The cosmic speed limit

Question

We already know by the postulates of Relativity, that the maximum theoretical possible velocity of a body is c or 299,792,458 m/s to be exact.

Also, if we think that the relative velocity would be exceeding c, then we are quite wrong.

Suppose, two bodies travelling in antiparallel directions have velocity of 0.75 each. Then only from my reference frame they would have $V_r$ of 1.50c . However, in their respective frames of reference, they have a $V_r$ of 0.96 c by the Lorentz transformation.

This is all very good. However, why is it exactly that no object in the universe can exceed the speed of light?

The Lorentz Transformation prevents this. Is there an experimental proof of this that no body can move faster than c? If there is, please do. Please inform me about this also. What is the explanation of this or I am getting something wrong?

Answer 1

However, why is it exactly that no object in the universe can exceed the speed of light? What is preventing us?

Mass. Objects with mass travel slower than light.

In an inertial frame, the relationship between mass, energy, and momentum is $$m^2 c^2 = E^2/c^2 - p^2$$ and the relationship between energy, momentum, and velocity is $$\vec v = \frac{\vec p c^2}{E}$$

We can combine the above to get $$v^2 = c^2 - \frac{m^2 c^6}{E^2}$$ from which if $0<m$ we can easily see that $v^2<c^2$ and we can find $$\lim_{E \rightarrow \infty} v = c$$

if it impossible why are photons capable of moving at velocity c?

Photons do not have mass. Things that do not have mass must travel at the speed of light. For $m=0$ we can easily obtain $v=c$ from the same equations.

Answer 2

$c$ is the speed of causality. It is the constant which connects the dimensions of time and space. Indeed, all objects are always moving through spacetime at $c$ - this is easily shown from calculating the quantity $$ u^\mu u_\mu = -c^2 $$ Where $u^\mu$ is the worldline tangent of a particle. It follows that if the particle traverses a certain distance in space, the corresponding distance traversed in time is reduced - this is the world's handwaviest explanation of time dilation. However, we also see that we can at most traverse the same distance in time and space. If we were to somehow traverse more distance in space than in time, our physical definition of velocity as "space interval traversed per time interval" completely falls apart.

This can also be seen from a Minkowski diagram: The faster one goes, the smaller the angle between the axes becomes, until they finally both collapse onto the diagonal for a particle moving on a null worldline. If one were to rotate them further, time and space would "switch places", again breaking our physical definition of velocity. As a side note, the mathematical framework of general relativity predicts such a "switching of places" for time and space beyond the event horizon of a Schwarzschild black hole, beyond which the escape velocity indeed exceeds $c$.

Answer 3

You can't move at $c$ because no matter how fast you go, you're not moving. You leave Earth at $0.99c$ and draw your Minkowski diagram: $(x, t)$ are two perpendicular axes (in the Euclidean sense, on the paper).

If you send out a light pulse: it moves away from you at $c$ is all directions, so if you have more rocket fuel: which way to point to reach $c$?

Forward seems like a good bet, but forward is equivalent to up to left to backwards, and backwards would put you right back in the rest-frame of Earth.

Also: right now, consider yourself in the frame of an ultra high energy cosmic ray (UHECR):

$$ \gamma =\frac E {m_pc^2} \approx 10^9 $$

so

$$ v \approx 1-\frac 1 2\frac 1{\gamma^2} =0.9999999999999999995c $$

In the UHECR-frame you are $5\,{\rm mm/year}$ slower than light--are you close to reaching $c$? If so, how do you get there?

As explained in the prior answer: this is the hyperbolic geometry of Minkowski space.

tl;dr You can't move relative to space.

Regarding the speed of light: don't get hung up on photons and mass. Look at Jefimenko's formulation of EM (equivalent to Maxwell's equations, https://en.wikipedia.org/wiki/Jefimenko%27s_equations).

Here, an non-zero field strength is cause by charges, currents, and their time derivatives on the past light cone, so that disturbance is a distance $ct$ from the source, after a time $t$. In this view, there is an effect propagating at the speed of causality.

Of course you can describe that as a wave, and quantize it to be a massless photon, but that gets us back to where we started. But really, the field I see right now at my location is a sum over my past light cone from me to the CMB, and all those "causes" and just effecting me "now", any earlier and you have all the time-ordering of space-like separated events cropping up.

Answer 4

If we expend a lot of energy to accelerate a particle to $0.99999999c$ relative to observer $\text{A}$, then from the point of view of observer $\text{B}$, who is co-moving with the particle, the particle is still going slower than a passing photon by a factor of $c$. In other words, no matter how much energy we use to accelerate an object, we are still no closer to moving at $c$ than before we accelerated.

Yes, I know that, but I guess my question was not clear enough. I want to know how we define energy limits. – Ritzthephysibeast

From the above, it should be clear there is a finite energy limit. The kinetic energy of a particle is given by $$\text{KE} = \frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2.$$

Mathematically, if we set $v=c$ we get: $$\text{KE}= \frac{mc^2}{\sqrt{1-c^2/c^2} }-mc^2 = \frac{mc^2}{0} - mc^2,$$ which is undefined. You might argue that as $v$ tends to $c$, the amount of energy required to accelerate it tends to infinity. From my first argument, $v$ does not even tend towards $c$, and there is no finite answer as to how much energy is required, even in principle, to accelerate a particle to $c$.

Answer 5

It is a property of the geometry of spacetime. As a vague analogy, imagine you are travelling on the surface of a sphere- no matter how much you accelerate, you never get further from the centre of the sphere or nearer to the horizon. Energy has nothing to do with it.