Constant Acceleration: Thought Experiment & Answers

In summary: The constant acceleration of the dime is equivalent to a force that needs to increase exponentially - in order for the acceleration to continue indefinitely, the force would need to be exponential in nature.2) As the velocity of the dime increases, the effect of the acceleration becomes more pronounced - the leading and trailing surfaces of the dime must be accelerating at different rates in order for the object to remain rigid.3) It's worth noting that there are a lot of subtle issues lurking in the weeds when it comes to discussing the effects of an accelerating object, because there are a lot of complicated interactions between the surrounding frames.
  • #1
Gulliver
9
0
I have this "thought" experiment. Imagine a large rigid object, a thin disc (dime) with the ("heads") surface area of the earth, accelerating in the "heads" direction for many years such that individuals on the surface (of the "head" side) experience the normal sensation of "gravity". This object is the only object in the Universe.

How long can the disc undergo this constant "felt" acceleration? (I believe the answer is "forever".) Does the force required to maintain the constant "felt" acceleration ever need to increase? (I believe the answer is "no".) Does the acceleration of the leading surface need to differ from the trailing surface in order for the disc to remain rigid at relativistic velocities? (I believe the answer is "yes".)

Thanks so much!
 
Physics news on Phys.org
  • #2
Yes, yes, and yes, assuming you are measuring the acceleration with a local accelerometer mounted on the leading and tailing edges.
 
  • #3
Ditto pervect (who probably meant "in principle forever, no, yes", i.e. agreeing with your guesses), except that I'd add:

1. The acceleration must produced by some force via the expenditure of energy, which presumably limits how long you can keep up constant acceleration,

2. For a thin disk accelerated parallel to its axis of symmetry, the variation in acceleration in the direction of motion would be very small; the effect is more pronounced if you consider a thin rod accelerated parallel to its axis of symmetry, which you can imagine being pulled with constant tension by a string,

3. This kind of discussion can quickly become surprisingly technical, since there are a lot of subtle issues lurking in the weeds. For example, in the rod, I am assuming the force is being measured at the pulled end of the string by an observer comoving with the pulled end.
 
Last edited:
  • #4
I'd better clarify the force question. In a frame comoving with the accelerating dime, the force would be constant.
 
  • #5
pervect said:
I'd better clarify the force question. In a frame comoving with the accelerating dime, the force would be constant.
Thanks. But here's my problem: Since the dime is accelerating, that frame is not valid for inertially sound measurements. So... in reality this constantly accelerating dime needs exponentially increasing force (and therefore energy), right?
 
  • #7
One of the best ways to describe the situation is to use four-vectors. On the minus side it's a bit technical, on the plus side ambiguities of language don't affect the answers.

Using the 4-vector approach, geometric units, and setting the constant acceleration to unity, the results from for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] would appear as below.

Note that I'm only writing out two of the components of the 4-vector, the other two components are identically zero.

[itex]\tau[/itex] = proper time

The position vector can be written as a function of proper time as follows:

[itex]x^i(\tau)[/itex] :

[tex]t = x^0 = sinh \, \tau \hspace{.5 in} d = x^1 = cosh \, \tau[/tex]

I've chosen this as the simplest way of writing down the equations of motion, note however that [itex]x^1(0)[/itex] is not equal to zero, but is rather equal to 1.

The 4-velocity vector can be written as a function of proper time as follows:

[tex]v^0 = \frac{d x^0}{d \tau} = cosh \, \tau \hspace{.5 in} v^1 = \frac{d x^1}{d\tau} = sinh \, \tau[/tex]

The energy-momentum 4-vector is just
[tex]p^0 = m v^0 = m \, cosh \, \tau \hspace{.5 in} p^1 = m v^1 = m \, sinh \, \tau [/tex]

Here m is the invariant mass of the dime. The simplest case to analyze is where the invariant mass m is constant. This requires the dime to be propelled by an external source, for instance the 'dime' might really be a laser powered light sail.

Note that [itex]p^0[/itex] can be interpreted as the energy of the dime (it's written here as a function of proper time), and [itex]p^1[/itex] can be interpreted as the momentum of the dime.

Note that the 4-velocity of the dime is the rate of change of the position of the dime with regard to proper time tau, rather than the coordinate time t.

The 4-force on the dime is
[tex]f^0 = \frac{d p^0}{d \tau} = m \, sinh \, \tau \hspace{.5 in} f^1 = \frac{d p^1}{d \tau} = m \, cosh \, \tau[/tex]

The 4-acceleration of the dime is
[tex]a^0 = \frac{d v^0}{d \tau} = sinh \, \tau \hspace{.5 in} a^1 = \frac{d v^1}{d \tau} = cosh \, \tau[/tex]

Some comments here are in order:

0) There's no problem in transforming the 4-vectors into an instantaneous comoving frame - we can transform both the 4-acceleration and the 4-force in this manner, because 4-vector transformations depend only on the relative velocity, i.e. all 4-vectors transform according to the Lorentz transform, which is indepenent of acceleration and depends only on velocity.

1) using the 4-vector approach, we can say that that the 4-force is directly proportional to the 4-acceleration, and the constant of proportionality is the invariant mass m of the dime.

2) In the frame of the dime, the 4-acceleration is constant. If you do the appropriate Lorentz boost, the 4-acceleration in an instantaneously comoving frame is just (0,1).

3) In the frame of the dime, the 4-force is also constant, and equal to m*a, and since a=1, the 4-force is equal to m.
5) We could write the 4-velocity and energy -momentum 4-vector in the instantaneous frame of the dime just as we did for the 4-acceleration and the 4-force, but they aren't very interesting.

6) The idea of constant proper acceleration is the same as the idea that the 4-acceleration in the instantaneously comoving frame is just (0,a). I've made a=1 for simplicity in writing the above expressions down.
 
Last edited by a moderator:
  • #8
Another Great Job.

pervect said:
One of the best ways to describe the situation is to use four-vectors. On the minus side it's a bit technical, on the plus side ambiguities of language don't affect the answers.

Using the 4-vector approach, geometric units, and setting the constant acceleration to unity, the results from for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] would appear as below.

Note that I'm only writing out two of the components of the 4-vector, the other two components are identically zero.

[itex]\tau[/itex] = proper time

The position vector can be written as a function of proper time as follows:

[itex]x^i(\tau)[/itex] :

[tex]t = x^0 = sinh \, \tau \hspace{.5 in} d = x^1 = cosh \, \tau[/tex]

I've chosen this as the simplest way of writing down the equations of motion, note however that [itex]x^1(0)[/itex] is not equal to zero, but is rather equal to 1.

The 4-velocity vector can be written as a function of proper time as follows:

[tex]v^0 = \frac{d x^0}{d \tau} = cosh \, \tau \hspace{.5 in} v^1 = \frac{d x^1}{d\tau} = sinh \, \tau[/tex]

The energy-momentum 4-vector is just
[tex]p^0 = m v^0 = m \, cosh \, \tau \hspace{.5 in} p^1 = m v^1 = m \, sinh \, \tau [/tex]

Here m is the invariant mass of the dime. The simplest case to analyze is where the invariant mass m is constant. This requires the dime to be propelled by an external source, for instance the 'dime' might really be a laser powered light sail.

Note that [itex]p^0[/itex] can be interpreted as the energy of the dime (it's written here as a function of proper time), and [itex]p^1[/itex] can be interpreted as the momentum of the dime.

Note that the 4-velocity of the dime is the rate of change of the position of the dime with regard to proper time tau, rather than the coordinate time t.

The 4-force on the dime is
[tex]f^0 = \frac{d p^0}{d \tau} = m \, sinh \, \tau \hspace{.5 in} f^1 = \frac{d p^1}{d \tau} = m \, cosh \, \tau[/tex]

The 4-acceleration of the dime is
[tex]a^0 = \frac{d v^0}{d \tau} = sinh \, \tau \hspace{.5 in} a^1 = \frac{d v^1}{d \tau} = cosh \, \tau[/tex]

Some comments here are in order:

0) There's no problem in transforming the 4-vectors into an instantaneous comoving frame - we can transform both the 4-acceleration and the 4-force in this manner, because 4-vector transformations depend only on the relative velocity, i.e. all 4-vectors transform according to the Lorentz transform, which is indepenent of acceleration and depends only on velocity.

1) using the 4-vector approach, we can say that that the 4-force is directly proportional to the 4-acceleration, and the constant of proportionality is the invariant mass m of the dime.

2) In the frame of the dime, the 4-acceleration is constant. If you do the appropriate Lorentz boost, the 4-acceleration in an instantaneously comoving frame is just (0,1).

3) In the frame of the dime, the 4-force is also constant, and equal to m*a, and since a=1, the 4-force is equal to m.



5) We could write the 4-velocity and energy -momentum 4-vector in the instantaneous frame of the dime just as we did for the 4-acceleration and the 4-force, but they aren't very interesting.

6) The idea of constant proper acceleration is the same as the idea that the 4-acceleration in the instantaneously comoving frame is just (0,a). I've made a=1 for simplicity in writing the above expressions down.

Aha! The key is that to the observer in an inertial frame of reference the force would remain the same, the mass would increase, and the acceleration would decrease. However to the observer of the dime, all three would be constant.

You really helped. That math was great and simple. Very enlightening. You don't know how many pages I've read (both on the web and in my "old" textbooks) and you manage to explain in so well. Good job.

Now... about the rigid body problem... I probably need to tell you where I'm coming from first.

Please don't think I'm lame here, but I'm on a mission. My personal hero is James Randi, randi.org. He has attempted to debunk the Flat Earth Society, theflatEarth'society.org. I have spent weeks and thousands of dollars to make sure that young minds get the proper scientific background to make a solid conclusion. I'm setting up a "counter" website and posting on their forums. They hold that the Earth is like a dime with the surface on the "heads" side, accelerated through space at 1g.

I'm convinced now, thanks to you and others here, that arguing limitation on acceleration, velocity, and force are unwarranted. The force is applied solely to the "tails" side (by dark energy, they claim). I suspect that I may be able to debunk this model by examining the effect of a constant acceleration to the trailing end of a rigid body. I believe that we have a large acceleration (1g) and such a thickness (of miles) that the relativistic effects would create too great of stress in the dime.

Forgive me if you think I'm wasting your time. I truly wish to help others by fighting those who don't apply scientific methods in their decision-making and then advocate their beliefs in an attempt to influence others.

(BTW, anyone who has the time and interest to join me on any such endeavor is most welcome. I really could use the help.)

Thank you so much.
 
Last edited by a moderator:
  • #9
Threads on this "flat Earth socieity" have appeared before, see for instance https://www.physicsforums.com/showthread.php?p=1067430

These threads tend to get locked for terminal silliness, however, as the thread above was. The consensus appears to be that the people arguing about this are just arguing for the sake of argument. So while we don't necessarily mind discussing the physics of an accelerated observer in SR, we'd rather avoid getting sucked into this "flat earth" debate.

Aha! The key is that to the observer in an inertial frame of reference the force would remain the same, the mass would increase, and the acceleration would decrease. However to the observer of the dime, all three would be constant.

The relativistic mass would increase, but note that the invariant mass m I use in the formulas I use above would not increase.

Note also that four-forces are [tex]\frac{dp}{d \tau}[/tex], the rate of change of the energy-momentum with respect to proper time.

"Standard" forces are defined as [tex]\frac{dp}{dt} [/tex], the rate of change of energy momentum with respect to coordinate time.

Four-forces are not compatible with the rather old-fashioned notion of relativistic mass, but are compatible with invariant mass.
 

1. What is constant acceleration?

Constant acceleration is the rate at which an object's velocity changes over time, while moving in a straight line. It can be represented by a constant value, such as meters per second squared (m/s²).

2. How does constant acceleration differ from uniform motion?

Uniform motion is when an object moves at a constant velocity, meaning there is no change in its speed or direction. Constant acceleration, on the other hand, involves a change in velocity over time, resulting in a curved motion instead of a straight line.

3. Can an object have both constant acceleration and constant velocity?

No, an object cannot have both constant acceleration and constant velocity at the same time. This is because constant acceleration requires a change in velocity, while constant velocity implies that there is no change in velocity.

4. How is constant acceleration calculated?

Constant acceleration can be calculated using the formula a = (vf - vi)/t, where a represents the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. This formula is also known as the average acceleration formula.

5. What is the significance of constant acceleration in physics?

Constant acceleration is a fundamental concept in physics and plays a crucial role in understanding the motion of objects. It helps explain various phenomena, such as free fall, projectile motion, and circular motion. It is also essential in the development of theories, such as Newton's laws of motion and the equations of motion.

Similar threads

  • Special and General Relativity
2
Replies
50
Views
2K
  • Special and General Relativity
2
Replies
69
Views
4K
  • Special and General Relativity
Replies
12
Views
884
  • Special and General Relativity
2
Replies
40
Views
2K
  • Special and General Relativity
3
Replies
75
Views
3K
  • Special and General Relativity
3
Replies
80
Views
12K
  • Special and General Relativity
Replies
11
Views
1K
  • Special and General Relativity
Replies
15
Views
1K
  • Special and General Relativity
Replies
15
Views
991
  • Special and General Relativity
Replies
27
Views
4K
Back
Top