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Can anyone provide me with a complete mathematical proof about why an observer frame cannot surpass the vacuum speed of light?

I have looked for answers in Quora and FB groups but no one is really convincing. In fact the majority assume that the constancy of vacuum speed of light in all referentials without any proof, other people "only" say that it can be proved using the causality principle but without any proof, which is really frustrating!

I hope my curiosity will be satisfied in PhysicsStackexchange, have a nice day!

Dale
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Youssef Benhachem
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That c is a limiting speed cannot be mathematically proved from first principles. Einstein did not calculate that it was true, he postulated that it was a constant for all observers and then calculated the consequences (such as its limiting characteristic and e = mc2). It is a postulate which has been borne out by innumerable scientific experiments, just like Newton's postulate that an object remains in steady motion until some force acts on it. Such cast-iron physical postulates eventually become elevated to laws of nature, not laws of mathematics.

However c as an upper limit may not be quite there yet. Einstein's equations allow superluminal particles - tachyons - for which c is a lower limit. An observer in such a particle's reference frame might, we suspect, see time running backwards. There is no theorem which forbids such antics, just a total lack of experimental evidence - to date.

Guy Inchbald
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    Where did Einstein postulate that $c$ is a limiting speed? – WillO Nov 23 '20 at 00:21
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    He didn't postulate it as a limiting speed. He postulated that all inertial observers would observe the same speed. The mathematics show that postulate results in c being a limiting speed. – Bill N Nov 23 '20 at 01:28
  • @BillN Thanks. I have taken you word for it and updated my answer acordingly. – Guy Inchbald Nov 23 '20 at 05:22
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    @GuyInchbald: Let me gently suggest that if you need to "take Bill N's word for it", then maybe you're not the person who should have been answering this question in the first place. – WillO Nov 23 '20 at 07:05
  • @WillO I am just lazy enough not to confirm from original sources. Courtesy seems a dying art, I am sure you will agree. – Guy Inchbald Nov 23 '20 at 07:17
  • @WillO That's a history of science thing, but this is a physics question. – wizzwizz4 Nov 25 '20 at 19:23
  • @wizzwizz4: The question asks for a proof of something. That question only makes sense if you know what axioms we're starting from. That in turn is a historical question. So you can't separate the question from the history. – WillO Nov 25 '20 at 20:08
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Can anyone provide me with a complete mathematical proof about why an observer frame cannot surpass the vacuum speed of light?

No, there is no such mathematical proof. It is perfectly mathematically acceptable to have $v>c$. Newton’s laws are mathematically valid and allow $v>c$.

The evidence that $v>c$ is not possible is experimental, not mathematical. My favorite summary of the experimental evidence is here: https://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

Section 6 in particular has several experiments directly addressing this fact. My favorites are the particle accelerators where kinetic energy is unbounded as $v$ approaches $c$

In fact the majority assume that the constancy of vacuum speed of light in all referentials without any proof

Note that assuming the invariance of $c $ is not the same thing as assuming that $c $ is the limiting speed. So in responding to a question on the latter it is not circular to assume the former. My above comments and reference apply to this assumption as well, but the relevant section is section 3.

Dale
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The constancy of the speed of light was a postulate of relativity that has been proven experimentally. That light speed is constant for all observers then implies that it is a limiting speed. See for example @CR Drost's Race a Light Pulse thought experiment here: From the speed of light being an invariant to being the maximum possible speed

As for the causality principle, see @John Rennie's answer here: Signal travels with speed greater than light breaks causality

Hope this helps.

Not_Einstein
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You are asking this question because you have learned the wrong units for velocity! Your units [m/sec] have two properties that don't agree with reality (and prompt you to ask your question). 1) There are numbers greater than $c=3 \times 10^8$ m/sec, so why can't I reach them? 2) In particular, why can't I keep adding up smaller velocity boosts until I get greater than $c=3 \times 10^8$ m/sec (ie: velocities are not additive to give the total velocity)?

If you had originally learned that velocity is measured by the Lorentz Group boost parameter $\lambda$ radians, where 1) $-\infty \lt \lambda \lt +\infty$ and 2)$\lambda s$ are additive to get the total boost, then there is no maximum $\lambda$ that you would be wondering about! Many physical transformations are Lie Group transformations (eg: rotations, boosts, strains) done by a number of radians.

The conversion from physically real $\lambda$ units to "historical" velocities $v$ is: $$ v=c\ \tanh(\lambda) $$ There is a need for a constant $c$ to give our historical units $v$ the dimensions of [m/sec]. This $c$ is analogous to the constant $\frac{180}{\pi}$ used to convert rotation angles from [radians] to historical [degrees].

Gary Godfrey
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  • The $\lambda$ you quote is not velocity, but rapidity. It is simply not true that the notion of velocity is a historical artifact - it is an important and relevant quantity, and to suggest otherwise is silly. There is nothing unphysical about the units $m/s$, and they don't lead to the contradictions you claim because units imply nothing (e.g. whether there is an ultimate limit) about the physical quantities to which they are associated. – J. Murray Nov 23 '20 at 01:59
  • @ J. Murray: Yes, the Lorentz Boost parameter is also called rapidity, and m/sec is still useful as an introduction to understanding boost. When boosts were discovered to be non-abelian, we learned that Lie Group parameters of boosts are in radians. Radians are not an arbitrary unit like [m/sec] that "implies nothing". Radians are special. They are dimensionless and you can't do dimensional analysis with them, cosh(radians) makes sense while cosh(m/sec) doesn't. – Gary Godfrey Nov 23 '20 at 10:28
  • Also, since the Lorentz Group, a standard velocity (ie both a standard meter and a standard second) is not needed anymore in the Bureau of Standards since we boost by radians. In fact, the removal of comparison objects to define units from the Bureau of Standards may be the arc of physics. If space-time translations are found to be non-abelian, it will be understood how to do space-time translations by radians, and the final standard meter or standard second will no longer be needed. – Gary Godfrey Nov 23 '20 at 10:48
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In special relativity the total energy of a massive particle moving at velocity $\mathbf{v}$ in some frame of reference is given by :

$$E=\frac{m_0c^2}{\sqrt{1-\frac{|\mathbf{v}|^2}{c^2}}}$$

So if the velocity exceeds the speed of light then that square root is operating on a negative number. The value moves out of the physically realistic range and that tells us we have passed the limits the theory works at.

So the special theory of relativity simply does not work if you try and apply it to velocities beyond the speed of light. And you asked about the special theory of relativity.

Put another way the theory predicts we won't see a particle travel at faster than the speed of light and we don't so we consider the theory valid and hence that prediction valid.

But the formula for the particle's energy also tells us that no matter how much energy we pump into a particle it won't travel faster than the speed of light because :

$$\frac {|\mathbf{v}|} c = \sqrt{1 - \left(\frac{m_oc^2}{E}\right)^2}$$

So the theory tells us there is no way to make a massive particle exceed the speed of light.

Does that mean the universe works that way ? Maybe yes, maybe no. The special theory of relativity does not really work on the scale of the entire universe - it's for locally flat spaces where it works well and experiments say it matches what the universe does on that scale. That's why we need the general theory of relativity. What a physicist would say is that the special theory of relativity is valid to within the experimental error we can measure and within the scope of the theory itself.

StephenG - Help Ukraine
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It all has to do with time dilation. A formula that proves it at the speed of light is:

Proper Time = Observer time / (1 - (v/c)^2)^.5

In the numerator, as he gets closer and closer to c, the observer time drops lower and lower, so the numerator becomes 0 at c. In the denominator, v/c gets closer and closer to 1, so 1-1=0 We end up with a formula 0/0, which just doesn't make sense. Infinity and such.

But even if we want to test a number like c+1, then the numerator will never change from 0, so no matter what, (other than in Star Trek) the proper time that the man faces will never change from 0 at any speed above c.

Since time has stopped for the man, and any motion requires time, this is mathematical proof that he can't go faster than c. That is your mathematical proof.

foolishmuse
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As we look at events in the world from different reference frames, we can relate the positions and times of those events by rotating the coordinates. If one considers a one-dimensional Minkowski space, the transform (rotation) operation for position change of a particle $\left( x_2-x_1=\Delta x \right)$ and time difference in those two positions $\left( t_2-t_1=\Delta t \right)$ looks like $$\pmatrix{\Delta x'\\ \Delta t'}=\pmatrix{~~\cosh \alpha &-\sinh \alpha \\ -\sinh \alpha &~~\cosh \alpha }\pmatrix{\Delta x\\ \Delta t}$$

The quantity $\alpha$ is related to the relative velocity of the new prime frame to the old frame. We will see that relationship later.

The prime coordinates represent the new reference frame quantities. If we define $\frac{\Delta x}{\Delta t}=V$ and $\frac{\Delta x'}{\Delta t'}=V'$ (the particle's velocities observed in the old and new frames) we get $$V'=\frac{V-\tanh \alpha}{1-V \tanh \alpha}$$ If the particle velocity in the new frame is zero ($0$), then, conceptually, the new frame relative velocity must equal the particle velocity in the old frame. Let's call that relative velocity $\beta$. But according to our transformation, with $V'=0$, $V=\tanh\alpha$. So $$\tanh\alpha = \beta.$$ There is no reason to restrict the value of $\alpha$, but the largest magnitude for hyperbolic tangent is $1$. From this we can reason that there is a maximum relative frame speed, $\beta=1$.

One can use differential calculus determine whether there is a maximum $V'$ for any particular $\alpha$. One will see this maximum to be $V'=1$. One will also see that if $V'=1$, then $V$ must also be 1, even for $|\beta|<1.$

This implies that there is a maximum speed for anything in a Minkowski space. What have observed so far gives us good reason to believe that space and time are modeled very well by Minkowski space.

If we postulate that light is measured to have the same value in all inertial reference frames, and in Minkowski space the only speed which has this property is the maximum speed, then light must travel at that maximum speed, and all speeds rescale by a factor c.

You should fill in the algebraic and calculus details to satisfy your curiosity and to get some good math practice.

Added Note With a little more work we can show that $\cosh\alpha=\gamma$ and $\sinh\alpha=\beta\gamma$, where $\gamma=\left(1-\beta^2)^{-1/2}\right)$

Bill N
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Anyone who works with a high energy particle accelerator can tell you that the thing has to be designed to deal with the fact that as the speed of the particles get near the speed of light, the inertial mass increases, but the speed does not. Then again it depends on who is doing the observing. We know that the nearest star is four light years away. If you could travel at near the speed of light (not possible with known science), you could get there in less than a year (as measured by a clock on your ship). But if you return home within another year, your dad will say that you have been gone for over eight years.

R.W. Bird
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Can anyone provide me with a complete mathematical proof about why an observer frame cannot surpass the vacuum speed of light?

Every proof (which of course sounds philosophical and experimental) that you get for this will be based on the postulate of special relativity. One of them says,

The speed of light in a vacuum is the same for all observers, regardless of the motion of the light source or observer.

Relativistic mass is one of the consequences of special relativity which is given by $$m_{rel}=\gamma \ m_0$$

As $v\rightarrow c$ so $m_{rel}\rightarrow \infty$ i.e. The faster something travels, the more massive it gets, and the more time slows – until you finally reach the speed of light, at which point time stops altogether.

In 1964, Bill Bertozzi at MIT accelerated electrons to a range of speeds. He then measured their kinetic energy and found that as their speeds approached the speed of light, the electrons became heavier and heavier – until the point they became so heavy it was impossible to make them go any faster. The maximum speed he could get the electrons to travel before they became too heavy to accelerate further? The speed of light. More on this here.

In another crucial test, physicists Joseph Hafele and Richard E. Keating flew synchronized, super-accurate cesium atomic clocks on various trips around the world on commercial airliners. After the journeys, all the moving clocks disagreed with each other and the reference clock back in the lab. Time ran slower for the moving clocks just as Einstein predicted. So the faster something travels, the more massive it gets, and the more time slows – until you finally reach the speed of light, at which point time stops altogether. And if time stops, well then, so does speed. And so nothing can travel faster than the speed of light.

although it's not rigorous as other proofs are in physics. Maybe it sounds like a philosophical one. The fact if you put $v=c$, everything becomes non-sense that says it can't be possible.

Young Kindaichi
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    I would avoid using the term relativistic mass. I know, I know Feynman used it in his lectures but still the concept is rather misleading and therefore not really used anymore. – AlmostClueless Nov 22 '20 at 20:20
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    What AlmostClueless said. See https://physics.stackexchange.com/a/133395/123208 But anyway, relativity doesn't place a limit on how much you can accelerate a particle (or a body composed of many particles), or how much momentum & kinetic energy you can give a particle, it just limits the velocity. However, the equipment you use to accelerate a particle will have a limitation in how much KE it can transfer to the particle. – PM 2Ring Nov 22 '20 at 21:45
  • This is the only correct answer. – Helen Dec 14 '20 at 19:21
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The causality argument is explained here. It follows from the theory of special relativity that if you can send a signal to someone at a speed of $a c$ with $a>1$, then doing so to someone who moves away from you at a speed larger than

$$v = \frac{2 a}{1+a^2} c$$

(note that $v < c$, you can put $a = 1+t$ and expand the expression out to see this easily), and letting the person send back the signal to you at a speed of $a c$ relative to his/her rest frame, will cause you to receive the signal back before you send it to the person.

This means that the existence of any sort of signal that can travel faster than light, allows you to build a device that you can use to communicate with yourself at some other time. A paradox can then be constructed by building a device that sends a signal into its own past that is programmed such that it will send a signal into a set time into its past if it did not receive such a signal and will not send a signal if it did receive such a signal from the future.

Count Iblis
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If you’re happy with the idea of wave/particle duality and recognise that at c a quantum must be in a purely wave state then you must either accept that c is limiting or propose some ‘tachyon’ state. None of that is in any way a proof though, especially if you’re not on board with the whole wave/particle thing, which I’d infer you’re not.

Frog
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