Question About Relativistic Acceleration

[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 criticize a line of reasoning, which leads to the conclusion that the answer is yes, not offend anyone.

:)

 

[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.

 

[Physicsguru]

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.

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

Kind regards,

Guru

 

[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.

 

[jcsd]

Any particle traveling 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.

 

[Physicsguru]

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.

He answered question one, and he failed to answer question two.

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

 

[Physicsguru]

Any particle traveling 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.

Firstly, how does knowing that any particle traveling 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 traveling at c has zero mass?

Kind regards,

Guru

 

[JesseM]

He answered question one, and he failed to answer question two.

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

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?

 

[jcsd]

Firstly, how does knowing that any particle traveling 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 traveling at c has zero mass?

Kind regards,

Guru

Actually particles traveling 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.

 

[Physicsguru]

Actually particles traveling 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.

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 traveling at c, must be zero?)

 

[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.

 

[JesseM]

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

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.

 

[marlon]

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

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

 

[Physicsguru]

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.

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

 

[JesseM]

How do you respond to this:

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

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

 

[dextercioby]

How do you respond to this:

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

Therefore:

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

Suppose that m=0, and v=c.

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.

 

[Physicsguru]

This equation is not correct in relativity, where if $$m_0$$is the rest mass,

$$p = m_0 v/\sqrt{1 - v^2/c^2}$$

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

 

[JesseM]

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

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.

 

[Physicsguru]

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.

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

Kind regards,

Guru

 

[JesseM]

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

Kind regards,

Guru

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.

 

[dextercioby]

*For light waves:
$$\lambda=\frac{c}{\nu}$$
*For matter waves (de Broglie) associated to relativistic massive particles it can be deduced from the equality u posted...

What are u trying to prove/learn here?

Daniel.

 

[Physicsguru]

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.

Would a relativistic analysis yield the following formula:

$$\lambda = \lambda_0 \sqrt{1-v^2/c^2}$$

?

Regards,

Guru

 

[Physicsguru]

*For light waves:
$$\lambda=\frac{c}{\nu}$$
*For matter waves (de Broglie) associated to relativistic massive particles it can be deduced from the equality u posted...

What are u trying to prove/learn here?

Daniel.

In this post, I am interested in "matter waves," but in this "side issue" i am leaving things up to Jesse, since he is the individual who asserted that any particle which moves with speed c, has a rest mass of zero. That's why I am asking him a few questions, since he is the one who made the assertion.

As for what I am trying to prove/learn here, I am trying to learn whether or not anyone here knows whether or not it is possible to accelerate a large body from rest to the speed of light in 24 hours.

Kind regards,

Guru

 

[dextercioby]

Who's $\lambda_{0}$ ?


Daniel.

 

[dextercioby]

As for what I am trying to prove/learn here, I am trying to learn whether or not anyone here knows whether or not it is possible to accelerate a large body from rest to the speed of light in 24 hours.

Kind regards,

Guru

Lemme join marlon & jcsd in telling you that accelerating a large body from rest to the speed of light in 24 hours is IMPOSSIBLE WITH THE PHYSICAL KNOWLEDGE MANKIND HAS PRODUCED BETWEEN 1905 AND 2005...

Daniel.

 

[JesseM]

Would a relativistic analysis yield the following formula:

$$\lambda = \lambda_0 \sqrt{1-v^2/c^2}$$

Is $$\lambda_0$$ supposed to be some sort of "rest wavelength"? That doesn't make sense, because as you can see by examining the equation for the DeBroglie wavelength $$\lambda = h/p$$, the wavelength goes to infinity as the momentum goes to zero.

 

[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.

Actually come to think of it I'm not completely sure that this equation is correct, because it's mixing an equation from relativity, $$p = m_0 v / \sqrt{1 - v^2/c^2}$$, with an equation from nonrelativistic QM, $$\lambda = h/p$$. Does anyone know if the equation for the DeBroglie wavelength also makes sense in relativistic QM/quantum field theory?

 

[russ_watters]

To put it more simply, you're using the Newtonian physics equation for momentum in a way that it was never intended and quite simply doesn't work (those are two separate flaws in your line of reasoning): it only works for objects with mass and it only works at low velocity.

edit: that equation is irrelevant anyway. Your question asks about accelerating an object with mass (presumably...) to C. Acceleration of a massive object to C has been predicted theoretically and proven experimentally to be impossible. And objects without mass (photons) do not "accelerate to c."

 

[jcsd]

In this post, I am interested in "matter waves," but in this "side issue" i am leaving things up to Jesse, since he is the individual who asserted that any particle which moves with speed c, has a rest mass of zero. That's why I am asking him a few questions, since he is the one who made the assertion.

As for what I am trying to prove/learn here, I am trying to learn whether or not anyone here knows whether or not it is possible to accelerate a large body from rest to the speed of light in 24 hours.

Kind regards,

Guru

1) matter waves are uninteresting in this context as if you treat them as classical plane waves then they are incompatible with both relativity and infact non-relativstic mechanics. That is to say you cannot treat them as actual waves and always expect to get meaningful results especially within relativity. In this case it is esepcially unintersing as soon as you consider the effects of time dialtion/length contraction on matter waves I don't see how you cannot get the wrong answer if you naively treat them ason the same footing as classical EM waves rather than giving them the fall QM treatment.

2) I have already shown that a particle that moves with speed c (i.e. has a null worldline) has a rest mass of zero

3) It's already been shown that in any frmae thta it is impossible to accelerate a particle (or indeed any body) from rest to c within any finite period of time.

 

[Physicsguru]

Here is where I was going with this. Suppose that:

$$\lambda = \lambda_0 \sqrt{1-v^2/c^2}$$

Therefore, the following equation would be a true statement:

$$Mv = \frac{m_0v}{\sqrt{1-v^2/c^2}} = \frac{h}{\lambda_0 \sqrt{1-v^2/c^2} }$$

From which it would follow that:

$$Mv = m_0v = \frac{h}{\lambda_0 }$$

Suppose now, that v=c. Therefore, it would follow that:

$$m_0c = \frac{h}{\lambda_0 }$$

Since not (c=0), it would therefore follow that:

$$m_0 = \frac{h}{c \lambda_0 }$$

Now, suppose that if v=c then $$m_0=0$$. It therefore follows that:

$$0 = \frac{h}{c \lambda_0 }$$

Since wavelength cannot be infinite, it therefore follows that Planck's constant is equal to zero, and that's known to be false. Therefore, it is not the case that if v=c then $$m_0=0$$.

Therefore, granted that v=c, it necessarily follows that not($$m_0=0)$$, which is the contrary of the assertion that if the speed of a particle is c, then it necessarily has a zero rest mass.

Obviously, you must end up disagreeing with some step which I have made, so I now ask you, which one?

Regards,

Guru

 

[russ_watters]

$$Mv = \frac{m_0v}{\sqrt{1-v^2/c^2}} = \frac{h}{\lambda_0 \sqrt{1-v^2/c^2} }$$

Obviously, you must end up disagreeing with some step which I have made, so I now ask you, which one?

Step 2, above, is false, for reasons already stated.

 

[Physicsguru]

Step 2, above, is false, for reasons already stated.

Russ, Energy=hf, therefore h has units of Kg m^2/s. Therefore, h divided by a quantity with units of length has units of classical momentum. How can you simply dismiss setting Mv equal to $$\frac{h}{\lambda}$$? You have said they are unequal, but how do you know that?

Kind regards,

Guru

 

[JesseM]

Here is where I was going with this. Suppose that:

$$\lambda = \lambda_0 \sqrt{1-v^2/c^2}$$

It was already pointed out to you that this equation makes no sense--what is $$\lambda_0$$? If you know that p=0 exactly (which in classical mechanics is what is meant by an object's rest frame), then the DeBroglie wavelength $$\lambda$$ is infinite.

 

[Physicsguru]

It was already pointed out to you that this equation makes no sense--what is $$\lambda_0$$? If you know that p=0 exactly (which in classical mechanics is what is meant by an object's rest frame), then the DeBroglie wavelength $$\lambda$$ is infinite.

How have you drawn the conclusion that the equation makes no sense? I agree with you, that without an interpretation for $$\lambda_0$$, the equation will never "make sense."

Kind regards,

Guru

 

[russ_watters]

Russ, Energy=hf, therefore h has units of Kg m^2/s. Therefore, h divided by a quantity with units of length has units of classical momentum. How can you simply dismiss setting Mv equal to $$\frac{h}{\lambda}$$? You have said they are unequal, but how do you know that?

Like I said before, we know they are unequal from both theory and experimentation. For the theory, Newton did not intend for momentum to be used that way when he wrote his momentum equation, and Einstein didn't intend for it to be used that way when he wrote his part. You're mixing classical mechanics with Relativity. In addition, it is well known that classical mechanics is flawed.

For the experimentation, well, there are lots of examples. Particle accelerators, for a start.

Why do you think that simply having the same units makes them equal?