Question About Relativistic Acceleration

[Physicsguru]

This calculation, of course, isn't valid. It's one of the traps a good physics instructor, teaching a class in SR, would lay onto the students.

To be able to use that kinematical equation, one has made an explicit assumption that the force (or dp/dt) applied to the object is a constant. But we know this isn't true in the observer's reference frame (the one who is observiing and measuring this vf and vi). As the velocity of the object increases, the observer is also seeing an increase in the mass (relativistic mass) of the object. Thus, to maintain a constant acceleration, the applied force has to increase. Immediately, that simply, first-year kinematic equation is no longer valid. And if, instead, one maintains that constant force, then the acceleration is no longer a constant (again due to the increasing mass) and you again can't use that kinematical equation.

Zz.

It was stipulated that the acceleration is constant, therefore the equation is valid provided SR is irrelevant to the question. SR self contradicts, so I fail to see its relevancy here.

Guru

 

[quantumdude]

It was stipulated that the acceleration is constant, therefore the equation is valid provided SR is irrelevant to the question.

Your provision isn't granted, because you are talking about a massive object moving at speed 'c'. Relativity applies.

SR self contradicts, so I fail to see its relevancy here.

No, SR is not self contradictory. It is not possible to derive two conflicting predictions from SR.

 

[jcsd]

It was stipulated that the acceleration is constant, therefore the equation is valid provided SR is irrelevant to the question. SR self contradicts, so I fail to see its relevancy here.

Guru

No you've used a Newtonian equation and expected to get an answer that is relvant to relativty. Relavity specifcally says that (extrinsic) accelaration cannot be constant indefintely as otherwise you soon find the force needed is infinite. Where does SR self contradict pray tell?

 

[JesseM]

It was stipulated that the acceleration is constant, therefore the equation is valid provided SR is irrelevant to the question. SR self contradicts, so I fail to see its relevancy here.

In SR, "acceleration is constant" doesn't mean that in a given inertial frame, I will see the velocity increase by the same amount each second. Rather, it means that if you look at the frame in which the ship is at rest at any given moment, and if one second later you see the velocity has increased by v in that frame, then if you switch to the ship's new rest frame at that moment, then one second later the velocity will have increased by v in that frame. Note that if you want "constant acceleration" to be synonymous with "constant g-force experienced by the ship's crew", you must use this definition.

Another way of thinking of this: suppose I am on a ship which has a gun onboard that will shoot out a smaller ship moving at 9.8 m/s relative to my ship. This ship in turn has a gun that shoots out an even smaller ship at 9.8 m/s relative to itself. If I shoot my gun, and the smaller ship shoots its gun, then according to Newtonian mechanics the smallest ship will now be moving at 19.6 m/s relative to me. But in relativity velocities don't add in this simple way: if you see a ship moving at velocity v in your frame, and I see that you are moving at velocity u in my frame, then in my frame the ship's velocity will not be u+v but $$(u+v)/(1 + uv/c^2)$$ (see relativistic velocities). This formula can in turn be derived from the "Lorentz transformation", which specify how coordinates (x,y,z,t) in frame S are mapped to coordinates (x',y',z',t') in frame S'.

 

[ZapperZ]

It was stipulated that the acceleration is constant, therefore the equation is valid provided SR is irrelevant to the question. SR self contradicts, so I fail to see its relevancy here.

Guru

Then you fail even basic, Newtonian physics.

dp/dt = m dv/dt + v dm/dt

If you say a is a constant, then all you are specifying is that dv/dt is a constant. The ONLY way in which you can use vf = vi + at is if you completely ignore the v dm/dt term. While this is perfectly valid in many cases in Newtonian physics where m doesn't change, it isn't valid HERE! The object's apparent mass increases as v increases!

You are welcome to come visit any particle accelerators and prove to yourself that your "derivation" here is faulty.

Zz.

 

[russ_watters]

How is classical mechanics flawed?

The answer has been given to you a number of times already and you're ignoring it. Classical mechanics is flawed because it does not adequately describe our experiments. As already said half a dozen times by now, it can't, for example, predict what goes on inside a particle accelerator. It also can't explain the observed constancy of the speed of light.

 

[Physicsguru]

No, SR is not self contradictory. It is not possible to derive two conflicting predictions from SR.

Sure it's possible, perhaps that's why I'm meeting with so much resistance.

Guru

 

[JesseM]

Sure it's possible, perhaps that's why I'm meeting with so much resistance.

OK, derive two contradictory conclusions from SR. But keep in mind what I said about the definition of "constant acceleration" in SR in my last post.

 

[Physicsguru]

In SR, "acceleration is constant" doesn't mean that in a given inertial frame, I will see the velocity increase by the same amount each second. Rather, it means that if you look at the frame in which the ship is at rest at any given moment, and if one second later you see the velocity has increased by v in that frame, then if you switch to the ship's new rest frame at that moment, then one second later the velocity will have increased by v in that frame. Note that if you want "constant acceleration" to be synonymous with "constant g-force experienced by the ship's crew", you must use this definition.

Another way of thinking of this: suppose I am on a ship which has a gun onboard that will shoot out a smaller ship moving at 9.8 m/s relative to my ship. This ship in turn has a gun that shoots out an even smaller ship at 9.8 m/s relative to itself. If I shoot my gun, and the smaller ship shoots its gun, then according to Newtonian mechanics the smallest ship will now be moving at 19.6 m/s relative to me. But in relativity velocities don't add in this simple way: if you see a ship moving at velocity v in your frame, and I see that you are moving at velocity u in my frame, then in my frame the ship's velocity will not be u+v but $$(u+v)/(1 + uv/c^2)$$ (see relativistic velocities). This formula can in turn be derived from the "Lorentz transformation", which specify how coordinates (x,y,z,t) in frame S are mapped to coordinates (x',y',z',t') in frame S'.

Your logic is simultaneously atrocious and astounding at the same time. There is a problem with the Lorentz transformations.

 

[jcsd]

Sure it's possible, perhaps that's why I'm meeting with so much resistance.

Guru

Can you name one self-contadictions (and don't name one of the so-called paradoxes of relativity as they are not true paradoxes)?

 

[Physicsguru]

Then you fail even basic, Newtonian physics.

dp/dt = m dv/dt + v dm/dt

If you say a is a constant, then all you are specifying is that dv/dt is a constant. The ONLY way in which you can use vf = vi + at is if you completely ignore the v dm/dt term. While this is perfectly valid in many cases in Newtonian physics where m doesn't change, it isn't valid HERE! The object's apparent mass increases as v increases!

You are welcome to come visit any particle accelerators and prove to yourself that your "derivation" here is faulty.

Zz.

This response is incorrect Zapper.

$$a = acceleration = dv/dt$$

Therefore:

dv = adt

Therefore:

$$\int dv = \int a dt$$

Since a is constant by stipulation, you can pull it out of the integral to obtain:

$$\int dv = a \int dt$$

Which leads to

$$Vf-Vi = a (t2-t1)= a \Delta t$$

Which is the kinematical formula for constant acceleration that was used.

QED

 

[JesseM]

Your logic is simultaneously atrocious and astounding at the same time. There is a problem with the Lorentz transformations.

No there isn't. For reference, the Lorentz transformation looks like this:

$$x' = \gamma (x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma (t - vx/c^2)$$
with $$\gamma = 1/\sqrt{1 - v^2/c^2}$$

The Lorentz transform makes sense for a few different reasons:

1. If a frame S assigns an event the coordinates (x,y,z,t), and you use the Lorentz transformation to map these coordinates to frame S', getting (x',y',z',t'), then if S' also uses the Lorentz transformation to map (x',y',z',t') back into S, he will get back the original coordinates (x,y,z,t).

2. Length in each observer's frame is just the distance in his coordinates from one end of an object to another

3. The time between two events in each observer's frame is just the time coordinate of the second minus the time coordinate of the first.

4. Velocity in each observer's frame is just distance/time in his coordinates

5. All the most accurate known laws of physics are invariant under the Lorentz transformation--in other words, if you have some physics equation expressed in terms of x',y',z',t' coordinates in frame S, and then you substitute in $$x' = \gamma (x - vt)$$, $$y' = y$$, $$z' = z$$, and $$t' = \gamma (t - vx/c^2)$$, then simplify, you will get back exactly the same equation but expressed in terms of x,y,z,t coordinates. The most accurate known laws of physics are not invariant under a "Galilei transformation", or:

$$x' = x - vt$$
$$y' = y$$
$$z' = z$$
$$t' = t$$

The Galilei transform is the one used in Newtonian mechanics (Newtonian laws such as $$F = GMm/[(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2]$$ are invariant under the Galilei transformation), and it's from the Galilei transform that we get the idea that if A is moving at velocity v relative to B, and B is moving at velocity u relative to C, then A would just be moving at u+v relative to C. Note that the Galilei transform also has properties 1-4 above, so again, the physical reason for preferring the Lorentz transform is just that the most accurate known laws are invariant under the Lorentz transform, but not under the Galilei transform.

 

[ZapperZ]

This response is incorrect Zapper.

$$a = acceleration = dv/dt$$

Therefore:

dv = adt

Therefore:

$$\int dv = \int a dt$$

Since a is constant by stipulation, you can pull it out of the integral to obtain:

$$\int dv = a \int dt$$

Which leads to

$$Vf-Vi = a (t2-t1)= a \Delta t$$

Which is the kinematical formula for constant acceleration that was used.

QED

Nope! Again, you are using a "special case" of a non-changing mass! The MOST GENERAL form of Newton's law is

F = dp/dt.

which can be expressed as

F = d(mv)/dt.

This is the most general expression since it takes into account not only a time varying velocity, but also a time-varying mass, as in the case of, for example, a rocket burning fuel during lift-off!

That is why this then can be written as

F = m dv/dt + v dm/dt.

It is ONLY for when F is a constant and dm/dt =0 can you then write

m dv/dt = constant, of dv/dt = constant, which means that the acceleration is a constant.

But how do you do this if dm/dt isn't zero? Especially considering the fact that dm/dt isn't even a constant in SR (rate of mass increase isn't linear)! Sure, you can write the equation

m dv/dt = F - v dm/dt

You can't say "Oh, I am forcing this equality (i.e. the RHS of the equation) to be a constant" when one quantity, "m" is blowing up to infinity! Your "stipulation" is physically unreal!

Zz.

 

[quantumdude]

Sure it's possible,

No, it is not possible. Far better men than you have tried.

I have an idea, why don't you post what you think is a self-contradiction in SR, and we will explain to you why you are wrong.

 

[quantumdude]

I think 5 pages is long enough for this nonsense.

Since the question in the opening post has been answered satisfactorily and repeatedly, this thread is done.