1. Feb 2, 2005

### Physicsguru

Hello everyone,

I regard myself as the smartest man in the world, and I have two questions for everyone here:

Question 1: Is it possible to accelerate to the speed of light in 24 hours.

Yes or No?

Question 2: If your answer to question one is yes, why is the answer yes; and if your answer to question one is no, why is it no?

Kind regards,

Guru

P.S.: The opening statement is to get the best minds here to criticise a line of reasoning, which leads to the conclusion that the answer is yes, not offend anyone.

:)

2. Feb 2, 2005

### jcsd

No it is impossible in any frame for an object to accelerate from rest to c, any other line of reasoning that suggests otherwise is a priori wrong.

3. Feb 2, 2005

### Physicsguru

You failed to answer question two, the more important of the two questions.

Kind regards,

Guru

4. Feb 2, 2005

### dextercioby

What do you mean he failed to answer the question??Maybe he didin't say specifically what theory forbids it (theory of relativity),but that was a good & correct answer...

Daniel.

5. Feb 2, 2005

### jcsd

Any particle travelling at c has zero mass and any particle with zero mass travels at c (this is easily shown by examing the four momentum of particles that travel at c), so a massive particle can never travel at c and a massless particle always travels at c. This is a widely knoiw and basic result of relativity.

6. Feb 2, 2005

### Physicsguru

Saying that the theory of relativity dictates the answer, is like saying the sky is green, because you read it in the Enquirer.

Kind regards,

Guru

7. Feb 2, 2005

### Physicsguru

Firstly, how does knowing that any particle travelling at c has an inertial mass of zero imply that the answer to question one is no, and secondly, can you prove that any particle travelling at c has zero mass?

Kind regards,

Guru

8. Feb 2, 2005

### JesseM

The difference is that there is an awful lot of evidence that relativity is correct in all its various predictions, including the equation for how energy increases with velocity, which implies an object with finite rest mass would need infinite energy to move at c.

Why don't you just explain your idea for how it is possible to reach the speed of light through acceleration, so people can criticize it?

9. Feb 2, 2005

### jcsd

Actually particles travelling at c don't strictly have zero 'inertial mass' (though some may argue what exactly inertial mass is), they do however have have a zero (rest) mass, the reason I mentioned this is to exclude the case of zero rest mass particles accelarting from rest to c.

The four moemntum of a particle is parallel to a particles worldline, this comes from the definition P = mU. The magnitude of the four moemntum is it's mass, if that mass is zero then the magnitude of the four momentum is non-zero and hence it is not a null vector which means the worldline is not null and the particle does not travel at c.

10. Feb 2, 2005

### Physicsguru

In question one, I made no reference to the kind of object being accelerated, be it particle or large body. For the sake of definiteness, let me rephrase question one as follows:

Question one: Is it possible to accelerate a large body from rest to the speed of light in 24 hours?

(As an aside, can you prove that the rest mass of any particle travelling at c, must be zero?)

11. Feb 2, 2005

### jcsd

Whether the body is spatially extended or not makes no difference (plus all bodies are made of particles anyway).

I suppose you could have some hypothetical body of zero rets mass mving at less than c, but it woldn't have any phsyical properties as it has an enrgy of zero, so most people would be more inclined to call it empty space.

12. Feb 2, 2005

### JesseM

Because the energy of an object with rest mass $$m_0$$ moving at velocity $$v$$ is $$E = m_0 c^2 / \sqrt{1 - v^2 / c^2}$$...if $$m_0$$ is nonzero, then as $$v$$ approaches $$c$$, the energy approaches infinity.

13. Feb 2, 2005

### marlon

It would take an infinite amount of time to accelerate a massive body to the speed of light.

The posts of JCSD are completely CORRECT.

regards
marlon

14. Feb 2, 2005

### Physicsguru

How do you respond to this:

$$P = mv = \frac{h}{\lambda}$$

Therefore:

$$m = \frac{h}{\lambda v}$$

Suppose that m=0, and v=c. Therefore:

$$0 = \frac{h}{\lambda c}$$

Therefore:

$$0 = \frac{1}{\lambda}$$

From which it follows that lambda is infinite. Since nothing can have an infinite wavelength, either not (m=0) or not (v=c). Since you are stipulating that v=c, it must be the case that not (m=0), contrary to your conclusion.

Regards,

Guru

15. Feb 2, 2005

### JesseM

This equation is not correct in relativity, where if $$m_0$$ is the rest mass, $$p = m_0 v/\sqrt{1 - v^2/c^2}$$

16. Feb 2, 2005

### dextercioby

I'm afraid your line of argument is not correct.The first formula u posted (interpreted in the assumption m=0) would indicate that "m" is the rest mass and that the formula P=mv would be purely NONRELATIVISTIC.However,it's easy to see that in nonrelativistic physics the "m=0" does not make any sense (the concept of REST MASS doesn't make sense,as it is simply absolute)...

Daniel.

17. Feb 2, 2005

### Physicsguru

Let m0 = rest mass, and let M = relativistic mass. Let v = speed of center of mass, in some reference frame. let h = planck's constant of nature, and let lambda denote 'wavelength.'

Definition: P= momentum = Mv

Therefore we have:

$$Mv = \frac{m_0v}{\sqrt{1-v^2/c^2}} = \frac{h}{\lambda}$$

I now ask you this, can the wavelength undergo length contraction or not?

Regards,

Guru

Last edited: Feb 3, 2005
18. Feb 2, 2005

### JesseM

the left and middle part of your equation becomes undefined if the rest mass is zero and the velocity is c. But sure, for a particle with nonzero rest mass moving at velocity less than c, the wavelength becomes smaller the higher its velocity.

19. Feb 2, 2005

### Physicsguru

What is the formula for the wavelength in terms of velocity?

Kind regards,

Guru

20. Feb 2, 2005

### JesseM

For a particle moving at a velocity slower than light, it's just $$\lambda = (h\sqrt{1 - v^2/c^2})/m_0v$$. But this equation has no well-defined limit as you let $$m_0$$ approach 0 and let v approach c.