# Accelration & special relativity.

I have a couple of questions, I'd be most grateful if someone could help me? Assume that you are accelerating at ate 'a' relative to a rest frame for distance 'D' which is also measued relative to the rest frame.

1) Is it right to use v=$$\sqrt{2aD}$$ to find the find velocity, and D=1/2at$$^{2}$$. to find final velocity and time elapsed relative to the rest frame. I'm fairly sure this is correct.

2) what is the final speed(v'), time elapsed(t'), acceleration(a') & distance(D') relative the moving reference frame.

I'd be very happy you someone show a derivation as well. Thanks

I have a couple of questions, I'd be most grateful if someone could help me? Assume that you are accelerating at ate 'a' relative to a rest frame for distance 'D' which is also measued relative to the rest frame.

1) Is it right to use v=$$\sqrt{2aD}$$ to find the find velocity, and D=1/2at$$^{2}$$. to find final velocity and time elapsed relative to the rest frame. I'm fairly sure this is correct.

No, it is not correct. The above is valid only in Newtonian mechanics, where F=m*dv/dt.

2) what is the final speed(v'), time elapsed(t'), acceleration(a') & distance(D') relative the moving reference frame.

m=rest mass
t=coordinate time
v=v(t)

If F is constant, you will end up with the equations of "Hyperbolic motion".
Here is a sketch for the derivation:

F=m*d/dt(v/sqrt(1-(v/c)^2)

so,

F/m=d/dt(v/sqrt(1-(v/c)^2)

Since F/m is constant(we call this "coordinate acceleration", a), the above becomes a very simple differential equation with the solution:

v=at/sqrt(1+(at/c)^2)

For at<<c, you recover the Newtonian equation v=at

If you integrate one more time, you will get x as a function of a and t. Indeed:

dx/dt=at/sqrt(1+(at/c)^2)

x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)

Again, for at<<c, you recover the Newtonian formula x(t)=at^2/2

Last edited:
bcrowell
Staff Emeritus
Gold Member
Starthaus, you're answering Mononoke's question as if the proper acceleration were constant. Mononoke posed the question by saying that it's the acceleration measured in the rest frame that's constant. If all quantities are measured in the rest frame, then the standard Newtonian equations like x=(1/2)at^2 are all correct.

Mononoke, you need to realize that under the conditions you've posed, (1) the person who's accelerating feels an acceleration that diverges to infinity as v approaches c, and (2) the accumulated energy expenditure also diverges to infinity.

Starthaus, you're answering Mononoke's question as if the proper acceleration were constant.

No, it is for constant coordinate acceleration. Please read the derivation carefully. Please look at the definition of F/m in the text.

Starthaus, what have I done wrong with rest frame calculation. And isn't bcrowell correct since i'm measuring in a rest frame. I thought this question was conceptually a little difficult for me to grasp because the object is moving in non inertial frame and lambda is difficult to define.

bcrowell
Staff Emeritus
Gold Member
No, it is for constant coordinate acceleration. Please read the derivation carefully. Please look at the definition of F/m in the text.

No, constant coordinate acceleration would be defined as d^2x/dt^2=constant, where x and t are measured in the rest frame. Constant F/m (where m is the invariant mass and F is the force measured in the rest frame) does not produce constant coordinate acceleration, since Newton's second law is only a low-velocity approximation.

In your #2 you're claiming to derive equations that simultaneously represent constant coordinate acceleration and hyperbolic motion. That's incorrect. Hyperbolic motion corresponds to constant proper acceleration, not constant coordinate acceleration in a rest frame: http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

Fredrik
Staff Emeritus
Gold Member
No, it is not correct. The above is valid only in Newtonian mechanics, where F=m*dv/dt.
...
If F is constant, you will end up with the equations of "Hyperbolic motion".
I didn't read your derivation, but what you said in the quote is wrong. The world line is a hyperbola when the proper acceleration is constant. When the coordinate acceleration is constant, the usual Newtonian stuff applies. It can't be constant for very long, because the coordinate speed can't reach c, but as long as it is, there's no need to mention force, momentum, etc.

bcrowell
Staff Emeritus
Gold Member
what is the final speed(v'), time elapsed(t'), acceleration(a') & distance(D') relative the moving reference frame.

Lorentz boosts are symmetric in the sense that if A sees B as moving at velocity v, B sees A as moving at -v. Therefore you have v'=-v. To relate D' to D, you can simply use the Lorentz length dilation factor. I believe the relationship between a and a' is also simply a factor of gamma, since the longitudinal component of force and momentum transforms that way. To get the elapsed proper time:
$$\int ds = \int \sqrt{dt^2-(atdt)^2}=\int\sqrt{1-a^2t^2}dt$$,
which comes out to be (t*Sqrt[1 - a^2*t^2])/2 + ArcSin[a*t]/(2*a), according to integrals.com.

The physical interpretation is a whole different matter. I doubt that this mathematical result corresponds to anything interesting or useful in the real world. It does *not* correspond to motion under the influence of a constant force -- the force that produces this motion is a varying one, both in the rest frame and in the accelerating frame.

bcrowell
Staff Emeritus
Gold Member
I didn't read your derivation, but what you said in the quote is wrong. The world line is a hyperbola when the proper acceleration is constant. When the coordinate acceleration is constant, the usual Newtonian stuff applies. It can't be constant for very long, because the coordinate speed can't reach c, but as long as it is, there's no need to mention force, momentum, etc.

Yes, I agree with this.

Lorentz boosts are symmetric in the sense that if A sees B as moving at velocity v, B sees A as moving at -v. Therefore you have v'=-v. To relate D' to D, you can simply use the Lorentz length dilation factor. I believe the relationship between a and a' is also simply a factor of gamma, since the longitudinal component of force and momentum transforms that way. To get the elapsed proper time:
$$\int ds = \int \sqrt{dt^2-(atdt)^2}=\int\sqrt{1-a^2t^2}dt$$,
which comes out to be (t*Sqrt[1 - a^2*t^2])/2 + ArcSin[a*t]/(2*a), according to integrals.com.

The physical interpretation is a whole different matter. I doubt that this mathematical result corresponds to anything interesting or useful in the real world. It does *not* correspond to motion under the influence of a constant force -- the force that produces this motion is a varying one, both in the rest frame and in the accelerating frame.

Thanks, but how did you get that integral

No, constant coordinate acceleration would be defined as d^2x/dt^2=constant, where x and t are measured in the rest frame. Constant F/m (where m is the invariant mass and F is the force measured in the rest frame) does not produce constant coordinate acceleration, since Newton's second law is only a low-velocity approximation.

In your #2 you're claiming to derive equations that simultaneously represent constant coordinate acceleration and hyperbolic motion. That's incorrect. Hyperbolic motion corresponds to constant proper acceleration, not constant coordinate acceleration in a rest frame: http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

I'll have to respectfully disagree. See here for an identical derivation. I am quite sure the paper was peer-reiviewed and published. Perhaps we are disagreeing on the terminology of "hyperbolic motion" but the math is correct.

bcrowell
Staff Emeritus
Gold Member
I'll have to respectfully disagree. See here for an identical derivation. I am quite sure the paper was peer-reiviewed and published. Perhaps we are disagreeing on the terminology of "hyperbolic motion" but the math is correct.

Nobody said anything was wrong with your math. Mononoke asked about motion with constant coordinate acceleration. The Iorio paper isn't describing motion with constant coordinate acceleration, and it doesn't claim to describe motion with constant coordinate acceleration.

Thanks, but how did you get that integral

sorry. I want to know where you get this from:
$$\int ds = \int \sqrt{dt^2-(atdt)^2}$$

bcrowell
Staff Emeritus
Gold Member
sorry. I want to know where you get this from:
$$\int ds = \int \sqrt{dt^2-(atdt)^2}$$

We're in flat spacetime, using Minkowski coordinates, so $ds^2=dt^2-dx^2$, where ds is interpreted as the proper time. Since the coordinate acceleration a=dv/dt is constant, we have v=at, and dx=(dx/dt)dt=at dt.

We're in flat spacetime, using Minkowski coordinates, so $ds^2=dt^2-dx^2$, where ds is interpreted as the proper time. Since the coordinate acceleration a=dv/dt is constant, we have v=at, and dx=(dx/dt)dt=at dt.

Sweet I got it

bcrowell
Staff Emeritus
Gold Member
Sweet I got it

Cool :-)

By the way, welcome to Physics Forums.

...which is precisely what my mathematical proof is about.

The Iorio paper isn't describing motion with constant coordinate acceleration,

I will have to respectfully disagree with you again. Please look at the opening sentence of paragraph 2 in the Iorio paper where he describes the basic notions.

No, constant coordinate acceleration would be defined as d^2x/dt^2=constant,

....which is precisely what you will obtain if you differentiate twice wrt the coordinate time t the expression I have derived in a previous post, i.e. x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)

where x and t are measured in the rest frame.

Yes, the rest frame of an inertial observer. See also the start of paragraph 2 of the Iorio paper.

Constant F/m (where m is the invariant mass and F is the force measured in the rest frame) does not produce constant coordinate acceleration, since Newton's second law is only a low-velocity approximation.

Nothing to do with Newton's second law, it is basic math. If F is constant in an inertial frame (I really don't know what you call the "rest frame", you will need to explain) and m is the proper mass, then F/m is a constant that is generally called coordinate acceleration, as opposed to "proper acceleration" (the acceleration felt by a comoving observer).

In your #2 you're claiming to derive equations that simultaneously represent constant coordinate acceleration and hyperbolic motion. That's incorrect. Hyperbolic motion corresponds to constant proper acceleration, not constant coordinate acceleration in a rest frame: http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

The wiki page is very badly written. Just take my final equation:

x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)

and you can easily obtain the standard equation of hyperbolic motion expressed in coordinate distance (x) and coordinate time (t), where a, as a parameter, is the coordinate acceleration:

((ax/c^2)+1)^2-(at/c)^2=1

I didn't read your derivation, but what you said in the quote is wrong. The world line is a hyperbola when the proper acceleration is constant. When the coordinate acceleration is constant, the usual Newtonian stuff applies. It can't be constant for very long, because the coordinate speed can't reach c, but as long as it is, there's no need to mention force, momentum, etc.

Dear Fredrik,

I would have to respectfully disagree with you. Just take the final equation from my derivation:

x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)

and you can easily obtain the standard equation of hyperbolic motion expressed in coordinate distance (x) and coordinate time (t), where a=F/m, as a parameter, is the coordinate acceleration:

((ax/c^2)+1)^2-(at/c)^2=1

As a mathematician, I am quite sure that you will appreciate the above proof, even if it is somewhat terse.

bcrowell
Staff Emeritus
Gold Member
I will have to respectfully disagree with you again. Please look at the opening sentence of paragraph 2 in the Iorio paper where he describes the basic notions.

I don't see anything in the paper that says what you seem to think it says.

bcrowell said:
No, constant coordinate acceleration would be defined as d^2x/dt^2=constant,

....which is precisely what you will obtain if you differentiate twice wrt the coordinate time t the expression I have derived in a previous post, i.e. x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)
This doesn't make any sense to me. Are you intending a to be treated as a constant in taking this derivative? If so, then you clearly aren't going to get a constant by taking the second derivative of this expression with respect to t.

I don't see anything in the paper that says what you seem to think it says.

"The equation of motion of a particle of mass m acted upon by a force F,
as viewed in an inertial frame I,"

"Note that t denotes the proper time of a standard clock located at the origin
of I."

Now, it is pretty obvious that I is not the frame comoving with the accelerated object.

This doesn't make any sense to me. Are you intending a to be treated as a constant in taking this derivative? If so, then you clearly aren't going to get a constant by taking the second derivative of this expression with respect to t.

dx/dt=at/sqrt(1+(at/c)^2)

d^x/dt^2=d/dt(dx/dt)=a

bcrowell
Staff Emeritus
Gold Member
Now, it is pretty obvious that I is not the frame comoving with the accelerated object.
Nobody said that it was. What you're missing here is that the motion described by the Iorio paper is not motion with constant coordinate acceleration.

dx/dt=at/sqrt(1+(at/c)^2)

d^x/dt^2=d/dt(dx/dt)=a
Here is the derivative, as evaluated by the computer algebra system Maxima. It does not equal a. Either we're miscommunicating, or you need to review your calculus. The most general function whose second derivative with respect to t is a constant a is a function of the form (1/2)at^2+c1t+c2.
Code:
(%i5) diff(a*t/sqrt(1+(a*t/c)^2),t);
3  2
a                a  t
(%o5)                 --------------- - -----------------
2  2             2  2
a  t          2  a  t      3/2
sqrt(----- + 1)   c  (----- + 1)
2                2
c                c

Nobody said that it was. What you're missing here is that the motion described by the Iorio paper is not motion with constant coordinate acceleration.

It is motion with F/m=constant. I have been telling you this from the first post. I made the unfortunate mistake of calling F/m coordinate acceleration. You may choose the call that "proper" acceleration, this is fine by me. This is the origin of our disagreement, a naming convention. My math is correct.

Here is the derivative, as evaluated by the computer algebra system Maxima. It does not equal a.

Sorry, this was an obvious typo, it is clear from post #2 that

F/m=d/dt(v/sqrt(1-(v/c)^2)

so,

a=d/dt(v/sqrt(1-(v/c)^2) (exactly as in the wiki page)

Either we're miscommunicating, or you need to review your calculus.

You are getting rude. Good bye.

Last edited: