# B Breaking the light barrier?

1. Apr 1, 2016

### Clever boy

according to einstien, I order for an object to move faster than the speed of light, we need an infinite amount of energy. Assuming that we could somehow generate this much energy, after moving faster than the speed of light, would the hypothetical particle generate a build up of light waves behind it in a similar fashion to sound waves and a sonic boom? I made the assumption that this would happen and stumbled across the thought at this "photonic boom" could have been the origin of the universe and the cause of the Big Bang. Please forgive me if this is far fetched and illogical. I'm only in grade nine.

2. Apr 1, 2016

### Ibix

I'm afraid you can't accelerate to faster than the speed of light.

Einstein based his theory on the postulate that the speed of light is always the same (it turns out to be about 3x108m/s, or 186,000 miles per second). You know how, if you look out of the window of a car travelling at 60mph, another car can stay right next to you? According to someone standing beside the road, that car is doing 60mph. According to you, both you and the other car are stationary, and the road is zipping backwards at 60mph. But everybody agrees that light passes them at 3x108m/s, whether they are standing by the road, in a car, or in a spaceship heading to the moon.

That means that passing the speed of light is impossible - because at one instant you would have to be travelling at the same speed as light and it would be stationary next to you. But it would also have to be doing 3x108m/s. That's a contradiction. So either the theory is wrong (and it's passed every test we've flung at it in over a century - and we've never seen anything travelling faster than light, either) or we can't exceed the speed of light.

So I can't answer your question. You are asking what would happen if we did something impossible. It's like asking what the world would be like if the colour blue were actually apples - the question makes no sense.

3. Apr 1, 2016

### Orodruin

Staff Emeritus
It seems you have not really grasped the concept of infinity. It does not just mean "very very large".

I am going to relabel this thread to "B".

4. Apr 1, 2016

### Staff: Mentor

As other posters mentioned it is not possible to go faster than c, which is the speed of light in vacuum. However, it is possible to go faster than the speed of light in a transparent medium, like water. This produces light which is the electromagnetic equivalent of a sonic boom. It is called Cherenkov radiation. It is responsible for the blue glow seen in water around a nuclear reactor.

5. Apr 7, 2016

### peety

Presumably there is a reason why this qualification doesn't suggest the possibility of transmitting information faster than light. If this is so, why? Would information break down? Would quantum aspects come into play?
Should we say that nothing can go faster than light but light can travel faster than c?

6. Apr 7, 2016

### jbriggs444

This qualification [that the speed of light in a medium can be less than the speed of light in a vacuum] allows for transmitting information faster than the slowed-down light, but not faster than the universal speed limit, c.
Nothing can go faster than c. But light can travel slower than c.

7. Apr 7, 2016

### peety

Thanks, I must have misundertood Dale's post.

8. Apr 7, 2016

### Staff: Mentor

Strictly from a geometry point of view (setting aside the impossibility of the objectvitself), there is no reason you couldn't see such an object, as long as you are inside the cone. Just like you can hear a plane traveling faster than the speed of sound if you are inside the Mach cone.

9. Apr 7, 2016

### Staff: Mentor

I guess the right way to say it is "nothing can travel faster than c, but light can travel slower than c in a transparent medium."
Edit: I see jbriggs444 said almost exactly the same thing.

10. Apr 8, 2016

### vanhees71

Well, that's a bit more complicated, because you have to specify which "speed" you are referring to when talking about "travel speed" of waves in a dispersive medium.

First you look at stationary states of single-frequency plane waves travelling in $x$ direction, which are of the form
$$A(t,\vec{x})=A_0 \exp(-\mathrm{i} \omega t + \mathrm{i} n(\omega) \omega x).$$
Here $A$ is some arbitrary field (e.g., a component of the electromagnetic field). If it's a real quantity, we silently take the real part of the exponential form, which is just more convenient to calculate with. Then $n(\omega)$ is a frequency dependent diffraction index, which in general is complex, describing both dispersion (real part) and damping (imaginary part) of the waves in the medium. If we have weak damping, i.e., if we can neglect the imaginary part of $n(\omega)$ we can define the phase velocity, defined by setting the phase of the wave constant
$$\omega t-\omega n(\omega) x=\text{const} \; \Rightarrow \; c_{\text{phase}}= \mathrm{d}x/\mathrm{d} t=1/n(\omega).$$
There's nothing preventing $n(\omega)<1$, and thus $c_{\text{phase}}$ can be $>1$ (I'm setting the speed of light in vacuo to 1 of course). This doesn't violate Einstein causality, because it's a "speed" of a stationary plane-wave state, which in nature can only be reached approximately in a limited region of space by switching on such a harmonic signal for a sufficiently long time. The speed doesn't refer to any cause-and-effect relationship between "events". It's just defining the speed of the propagation of the phase of a stationary plane wave.

Another question is, how fast signals can travel, i.e., the speed of a wave packet of limited spatial width. Such a wave packet is constructed by a Fourier transform, i.e., the decomposition of the wave in the plane-wave modes
$$A(t,x)=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} A_{\omega_0}(\omega) \exp [-\mathrm{i} \omega t + \mathrm{i} k(\omega) x], \quad k(\omega)=\omega n(\omega).$$
Suppose now that $A_{\omega_0}$ is quite narrowly peaked around a single frequency. Then we can approximate the integral as
$$A(t,x) \simeq \exp[-\mathrm{i} \omega_0 t+\mathrm{i} k(\omega_0) x] \int_{\mathbb{R}} \mathrm{d} \omega' \frac{A_{\omega_0}(\omega_0+\omega)}{2 \pi} \exp \left [\mathrm{i} \frac{\omega'}{v_g} (x-v_g t) \right]$$
with
$$v_g=\left (\frac{\mathrm{d} k}{\mathrm{d} \omega} \right)^{-1}_{\omega=\omega_0}.$$
Thus the "envelope" of the wave packet (characterized, e.g., by its peak) goes with the group velocity $v_g$. Also this quantity can take values $>c$ (in the region of anomalous dispersion).

The only speed that can never exceed the speed of light is the front velocity of a wave of finite spatial extension. This has been shown already around 1910 by Sommerfeld in answer to a question by W. Wien, how the faster-than-light phase and group velocities well known in optics are compatible with the special theory of relativity.

11. Apr 8, 2016

### Staff: Mentor

You are, of course, correct. But the thread is marked "B", so I was leaving out the complications.

12. Apr 8, 2016

### pervect

Staff Emeritus
The simplest, accurate way I can think of to describe the situation doesn't involve energy at all. Instead, it involves a large number of rockets, each of which moves faster than the last. We give them each a number, $r_0$, $r_1$, $r_2$, etc - we call the n'th rocket $r_n$. I hope this isn't too advanced, as we are using some algebra, but I don't know how to write a good response without at least algebra.

Now, we can imagine that the relative velocity between $r_0$ and $r_1$ has some fixed value, say 1 meter/second. And the same is true for $r_2$ and $r_1$, so each rocket in the chain moves 1 meter per second faster than the previous chain.

The question becomes - what is the relative velocity between the n'th rocket, and the first rocket, $r_0$? If the velociites of the rockets just added, it would be infinite. But they don't "just add" in relativity, there is a formula for how they add, called the velocity addition formula. Subtracting velocities works the same as adding a negative velocity (this is algebra, again).

To go much further than this requires more math. But the key point is that although you have an unbounded or infinite number of rockets, and although each rocket moves 1 meter/second faster than the previous rocket, the relative vlocity between the first rocket and the last remains lower than the speed of light.

The "energy" explanation is common in popularizations, and it's not really wrong - it just doesn't address the heart of the problem.

I hope this helps.

13. Apr 8, 2016

### jbriggs444

Being picky here. If you label your rockets 0, 1, 2, ... and continue that sequence infinitely, you will not have a last rocket. However you are correct that every rocket in that infinite sequence will have a finite velocity strictly less than c.

14. Apr 9, 2016

### pervect

Staff Emeritus
Formally, I should have said that if you take the limit of the difference between the n-th rocket in the sequence and the 0-th rocket, the limit approaches a finite value, "c", equal to the speed of light. But I was trying to avoid the math, so I was rather sloppy. The intent was to be more understandable to the OP, who has likely not been exposed to the concept of a limit yet, as he is in the 9th grade.