# V > c for beta-rays?

1. May 11, 2012

### elias2010

Hello, I repeat my question
As is well known that radioactive decay is a random, automatic phenomenon that nothing can stop it. This is unaffected by any variable such as temperature, pressure, acceleration. Each radioactive element has a specific fixed as half-life, wavelength and period, which is fully stable and unaffected.
Imagine a spaceship made of depleted uranium in the earth is moving at a speed approaching that of light.
B-rays emitted in the direction that moves beyond the speed of light?
No, says the special theory. But then is the paradox. A Geiger counter in the spacecraft detects radiation and another mounted on the outside 'in front of the window' was not detected. Both are stationary observers on the craft. To illustrate this, suppose that the craft is made of non-radioactive materials and moving towards earth with a constant speed. Inside there's a piece of uranium, a Geiger counter in front of A is in relation to the direction of craft speed and a second B behind it. If the speed is low, both counters detect radiation. Recall here that the speed of light is constant independent of the observer and inaccessible. So if the speed is equal to that of light, A did not detect radiation. Which of the two would happen if there was no earth or other reference point to know the speed? That is, whether the rings A or not, depends on the observer. The theory introduces a new conception of reality, where a natural phenomenon does not happen by itself, but depends on the existence of observers.
But you say, nobody can approach the speed of light. But perhaps one core of radioactive isotope in a large accelerator can do. You just have to try it, if there at CERN read our articles. We will see then if the speed prevents the decay of the nucleus, which is, as we said, a random phenomenon. And how the core 'knows' that moving relatively to the observer to 'postpone' its breakdown of?

2. May 11, 2012

### Ich

Re: Quick question on Special Relativity

The postulates of relativity:

1) principle of relativity means that there is no preferred state of motion, which directly means that the outcome of your rocket experiment is independent of its velocity.
2) speed of ligth the same in every reference frame means that there is no reference frame where the speed of light is zero as in your "So if the speed is equal to that of light, A did not detect radiation.". Which means again that the rocket can't go at the speed of light.

So, even without any math, your conclusions directly contradict the postulates that the theory is built on. You sure you're talking about relativity? Where did you learn it from?

I don't understand what you're asking.
In the core's system, there is no time dilation. Time dilation is a relation between different frames, like core frame and laboratory frame. It has nothing to do with your notion of "slowing processes".

3. May 11, 2012

### elias2010

Re: Quick question on Special Relativity

I am not talking about time dilation,what I am asking for is that if the moving core decays,b-rays emmited beyond the speed of light?

4. May 11, 2012

### Ich

5. May 11, 2012

### elias2010

Re: Quick question on Special Relativity

Thank you,I'll see it,but what about the isotope core emission relative to lab?

6. May 11, 2012

### Ich

Re: Quick question on Special Relativity

Either the nucleus is going v<c, then you're talking about time dilation, which I already answered. Or it is going >=c, then you're talking about something impossible.

7. May 11, 2012

### HallsofIvy

Re: new question on Special Relativity

No one has said you cannot approach the speed- you just can't ever get there. And, no, the speed does not "prevent" decay of the nucleus. You are just using the wrong formula for combination of speeds. If an atom on a space ship, moving at, say, .9c relative to me, emits a beta ray moving at .9c relative to the atom, directly toward me, I would measure the beta ray as moving at
$$\frac{.9c+ .9c}{1+ \frac{(.9c)(.9c)}{c^2}}= \frac{1.8c}{1.81}= .99c$$
still less than the speed of light.

8. May 11, 2012

### Staff: Mentor

Beta radiation consists of electrons or positrons. It is always slower than c.

If you believe otherwise then please provide a mainstream scientific reference which states that.

9. May 11, 2012

### Whovian

I think what (s)he's saying is the very well known newtonian formula for adding velocities, which is invalid in this case.

10. May 11, 2012

### elias2010

Thank you all
What I ask in my first question is that if A counter rings,is this for an observator on earth means that b-rays beyond the speed of light?

11. May 11, 2012

### Whovian

Again, no, the formula isn't $\vec{v}_{1,3}=\vec{v}_{1,2}+\vec{v}_{2,3}$.

12. May 11, 2012

### Staff: Mentor

Also, a Geiger counter does not measure speed of a radioactive particle, it only detects its presence. To measure speed you will need something else, like a Doppler radar.

13. May 11, 2012

### Mentz114

In this setup, the detectors are stationary wrt the source. Therefore they will both count beta rays at all times. Furthermore, every observer will agree on the counts.

No, this is wrong. The counters operation does not depend on a relative velocity with some other frame.

14. May 12, 2012

### elias2010

Ok,but what about γ-rays also emitting by the nucleus?

15. May 12, 2012

### Whovian

Since those are light, they travel at c in all reference frames.

Remember that if we use the relativistic formula for adding speeds, if something's moving at c in one reference frame, it's moving at c in all reference frames (in fact, ignore the relativistic formula for adding speeds, this is what Special Relativity comes from), and if something's moving at <c in one reference frame, you get a speed of <c in all reference frames.

16. May 12, 2012

### Mentz114

As Whovian has said, γ-rays are light so they always travel at c in all frames.

If you placed your apparatus ( the emitter and the detectors ) in any non-accelerating lab it will give the same results. The laws of physics are the same in all inertial frames. You should be aware that there is no absolute motion. The lab may be moving close to c with respect to some other frame but that is irrelevant. In the lab frame there are no relativistic effects.

If the emitter was in motion with respect to the detectors, there would relativistic effects like time dilation, but the counts of the detectors would be the same in all frames, although the rate of detection might be different.

17. May 12, 2012

### elias2010

Forgive me but I have doubts until this could be experimentally assured.I asked for it:
"But perhaps one core of radioactive isotope in a large accelerator can do. You just have to try it, if there at CERN read our articles."
The core is moving close to c and emitts γ-rays.The frame is the lab.
Supposed that math produce physics you know Zenon's parodox.

18. May 12, 2012

### Vorde

More importantly, nothing emitted of any sort will appear to travel faster than light from any inertial reference frame. Either the radiation is EMR and therefore traveling at v=c in all reference frames, or it is traveling v<c in which case as per the relativistic velocity addition formula, no reference frame will observe it moving faster than light.

19. May 12, 2012

### Staff: Mentor

This has been experimentally validated for decades. Particle accelerators produce a plethora of unstable particles which radioactively decay. Their decay rates can be measured and compared to the decay rates of the same particles at rest. My favorite are the muon decay experiments of Bailey, but here is a broad overview:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Zeno's paradox is completely unrelated to this topic.

20. May 14, 2012

### elias2010

HallsoIvy wrote:

"No one has said you cannot approach the speed- you just can't ever get there. And, no, the speed does not "prevent" decay of the nucleus. You are just using the wrong formula for combination of speeds. If an atom on a space ship, moving at, say, .9c relative to me, emits a beta ray moving at .9c relative to the atom, directly toward me, I would measure the beta ray as moving at
.9c+.9c1+(.9c)(.9c)c2=1.8c1.81=.99c
still less than the speed of light."

If the spaceship moving at c (as its supposed in my statement;impossible?) from the formula above you would measure the beta ray as moving at c.That means that it could not reach counter A so A not rings,or the atom could not emit (speed prevent decay).
What would be happened in fact?