Yet another relativistic acceleration question

[starthaus]

The uncertainty of both of these quantities is given by $$\Delta$$x = $$\Delta$$(x₀/y) = (1/y)$$\Delta$$x₀

Nope, see previous post.

and $$\Delta$$p = $$\Delta$$(mv) = $$\Delta$$(ym₀v) = (ym₀)$$\Delta v$$

Also, no, see previous post. Even if you differentiated the exressions correctly, your approach is flawed, since it doesn't prove in any way that $$v$$ can ever exceed $$c$$.


The rest of the "derivation" is all messed up. The conclusion, as well.

 

[Dale]

To DaleSpam, for an object moving at any given velocity v as measured from the rest frame of a distant stationary observer, let p = mv be the momentum and x be the position of the object, as measured in that same frame. The uncertainty of both of these quantities is given by $$\Delta$$x = $$\Delta$$(x₀/y) = (1/y)$$\Delta$$x₀ and $$\Delta$$p = $$\Delta$$(mv) = $$\Delta$$(ym₀v) = (ym₀)$$\Delta$$v, where the subscript naughts describe the corresponding measurement values when the ship was at rest wrt to the observer (keep in mind that at a specific velocity, m₀, $$\Delta$$x₀, and the Lorentz factor, y, are all constant from the observer's perspective). The HUP thus states that $$\Delta$$x$$\Delta$$p = (1/y)$$\Delta$$x₀(ym₀)$$\Delta$$v = m₀$$\Delta$$x₀$$\Delta$$v $$\geq$$ [STRIKE]h[/STRIKE]/2. From this we see that $$\Delta$$v $$\geq$$ [STRIKE]h[/STRIKE]/2m₀$$\Delta$$x₀, i.e. the uncertainty in an observer's measurement of the ship's velocity is bounded by its rest mass and the uncertainty in it's position measurement while at rest. Because of this, there should be some velocity extremely close to c such that the uncertainty of the velocity measurement makes it theoretically impossible to say whether the ship is actually going slightly slower or slightly faster than c, from the observer's perspective.

The uncertainty relationship does not transform this way. Also, as others mentioned y is a function of v so you cannot just pull it out like that. Here is a short paper that describes how to transform the uncertainty relation:
http://docs.google.com/viewer?a=v&q...y1rw4r&sig=AHIEtbRsPyQakkAL9sKha_DVN7zS3XQnbA

 

[RexxXII]

@starthaus: For any given velocity, at that velocity y is constant (as I made sure to mention in my post), so it is being treated as such. Also, y is not dependent on the operator v, but on the magnitude squared of v, which is a number. Furthermore, what does differentiation have to do with an expectation value? If $$\Delta$$x represents a width on either side of an object (simplifying the concept to 1D), then why would this width not be Lorentz contracted like the rest of the ship, such that $$\Delta$$x = (1/y)$$\Delta$$x₀? And from a basic logic point of view, how does a misinterpretation of my result imply that my approach is flawed? Also, assuming that my derivation and results are correct (which I highly doubt you would or could ever do, even for simple didactic reasons) why would an observer measuring a ship going faster than the speed of light within their observational uncertainty not qualify as the ship going faster than the speed of light from their perspective? The very fact that this idea has not been fully refuted by anyone in this thread (namely you or DaleSpam) is why I'm asking the question.

@DaleSpam: That is an interesting paper which I haven't seen before. My question is whether or not those same equations still apply when one is working in a system of baryons rather than mesons. Because the derivations were based on the assumption that the particle was a harmonic oscillator bound state of two quarks (which must then be a quark and an antiquark to conserve color), it seems unlikely that the same results would be applicable to a system of three quarks that are bound in a much more complicated manner than a simple harmonic oscillator, as is the type of matter that this ship is made of. In fact, this article doesn't take into account at all the fact that one of the quarks is an antiparticle or the behavior of the interaction between the quark and antiquark, which certainly were not as well understood in the 70's when this paper was first published as they are today. Is there another source for this transformation relation that doesn't rely on meson based derivations?

 

[starthaus]

@starthaus: For any given velocity, at that velocity y is constant (as I made sure to mention in my post), so it is being treated as such. Also, y is not dependent on the operator v, but on the magnitude squared of v, which is a number.

This is a festival of mistakes. Please answer the following:

1. If the above is true, why did you differentiate $$v$$ ? According to your claim, $$v=constant$$ so $$\Delta v=0$$.

2. It is already known in maistream physics that $$v \gamma(v)$$ get differentiated as a function, i.e. together.
I have already shown that to you in a previous post. What entitles you to differentiate (incorrectly) only $$v$$? Are you trying to rewrite the basic rules of calculus and physics?
Can you try differentiating $$\frac{v}{\sqrt{1-(v/c)^2}}$$ correctly?

3. Why makes you think that $$v^2$$ is a number?


4. Dalespam has already given you a reference as to how this is done correctly, why do you insist in doing it incorrectly?

5. Your incorrect derivation shows $$v>c$$ despite the fact that several of us have shown you that this is impossible. Why persist in your errors?

Furthermore, what does differentiation have to do with an expectation value?

Correct differentiation has everything to do with obtaining correct results. Errors like yours lead to incorrect results.

Also, assuming that my derivation and results are correct (which I highly doubt you would or could ever do, even for simple didactic reasons)

Because your derivation is not correct. It has elementary errors of both calculus and basic physics. See above.

 

[DrGreg]

As RexxXII doesn't seem to understand how differentiation is involved, for small Δv, Δp approximates to

$$\Delta p \approx \frac{dp}{dv} \Delta v$$​

 

[Dale]

@DaleSpam: That is an interesting paper which I haven't seen before. My question is whether or not those same equations still apply when one is working in a system of baryons rather than mesons. Because the derivations were based on the assumption that the particle was a harmonic oscillator bound state of two quarks (which must then be a quark and an antiquark to conserve color), it seems unlikely that the same results would be applicable to a system of three quarks that are bound in a much more complicated manner than a simple harmonic oscillator, as is the type of matter that this ship is made of. In fact, this article doesn't take into account at all the fact that one of the quarks is an antiparticle or the behavior of the interaction between the quark and antiquark, which certainly were not as well understood in the 70's when this paper was first published as they are today. Is there another source for this transformation relation that doesn't rely on meson based derivations?

The wavefunction is for any pair of equal mass particles in a quadratic potential. It would probably apply pretty closely for a pair of unequal mass particles in a quadratic potential (e.g. hydrogen atom) using the reduced mass, but it would probably would not apply directly to more than two particles nor to particles in a non-quadratic potential. However, the exact wavefunction is not the key point I intended to convey, but rather the derivation itself. Also, there are a few general results that stem from the geometry of the Lorentz transform rather than the details of the wavefunction:
1) A minimum uncertainty relation in one frame will not necessarily transform to a minimum uncertainty relation in another frame, but in general a much larger uncertainty.
2) The uncertainty relation is only frame-invariant when expressed in light-cone coordinates.
3) Most importantly, the momentum uncertainty is always finite, which directly implies that the velocity is always less than c.

Btw, I would also highly recommend that you examine DrGreg's post in detail. dp/dv goes to infinity as v goes to c so for a finite delta p you immediately see that delta v must go to zero.

 

[starthaus]

As RexxXII doesn't seem to understand how differentiation is involved, for small Δv, Δp approximates to

$$\Delta p \approx \frac{dp}{dv} \Delta v$$​

Yes, this is a nice way of putting it.

 

[starthaus]

Btw, I would also highly recommend that you examine DrGreg's post in detail. dp/dv goes to infinity as v goes to c so for a finite delta p you immediately see that delta v must go to zero.

Very nice way of putting it, I guess you have made this point several times already.

 

[RexxXII]

This is a festival of mistakes. Please answer the following:

1. If the above is true, why did you differentiate $$v$$ ? According to your claim, $$v=constant$$ so $$\Delta v=0$$.

I honestly don't understand what you think I'm doing that involves differentiation. If you're referring to the expectation value of v, even though the magnitude of v is constant, v itself is being treated as an operator in this case, so it is not a constant.

2. It is already known in maistream physics that $$v \gamma(v)$$ get differentiated as a function, i.e. together.
I have already shown that to you in a previous post. What entitles you to differentiate (incorrectly) only $$v$$? Are you trying to rewrite the basic rules of calculus and physics?
Can you try differentiating $$\frac{v}{\sqrt{1-(v/c)^2}}$$ correctly?

Yes, it is certainly is mainstream physics to do so. But due to the fact that I'm not differentiating anything ($$\Delta$$v was not necessarily small in this case, which was the entire point of the question) this shouldn't matter.

3. Why makes you think that $$v^2$$ is a number?

Because it isn't $$v^2$$ that appears in the Lorentz factor, it's $$|v|^2$$. One is an operator, and one is a magnitude squared of a different operator, which happens to be a scalar.

4. Dalespam has already given you a reference as to how this is done correctly, why do you insist in doing it incorrectly?

If you took the time to read my response, which DaleSpam had the common courtesy to do before responding, you'd know the answer to that.

5. Your incorrect derivation shows $$v>c$$ despite the fact that several of us have shown you that this is impossible. Why persist in your errors?

If you're so obviously correct and I'm so obviously erroneous, why are you wasting both of our time by responding to this thread at all? It's clear that no matter what I say, as long as it isn't what you think, I'm automatically wrong and you're intention is not to try and help me understand why but to point that out as rudely as possible.

Correct differentiation has everything to do with obtaining correct results. Errors like yours lead to incorrect results.

You're answer to "what does calculating expectation value have to do with differentiation?" is "everything"? Don't you see how supremely unhelpful of an answer that is? Thank you to DrGreg for actually giving me a decent answer so that I can see where my understanding is going astray. For your information, starthaus, I have sincere questions about relativity that nobody has been able to answer for me, and I'm attempting to look for those answers if they exist, and create new ones if they don't. If you believe that relativity is correct and that you understand it, why don't you try explaining to me why I'm wrong instead of yelling at me that I am wrong. What I'm saying should be wrong; that's the entire point of me asking why it is.

Because your derivation is not correct. It has elementary errors of both calculus and basic physics. See above.

Thank you for proving my point more clearly and ironically than I could have possibly done.

 

[diggy]

I can also record the time that has elapsed since I began to accelerate using my clock (the proper time of my reference frame).

Check -- everything you say is good up to here (I think).

Though I have no actual velocity in my constantly accelerating frame, using my calculator I should be able to calculate an "extrapolated instantaneous velocity", v, that some distant observer would see,

Check -- note that you have to use *their* clock and *their* distance for this. (Obviously they would be instantaneous times and distances, but still theirs, not yours -- you have your own reference frame). If you want to use your clock you have to convert it to *their* time.

After accelerating at 10 m/s2 for about 350 days I should finally reach an extrapolated velocity of 300,000,000 m/s (speed of light) and can even exceed this to reach any arbitrary velocity, provided I have enough fuel.

No --- your 10m/s2 is not going to be 10m/s2 in their frame anytime after t=0. The equivalence principle is about how to handle gravity, it doesn't mean that acceleration is uniform in every reference frame.

 

[diggy]

These are a more general form of the same equations posted before, which still define acceleration in a way differently than I have. In this set of equations acceleration is being measured by a distant observer which can then not be used in conjunction with the proper time of the traveller, as they do not refer to the same reference frame.

Right, and likewise you can't mix your acceleration with their clock. Different reference frames.

 

[diggy]

why can one not accelerate constantly to achieve any arbitrary velocity as calculated in their reference frame?

I think this is basically the heart of the misunderstanding. *Their* reference frame is no longer *theirs* once they leave it -- it becomes the observers reference frame. In *their* reference frame acceleration always equals 10 and velocity always equals 0.

Sorry for not posting this all together. Hopefully it is all clear.

 

[Janus]

To your second comment, in this scenario the momentum of the exhaust results from the mass of the fuel being converted to directly into energy. Because the momentum of the ship is unbounded, and the fuel is being propelled along with the ship, the momentum of the fuel is also unbounded. This means that it's mass increases to infinity as the ship approaches the speed of light. Because this mass is then converted into energy and then into exhaust momentum, we have a rocket exhaust with an infinite momentum that can be used to accelerate to c.

Let's try to put this differently. First consider a classical rocket without relativistic effects. As the rocket burns fuel, kinetic energy is released which expels the exhaust backwards, propelling the rocket forward.
As the rocket gains speed, it, along with any as of yet unused fuel, gains kinetic energy( as measured by someone the rocket is moving with respect to). But this gain of kinetic energy of the fuel cannot, in turn, be used to expel exhaust even faster and accelerate the ship even more efficiently.

Now to the relativistic rocket. The increase in mass of the fuel as measured in the rest frame is just a expression of the fuel's kinetic energy in that frame. (In fact, in many scientific circles, the very concept of "relativistic mass" has lost favor and it is just considered energy). You can no more tap that "mass" and convert it into energy to drive the ship than you can tap the kinetic energy of the fuel to propel the classical rocket in the first example.

 

[starthaus]

I honestly don't understand what you think I'm doing that involves differentiation. If you're referring to the expectation value of v, even though the magnitude of v is constant, v itself is being treated as an operator in this case, so it is not a constant.

...but $$\gamma(v)$$ is NOT constant. This is the starting point of your chain of mistakes.

Yes, it is certainly is mainstream physics to do so. But due to the fact that I'm not differentiating anything ($$\Delta$$v was not necessarily small in this case, which was the entire point of the question) this shouldn't matter.

How do you expect to calculate $$\Delta p$$? Certainly , not the way you have been trying. You seem to get the fact that $$p$$ varies directly as a function of $$v$$. Yet, you seem to persistently ignore that $$p$$ also varies as a function of $$v$$ through $$\gamma(v)$$.