# Does everything that has to do with velocity

1. Aug 13, 2008

### epkid08

Given that you have a constant force applied to any mass, what is the rate at which the mass' acceleration decreases as its velocity approaches c?

Does everything that has to do with velocity in some way, have a relativistic transformation i.e. t'=f(t)?

2. Aug 13, 2008

### Fredrik

Staff Emeritus
I would define a constant force as something that causes constant proper acceleration. The world line of an object that has constant proper acceleration is a hyperbola (-t2+x2=constant). I have to go to bed, so I'm not going to work out the details now, but maybe this thread will be helpful. (See the first question I asked and DrGreg's answer).

I'm not sure how to interpret that question. There is a coordinate system called an inertial frame associated with each velocity, and everything that happens in spacetime can be described using the coordinates of any inertial frame. A "relativistic transformation" (a Lorentz transformation) is a coordinate change from one inertial frame to another, i.e. if you know how something looks in one inertial frame, you would use a Lorentz transformation to find out how it looks in another.

3. Aug 13, 2008

### MeJennifer

Since force obviously decreases with distance could you explain what you mean when you say you are applying a constant force to a mass that is rapidly moving away from you?

4. Aug 13, 2008

### epkid08

Constant force applied to a mass causes the mass to have a constant acceleration.

5. Aug 13, 2008

### DaveC426913

Yah, I think you're misunderstanding the question. Pretend he's asking about a rocket with constant thrust.

6. Aug 13, 2008

### DaveC426913

Here is a http://www.1728.com/reltivty.htm" [Broken] you can play with and draw your own conclusions.

Last edited by a moderator: May 3, 2017
7. Aug 13, 2008

### MeJennifer

In that case the rocket acceleration will never descrease and the speed relative to something else will never be c.

8. Aug 13, 2008

### epkid08

To further my first question, if we have a relativistic transformation of t, t', as well as of x, x', why can't we take $$\frac{x'}{t'^2}$$, and have that be a relativistic transformation of a, which then simplifies to

$$a'=a\sqrt{1-(at/c)^2}$$?

Does the above equation correctly represent the boundary of a constant acceleration as velocity approaches c?

9. Aug 13, 2008

### epkid08

The acceleration has to decrease as the velocity approaches c.

10. Aug 13, 2008

### MeJennifer

No that is not true at all.

11. Aug 13, 2008

### Al68

by a factor of 1/sqrt[1 +(at/c)^2]. So delta v = at / sqrt[1 + (at/c)^2] where a is proper acceleration (ie force applied/proper mass), t is time in the rest frame in which the mass was initially at rest, and v is relative velocity between that rest frame and the accelerated mass.

Al

12. Aug 14, 2008

### Staff: Mentor

Acceleration is the force you feel when pushed against your chair in an accelerating rocket. This force is always present when the engines are firing. That's one of the main points of special relativity.

13. Aug 14, 2008

### DaveC426913

Explain.

The OP is asking why - if the first hour of his trip gets him from .98c to .99c (delta v=+.01c), but the second hour of his trip gets him from .99c to only .999c (delta v=+.001c) - why that doesn't constitute a decease in acceleration.

14. Aug 14, 2008

### epkid08

It might not constitute a decrease in acceleration in an objects inertial frame, but I think it obviously constitutes a decrease in observed acceleration of that object, if not, then I'm definitely missing something.

15. Aug 14, 2008

### Staff: Mentor

This is the difference between proper acceleration (loosely speaking, what the object itself "feels") and coordinate acceleration in some inertial reference frame.

16. Aug 14, 2008

### epkid08

17. Aug 14, 2008

### Fredrik

Staff Emeritus
The proper acceleration (what the object "feels", what it measures with an accelerometer) is the coordinate acceleration ($d^2x/dt^2$) in a co-moving inertial frame. So "constant proper acceleration" doesn't mean that $d^2x/dt^2$ is constant in one particular inertial frame. It means that if we calculate $d^2x/dt^2$ in the inertial frame that's moving with the same velocity as the object we get the same result at every point on the object's world line. (Note that this means that we're using a different inertial frame at each point).

When the proper acceleration is constant, the coordinate acceleration (in one specific inertial frame) is decreasing.

18. Aug 14, 2008

### epkid08

That's what I thought coming into this topic, the acceleration of an object as viewed by an observer decreases as the object's velocity approaches c, but my real question was, at what rate does it decrease? Is this the correct way to show the change from proper acceleration to coordinate acceleration?

$$a_c=a_p\sqrt{1-(\frac{a_pt}{c})^2}$$

Or is it more complicated than this? I got that equation using lorentz transformations, solving for x'/(t'^2).

19. Aug 14, 2008

### DrGreg

Acceleration, like almost everything else in relativity, is a relative concept. Although everyone agrees what zero acceleration is, they disagree on the magnitude of a non-zero acceleration.

Thus an acceleration that is constant according the object feeling the acceleration (constant "proper acceleration") would, according to an inertial observer, be a decreasing acceleration.

It turns out, surprisingly, that, for linear motion of a constant mass, the "force" is invariant, i.e. all inertial observers agree on its value. (The appropriate relativistic equation for force is $\textbf{f} = d\textbf{p}/dt$, not $\textbf{f} = m \, d^2\textbf{x}/dt^2$.)

Thus a constant force applied to a constant mass in the same direction as its velocity causes constant proper acceleration but decreasing "coordinate acceleration" relative to an inertial observer.

The two accelerations are linked by

$$a_{proper} = \gamma^3 \, a_{coord}$$
$$\gamma = 1/\sqrt{1-v^2/c^2}$$​

This is proved in this thread, posts #13,#14,#15 equation (12) (with a correction in post #28).

For an example of a force acting on a non-constant mass, consider friction. Friction may cause a body to heat up, and that gain in energy is accounted for as an increase in mass (i.e. rest mass). Another, more obvious, example is a rocket ejecting burnt fuel and therefore losing mass.

_______________

Proof that, for linear motion of a constant mass, force is invariant:

In the notation of the posts I referred to above

$$p = m \, dx/d\tau = mc \, \sinh\phi$$
$$\frac {dp}{d\tau} = mc \, \frac{d\phi}{d\tau} \, \cosh\phi$$
$$f = \frac {dp}{dt} = \frac {dp/d\tau}{dt/d\tau} = mc \, \frac{d\phi}{d\tau} = m \, a_{proper}$$​

20. Aug 14, 2008

### epkid08

I read your proof and understood it, but my question was more directed at the at the equation $$a_c=f(a_p)$$, not $$a_p=f(a_c)$$.

Before I go further let me ask you a simple question, in your equation, $$a_{p} = \gamma^3 \, a_{c}$$, isn't the velocity equal to $$\frac{x}{\tau}$$? So in that equation, you will need to know both proper time and coordinate time in order to execute the function, which doesn't really show $$\frac{da_c}{da_p}$$.

Back to what I was saying before, here are the relativistic transformations for x and $$\tau$$:

$$x'=x\gamma$$
$$t'=\tau\gamma$$

My question is why can't we use this to calculate coordinate acceleration, in terms of proper velocity/acceleration i.e. solve for x'/t'^2?

$$a_c = a' = \frac{x'}{t'^2} = \frac{x\gamma}{\tau^2\gamma^2} = \frac{x}{\tau^2\gamma} = a_p\sqrt{1-(\frac{a_p\tau}{c})^2}$$

21. Aug 15, 2008

### DrGreg

In my posts, x and t are coordinates of the "stationary" inertial observer, $\tau$ is the proper time of the accelerating particle (and I used x' and t' as coordinates of a co-moving inertial observer moving with speed $v_0 = c \tanh \phi_0$ relative to the "stationary" observer.) I'm not sure that your notation is the same as mine.

The equations

$$x'=x\gamma$$
$$t'=\tau\gamma$$​

can't be true whatever your symbols mean.

Maybe you meant

$$x'=\gamma_0(x - v_0t)$$
$$t'=\gamma_0(t - v_0x/c^2)$$​

in my notation?

22. Aug 15, 2008

### DrGreg

No, it's $dx/dt$ (which turns out be be $c^2 t / (x + c^2/\alpha)$). Do you understand calculus? $dx/dt$ means something very different to $x/t$.

If you want equations in terms of particle time $\tau$, you've already got them in the posts in the other thread I referred to earlier. If you want equations in terms of coordinate time t, they are:

$$x = \frac {c^2}{\alpha} \left[ \sqrt{1 + \left( \frac{\alpha t}{c} \right)^2 } - 1 \right]$$ .......(A)
$$v = \frac{\alpha t} {\sqrt{1 + \left( \frac{\alpha t}{c} \right)^2 }}$$ .......(B)
$$\gamma = \sqrt{1 + \left( \frac{\alpha t}{c} \right)^2 }$$ .......(C)
$$a = \frac{\alpha} {\gamma^3}$$ .......(D)​

where $\alpha = a_p$ is constant proper accceleration, $a = a_c$ is coordinate acceleration.

(A) follows from equations (16) and (17) in my original proof in the other thread (with a change-of-origin so x=0 when t=0), and the rest follows by differentiation.

And if the particle's mass is m,

$$p = m \alpha t$$ .......(E)
$$E = \gamma m c^2$$ .......(F)
$$f = dp/dt = m \alpha$$ .......(G)​

23. Aug 15, 2008

### epkid08

I think we had the same notation, I'm just going off about a different way of solving for coordinate acceleration of an object. My idea was to use the Lorentz transforms of (x,t), (x',t'), to find a', the coordinate acceleration. I would do this by taking x', and dividing it by t'^2. I was wondering why this leads to a wrong solving of coordinate acceleration, as it's different than the equation you had posted.

24. Aug 15, 2008

### Fredrik

Staff Emeritus
epkid, you are mistaken about the form of Lorentz transformations. You can't transform the time and space coordinates separately. A Lorentz transformation always mixes them up, unless we're talking about a rotation in space (which doesn't change the time coordinate at all). The general form in 1+1 dimensions (excluding translations, reflections and time-reversal, and using units such that c=1) is

$$\begin{pmatrix}t' \\ x'\end{pmatrix}=\gamma\begin{pmatrix}1 && -v\\ -v && 1\end{pmatrix}\begin{pmatrix}t \\ x\end{pmatrix}$$

If you don't like matrices, then look at DrGreg's post again. He said the same thing, but without using matrices.

25. Aug 18, 2008

### DrGreg

$a' = x'/t'^2$ would be wrong even for constant coordinate acceleration. You ought to know the equation $x' = u't' + a't'^2/2$ which has an extra factor of 1/2. But the coordinate acceleration a' is not constant, so you have to use calculus techniques like I did.

Please also remember that the accelerating object is not an inertial observer, so there are no inertial coordinates associated with the object. The best you can do is consider a co-moving inertial observer (x',t') who is momentarily travelling at the same speed as the object. But to examine the whole motion you need an infinite number of different co-moving inertial observer all at different points along the trajectory of the accelerating object. That's where the calculus comes in.