Reasons why lightspeed travel is impossible

[aexyl93]

So, as something with rest mass gets closer to the speed of light, it gains more mass, making it require greater force to accelerate it making it gain even more mass and so on.
So the reason we use to say nothing can travel at or faster than light is because it gains more mass, and requires more and more energy to accelerate at the same rate the closer its speed is to light. (That's the only reason I've seen for why faster than light travel is impossible)

Suppose a rocket is propelling itself by burning fuel. The rate of energy released from burning the fuel is constant from the rocket's point of view, and therefore has a constant force applied to it from its point of view. But from an observer's view, the rate of energy released from burning the fuel, and the force resulting from it, is less than that viewed by the rocket because time for the rocket is slower than time for the observer.

So...
1. Can the rocket detect its own change in mass from near lightspeed travel?

2. If it can't, then isn't its acceleration from its view still constant because the force applied by the fuel is also constant from its view?

3. Even though the rocket perceives itself as having a constant acceleration, from the observer's view it's acceleration is decreasing right?

4. As the rocket gets faster, time for it slows, and its acceleration due to the fuel burning will decrease from the observer's view right?

So my question is really that can't you say that lightspeed travel is impossible not (or also?) because it gains mass and requires an infinite amount of energy, but because time slows down by an exponentially increasing factor the closer it gets to the speed of light, which would make it require an infinite amount of time?

I've looked but I haven't seen anywhere that says requiring an infinite amount of time is a reason that faster than light travel is impossible

(Please correct me if I'm wrong about anything which I probably am, I may have misused the word force...)

 

[Passionflower]

So, as something with rest mass gets closer to the speed of light, it gains more mass, making it require greater force to accelerate it making it gain even more mass and so on.

So the reason we use to say nothing can travel at or faster than light is because it gains more mass, and requires more and more energy to accelerate at the same rate the closer its speed is to light. (That's the only reason I've seen for why faster than light travel is impossible)

I see this argument a lot and I have reservations about this argument.

How could one argue that the increase of relativistic mass, from a different frame nevertheless, has an impact on the rocket's ability to accelerate?

If a rocket accelerates with 1g for a long time it will actually lose rest mass instead of gain it as it needs to burn up a lot of fuel to keep accelerating. So in fact the rocket will accelerate easier in a later stage because it has converted mass to energy and that energy is dispatched in the opposite direction (in some amount, as seen from the observer's frame, related to the Doppler factor if I am not mistaken).

Perhaps when a rocket is pushed the argument is valid.

Who can shine some light on this?

 

[Dale]

1. Can the rocket detect its own change in mass from near lightspeed travel?

No.

2. If it can't, then isn't its acceleration from its view still constant because the force applied by the fuel is also constant from its view?

Yes.

3. Even though the rocket perceives itself as having a constant acceleration, from the observer's view it's acceleration is decreasing right?

Yes.

4. As the rocket gets faster, time for it slows, and its acceleration due to the fuel burning will decrease from the observer's view right?

Yes.

 

[starthaus]

So my question is really that can't you say that lightspeed travel is impossible not (or also?) because it gains mass and requires an infinite amount of energy, but because time slows down by an exponentially increasing factor the closer it gets to the speed of light, which would make it require an infinite amount of time?

If $$a$$ is the proper acceleration measured by the rocket, then, its speed wrt to an observer is known to be :

$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

The above explains why the rocket can never reach $$v=c$$

 

[Janus]

So, as something with rest mass gets closer to the speed of light, it gains more mass, making it require greater force to accelerate it making it gain even more mass and so on.
So the reason we use to say nothing can travel at or faster than light is because it gains more mass, and requires more and more energy to accelerate at the same rate the closer its speed is to light. (That's the only reason I've seen for why faster than light travel is impossible)

Suppose a rocket is propelling itself by burning fuel. The rate of energy released from burning the fuel is constant from the rocket's point of view, and therefore has a constant force applied to it from its point of view. But from an observer's view, the rate of energy released from burning the fuel, and the force resulting from it, is less than that viewed by the rocket because time for the rocket is slower than time for the observer.

So...
1. Can the rocket detect its own change in mass from near lightspeed travel?

2. If it can't, then isn't its acceleration from its view still constant because the force applied by the fuel is also constant from its view?

3. Even though the rocket perceives itself as having a constant acceleration, from the observer's view it's acceleration is decreasing right?

4. As the rocket gets faster, time for it slows, and its acceleration due to the fuel burning will decrease from the observer's view right?

So my question is really that can't you say that lightspeed travel is impossible not (or also?) because it gains mass and requires an infinite amount of energy, but because time slows down by an exponentially increasing factor the closer it gets to the speed of light, which would make it require an infinite amount of time?

I've looked but I haven't seen anywhere that says requiring an infinite amount of time is a reason that faster than light travel is impossible

(Please correct me if I'm wrong about anything which I probably am, I may have misused the word force...)

Another explanation use the addition of velocities theorem.

Consider point 2.

From the Rocket's point of view its acceleration is constant (or even increasing due to the changing fuel to payload ratio.)

Now assume that the rocket is dropping off space buoys as it travels, with each buoy retaining the velocity of the rocket at the moment it is released. Furthermore, let's say that these buoys are dropped off as the ship attains a fixed velocity difference from the last buoy as measured from the ship. If the ship drops the first buoy when it achieves a velocity of 0.1 c relative to its starting point it drops a buoy, and then when it achieves a velocity of 0.1c relative to that buoy, it drops another, etc.

The formula for velocity addition is

$$w= \frac{u+v}{1+\frac{uv}{c^2}}$$

Which means if u is the velocity of a buoy with respect to the starting point, and v is the velocity of the ship with respect to the buoy (as measured by the ship), then w is the velocity of the ship with respect to the starting point.

If the first buoy is dropped off at 0.1c and then we calculate the ship's velocity when it it moving at 0.1c relative to the buoy (and ready to drop the second buoy), we get an answer of:
0.198c
Which is also the velocity of the second buoy with respect to the starting point.

If the ship drops off a third buoy when it it moving at 0.1c with respect to the second buoy it and the third buoy will now have a velocity with respect to the staring point of:
0.292c

And here is a list of the velocities of the next 12 buoys.
0.381c
0.463c
0.538c
0.606c
0.666c
0.718c
0.763c
0.802c
0.835c
0.863c
0.886c
0.906c

Note that even though the ship, by just adding up its changes of velocity from buoy to buoy would calculate that it has accelerated by a total of 1.4c, it is actually only moving at 0.906c relative to where it started.

Furthermore, if you look at the velocity addition formula again:

$$w= \frac{u+v}{1+\frac{uv}{c^2}}$$

You will note that no matter how many buoys the ship drops off and accelerates from, it will never achieve a velocity of c relative to its starting point. As long as u and v are smaller than c, w will come out to be smaller than c, and since each subsequent calculation uses the answer from the previous one for u, you always get an answer smaller than c.

 

[bcrowell]

So my question is really that can't you say that lightspeed travel is impossible not (or also?) because it gains mass and requires an infinite amount of energy, but because time slows down by an exponentially increasing factor the closer it gets to the speed of light, which would make it require an infinite amount of time?

This seems fine to me, except for the misuse of the term "exponentially" -- time only occurs in the base of the relevant expressions, not the exponent.

To the observer in the Earth's frame, the rocket would need infinite energy in order to reach c, and if the rocket had a source of energy that allowed it to keep blasting its rockets with constant thrust (constant as measured in the rocket's frame), its velocity would approach c, but would never reach it. (I.e., it would take infinite time to reach c.)

To the observer in the rocket's accelerating frame, the rocket's kinetic energy is always zero. To this observer, the universe is permeated by a gravitational field (the equivalence principle at work), and this gravitational field is accelerating the earth. To hover in this gravitational field, the rocket engine is continually engaged in the business of spewing out high-energy exhaust gases. The Earth's velocity approaches c, but never reaches it. (I.e., it would take infinite time to reach c.)

 

[Passionflower]

To this observer, the universe is permeated by a gravitational field (the equivalence principle at work), and this gravitational field is accelerating the earth. To hover in this gravitational field, the rocket engine is continually engaged in the business of spewing out high-energy exhaust gases. The earth's velocity approaches c, but never reaches it. (I.e., it would take infinite time to reach c.)

What do you mean, I can't follow what you say.

 

[aexyl93]

Okay, so then requiring an infinite amount of time is another reason for why an object with rest mass can't travel at or faster than the speed of light right? I looked around some and I haven't seen that supplied as a reason. Does anyone know why it isn't normally suppplied as a reason?

Another explanation use the addition of velocities theorem.

Consider point 2.

From the Rocket's point of view its acceleration is constant (or even increasing due to the changing fuel to payload ratio.)

Now assume that the rocket is dropping off space buoys as it travels, with each buoy retaining the velocity of the rocket at the moment it is released. Furthermore, let's say that these buoys are dropped off as the ship attains a fixed velocity difference from the last buoy as measured from the ship. If the ship drops the first buoy when it achieves a velocity of 0.1 c relative to its starting point it drops a buoy, and then when it achieves a velocity of 0.1c relative to that buoy, it drops another, etc.

The formula for velocity addition is
$$w= \frac{u+v}{1+\frac{uv}{c^2}}$$

Thanks for giving another explanation. I've never heard of the addition of velocities theorem before, but I like it as an explanation. It easily shows mathematically why it's impossible.

This seems fine to me, except for the misuse of the term "exponentially" -- time only occurs in the base of the relevant expressions, not the exponent.

And sorry for misusing 'exponentially' I thought it would be suitable since the rate at which the factor changes increases the closer the speed is to c. So I guess would asymptotically be more appropriate then?

Also, its amazing how this post got 6 replies within a few hours while another post I made yesterday in this same forum has no replies. I guess some questions are more interesting or have more people that have an answer for it?

 

[DaveC426913]

Okay, so then requiring an infinite amount of time is another reason for why an object with rest mass can't travel at or faster than the speed of light right? I looked around some and I haven't seen that supplied as a reason. Does anyone know why it isn't normally suppplied as a reason?

It is important for you to understand that these are all nothing more than looking at the same thing from different points of view.

Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time - is the same thing as saying it would acquire infinite mass.

 

[qraal]

From the viewpoint of a spaceship with a clock and an accelerometer it can accelerate to speeds that are faster than light as measured by integrating the readings from the two instruments, but space-time as manifested by the view of the stars and a timing clock pulse from Earth does weird things, from the ship's perspective. But it depends on what we measure. A ship that continuously accelerates from rest to then stop at a distant destination will find that a clock signal from it's origin has counted out just a couple of years of flight, even if it travels millions of light-years (as measured by other means.) If the distance is instead measured by a series of radar pulses from the ship, then it'll get a quite significantly different measurement. From it's point of view the space-time displacement between it's origin and it's destination shrank compared to measurements taken by a stationary observer.

 

[bcrowell]

From the viewpoint of a spaceship with a clock and an accelerometer it can accelerate to speeds that are faster than light as measured by integrating the readings from the two instruments

You can integrate $\int a d\tau$, where a is the proper acceleration and $\tau$ is the proper time. However, that doesn't give your actual velocity relative to the stars as you pass by them. That's because an integral is basically a Riemann sum, but velocity increments don't add linearly.

 

[Janus]

I see this argument a lot and I have reservations about this argument.

How could one argue that the increase of relativistic mass, from a different frame nevertheless, has an impact on the rocket's ability to accelerate?

If a rocket accelerates with 1g for a long time it will actually lose rest mass instead of gain it as it needs to burn up a lot of fuel to keep accelerating. So in fact the rocket will accelerate easier in a later stage because it has converted mass to energy and that energy is dispatched in the opposite direction (in some amount, as seen from the observer's frame, related to the Doppler factor if I am not mistaken).

Perhaps when a rocket is pushed the argument is valid.

Who can shine some light on this?

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.

 

[JesseM]

Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time - is the same thing as saying it would acquire infinite mass.

I don't think it's quite the same. If you ignore dynamical issues like energy and just think in terms of kinematics, you can write down an equation describing the worldline of an object that reaches c in finite time (for example, x(t) = (a/2)*t^2, where x and t are coordinates of an inertial frame and a is some constant coordinate acceleration), and if you calculated proper acceleration as and proper time using the usual kinematical equations of SR, I believe you would find that at the moment the object reaches c the proper acceleration has gone to infinity but the proper time is still at some finite value.

 

[starthaus]

I don't think it's quite the same. If you ignore dynamical issues like energy and just think in terms of kinematics, you can write down an equation describing the worldline of an object that reaches c in finite time (for example, x(t) = (a/2)*t^2, where x and t are coordinates of an inertial frame and a is some constant coordinate acceleration), and if you calculated proper acceleration as and proper time using the usual kinematical equations of SR, I believe you would find that at the moment the object reaches c the proper acceleration has gone to infinity but the proper time is still at some finite value.

I don't think so, see post #4. DaveC426913 is correct.

 

[JesseM]

I don't think so, see post #4. DaveC426913 is correct.

What specific statement in my post are you disagreeing with? Again remember that I am separating dynamics from kinematics--of course it is impossible dynamically to accelerate to c in finite time since this would require infinite energy and force, but we can consider the possibility from a purely kinematical point of view, and applying the usual kinematical equations of SR the proper time would not go to infinity for the worldline I mentioned (which reaches c in finite coordinate time, the proper acceleration going to infinity at that time).

 

[starthaus]

What specific statement in my post are you disagreeing with?

The fact that you disagree with Dave's initial claim that finite proper acceleration for an infinite time produces the same effect as infinite proper acceleration for a finite time. His claim is correct, you can prove it trivially from math in post 4.

Again remember that I am separating dynamics from kinematics

You don't need to, the formula $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$ encapsulates both the kinematics and the dynamics.

 

[bcrowell]

I don't think it's quite the same. If you ignore dynamical issues like energy and just think in terms of kinematics, you can write down an equation describing the worldline of an object that reaches c in finite time (for example, x(t) = (a/2)*t^2, where x and t are coordinates of an inertial frame and a is some constant coordinate acceleration), and if you calculated proper acceleration as and proper time using the usual kinematical equations of SR, I believe you would find that at the moment the object reaches c the proper acceleration has gone to infinity but the proper time is still at some finite value.

Well, I can think of a few purely kinematical objections:

(1) If the object reached v=c at t=t_c, instantaneously comoving Minkowski frames would exist for all t<t_c, but not for $t \ge t_c$.

(2) There is causality violation at t>t_c.

(3) If the object has finite size, and its size as measured by comoving observers is constant (i.e., it's Born-rigid), then its longitudinal size as measured by inertial observers becomes zero at t=t_c. But if its front and back coincide at t=t_c, and its front and back both have v=c at t=t_c, then it's hard to imagine how any description of the time-evolution of the system could make its front and back *not* coincide at t>t_c; if the initial conditions are the same, how can the final conditions be different?

 

[JesseM]

Well, I can think of a few purely kinematical objections:

(1) If the object reached v=c at t=t_c, instantaneously comoving Minkowski frames would exist for all t<t_c, but not for $t \ge t_c$.

How is that a kinematical objection? You wouldn't object to the fact that light moves at c on kinematical grounds, in spite of the fact that light has no comoving Minkowski frame.

(2) There is causality violation at t>t_c.

Again, I don't understand how you are defining "kinematical objection". Tachyons are consistent with the axioms of SR, we can imagine a logically possible set of laws of physics where tachyons exist and causality is violated but the basic kinematical laws work the same as in SR. If someone said they could rule something out on purely kinematical grounds, I would normally interpret that to mean that they could show it was impossible using only the kinematical rules with no additional assumptions about the detailed laws of physics (for example, if the kinematical rules are those of SR, on purely kinematical grounds we can rule out the possibility that there would be some non-inertial path a clock could take between two events A and B such that it would have elapsed more time than a clock which moves inertially from event A to event B)

(3) If the object has finite size, and its size as measured by comoving observers is constant (i.e., it's Born-rigid), then its longitudinal size as measured by inertial observers becomes zero at t=t_c. But if its front and back coincide at t=t_c, and its front and back both have v=c at t=t_c, then it's hard to imagine how any description of the time-evolution of the system could make its front and back *not* coincide at t>t_c; if the initial conditions are the same, how can the final conditions be different?

Is it even possible for an object to remain Born rigid if each point on the object doesn't have uniform proper acceleration? (not a rhetorical question, I'm really not sure) Anyway I see no reason to impose Born rigidity as a condition if we're only talking about kinematics, not what is realistic according to dynamical laws dealing with solid objects.

 

[JesseM]

The fact that you disagree with Dave's initial claim that finite proper acceleration for an infinite time produces the same effect as infinite proper acceleration for a finite time.

Uh, I never disagreed with his "initial claim", I only objected to the claim of his I quoted, where he said "Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time - is the same thing as saying it would acquire infinite mass."

You don't need to, the formula $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$ encapsulates both the kinematics and the dynamics.

I think you need to look into the distinction between kinematics and dynamics more carefully, that equation is purely kinematical since it can be derived from the definitions of velocity and proper acceleration and the coordinate transformation of SR, without any consideration of the dynamical laws which predict what trajectory a given object will actually take in a specific physical situation (and the equation only deals with velocity as a function of time for special the case of constant proper acceleration, whereas I was talking about a kinematical analysis of a hypothetical object moving with ever-increasing proper acceleration)

 

[bcrowell]

How is that a kinematical objection? You wouldn't object to the fact that light moves at c on kinematical grounds, in spite of the fact that light has no comoving Minkowski frame.

Point taken, but I think it's important to keep a clear distinction between observers and objects that can't be observers. When we talk about a rocket, we implicitly assume that it's possible for an observer to be aboard. When we talk about a ray of light, we implicitly assume that the ray of light isn't an observer. In your example, the object suddenly makes the transition from being in the same class as the rocket (observer-compatible) to being the same class as the ray of light (observer-incompatible). This seems to me like a purely kinematic issue: can you tie the origin of a coordinate system to a particular world-line?

Again, I don't understand how you are defining "kinematical objection". Tachyons are consistent with the axioms of SR, we can imagine a logically possible set of laws of physics where tachyons exist and causality is violated but the basic kinematical laws work the same as in SR.

Yes, but again we have two classes of objects: those that move at less than c and those that move at more than c. A tachyon can't slow below c any more than a normal object can accelerate above c. I think you run into real kinematic difficulties when you try to posit objects that switch between the two classes.

Is it even possible for an object to remain Born rigid if each point on the object doesn't have uniform proper acceleration? (not a rhetorical question, I'm really not sure)

Actually I think if it's going to remain Born-rigid then the proper accelerations have to be *un*-equal. This is basically the Bell spaceship paradox.

Anyway I see no reason to impose Born rigidity as a condition if we're only talking about kinematics, not what is realistic according to dynamical laws dealing with solid objects.

Well, Born-rigidity is a purely kinematical condition: all it says is that the radar distance between nearby points remains constant. And when, e.g., we find that a Born-rigid disk can't have an angular acceleration, that is a purely kinematical prohibition, not a dynamical one; the problem is that if you wanted to have an angular acceleration, accelerations would have to occur simultaneously at all points around a circumference, but that simultaneity is kinematically impossible.

If you don't require Born-rigidity, then I think you have a problem, because without some sort of requirement of this type, you could have the object get infinitely distorted according to a comoving observer. I think it's hard to maintain the idea of an "object" when the object is distorted by a factor of *infinity* in its own comoving frame.

 

[JesseM]

Point taken, but I think it's important to keep a clear distinction between observers and objects that can't be observers. When we talk about a rocket, we implicitly assume that it's possible for an observer to be aboard. When we talk about a ray of light, we implicitly assume that the ray of light isn't an observer. In your example, the object suddenly makes the transition from being in the same class as the rocket (observer-compatible) to being the same class as the ray of light (observer-incompatible). This seems to me like a purely kinematic issue: can you tie the origin of a coordinate system to a particular world-line?

You can attach a non-inertial coordinate system to a hypothetical object that goes from subluminal to superluminal...you can't say it remains at the origin of an inertial frame, but they you can't say that for any non-inertial rocket, even one that remains slower-than-light forever. Still, I agree that my hypothetical rocket does have the special characteristic that it has an instantaneous comoving inertial frame for some points on its worldline but not others, whereas a photon or a tachyon never would and a sublight object always would.

Yes, but again we have two classes of objects: those that move at less than c and those that move at more than c. A tachyon can't slow below c any more than a normal object can accelerate above c. I think you run into real kinematic difficulties when you try to posit objects that switch between the two classes.

What do you mean by "kinematic difficulties" though? There's nothing kinematically inconsistent about this. Since kinematics deals purely with coordinate motions and coordinate transformations and not with the actual laws of physics, I would just point out that in a Newtonian universe with Galilei-symmetric laws, you could still do a kinematical analysis of how the motions of objects would look if you used the set of coordinate systems given by the Lorentz transformation, and in such a universe there'd be no reason an object couldn't accelerate from slower-than-c to faster-than-c relative to some such coordinate system.

Actually I think if it's going to remain Born-rigid then the proper accelerations have to be *un*-equal.

I think you misunderstood me, I didn't say anything about the proper accelerations of each point being equal to the proper acceleration of other points, I asked whether Born rigid acceleration implies the requirement that the the proper acceleration of each point is uniform, by which I meant each point's proper acceleration remains constant over time. I had thought this was a requirement for Born rigidity, which would mean that if part of the rocket had x(t) = (a/2)*t^2 as I suggested, then the rocket as a whole couldn't be Born rigid since this equation implies a non-uniform proper acceleration (increasing to infinity at the moment dx/dt = c)

Anyway I see no reason to impose Born rigidity as a condition if we're only talking about kinematics, not what is realistic according to dynamical laws dealing with solid objects.

Well, Born-rigidity is a purely kinematical condition: all it says is that the radar distance between nearby points remains constant. And when, e.g., we find that a Born-rigid disk can't have an angular acceleration, that is a purely kinematical prohibition, not a dynamical one; the problem is that if you wanted to have an angular acceleration, accelerations would have to occur simultaneously at all points around a circumference, but that simultaneity is kinematically impossible.

Yes, you can derive various kinematical conclusions about whether a given fact is consistent or inconsistent with Born rigidity, but there's no reason to say that a lack of Born rigidity shows that a given equation of motion is impossible on kinematical grounds (even if we take dynamics into account there are obviously plenty of physical scenarios where an object would accelerate in a non-Born-rigid way!)

 

[Dr Lots-o'watts]

So the reason we use to say nothing can travel at or faster than light is because it gains more mass, and requires more and more energy to accelerate at the same rate the closer its speed is to light. (That's the only reason I've seen for why faster than light travel is impossible)

Light doesn't care how fast anything goes, it goes its speed no matter what. If a car goes 0.9c, and it it turns on a headlight, that light goes c relative to the driver, and it goes c relative to a fire hydrant.

If a lamppost emits light, that light also goes c relative to the fire hydrant, and it also goes c relative to the driver in the 0.9c car.

See? No talk of that absurd infinite mass or energy. All I said was light speed was the same for everyone. 4 words : c is strictly constant. Straight to the point, simple, elegant. It's all you need.

 

[starthaus]

Uh, I never disagreed with his "initial claim", I only objected to the claim of his I quoted, where he said "Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time - is the same thing as saying it would acquire infinite mass."

His claim is correct. I have just showed you that.

I think you need to look into the distinction between kinematics and dynamics more carefully, that equation is purely kinematical since it can be derived from the definitions of velocity and proper acceleration and the coordinate transformation of SR, without any consideration of the dynamical laws which predict what trajectory a given object will actually take in a specific physical situation

This is also incorrect since $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$ is obtained from integrating the equation:

$$\frac{d}{dt}\frac{mv}{\sqrt{1-(v/c)^2}}=F$$

whereas I was talking about a kinematical analysis of a hypothetical object moving with ever-increasing proper acceleration)

So was I.

 

[JesseM]

His claim is correct. I have just showed you that.

The claim I objected to does not seem correct. He said "Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time", while I pointed out that if we ignore dynamical concerns about the need for infinite energy to achieve infinite acceleration, then there is a kinematically possible worldline that reaches c in finite coordinate time, and its proper acceleration goes to c but its proper time does not. Do you disagree with my claim, or do you agree with it but somehow disagree it contradicts the statement of Dave I quoted? Please give a specific answer if you want to continue to act like you are "disagreeing" with something I actually said (as opposed to strawmen like the idea that I was disagreeing with Dave's 'initial claim').

This is also incorrect since $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$ is obtained from integrating the equation:

$$\frac{d}{dt}\frac{mv}{\sqrt{1-(v/c)^2}}=F$$

OK, I suppose the equation can be either kinematical or dynamical depending on the context--if you are considering what the path of a rocket will be under the conditions of constant force, dynamically you can show that this means constant proper acceleration, which implies the function for v as a function of t that you wrote. However, you can show that constant proper acceleration implies that equation without ever considering what physical conditions would be required for constant proper acceleration, and thus there is a purely kinematical derivation of the idea that constant proper acceleration a implies $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$, a derivation which does not involve any consideration of "force" or other dynamical concerns.

whereas I was talking about a kinematical analysis of a hypothetical object moving with ever-increasing proper acceleration)

So was I.

Are you sure you read me correctly? Are you really claiming that you were also talking about the case of an object moving with ever-increasing proper acceleration? If so why did you invoke the equation $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$ which only applies under conditions of constant proper acceleration?

 

[starthaus]

The claim I objected to does not seem correct. He said "Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time", while I pointed out that if we ignore dynamical concerns about the need for infinite energy to achieve infinite acceleration, then there is a kinematically possible worldline that reaches c in finite coordinate time, and its proper acceleration goes to c but its proper time does not. Do you disagree with my claim, or do you agree with it but somehow disagree it contradicts the statement of Dave I quoted? Please give a specific answer if you want to continue to act like you are "disagreeing" with something I actually said (as opposed to strawmen like the idea that I was disagreeing with Dave's 'initial claim').

It is very simple, really. Like I said, start with the equation of motion:$$\frac{d}{dt}\frac{mv}{\sqrt{1-(v/c)^2}}=F$$

You can integrate it for arbitrary force by using the notation:

$$A=\int {\frac{F}{m}} dt$$

Then,

$$v=\frac{A}{\sqrt{1+(A/c)^2}}$$

So, for $$A->\infty$$ $$v->c$$ with $$v<c$$

The above encapsulates BOTH the case $$a->\infty$$ AND $$t->\infty$$, so, DaveC's claim is correct as stated and your objection to his claim is not.

OK, I suppose the equation can be either kinematical or dynamical depending on the context

It is, see above.

Are you sure you read me correctly? Are you really claiming that you were also talking about the case of an object moving with ever-increasing proper acceleration? If so why did you invoke the equation $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$ which only applies under conditions of constant proper acceleration?

It doesn't, I just tried to keep it simple. See the complete derivation above, it covers all cases , INCLUDING the variable force (and, implicitly, the variable acceleration).

 

[JesseM]

It is very simple, really. Like I said, start with the equation of motion:$$\frac{d}{dt}\frac{mv}{\sqrt{1-(v/c)^2}}=F$$

You can integrate it for arbitrary force by using the notation:

$$A=\int {\frac{F}{m}} dt$$

Why would the acceleration be obtained by integrating force/mass? That doesn't seem to make sense in terms of units, force/mass has units of acceleration so if you integrate with respect to time you should get something with units of velocity. Are you basing this on some textbook or other source, or is this your own original argument?

Then,

$$v=\frac{A}{\sqrt{1+(A/c)^2}}$$

That last equation is missing the factor of t that should be multiplied by each A, without it the units don't work out. And is A here supposed to represent the proper acceleration (the magnitude of the four-acceleration vector), or the coordinate acceleration? It seems that when you wrote down the equation for F, you were talking about the coordinate force obtained by taking the derivative of mass*coordinate velocity*gamma, rather than the four-force obtained by taking the derivative of mass times the energy-momentum four-vector...

So, for $$A->\infty$$ $$v->c$$ with $$v<c$$

The above encapsulates BOTH the case $$a->\infty$$ AND $$t->\infty$$, so, DaveC's claim is correct as stated and your objection to his claim is not.

Again, can you be specific on what aspect of my objection you are claiming is not correct? Are you disputing that kinematically, if we consider a hypothetical object with a worldline given by x(t) = (A/2)*t² where A is a constant with units of acceleration, then v(t) = A*t so the object will reach c at time coordinate t=c/A, and the proper acceleration reaches infinity at that time but the proper time is still finite?

If you just disagree that the proper time would be finite from t=0 to t=c/A in this case, remember that the proper time is from t=0 to t=c/A is given by $$\int_{t=0}^{t=c/A} \sqrt{1 - v(t)^2/c^2} \, dt$$, you can use the Mathematica integrator page to verify that with v(t) = A*t the integral works out to $$\frac{At \sqrt{1 - (At/c)^2 } + c * arcsin(At/c)}{2A}$$, so plugging in t=c/A and then subtracting t=0 we get $$\frac{c * arcsin(1)}{2A} - \frac{c * arcsin(0)}{2A} = \frac{c * (\pi/2)}{2A}$$

It doesn't, I just tried to keep it simple. See the complete derivation above, it covers all cases , INCLUDING the variable force (and, implicitly, the variable acceleration).

No, it's easy to show that for variable proper acceleration, it is not true that $$v=\frac{At}{\sqrt{1+(At/c)^2}}$$. Consider the simple case I mentioned where x(t) = (A/2)*t^2 with constant A, which implies v(t) = A*t and a(t) = A. Clearly v(t) = a(t) * t in this case, so if A is supposed to represent the coordinate acceleration in that equation it doesn't work here. On the other hand, if A is supposed to represent the proper acceleration it still doesn't work in this case, since as mentioned in this textbook the general relation between coordinate acceleration dv/dt and proper acceleration a is $$a = \frac{1}{(1 - v^2/c^2)^{3/2}} dv/dt$$, so if a(t) = dv/dt = A (a constant) for the case I am discussing here, then the proper acceleration a is given by $$\frac{A}{(1 - v^2/c^2)^{3/2}}$$, which implies A = a * (1 - v²/c²)^3/2. So, v(t) = A*t implies v = t * a * (1 - v²/c²)^3/2 for the worldline I am describing, which is inconsistent with the equation v = at / sqrt(1 + (at/c)²)...you can see that they're inconsistent if you substitute that last equation for v into the previous equation v = t * a * (1 - v²/c²)^3/2, if you do the right side of the equation becomes:

at * [1 - (at/c)²/(1 + (at/c)²)]^3/2 = at * [1/(1 + (at/c)²)]^3/2 = at / (1 + (at/c)²)^3/2

But if you substitute v = at / sqrt(1 + (at/c)²) into the left side of the same equation v = t * a * (1 - v²/c²)^3/2 you just get back at / sqrt(1 + (at/c)²) for the left side, which is clearly different than what we found when we substituted this expression for v into the right side. So, a worldline which satisfies v = t * a * (1 - v²/c²)^3/2 cannot also satisfy v = at / sqrt(1 + (at/c)²).

 

[starthaus]

Why would the acceleration be obtained by integrating force/mass?

It isn't. What gave you this idea? The fact that I labeled the integral with the letter "A"?

That last equation is missing the factor of t that should be multiplied by each A, without it the units don't work out.

It isn't missing anything, you fail to follow basic math."A" has dimensions of speed.

 

[JesseM]

It isn't. What gave you this idea? The fact that I labeled the integral with the letter "A"?

No, I assumed you were trying to derive the equation you had written in earlier posts, $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$. Have you now abandoned the argument that this previous equation is somehow relevant to demonstrating I am wrong?

Anyway, if you define F/m by the equation $$\frac{d}{dt}\frac{v}{\sqrt{1-(v/c)^2}}$$, then by definition integrating that with respect to t give $$\frac{v}{\sqrt{1-(v/c)^2}}$$. If you define A = v/sqrt(1 - (v/c)²) and then solve for v, then yeah, you do get $$v = A/\sqrt{1 + (A/c)^2 }$$. 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?

And still you haven't given me a straight answer to whether you actually disagree with any specific thing I have claimed, or if you are just fantasizing about something I never said as you like to do so often. Again, my claim was just that kinematically we can come up with a worldline, x(t) = (a/2)*t^2, which reaches c at finite time t=c/a, but while the proper acceleration reaches infinity at t=c/a, the proper time remains finite at t=c/a. Do you disagree with this or not? And if you don't disagree, are you for some reason disagreeing with my opinion that this conflicts with Dave's statement "Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time"?

 

[starthaus]

No, I assumed you were trying to derive the equation you had written in earlier posts, $$v=\frac{at}{\sqrt{1+(at/c)^2}}$$.

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

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

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

 

[starthaus]

I believe you would find that at the moment the object reaches c the proper acceleration has gone to infinity but the proper time is still at some finite value.

...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).
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.

 

[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?