Reasons why lightspeed travel is impossible

[JesseM]

For F=constant you get v=\frac{at}{\sqrt{1+(at/c)^2}} where a=F/m. This is trivial.

It's not actually trivial that constant proper acceleration is the same as constant F, since there are two different notions of "force" in relativity, as I mentioned earlier, and you seem to be talking about "force" as defined by the derivative w/respect to coordinate time of mass*coordinate velocity*gamma, as opposed to the four-force which is the derivative w/respect to proper time of the energy-momentum four-vector. Apparently it does work out that the notion of force you're using will be constant in the case of constant proper acceleration, despite the fact that this force involves coordinate velocity and time in a single inertial frame whereas proper acceleration deals with the coordinate acceleration in a series of instantaneously comoving frames (and the proper acceleration can also be understood as the magnitude of the acceleration four-vector), but this is a nontrivial fact which requires some proof.

Have you now abandoned the argument that this previous equation is somehow relevant to demonstrating I am wrong?

Not at all,

OK, then how do you think that equation was relevant? I was talking about a case involving increasing proper acceleration, whereas the previous equation v=\frac{at}{\sqrt{1+(at/c)^2}} applies only to the case of constant proper acceleration.

I didn't but in the process I have discovered that you made some other, more basic, errors.

My "error" was only in misunderstanding what you were trying to derive, because you didn't give any explanation and didn't mention that you were no longer talking about the equation v=\frac{at}{\sqrt{1+(at/c)^2}}. Once I realized your A was not supposed to be the same as the a in the previous equation you'd been talking about, and in fact was not supposed to refer to acceleration at all, I had no problem following what you were doing...but if it makes you feel good to crow about how you have caught me in an "error" go ahead. Can we get back to the actually relevant question though, namely how does this statement about the relation between v and A have any relevance whatsoever to showing an error in my comment to DaveC? Like I asked before: What is that supposed to prove? How is it even relevant that you defined A in terms of an integral of F/m, as opposed to just starting from the arbitrary definition A = v/sqrt(1 - (v/c)2)? How does this tell us anything about either the proper acceleration or the proper time needed to reach c?

 

[JesseM]

...which is precisely what DaveC posted. Read his post one more time and you'll hopefully understand. He's telling you that you get v->c for EITHER a->\infty (when t=bounded) OR t->\infty (when a=bounded).

That's not how I interpreted his post. He seemed to be saying that the conditions were all physically equivalent (that each one is 'the same thing'), meaning that in any scenario where one was satisfied the other two would be as well, not that they were a bunch of distinct conditions and that at least one (but not all) would have to be satisfied for a rocket to reach c.

I put his post in a mathematical form at post 25 by explaining that v->c for A->\infty where A=\int {\frac{F}{m}} dt. This puts DaveC's post in the most general form and includes his claim as particular cases.

Again I don't see how your derivation involving A is relevant. For example you say nothing about the relation between A and time to justify DaveC's "is the same thing as saying it would take an infinite amount of time" comment, nor do you say anything about the precise relation between your F and the proper acceleration (which is somewhat nontrivial as I mentioned above), nor do you show what conditions are needed to prove that F must go to infinity if A goes to infinity (I'm sure you'd agree that A can go to infinity in the limit as t goes to infinity without F needing to approach infinity, whereas if A goes to infinity in finite t then F must approach infinity at the same time, but your comments don't deal with these separate cases)

 

[starthaus]

That's not how I interpreted his post. He seemed to be saying that the conditions were all physically equivalent (that each one is 'the same thing'), meaning that in any scenario where one was satisfied the other two would be as well,

I think that you interpret it incorrectly. You should ask him rather than making guesses.

 

[DaveC426913]

I think that you interpret it incorrectly. You should ask him rather than making guesses.

Caught. Must confess. :blush:
It was my intention that these are all ways of expressing the same thing. The OP seemed to be under the impression that there were several distinct reasons why c could not be attained by anything massive. It was my contention that there is really only one causal element, and that these various infinities are simply different ways of rationalizing it.

 

[Drakkith]

I just read something that was saying that if you impart enough energy onto a particle, the gravitational force would become high enough to equal the other dominant forces. (Strong, Electromagnetic, and Weak)

Would this mean that at a certain velocity, your mass becomes so high you could collapse in on yourself, forming a black hole or something?

 

[Passionflower]

It's not actually trivial that constant proper acceleration is the same as constant F, since there are two different notions of "force" in relativity, as I mentioned earlier, and you seem to be talking about "force" as defined by the derivative w/respect to coordinate time of mass*coordinate velocity*gamma, as opposed to the four-force which is the derivative w/respect to proper time of the energy-momentum four-vector. Apparently it does work out that the notion of force you're using will be constant in the case of constant proper acceleration, despite the fact that this force involves coordinate velocity and time in a single inertial frame whereas proper acceleration deals with the coordinate acceleration in a series of instantaneously comoving frames (and the proper acceleration can also be understood as the magnitude of the acceleration four-vector), but this is a nontrivial fact which requires some proof.

These statements surprise me.
Isn't this basic special relativity?

 

[JesseM]

These statements surprise me.
Isn't this basic special relativity?

I'm not saying the proof would be particularly complicated, but I don't think it's so trivial that it requires no proof, or that someone who was familiar with basic SR but had never seen such a proof would find it immediately obvious how to prove it or even find it obvious that it's true in the first place.

 

[starthaus]

It's not actually trivial that constant proper acceleration is the same as constant F, since there are two different notions of "force" in relativity, as I mentioned earlier, and you seem to be talking about "force" as defined by the derivative w/respect to coordinate time of mass*coordinate velocity*gamma, as opposed to the four-force which is the derivative w/respect to proper time of the energy-momentum four-vector. Apparently it does work out that the notion of force you're using will be constant in the case of constant proper acceleration, despite the fact that this force involves coordinate velocity and time in a single inertial frame whereas proper acceleration deals with the coordinate acceleration in a series of instantaneously comoving frames (and the proper acceleration can also be understood as the magnitude of the acceleration four-vector), but this is a nontrivial fact which requires some proof.

It is very easy to show that \frac{dp}{dt}=F where F is the 3-force I'm using throughout this thread and t is coordinate time is the same as \frac{dp}{d\tau}=F_M where F_M is the Minkowski force and \tau is proper time. I did a short writeup that you can read https://www.physicsforums.com/blog.php?b=1911 .

 

[Austin0]

Classically, the rocket equation is:

V_f = V_e \ln \left ( \frac {M_i}{M_f} \right )

With Vf being the final velocity of the rocket,
Ve the exhaust velocity,
Mf the final mass of the rocket
and
Mi the intial mass of the rocket.

Factoring in Relativity gives us:

V_f = c \tanh \left (\frac{Ve}{c} \ln \left(\frac{M_i}{M_f}\right) \right )

Which gives us the result that any ratio of initial mass to final mass less than infinite results in a final velocity of less than c.

A rocket accelerates forward by throwing mass backwards, the efficiency of this process depends on the exhaust velocity.

So if I'm in the "stationary" frame watching the rocket, the efficiency of the rocket will be determined by the difference between the rocket's velocity and the exhaust velocity.

Assume that the rocket is already moving at 0.9c relative to me and the exhaust velocity relative to the ship is 0.1c. Relative to me, the exhaust velocity is

\frac{0.9c-0.1c}{1- \frac{0.9c(0.1c)}{c^2}} = 0.879c

and difference between rocket and exhaust velocity is 0.021c

When the rocket reaches 0.95c, the exhaust velocity with relative to me will be 0.939c for a difference of 0.011c , almost half that of the difference at 0.9c.

At a rocket velocity of 0.99c the difference drops off to 0.002c

From my frame, the rocket's efficiency drops off as it approaches c and the rocket's acceleration decreases the nearer it gets to c.

AM I correct in assuming that constant proper acceleration implies constant thrust and fuel consumption as measured within the accelerating system?
SO if propulsion was ion dirve of some kind and the thrust was measured by some finite number of ions per unit time, then it would seem to mean that not only would the momentum of the ions [per particle] decrease as you have shown here, but the number of ions/dt would also decrease by the gamma factor due to the internal dilation of clocks in the accelerated system.
SO the total combined observed decrease in thrust would be even greater. Is this at all correct??
I have just learned of the 2/\gamma³ coordinate acceleration falloff rate and am trying to understand why this would be the case.
Thanks