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- TL;DR Summary
- Does dv/dx=-Fv/mc^2(1-v^2/c^2)^2 describe it adequately?

I am trying to push the boundaries of special relativity with a self-imposed challenge problem. A common derivation of relativistic kinetic energy involves an object to which a constant force is applied. I want to consider a similar scenario, but instead of a point object we now have a uniform rod accelerating along its axis.

Let's say there is a constant force [itex]F[/itex] applied to the center of the rod. Its momentum is [itex]p=Ft[/itex], from which we get its velocity:

[tex]\frac{v_\mathrm{CM}}{c}=\frac{Ft/mc}{\sqrt{1+(Ft/mc)^2}}.[/tex]

The velocity of the CM of the rod increases, asymptotically approaching [itex]c[/itex]. That means its length contracts more and more, so the rod gets shorter and shorter. But that means the front of the rod must be traveling more slowly than the center, and the back end more quickly, because of the time-dependent Lorentz factor. This, in turn, means that the Lorentz factor at the back of the rod is greater than the Lorentz factor at the front of the rod.

This observation is not unique to the endpoints of the rod; it is true for every distinct pair of points along the rod. We thus reformulate the above statement as follows: The Lorentz factor decreases monotonically as we move along the rod from back to front. In particular, it is nonuniform: [itex]\gamma=\gamma(x)[/itex]. (There is also, obviously, a time-dependence, but we'll get to that later.)

The continuous variation of [itex]\gamma[/itex] with position suggests a differential equations approach. Knowing [itex]v_\mathrm{CM}(t)[/itex], I obtain

[tex]\gamma_\mathrm{CM}(t)=\sqrt{1+\left(\frac{Ft}{mc}\right)^2}.[/tex]

In particular, [itex]\gamma_\mathrm{CM}(t)[/itex] satisfies the differential equation

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=-\frac{(Ft/mc)(F/mc)}{[1+(Ft/mc)^2]^\frac{3}{2}}=-\frac{Fv}{mc^2}\frac{1}{\gamma^2}.[/tex]

A segment of rod of length [itex]\mathrm{d}x[/itex] will Lorentz-contract to a length [itex]\mathrm{d}x'=\mathrm{d}x/\gamma[/itex]. Therefore, the velocity of a point located at position [itex]x+\mathrm{d}x[/itex] relative to a point at position [itex]x[/itex] is

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\mathrm{d}x}{\gamma}\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]\,\mathrm{d}x=-\frac{Fv}{mc^2}\frac{\mathrm{d}x}{\gamma^2}=-\frac{Fv}{mc^2}\left(1-\frac{v^2}{c^2}\right)\,\mathrm{d}x.[/tex]

Let us call this velocity [itex]\mathrm{d}u[/itex] to reflect its differential nature. We add velocities relativistically to get the velocity [itex]v(x+\mathrm{d}x)[/itex] of the point at position [itex]x+\mathrm{d}x[/itex] in terms of the velocity [itex]v(x)[itex] of the point at position [itex]x[/itex] and the differential relative velocity [itex]\mathrm{d}u[/itex]:

[tex]v(x+\mathrm{d}x)=\frac{v(x)+\mathrm{d}u}{1+v(x)\,\mathrm{d}u/c^2}.[/tex]

Since [itex]\mathrm{d}u[/itex] is differential and [itex]v(x)<c[/itex], we can Taylor expand and throw away all terms of order [itex]\mathrm{d}u^2=0[/itex]:

[tex]v(x+\mathrm{d}x)=(v(x)+\mathrm{d}u)\left(1-\frac{v(x)\,\mathrm{d}u}{c^2}\right)=v(x)+\mathrm{d}u-\frac{v(x)^2}{c^2}\,\mathrm{d}u[/tex]

[tex]\Rightarrow v(x+\mathrm{d}x)-v(x)=\left(1-\frac{v(x)^2}{c^2}\right)\,\mathrm{d}u

=-\left(1-\frac{v(x)^2}{c^2}\right)\frac{Fv(x)}{mc^2}\left(1-\frac{v(x)^2}{c^2}\right)\mathrm{d}x.[/tex]

Therefore

[tex]\frac{v(x+\mathrm{d}x)-v(x)}{\mathrm{d}x}\equiv\frac{\mathrm{d}v}{\mathrm{d}x}=-\frac{Fv}{mc^2}\left(1-\frac{v^2}{c^2}\right)^2.[/tex]

This is a differential equation, and to determine the unique solution we specify the initial condition that [itex]v(0)=v_\mathrm{CM}[/itex]. Note that the slope field of this expression is as we would expect.

Does this approach look good? Qualitatively, the numerical solutions to this ODE (and the analytic inverse function [itex]x(v)[/itex]) behave exactly as they should; as [itex]v[/itex] increases, [itex]x[/itex] decreases until [itex]\lim_{v\rightarrow c}x(v)=-\infty[/itex]. Thus, although points further back along the rod move more quickly, they never reach [itex]c[/itex]. Similarly, [itex]\lim_{v\rightarrow 0}x(v)=+\infty[/itex], meaning that although points further forward along the rod move more slowly, the velocity never reaches [itex]0[/itex]. If we integrate the Lorentz-contracted length from the point where [itex]v=v_\mathrm{CM}[/itex] to the ``point'' (infinitely far away along the uncontracted rod) where [itex]v=c[/itex], we do indeed obtain an infinite length -- again, this is good, as any finite length would mean the ``end'' of the massive rod would be traveling superluminal velocity.

I want to check this solution, not only mathematically but also physically. Obviously [itex]x(v)[/itex] can be converted to [itex]x(1/\gamma)[/itex] by substituting [itex]\frac{v}{c}=\sqrt{1-(\frac{1}{\gamma})^2}[/itex]. Now, I cannot get the total contracted length of the rod since I don't know how fast its endpoints are moving. However, given that some point has velocity [itex]v[/itex], I can find its [itex]x[/itex]-coordinate (and, thus, whether or not there is a point on my rod with this velocity, using the condition that [itex]|x|<\frac{L}{2}[/itex]). I am confident in my reasoning everywhere except for one step:

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=-\frac{Fv}{mc^2}\frac{1}{\gamma^2}.[/tex]

I know that this is true for [itex]\gamma_\mathrm{CM}[/itex]. But is it logical to assert that it is true for all positions along the rod? Does this assertion, in conjunction with the function [itex]\gamma(x)[/itex], imply that the ratio of [itex]\gamma[/itex] at time [itex]t[/itex] to [itex]\gamma[/itex] at time [itex]0[/itex] must be this same function that was found for [itex]\gamma_\mathrm{CM}[/itex] early on? Because that can't be true -- [itex]\gamma(x)\equiv 1[/itex] for all [itex]x[/itex] at time [itex]t=0[/itex].

But then how should I find the velocity profile for the uniform rod? It looks like the space- and time-dependence of the rod are coupled, so I can't treat them as obeying separate ordinary differential equations. Should I treat [itex]\gamma[/itex] as a function of two variables, [itex]\gamma(x,t)[/itex], and derive a partial differential equation instead? My gut says the derivation would go something like this:

[tex]\Rightarrow v(x+\mathrm{d}x)-v(x)=\left(1-\frac{v(x)^2}{c^2}\right)\,\mathrm{d}u[/tex]

where

[tex]\mathrm{d}u=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]\,\mathrm{d}x[/tex]

which gives

[tex]\frac{\mathrm{d}v}{\mathrm{d}x}=\left(1-\frac{v^2}{c^2}\right)\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right].[/tex]

But

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left[\sqrt{1-\frac{v^2}{c^2}}\right]

=-\frac{v}{\sqrt{1-v^2/c^2}}\frac{\mathrm{d}v}{\mathrm{d}t},[/tex]

so

[tex]\frac{\mathrm{d}v}{\mathrm{d}x}=-\left(1-\frac{v^2}{c^2}\right)\frac{v}{\sqrt{1-v^2/c^2}}\frac{\mathrm{d}v}{\mathrm{d}t},[/tex]

i.e.,

[tex]\frac{\partial v}{\partial x}=-v\sqrt{1-\frac{v^2}{c^2}}\frac{\partial v}{\partial t}.[/tex]

This looks like a demented version of the advection equation. I'm not even totally sure what boundary conditions apply here; for sure, [itex]v(x=0,t)=(Ft/m)/\sqrt{1+(Ft/mc)^2}[/itex] and [itex]v(x,t=0)=0[/itex] must be true, but I doubt that will be enough to solve this PDE.

What do you guys think? Was my initial approach right? Did I underestimate this problem, and now have to solve a PDE? Or did I overthink it, and actually the solution is simple?

I acknowledge that special relativity is only really supposed to be about inertial reference frames and so what I'm doing here is a bit of a stretch. Professors I talk to agree that this is a teaser for general relativity. I'm prepared to bring mathematical machinery to bear here. I really just want to know two things: First, can anything be said about the velocity as a function of position (and time) for a rod with constant force applied to its CM? And second, what approach will get me there in a way that is consistent with relativistic kinematics? From the qualitative argument I laid out in the beginning, I simply do not buy that an accelerated uniform rod Lorentz-contracts uniformly.

This is not a homework-style problem, though like my previous questions it is rather phrased like one. Your feedback is appreciated.

Cheers,

QM

Let's say there is a constant force [itex]F[/itex] applied to the center of the rod. Its momentum is [itex]p=Ft[/itex], from which we get its velocity:

[tex]\frac{v_\mathrm{CM}}{c}=\frac{Ft/mc}{\sqrt{1+(Ft/mc)^2}}.[/tex]

The velocity of the CM of the rod increases, asymptotically approaching [itex]c[/itex]. That means its length contracts more and more, so the rod gets shorter and shorter. But that means the front of the rod must be traveling more slowly than the center, and the back end more quickly, because of the time-dependent Lorentz factor. This, in turn, means that the Lorentz factor at the back of the rod is greater than the Lorentz factor at the front of the rod.

This observation is not unique to the endpoints of the rod; it is true for every distinct pair of points along the rod. We thus reformulate the above statement as follows: The Lorentz factor decreases monotonically as we move along the rod from back to front. In particular, it is nonuniform: [itex]\gamma=\gamma(x)[/itex]. (There is also, obviously, a time-dependence, but we'll get to that later.)

The continuous variation of [itex]\gamma[/itex] with position suggests a differential equations approach. Knowing [itex]v_\mathrm{CM}(t)[/itex], I obtain

[tex]\gamma_\mathrm{CM}(t)=\sqrt{1+\left(\frac{Ft}{mc}\right)^2}.[/tex]

In particular, [itex]\gamma_\mathrm{CM}(t)[/itex] satisfies the differential equation

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=-\frac{(Ft/mc)(F/mc)}{[1+(Ft/mc)^2]^\frac{3}{2}}=-\frac{Fv}{mc^2}\frac{1}{\gamma^2}.[/tex]

A segment of rod of length [itex]\mathrm{d}x[/itex] will Lorentz-contract to a length [itex]\mathrm{d}x'=\mathrm{d}x/\gamma[/itex]. Therefore, the velocity of a point located at position [itex]x+\mathrm{d}x[/itex] relative to a point at position [itex]x[/itex] is

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\mathrm{d}x}{\gamma}\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]\,\mathrm{d}x=-\frac{Fv}{mc^2}\frac{\mathrm{d}x}{\gamma^2}=-\frac{Fv}{mc^2}\left(1-\frac{v^2}{c^2}\right)\,\mathrm{d}x.[/tex]

Let us call this velocity [itex]\mathrm{d}u[/itex] to reflect its differential nature. We add velocities relativistically to get the velocity [itex]v(x+\mathrm{d}x)[/itex] of the point at position [itex]x+\mathrm{d}x[/itex] in terms of the velocity [itex]v(x)[itex] of the point at position [itex]x[/itex] and the differential relative velocity [itex]\mathrm{d}u[/itex]:

[tex]v(x+\mathrm{d}x)=\frac{v(x)+\mathrm{d}u}{1+v(x)\,\mathrm{d}u/c^2}.[/tex]

Since [itex]\mathrm{d}u[/itex] is differential and [itex]v(x)<c[/itex], we can Taylor expand and throw away all terms of order [itex]\mathrm{d}u^2=0[/itex]:

[tex]v(x+\mathrm{d}x)=(v(x)+\mathrm{d}u)\left(1-\frac{v(x)\,\mathrm{d}u}{c^2}\right)=v(x)+\mathrm{d}u-\frac{v(x)^2}{c^2}\,\mathrm{d}u[/tex]

[tex]\Rightarrow v(x+\mathrm{d}x)-v(x)=\left(1-\frac{v(x)^2}{c^2}\right)\,\mathrm{d}u

=-\left(1-\frac{v(x)^2}{c^2}\right)\frac{Fv(x)}{mc^2}\left(1-\frac{v(x)^2}{c^2}\right)\mathrm{d}x.[/tex]

Therefore

[tex]\frac{v(x+\mathrm{d}x)-v(x)}{\mathrm{d}x}\equiv\frac{\mathrm{d}v}{\mathrm{d}x}=-\frac{Fv}{mc^2}\left(1-\frac{v^2}{c^2}\right)^2.[/tex]

This is a differential equation, and to determine the unique solution we specify the initial condition that [itex]v(0)=v_\mathrm{CM}[/itex]. Note that the slope field of this expression is as we would expect.

Does this approach look good? Qualitatively, the numerical solutions to this ODE (and the analytic inverse function [itex]x(v)[/itex]) behave exactly as they should; as [itex]v[/itex] increases, [itex]x[/itex] decreases until [itex]\lim_{v\rightarrow c}x(v)=-\infty[/itex]. Thus, although points further back along the rod move more quickly, they never reach [itex]c[/itex]. Similarly, [itex]\lim_{v\rightarrow 0}x(v)=+\infty[/itex], meaning that although points further forward along the rod move more slowly, the velocity never reaches [itex]0[/itex]. If we integrate the Lorentz-contracted length from the point where [itex]v=v_\mathrm{CM}[/itex] to the ``point'' (infinitely far away along the uncontracted rod) where [itex]v=c[/itex], we do indeed obtain an infinite length -- again, this is good, as any finite length would mean the ``end'' of the massive rod would be traveling superluminal velocity.

I want to check this solution, not only mathematically but also physically. Obviously [itex]x(v)[/itex] can be converted to [itex]x(1/\gamma)[/itex] by substituting [itex]\frac{v}{c}=\sqrt{1-(\frac{1}{\gamma})^2}[/itex]. Now, I cannot get the total contracted length of the rod since I don't know how fast its endpoints are moving. However, given that some point has velocity [itex]v[/itex], I can find its [itex]x[/itex]-coordinate (and, thus, whether or not there is a point on my rod with this velocity, using the condition that [itex]|x|<\frac{L}{2}[/itex]). I am confident in my reasoning everywhere except for one step:

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=-\frac{Fv}{mc^2}\frac{1}{\gamma^2}.[/tex]

I know that this is true for [itex]\gamma_\mathrm{CM}[/itex]. But is it logical to assert that it is true for all positions along the rod? Does this assertion, in conjunction with the function [itex]\gamma(x)[/itex], imply that the ratio of [itex]\gamma[/itex] at time [itex]t[/itex] to [itex]\gamma[/itex] at time [itex]0[/itex] must be this same function that was found for [itex]\gamma_\mathrm{CM}[/itex] early on? Because that can't be true -- [itex]\gamma(x)\equiv 1[/itex] for all [itex]x[/itex] at time [itex]t=0[/itex].

But then how should I find the velocity profile for the uniform rod? It looks like the space- and time-dependence of the rod are coupled, so I can't treat them as obeying separate ordinary differential equations. Should I treat [itex]\gamma[/itex] as a function of two variables, [itex]\gamma(x,t)[/itex], and derive a partial differential equation instead? My gut says the derivation would go something like this:

[tex]\Rightarrow v(x+\mathrm{d}x)-v(x)=\left(1-\frac{v(x)^2}{c^2}\right)\,\mathrm{d}u[/tex]

where

[tex]\mathrm{d}u=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]\,\mathrm{d}x[/tex]

which gives

[tex]\frac{\mathrm{d}v}{\mathrm{d}x}=\left(1-\frac{v^2}{c^2}\right)\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right].[/tex]

But

[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left[\sqrt{1-\frac{v^2}{c^2}}\right]

=-\frac{v}{\sqrt{1-v^2/c^2}}\frac{\mathrm{d}v}{\mathrm{d}t},[/tex]

so

[tex]\frac{\mathrm{d}v}{\mathrm{d}x}=-\left(1-\frac{v^2}{c^2}\right)\frac{v}{\sqrt{1-v^2/c^2}}\frac{\mathrm{d}v}{\mathrm{d}t},[/tex]

i.e.,

[tex]\frac{\partial v}{\partial x}=-v\sqrt{1-\frac{v^2}{c^2}}\frac{\partial v}{\partial t}.[/tex]

This looks like a demented version of the advection equation. I'm not even totally sure what boundary conditions apply here; for sure, [itex]v(x=0,t)=(Ft/m)/\sqrt{1+(Ft/mc)^2}[/itex] and [itex]v(x,t=0)=0[/itex] must be true, but I doubt that will be enough to solve this PDE.

What do you guys think? Was my initial approach right? Did I underestimate this problem, and now have to solve a PDE? Or did I overthink it, and actually the solution is simple?

I acknowledge that special relativity is only really supposed to be about inertial reference frames and so what I'm doing here is a bit of a stretch. Professors I talk to agree that this is a teaser for general relativity. I'm prepared to bring mathematical machinery to bear here. I really just want to know two things: First, can anything be said about the velocity as a function of position (and time) for a rod with constant force applied to its CM? And second, what approach will get me there in a way that is consistent with relativistic kinematics? From the qualitative argument I laid out in the beginning, I simply do not buy that an accelerated uniform rod Lorentz-contracts uniformly.

This is not a homework-style problem, though like my previous questions it is rather phrased like one. Your feedback is appreciated.

Cheers,

QM