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Finding v(t) at high speeds with F constant

by James Brady
Tags: relativity special
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James Brady
#1
Jan1-13, 05:20 AM
P: 31
I'm trying to find out what would happen to a mass if it had a constant force applied to it for a very long time. It eventually approaches the speed of light, I want to plot it's velocity with respect to time as it gets harder and harder to push. Here's what I came up with.

From Newton's 2nd Law: dv/dt = F/M.
Making this relativistic: dv/dt = [itex]\frac{F}{M}[/itex] *[itex]\sqrt{1 - v^2/c^2}[/itex]

Setting up the differential equation:

[itex]\frac{dv}{\sqrt{1 - v^2/c^2}}[/itex] = [itex]\frac{F}{M}[/itex]*dt

I multiply both sides of the equation by 1/c to make it integrateable.

[itex]\frac{dv}{C * \sqrt{1 - v^2/c^2}}[/itex] = [itex]\frac{F}{C * M}[/itex]*dt

And solving for the integral of both sides I get...

sin^-1(v/c) = [itex]\frac{F * t}{M*C}[/itex]

v(t) = sin([itex]\frac{F * t}{M*C}[/itex]) * C

I'm pretty happy with this answer , the units check out alright, and it looks good. But once you get close to the speed of light, the velocity turns around and starts going back down because sine is one of those cyclic functions. I know of course you will not all of a sudden reverse acceleration once you near the speed of light though, so I know this isn't right.

Any help would much be appreciated, I will give you an e-high-five. Or an e-hug if you're into that sort of thing. Seriously, I'm that lonely.
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Mentz114
#2
Jan1-13, 05:38 AM
PF Gold
P: 4,087
I have to tell you that your result is not right because the sin should be a sinh. The hyperbolic function approaches an asymptote so the velocity gets closer and closer to c without reaching it in any coordinates. Have a look at this

http://gregegan.customer.netspace.ne...erHorizon.html



Happy new year, but no hugs, please !
Bill_K
#3
Jan1-13, 03:58 PM
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I'm trying to find out what would happen to a mass if it had a constant force applied to it for a very long time.
Depends first of all what you mean by constant force. Usually it is taken to mean constant in the rest frame of the particle. This would be true, for example, of a rocket ship with an onboard engine exerting a constant thrust. In that case, as Mentz114 says, the position and velocity will be hyperbolic functions.

But it sounds like what you mean instead is a constant external force, i.e. constant in the original rest frame, and then you do get something different.

Newton's Law is F = dp/dt. (Yes, that's coordinate t.) In special relativity, p = γmv, so the equation of motion is F dt = d(γmv) where F is constant. This is simply integrated to give Ft = γmv, which can then be solved algebraically for v: v2/c2 = (Ft/m)2/(1 + (Ft/m)2). Note that as t → ∞, v approaches c.

andrien
#4
Jan2-13, 01:21 AM
P: 1,020
Finding v(t) at high speeds with F constant

check out,your newton's second law is wrong.You should proceed from the definition
F=dp/dt.
Bill_K
#5
Jan2-13, 06:32 PM
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Yes, F=dp/dt. So F dt = dp = d(γmv), and since F is constant, Ft = γmv. What's wrong?
James Brady
#6
Jan3-13, 12:29 AM
P: 31
Excellent, I started with the correct solved force differential equation: Ft = γmv, but I couldn't get to where you got with algebra, I kept on getting v^2/C^2 = 0...

Ft = [itex]\frac{mv}{\sqrt{1 - v^2/c^2}}[/itex] I figured out that v = Ft/m so I got

(Ft/m) = [itex]\frac{(Ft/m)}{\sqrt{1 - v^2/c^2}}[/itex]

(Ft/m)2 = [itex]\frac{(Ft/m)^2}{1 - v^2/c^2}[/itex]

(Ft/m)2[1 - v2/c2] = (Ft/m)2

1- v2/c2 = [itex]\frac{(Ft/m)^2}{(Ft/m)^2}[/itex]
-v2/c2 = 1 - 1 = 0
James Brady
#7
Jan3-13, 12:31 AM
P: 31
Mentz114, here's a non-threatening high five *high-five*
andrien
#8
Jan3-13, 05:37 AM
P: 1,020
Quote Quote by Bill_K View Post
Yes, F=dp/dt. So F dt = dp = d(γmv), and since F is constant, Ft = γmv. What's wrong?
well,I was not saying to you but to OP.i.e.
dv/dt=F(√(1-v2/c2)/m0 is wrong.Proceed with F=dp/dt,it will give something different.
andrien
#9
Jan3-13, 05:42 AM
P: 1,020
Quote Quote by James Brady View Post
Excellent, I started with the correct solved force differential equation: Ft = γmv, but I couldn't get to where you got with algebra, I kept on getting v^2/C^2 = 0...

Ft = [itex]\frac{mv}{\sqrt{1 - v^2/c^2}}[/itex]
just solve for v in it.
vanhees71
#10
Jan3-13, 08:18 AM
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In relativistic dynamics you have to be careful not to spoil relativistic covariance. The closest thing to "constant acceleration" is the motion of a charged particle in a homogeneous electric field. The equation of motion in covariant form reads (setting [itex]c=1[/itex]):
[tex]m \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=q F^{\mu \nu} u_{\nu}.[/tex]
Since [itex]F^{\mu \nu}=-F^{\nu \mu}[/itex]=0[/itex] this equation of motion is consistent with the constraint [itex]u_{\mu} u^{\mu}=1=\text{const}[/itex], because
[tex]\dot{u}^{\mu} u_{\mu}=\frac{q}{m} F^{\mu \nu} u_{\nu} u_{\mu}=0.[/tex]
For a constant electric field in [itex]x[/itex]-direction the equations of motion read
[tex]m \dot{u}^0=q E u^1, \quad m \dot{u}^1=q E u^0.[/tex]
The easiest way to solve this is to add the two equations, leading to
[tex]\frac{\mathrm{d}}{\mathrm{d} \tau}(u^0+u^1)=\frac{q E}{m} (u^0+u^1).[/tex]
This gives
[tex]u^0+u^1=C \exp \left(\frac{q E}{m} \tau \right).[/tex]
Assuming that we start with [itex]\vec{u}=0[/itex], i.e., [itex]u^0(0)=1, \quad u^1(0)=0[/itex] we have
[tex]u^0+u^1=\exp \left (\frac{q E}{m} \tau \right ).[/tex]
From the constraint we have
[tex]1=(u^0)^2-(u^1)^2=(u^0+u^1)(u^0-u^1)=\exp \left (\frac{q E}{m} \tau \right ) (u^0-u^1).[/tex]
This gives
[tex]u^0-u^1=\exp \left (-\frac{q E}{m} \tau \right ).[/tex]
Thus we find
[tex]u^0=\cosh\left (\frac{q E}{m} \tau \right ), \quad u^1=\sinh \left (\frac{q E}{m} \tau \right ).[/tex]
The three-velocity is
[tex]v=\frac{u^1}{u^0}=\tanh \left (\frac{q E}{m} \tau \right ) \in (-1,1).[/tex]
Assuming further that [itex]x^0(0)=t(0)=0, \quad \vec{x}(0)=0[/itex], we can integrate this once more to get
[tex]t=\frac{m}{q E} \sinh \left (\frac{q E}{m} \tau \right ), \quad x=\frac{m}{q E} \left [\cosh \left (\frac{q E}{m} \tau \right )-1 \right].[/tex]
We get the trajectory in terms of the coordinate times simply by using [itex]\cosh^2 \alpha-\sinh^2 \alpha=1[/itex]:
[tex]x=\frac{m}{q E} \left [\sqrt{\left (\frac{q E}{m} t \right)^2+1}-1 \right].[/tex]
For early times, where
[tex]\left (\frac{q E}{m} t \right)^2 \ll 1[/tex]
we find
[tex]x \approx \frac{q E}{2m} t^2,[/tex]
i.e., the non-relativistic limit, corresponding to the motion in the constant electric field as it should be.
Bill_K
#11
Jan3-13, 08:54 AM
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In relativistic dynamics you have to be careful not to spoil relativistic covariance. The closest thing to "constant acceleration" is the motion of a charged particle in a homogeneous electric field.
You also have to carefully read the question, and not assume it is one you have seen before. Neither relativistic covariance nor constant acceleration were specified in this case.
vanhees71
#12
Jan3-13, 09:59 AM
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Sorry that I misunderstood the question, but the original posting doesn't make sense. I've my quibbles with what you mean by "force" in the relativistic case. Usually one defines as "force" the non-covariant quantity [tex]\mathrm{d} \vec{p}/\mathrm{d} t[/tex]. Then I concluded from the equation of motion given in this posting that what was in the mind of the poster was the case of a constant electric field, leading to the constant "force" [itex]q E[/itex]. It's of course not constant acceleration in a strict sense but constant acceleration in the non-relativistic limit only. Translating "constant force" into "constant electric field" was the closest which came to my mind.

Of course you can as well start in the non-covariant formalism, but you still have to be carefull, whether "the force" makes sense. In this case I think I solved what was intended by the original poster since by chance he has guessed a valid force somehow. In the non-covariant formalism you have
[tex]m \frac{\mathrm{d}}{\mathrm{d} t} \frac{v}{\sqrt{1-v^2}}=q E,[/tex]
where I already specialized to one-dimensional motion.

This can be integrated immediately once. Using the initial condition [itex]v(0)=0[/itex] you obtain
[tex]\frac{v}{\sqrt{1-v^2}}=\frac{q E}{m} t.[/tex]
I don't understand, what the original poster has done in his calculation, but his result is not a solution of the equation of motion. Of course we can solve the above for [itex]v[/itex]. We only have to be a bit careful with the sign, but that's not a real problem, because obviously the sign of [itex]v[/itex] is given by the sign of [itex]q E[/itex], which is physically clear anyway.

So we just square the above equation, leading to
[tex]v^2=\left (\frac{q E}{m} t \right )^2 (1-v^2).[/tex]
This gives
[tex]v=\frac{q E t}{m \sqrt{1+\left (\frac{q E}{m} t \right )^2}} = \frac{q E t}{\sqrt{m^2+(q E t)^2}}.[/tex]
This is equivalent to my previous solution, as can be seen by taking the derivative of [itex]x(t)[/itex] wrt. the coordinate time.

Anyway, to answer the original question, may take the limit [itex]t \gg m/(q E)[/itex]. Then we get
[tex]v=\frac{q E t}{|q E t| \sqrt{1+\frac{m^2}{(q E t)^2}}} \approx \text{sign}(q E) \left [1-\frac{m^2}{2(q E t)^2} \right].[/tex]
This answers the original question, how the velocity behaves for large times or (equivalently) if the particle's speed comes close to [itex]1[/itex], i.e., the speed of light.
James Brady
#13
Jan4-13, 02:45 AM
P: 31
Oh you know what, that last entry was just an algebra fail. I figured out how to solve for v. I got it now, thanks everyone for the help.

Vanhees71, Yeah I never specified the type of force, but I was actually picturing in my mind some kind of gigantic space railroad which produced an electric field which would force the particle to accelerate. So I understood your thinking.


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