 France

For what is worth: I have been refreshing up a bit on the subject (it had been years since I last studied this, having left physics for computer science) and while the core of the subject matter was still intact, the more subtle aspects were completely missing. I now see where I was wrong, and I concede the point entirely. I also would like to apologise for acting like an a$$. I always had a tendency to act this way, and it seems to be getting worse with age despite my efforts to suppress it...

Marzo Junior

Posted: 6/30/2018 6:49 PM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

Could someone help me understand, or even get a vague mental picture and understanding, of the (current model of the) geometry of the universe? It is my understanding (and correct me if I'm wrong) that the current model of the universe is that it has finite volume, but has no edge. You can't just traverse to the "edge" of the universe and go outside of it (or be stopped by some strange horizon, where the universe stops existing). On the other hand, at the same time, the geometry of the universe is postulated to be "flat". Whatever that means. As a programmer, I have difficult time undrestanding anything else than Euclidean geometry. All of those other geometries go above my head. Let's assume that the universe, for some incomprehensibly strange reason, just stopped expanding, and becomes completely steady-state. It just stops expanding or contracting, and just stays as it is at this exact moment. Let's also assume you start traveling away from Earth, and you are able to do it for as long as needed, as fast as needed. What happens, given enough time? If there is no "edge" to be reached, what exactly happens if you just keep going?

Posted: 6/30/2018 10:51 PM

Quote

Amaraticando

It/Its

Editor, Player (159)

Joined: 1/10/2012
Posts: 673
Location: Brazil

The current model does not assume that the universe has a finite volume. The best data we have points to a flat model within the margin of error ( Ω = 1). That means, it's consistent with a infinite universe or at least one which is much larger than the observable universe if the curvature is near. If Ω is slightly less than 1, the total universe is quite big but finite. If 1, it's flat and infinite. If more than 1, it's infinite and hyperbolic. Were the universe finite, you would end up in the opposite side from when you started traveling, like Mario does in vertical levels of SMB3. Except that it would be a sphere, not a rectangle. Actually, there's a possibility that the total universe is smaller than the observable universe: in this case if you see distant galaxies in opposite directions, you might be looking at the same one.

Posted: 7/1/2018 10:23 AM

Quote

rhebus

Joined: 2/19/2010
Posts: 248

Warp wrote:

It is my understanding (and correct me if I'm wrong) that the current model of the universe is that it has finite volume, but has no edge. You can't just traverse to the "edge" of the universe and go outside of it (or be stopped by some strange horizon, where the universe stops existing). On the other hand, at the same time, the geometry of the universe is postulated to be "flat". Whatever that means.

I can't speak to what our actual universe is shaped like, but if you want an example of a finite-volume universe with no boundary and flat curvature, you can think of games like Asteroids, where you warp to the other side of the screen when you hit the edge. If you still think asteroids has a boundary, you could also consider Defender, where the camera follows the player and there is no left/right boundary at all, not even a "warping" one. This sort of model easily generalizes to three dimensions. (Again, I'm not saying this is an actual viable model for our universe - I seem to recall that theories of our universe have curvature and finiteness linked together, such that the only viable finite universes have positive curvature (like the positively-curved surface of a sphere is finite but boundless), and the only infinite universes have flat or hyperbolic curvature. In other words, I'm not sure anyone is postulating a finite, flat universe, even if asteroids shows that such a thing is mathematically consistent.)

Posted: 7/1/2018 4:02 PM

Quote

OmnipotentEntity

He/Him

Player (36)

Joined: 9/11/2004
Posts: 2630

Sort of a quick primer on curved space: http://scottburns.us/vision-in-curved-space/

Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.

Posted: 7/2/2018 5:35 AM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

Amaraticando wrote:

The current model does not assume that the universe has a finite volume. The best data we have points to a flat model within the margin of error ( Ω = 1). That means, it's consistent with a infinite universe or at least one which is much larger than the observable universe if the curvature is near. If Ω is slightly less than 1, the total universe is quite big but finite. If 1, it's flat and infinite. If more than 1, it's infinite and hyperbolic.

Isn't even the notion that the universe might be infinite in contradiction with the hypothesis that at one point the universe, in its entirety, was a singularity that expanded? I don't think you can have both. That would be impossible.

Posted: 7/2/2018 1:03 PM

Quote

OmnipotentEntity

He/Him

Player (36)

Joined: 9/11/2004
Posts: 2630

Warp wrote:

Isn't even the notion that the universe might be infinite in contradiction with the hypothesis that at one point the universe, in its entirety, was a singularity that expanded? I don't think you can have both. That would be impossible.

It's not a contradiction. Space and time did not exist before the universe existed. So it's entirely possible that at the instant the big bang occurs and all space comes into existence, it comes into existence as infinite.

Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.

Posted: 7/2/2018 7:45 PM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

OmnipotentEntity wrote:

It's not a contradiction. Space and time did not exist before the universe existed. So it's entirely possible that at the instant the big bang occurs and all space comes into existence, it comes into existence as infinite.

I don't really understand why it would expand, if it already became into existence as infinite. The place where it expanded from would have been somehow different from the entirety of everywhere else. (Unless I'm completely talking out of my posterior here, it's not like it was just that all energy was at this initial location and then started expanding to its surroundings, but that space itself experienced, and still experiences, a metric expansion. Space itself becomes larger, and matter/energy just moves with it, like a conveyor belt moves material.)

Posted: 8/10/2018 7:40 AM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

A very old method (which is in fact still used in some situations today) for getting color photographs when all you have is a black&white (well, monochromatic) camera is to take three photos with a red, a green and a blue filter in front of the camera, and then combine them. (This method is used even today in some situations. IIRC the Hubble Space Telescope uses essentially this method to get color photos.) There is one thing I don't really understand about this method though. More precisely, I don't understand why violet colors show up in these photos just fine. Why is this a problem? Violet is above blue in the frequency range. Since we have receptor cells for only red, green and blue, but not for violet, in theory we shouldn't be able to see violet at all. The reason why we see violet is complicated. Our blue receptor cells perceive violet frequencies to a degree. This in itself ought to mean that we should see violet colors as dark blue. However, curiously, or red receptor cells also perceive violet, even though red is on the opposite end of the color spectrum. It is my understanding that this is because the frequency of violet is twice that of (low) red, and there's a resonance effect going on, where red cells not only perceive red frequencies but also frequencies that are twice as large (in this case violet). Thus our eyes see violet as a combination of blue and red. This is also the reason why violet can be displayed on TVs and computer monitors: They can be likewise represented as a combination of red and blue, which causes the same illusory violet color perception for our brains. But now enters the color photography using color filters problem: We have a monochromatic camera, and we take three photos with it, with a red, a green and a blue filter, and we combine them and we get a color photograph. And violet colors show up just fine! How? The red filter should be blocking violet frequencies out completely. There should be no violet at all (unless, somehow, the object is emitting both blue and red light). Yet somehow they show up. Do red filters actually let violet frequencies through (perhaps due to the fact that they are twice the frequency of red)?

Posted: 8/12/2018 9:15 AM

Quote

FractalFusion

Editor, Expert player (2072)

Joined: 6/15/2005
Posts: 3282

Warp wrote:

But now enters the color photography using color filters problem: We have a monochromatic camera, and we take three photos with it, with a red, a green and a blue filter, and we combine them and we get a color photograph. And violet colors show up just fine! How?

I assume that by "violet colors" you are referring to violet light (400-440nm) and not digital (or otherwise) renderings of purple colors. Actually, it depends on the colors that the filters let through (since the red/green/blue filters only form an approximation when combined together). Perhaps violet colors show up just fine because the "blue" filter is actually closer to violet than it first appears (maybe it filters 400-440nm light, rather than the 470-500nm that is typical of blue). Or maybe the blue filter allows a wider range than it appears at first glance.

Warp wrote:

It is my understanding that this is because the frequency of violet is twice that of (low) red, and there's a resonance effect going on

p4wn3r wrote:

When it receives violet, the blue cone is stimulated because its frequency is not far below violet, and the red one as well, since it's stimulated by the second harmonic

I've never heard of a resonance effect or harmonic for electromagnetic waves before, but I don't know about it. I only know about resonance and harmonics in sound waves (e.g. vibrating string). The typical human sees color using three cones, often called "red", "green", and "blue". The typical sensitivities of the cones are shown in the image below, taken from the Color vision page on Wikipedia:

In this model, violet light mostly stimulates the blue cone, compared to the red and green cones. Note that the peak of the blue cone is in the violet range. I have seen some other models where the red cone stimulation actually increases in the violet range, but I don't know if they are accurate or not. Searching the internet reveals that there is a disagreement in this area.

Posted: 8/12/2018 11:34 AM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

FractalFusion wrote:

I've never heard of a resonance effect or harmonic for electromagnetic waves before, but I don't know about it. I only know about resonance and harmonics in sound waves (e.g. vibrating string).

"Resonance" was probably the completely wrong word to use in this case, but I didn't/don't know the technically proper term (which is probably that "harmonic" instead). I have never actually seen an explanation of why the red receptor cells in our eyes are stimulated by violet light, but I have noticed that the frequency of violet is exactly twice that of low red, and I suspect that's exactly the reason.

Or maybe the blue filter allows a wider range than it appears at first glance.

I suppose that if the blue filter allows violet to pass through, and this is then somehow replicated when composing the end result, and this end result somehow now emanates violet light, it will work. However, if we are digitally composing an RGB image out of it, some kind of algorithm would need to be used for the blue-filter-image to produce purple colors out of it (ie. it would affect also the red component of the pixels). OTOH, I'm assuming that in typical digital compositions RGB isn't the color model used during the process (only as the very end result), so there might be more about it there.

Posted: 8/12/2018 12:37 PM

Quote

p4wn3r

Player (42)

Joined: 12/27/2008
Posts: 873
Location: Germany

Resonance is not the right term for detection/generation of further harmonics. It's nonlinear optics. If you make monochromatic light pass through a nonlinear filter, it's still periodic, but you get distortions arising from integer multiples/divisors of the base frequency. Commercial spectrometers have been built carefully to avoid nonlinear optical effects, but the eye doesn't have this mechanism. You have to see how the graph supplied was measured. If the scientists isolated the cells and passed light through a linear medium, indeed violet will not stimulate the red cones. But if they are operating in their natural environment, which is nonlinear, it should stimulate a little bit. The devil is in the experimental details, they are important to interpret the data.

Posted: 12/17/2018 1:06 PM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

Wikipedia defines Joule as "the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second." This seems to imply that a resistor of 1 ohm will always dissipate the exact same amount of heat regardless of what material it's made of. That feels surprising and counter-intuitive. Or, perhaps, it could be restated as: It seems to imply that there's a tight unique relationship between resistance and heat dissipation, regardless of the material.

Posted: 12/18/2018 6:36 AM

Quote

OmnipotentEntity

He/Him

Player (36)

Joined: 9/11/2004
Posts: 2630

Warp wrote:

Wikipedia defines Joule as "the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second." This seems to imply that a resistor of 1 ohm will always dissipate the exact same amount of heat regardless of what material it's made of. That feels surprising and counter-intuitive. Or, perhaps, it could be restated as: It seems to imply that there's a tight unique relationship between resistance and heat dissipation, regardless of the material.

That's not strictly true though. For instance, if you have an LED, the energy is dissipated as light. If you have a speaker, it's dissipated as sound. If you have a motor, the energy becomes mechanical work. And heat makes up only the rest of the energy. However, if we're talking about a passive resistive component that does no external work, then Wikipedia is correct. This comes simply from the conservation of energy. You can think of it in reverse. There isn't any reason for two materials to have the same heat dissipation when current is ran through it. Instead the amount material and its geometry determines how much energy is lost, and from this, the value of its resistance may be determined.

Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.

Posted: 12/18/2018 7:38 AM

Quote

upthorn

He/Him

Emulator Coder, Active player (391)

Joined: 3/24/2006
Posts: 1802

Warp wrote:

Wikipedia defines Joule as "the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second." This seems to imply that a resistor of 1 ohm will always dissipate the exact same amount of heat regardless of what material it's made of. That feels surprising and counter-intuitive. Or, perhaps, it could be restated as: It seems to imply that there's a tight unique relationship between resistance and heat dissipation, regardless of the material.

I think your confusion here comes from an incorrect understanding of "heat dissipation." The amount of energy lost to heat (converted from electrical energy to quasi-randomly directed kinetic energy) is only dependent on the amount of current and the amount of resistance. How many degrees of environmental temperature change results from this heat-generation is dependent on the material of the resistor as different materials have different specific heat capacity, and its heat transfer coefficient. The specific heat capacity of a material is the number of joules required to raise one gram of the material by 1 Kelvin. The heat transfer coefficient of a material is a term in the formula which relates the rate of temperature equalization to the difference in temperature between the material and the surrounding air. So the basic concept is that the scientific definition of "heat" (undirected kinetic energy) is not the same as the common usage of the word "heat" by laymen (temperature level). So, while, the temperature change (laymen's definition of heat) caused by running a 1 ampere current through 1 ohm of resistance is dependent on the material causing the resistance, the amount of electrical energy transformed into undirected kinetic energy (physical science definition of heat) is not.

How fleeting are all human passions compared with the massive continuity of ducks.

Posted: 12/24/2018 10:10 AM

Quote

Warp

Banned User

Joined: 3/10/2004
Posts: 7698
Location: Finland

Every single person knows the famous equation E=mc², but not many people know what it actually means. If I understand correctly, it simply establishes the equivalence of (rest) mass and energy, and the formula is simply a conversion formula between the two. In fact, the relationship is completely linear, using a constant conversion factor (of c²). This means that 1 kg of mass has (by definition) exactly 89875517873681764 joules of energy. Or, conversely, if there are 89875517873681764 joules of energy in something, that something has exactly 1 kg of mass. (Hmm, wouldn't that be a nifty definition of the kilogram? Although I suspect that would be a circular definition, if the joule is defined in terms of the kilogram, which would mean that you can't define the kilogram in terms of joules. But I digress.) One nifty thing about the equation is that the units on both sides of the equal sign match. Of course this is a necessity, else it would be an invalid equation. It still feels nifty, though, given that it's not immediately obvious how a joule is equal to kg*(m/s)². But the question arises: Why precisely c, and not some other velocity? How did Einstein come up with c in particular? Why couldn't it have been some other velocity? It is my understanding that the c isn't there willy-nilly, or as a wild guess, but it's actually derived. It can't be anything else than c, and it's derived from what we know about physics. Does anyone have a simple explanation of how you can derive that the velocity that has to be used in that formula is precisely c, and not something else?

Posted: 12/24/2018 2:08 PM

Quote

p4wn3r

Player (42)

Joined: 12/27/2008
Posts: 873
Location: Germany

There are two ways to derive E=mc², or the more general E=gamma*mc², where gamma, the Lorentz factor, becomes 1 if the particle is at rest, v=0. The one that was used by Einstein, and is actually what's used to test relativity everyday in particle colliders, is to look at what happens in collisions. It turns out that if you require that a theory conserves energy and momentum, and analyze the result of a collision from different frames where space and time transform according to the postulates of relativity, you find that the only consistent choice of energy is E=gamma*mc2 plus an unobservable constant. In particular, if the masses of the particles are different in initial and final states (a particle "becomes" another with different mass), the mass energy converts to kinetic energy and vice-versa. Actually, you are right that we can measure energy in terms of mass. If you include quantum mechanics, you can also measure length and time in terms of energy. This is done in particle physics all the time. An example: if you want to produce a Higgs boson, with mass of 126 GeV by colliding an electron and a positron, their kinetic energies should be at least 63 GeV (you can neglect the electron mass energy because it's much smaller than that) so that the sum can produce the Higgs at rest. The second way to deduce is more mathematical but you can get the answer with less calculation. One way to derive physical theories is to use the principle of least action, which states a particle's path should minimize a given quantity. It turns out that, from the postulates of relativity, you can deduce that a free particle maximizes its proper time when moving. Then, using techniques from analytical mechanics you find, using a few derivatives that, a particle that maximizes proper time necessarily has an energy of gamma*mc2. Finally, the c that appears in the formula is the velocity that all inertial frames agree. That means, if an object is traveling at c in a given inertial frame, it necessarily travels at the same speed at all other inertial frames. That uniquely fixes c, so the velocity in the formula cannot be any value. Of course, nothing stops you from creating a theory where c is replaced by the velocity of sound, for example. You would get a consistent theory where energy is proportional to the square of the speed of sound. However, you would get bizarre results which are not observed experimentally, like massive things unable to break the speed of sound.

Forum Off topic Physics questions

Forum

Off topic

Physics questions