Question About Relativistic Acceleration

[Physicsguru]

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.

How is classical mechanics flawed?

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

I don't think that they are equal, but I do think that they are proportional.

Kind regards,

Guru

 

[Physicsguru]

Infinite time. Acceleration doesn't work the same way in relativity that it does in Newtonian mechanics, your velocity at time t won't just be (acceleration rate)*(time since your velocity was zero).

Let me try this a different way JesseM. Suppose that you are in a ship which is accelerating at 9.8 m/s^2. If you start from rest, how fast will your ship be moving after 24 hours?

Regards,

Guru

 

[JesseM]

Let me try this a different way JesseM. Suppose that you are in a ship which is accelerating at 9.8 m/s^2. If you start from rest, how fast will your ship be moving after 24 hours?

24 hours according to the onboard clock, or according to clocks in the inertial frame that the ship started out at rest in?

 

[jcsd]

Physics guru the answer to you question s have been given, is this thread going anywhere? The only challenge I see are your misconceptions onn relativity.
As for accelartion if you mean extrinsic acceleration (i.e. the coordinate accelartion as measured form some inertial frame) then it is impossible to maintain an extrinsic accelartion of 9.8 m/s^2 indefintely as you soon find the force required to keep up that accelartion is infinite. If you mean intrinsic acceleration, then yes you can maintin an instrinsic accelartion of 9.81 m/s^2 indefintely, but you will never reach a velocity of c or greater in any inertial frame.

 

[Physicsguru]

24 hours according to the onboard clock, or according to clocks in the inertial frame that the ship started out at rest in?

Give the answer in both frames.

(I would also suggest that you cover both cases case 1) relativity correct, case 2) relativity incorrect).

Regards,

Guru

 

[Physicsguru]

Physics guru the answer to you question s have been given, is this thread going anywhere? The only challenge I see are your misconceptions onn relativity.
As for accelartion if you mean extrinsic acceleration (i.e. the coordinate accelartion as measured form some inertial frame) then it is impossible to maintain an extrinsic accelartion of 9.8 m/s^2 indefintely as you soon find the force required to keep up that accelartion is infinite. If you mean intrinsic acceleration, then yes you can maintin an instrinsic accelartion of 9.81 m/s^2 indefintely, but you will never reach a velocity of c or greater in any inertial frame.

JCSD the answer is not as simple as the question appears. To put this another way, no one is addressing the question in an epistemologically correct manner, since no one is actually certain that E=Mc^2. This is supposed to be a hard question, which eventually leads to the Meissner effect, but this thread is nowhere near that point yet.

Regards,

Guru

 

[quantumdude]

JCSD the answer is not as simple as the question appears.

Yes, it is. According to the best information we have, it is not possible to accelerate any massive object to a speed of c in any finite amount of time.

To put this another way, no one is addressing the question in an epistemologically correct manner, since no one is actually certain that E=Mc^2.

You're asking the members of this Forum to make a scientific prediction. No such prediction is ever certain. Furthermore, no one is relying on E=mc². That equation describes a particle at rest. The respondents to your question are referring to relativistic kinematics.

But in any case, we are as sure of both E=mc² and of relativistic kinematics as we are of anything else in science.

This is supposed to be a hard question, which eventually leads to the Meissner effect, but this thread is nowhere near that point yet.

Why don't you just make your point?

 

[jcsd]

No it's a rather simplistic question that has alreday been answered and as for the relevance of the Meissner effect I really can't see where you're going. If you think there is something being missed it is best you actually say what it is rather than leading everyone on a wild goose chase as the answer already given is undeniably corrcet (within the context of relativty, if you're talking baout any other context then why post it in this forum?) so if you've got another answer you have made a mistake.

 

[JesseM]

Give the answer in both frames.

(I would also suggest that you cover both cases case 1) relativity correct, case 2) relativity incorrect).

OK, you can see the equations for velocity as a function of time (both proper time and coordinate time) for constant acceleration according to relativity on http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page. For onboard time T, the equation is:
v = c*tanh(aT/c)
It's easier to evaluate this equation if you use units of years and light-years instead of meters and seconds, since c=1 in these units; the page mentions that an acceleration of 9.8 m/s^2 is approximately equal to 1.03 lyr/yr^2. Meanwhile, 1 day = 1/365 years, or about 0.00274 years, so aT/c will be about 0.00282 in these units. Here is a graphing calculator applet that can do the tanh function--in this case, if I type tanh(0.00282) and click the "Eval" button, I get back a number which is still approximately equal to 0.00282, I guess because tanh(x) is close to x when x is close to zero. So, this means that after 24 hours of onboard time, the velocity would be about 0.00282c.

For time in the reference frame where the velocity is being measured, the equation given is:
v = at/\sqrt{1 + (at/c)^2}
So, with a = 1.03 and t=0.00274, this will also be very close to 0.00282c. I guess you'd need a significantly larger time or acceleration for there to be any noticeable difference.

As for covering the case 2), "relativity incorrect", there would of course be an infinite number of possible answers depending on what alternate theory you choose. For example, one of these theories would be that anyone who accelerates at 1G for exactly 24 hours is instantly transported away from our universe and into smurfworld, where they live smurfilly ever after moving at velocity smurf. Did you have a specific theory in mind?

 

[Physicsguru]

No it's a rather simplistic question that has alreday been answered and as for the relevance of the Meissner effect I really can't see where you're going. If you think there is something being missed it is best you actually say what it is rather than leading everyone on a wild goose chase as the answer already given is undeniably corrcet (within the context of relativty, if you're talking baout any other context then why post it in this forum?) so if you've got another answer you have made a mistake.

The question isn't simplistic at all. No one here has been handling accelerating reference frames very well.

Regards,

Guru

 

[jcsd]

The question isn't simplistic at all. No one here has been handling accelerating reference frames very well.

Regards,

Guru

You don't need to use accelerated frames here (if you are talking about accelerated frames you need to specifcally say so), infact I've specifcally been careful to always make it clear thta I'm talking about inertial frames only as bodies can have coordinate speeds of greater than c in non-inertial frames (that fact is pretty trivial).

 

[quantumdude]

You don't need to use accelerated frames here (if you are talking about accelerated frames you need to specifcally say so),

I think he means the rocket, which is accelerating.

 

[jcsd]

I think he means the rocket, which is accelerating.

that's what I thought possibly, but by defintion the coordinate speed of the accelarted rocket in it's own frame is zero (it's intrinsic accelartion is it's accelartion relative to it's momentarily comoving inertial frame, however the mometarily comoving inertial frame is different at each instant).

 

[Physicsguru]

24 hours according to the onboard clock, or according to clocks in the inertial frame that the ship started out at rest in?

Give me the answer in both frames.

Regards,

Guru

 

[JesseM]

Give me the answer in both frames.

I already did--did you miss my post on the last page?

 

[Physicsguru]

I already did--did you miss my post on the last page?

Yes I did miss it, and I just went back and read it, and of course you arrived at what I expected... namely that you would need a far larger acceleration to be anywhere near c, after 24 hours.

So now, let me ask you this. What would the rate of accleration have to be, in order for you to be near the speed of light after one day?

 

[JesseM]

So now, let me ask you this. What would the rate of accleration have to be, in order for you to be near the speed of light after one day?

Just solve the two equations I gave for a and you should be able to find the answer. Have a stab at it yourself, and if you have trouble I'll help you out.

 

[Physicsguru]

Just solve the two equations I gave for a and you should be able to find the answer. Have a stab at it yourself, and if you have trouble I'll help you out.

It's a trivial problem.

initial speed=0
final speed = 299792458 m/s
t = 24 hours = 24*3600 seconds

Vf=Vi + at

299792458 = a 24*3600

a = 299792458/(24*3600) m/s^2

Guru

 

[JesseM]

It's a trivial problem.

initial speed=0
final speed = 299792458 m/s
t = 24 hours = 24*3600 seconds

Vf=Vi + at

No, this equation would only be correct in Newtonian physics. Like I said before, in relativity your velocity after accelerating from rest for time t is not at. I meant you should try solving the equations I gave in my previous post for a to see how this works in relativity.

 

[ZapperZ]

It's a trivial problem.

initial speed=0
final speed = 299792458 m/s
t = 24 hours = 24*3600 seconds

Vf=Vi + at

299792458 = a 24*3600

a = 299792458/(24*3600) m/s^2

Guru

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.

 

[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.