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The (relativistic) mass of an object measured by an observer in the $xyz$-frame is given by $$m = \frac{m_{rest}}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}.$$ Mathematically $v$ could be greater than the speed of light, but the mass $m$ would become imaginary. Physically we would have to get to the speed of light first i.e. $v = c$, which gives us an undefined value for $m$. So we believe that nothing moves faster than the speed of light because we do not like observables to be imaginary?

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One simple answer is that there is not physical mechanism that will accelerate particle to velocities beyond the speed of light, because it would require an infinitely large amount of energy to accelerate particle arbitrarily close to the speed of light. – Prathyush Oct 29 '12 at 1:23
True, but this argument does not exclude particles that 'started off' moving faster than the speed of light. – Mark Mitchison Oct 29 '12 at 1:25
Why downvote? This is a perfectly valid question. – glebovg Oct 29 '12 at 18:05
The issue isn't that we don't want observables to be imaginary. What you get when you measure the weight or inertia of an object is its mass-energy, not its mass, and the mass-energy of a tachyon is real, not imaginary. – Ben Crowell Apr 5 '13 at 23:19
Related: – Qmechanic Jun 5 '13 at 13:21

4 Answers 4

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Physically, you math guys aren't allowed to cross near the boundary $c$ (speed of light). Special Relativity does that. SR says that it would be impossible for a particle to be accelerated to $c$ because the speed of light (maximum possible measured velocity) is constant in vacuum for all inertial observers (i.e.) Observers in all inertial frames would measure the same value for $c$. Not only the fact that infinite energies are required to accelerate objects to speed of light, (but) an observer would see things going crazy around the guy (or an object) traveling at $c$ such as length contraction (length would be contracted to zero), time dilation (time would freeze around him) & infinite mass. You can't enjoy anything when you travel at $c$. But, the stationary observer who's measuring your speed (relative to his frame) would definitely suffer..!

Note: But, there are some quantum mechanical solutions that allow negative masses like the expression for relativistic energy-momentum. Let's try not to make the subject more complicated. $$E^2=p^2c^2+m^2c^4$$

There are hypothetical particles (having negative mass squared (or) imaginary mass) always traveling faster than the speed of light called Tachyon. This was assumed by Physicists in order to investigate the faster than light case. So When $v>c$, the denominator becomes a imaginary. But, Energy is an observable. It should be some integer. A consistent theory could be made if their mass is made to be imaginary and Energy to be negative. Using these data in the E-p relation, we would arrive at a point $p^2-E^2=m^2$, where $m$ is real. This makes Tachyons behave a kind of opposite to that of ordinary particles. When they gain energy, their momentum decreases (which strongly disproves all our assumptions).

The first reason that this investigation blown off is Cherenkov radiation where particles traveling faster than light emit this kind of radiation. As far as now, No such radiation has been observed in vacuum proving the existence of these..! It's like making a pencil to stand at its graphite tip. If it would stand, physicists would've to blow up their heads :-)

There are tougher stories on the topic when you Google it out...

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Suppose a particle starts traveling faster than light without getting to the speed of light first. Just because such beasts would cause strange things to happen does not mean it is impossible. – glebovg Oct 29 '12 at 2:52
Unless you can come up with an operational definition of "negative" or "imaginary" mass, then you are just doing maths, not physics. You need to define what such a negative mass would look like physically (e.g. repelling "positive mass" objects by gravity, a behaviour that has never been observed to my knowledge). – Mark Mitchison Oct 29 '12 at 3:16
no worries, actually it was intended as a comment for glebovg, but I'm glad if it helped you sharpen up your answer :) – Mark Mitchison Oct 29 '12 at 17:16
There are at least three different questions one could ask, which some of the comments and answers seem to be assuming incorrectly are the same question. (1) Does nothing move faster than light? (2) Do known forms of matter such as atoms never move faster than light? (3) Can we have observers and frames of reference that move faster than light? The OP asked about 1. Prathyush's comment relates to 2. Crazy Buddy's answer seems to be about 3. The simple answer to 2 is that rest mass is observed to be a fixed property of particles, and you can't [...] – Ben Crowell Apr 5 '13 at 23:25
[...] accelerate a particle past c without changing its rest mass. This is why we talk about tachyons as a separate type of particle. Question 3 has a somewhat technical answer; see V. Gorini, "Linear Kinematical Groups," Commun Math Phys 21 (1971) 150. – Ben Crowell Apr 5 '13 at 23:27

Actually, a quick search on Wikipedia shows that you have misinterpreted this formula: imaginary-mass particles do not propagate faster than the speed of light when you take quantum mechanics into account. A much better reason not to believe in faster-than-light particles is that they have never been observed to exist. Furthermore, if they were to exist, I could in principle do rather confusing things like cause the death of my own grandmother before my mother was even born. Generally, if something does not appear to exist, and it would cause everyone a massive headache if that something did exist, it is easier to assume that it doesn't! Likewise, I do not believe that an enormous colony of pixies living on the dark side of the moon is planning a surprise birthday party for me next year. Of course, someone, somewhere probably does believe that :)

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If something has not been observed it does not mean it does not exist. Only a bad physicist can think this way! According to your argument, quantum mechanics is a complete nonsense and we should not believe in it because it makes your head hurt. Overall, a pointless answer. – glebovg Oct 29 '12 at 2:45
@glebovg Let me reiterate, or perhaps rephrase, my point. I said (or should have said) if there is no evidence for something then there is no reason to believe in it if it seems to disagree with other things you know. I apologise if you were confused by my flippant tone of voice: the comment about headaches was intended to amuse, not to be taken literally. Science is about drawing rational conclusions based on evidence. Quantum mechanics does not "make my head hurt", to carry on my (admittedly poor) analogy. On the contrary, it explains things I wouldn't understand otherwise. – Mark Mitchison Oct 29 '12 at 2:52
Quantum mechanics is therefore more like paracetamol :) And just to ram the point home, I would contend that only a bad physicist believes in things just because they haven't been proved beyond all reasonable doubt. That doesn't mean you can't maintain an open mind, but ideally, we should be skeptical about everything for which there is not a significant body of evidence. Your suggestion seemed to be that we don't believe in tachyons because a certain equation gives unsavoury results. My point was that such an aesthetic consideration is never valid. Only evidence and logic matter. – Mark Mitchison Oct 29 '12 at 2:56
@glebovg First of all, you are speaking for yourself there: I would not say I "strongly believe" in black holes. Second, this is not a very good example, since black holes are a natural prediction of general relativity. Other predictions of general relativity (gravitational lensing and time dilation) have been shown to be true, therefore it is perfectly reasonable to assume, in the absence of any better evidence, that other predictions of GR are correct. – Mark Mitchison Oct 29 '12 at 3:05
On the other hand, the proposition that particles can travel faster than the speed of light is not based on either evidence nor logical deduction from existing theory (like black holes), there is no proposed mechanism for how this could happen, and it would not explain any as-yet unexplained phenomena. I think the difference is pretty clear. – Mark Mitchison Oct 29 '12 at 3:07

The use of relativistic mass is deprecated in modern physics, which means that we can explain why nothing moves faster than the speed of light (in the contest of special relativity) without even mentioning relativistic mass.

The special relativistic energy for a massive particle is $E^2=p^2c^2+m^2c^4$, where $m$ is mass [1]. Solving the Hamilton equation we obtain the velocity $v = pc^2/E$. It is evident that $v = c$ only when $E=|p|c$, which implies that $p$ has to be infinite: $\lim_{p \rightarrow \infty} (p^2c^2+m^2c^4) = p^2c^2$. Therefore you need infinite energy for accelerating a massive particle at the speed of light. The energy of all the observable universe is finite. You cannot break the c limit.

For a massless particle, special relativistic energy is $E^2=p^2c^2$. Solving the Hamilton equation we obtain the velocity $|v| = c$, which is a constant. Therefore no matter what energy you give to a massless particle, it will be always moving at the speed of light. Again you cannot break the c limit.

[1] This is standard notation but disagree with your notation. Relativistic mass is often denoted by $m_{rel}.$

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according to sudarsan, when particeles travel with velocity greater than that of light, the mass becomes negative or imaginary. they are called Tachons. I have shown in an earlier article that the mass becomes imaginary or negative. According to the method used in quatum mechanics, by multiplying by a complex conjugate [that is chaning -i to +i] one gets a positive mass but slightly less than the orignal

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I was wondering, what do you mean by slightly less? The modulus will still be the same... And the method is not used only in quantum mechanics. if you have any number fraction with complex number in the denominator, you can always multiply top and the bottom by a complex conjugate and then the complex part remains only in the numerator. This is, as far as I understand, some basic complex algebra. – gns-ank Mar 23 '13 at 18:33

protected by Qmechanic Mar 23 '13 at 16:49

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