The Starship Engineer's Notebook: Universal Speeding: Using The Four-Force

There are two important, well-known rules about interstellar travel:

Nothing can travel faster than light.
If you want to get anywhere interesting in a reasonable amount of time then you need to go faster than light.

This leads to three options:

You just can't go anywhere.
You can go someplace nearby, but you have to stay on your ship for a very long time.
You must break rule one.

Option 1 is boring, so we'll ignore it. Option 2 is something we're already looking at. For now, let's talk about Option 3: If the speed of light is the universal speed limit then let's start speeding.

Before we can exceed the speed limit we need to know what the speed limit is, and why it's the limit. So why can't anyone go faster than the speed of light? Let's begin with some very basic concepts in Special Relativity, because we'll need them later. We've already shown how the Lorentz Factor γ can be derived from Galilean Relativity and Lorentz Invariance. While this approach is useful and instructive for discussing ideas like time dilation and length contraction, it only tells us that going faster than light should be strange, not why we simply can't get there.

Four-Vectors and Boosts

Define a four-vector describing an event, a specific point in space and a specific time:

$\mathbf{v}=\begin{pmatrix}v_0\\v_1\\v_2\\v_3\end{pmatrix}=\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}$

And we understand that relative velocity causes observers' perception of time to differ.

Lorentz boost x direction standard configuration.svg

By Maschen - Own work, CC0, Link

To describe the change in their perception of the same moment in both frames in four dimensions, we'll introduce the matrix form of a Lorentz Boost:

$x'=\begin{pmatrix}ct'\\x'\\y'\\z'\end{pmatrix}=\Lambda x=\Lambda\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}$

Where a boost in the +x direction looks like:

$\Lambda=\begin{bmatrix}\gamma&-\gamma\frac{v}{c}&0&0\\-\gamma\frac{v}{c}&\gamma&0&0\\0&0&1&0\\0&0&0&1\\\end{bmatrix}$

Proper Time

We want to find a quantity that remains unchanged during a boost, so that both frames can have something that remains consistent between them. Traditional 3 dimensional vector lengths are unchanged by a change in coordinate system. Vector lengths can be found from their components by the Pythagorean Theorem. Let's take a hint from Pythagoras and find the square of both t' and x' :

$\begin{align}(ct')^2=\gamma^2(ct-x)&=\gamma^2(c^2t^2-2vxt+\frac{v^2}{c^2}x^2)\\(x')^2=\gamma^2(x-vt)&=\gamma^2(x^2-2vxt+v^2t^2)\\\rightarrow(ct')^2-(x')^2&=\gamma^2(c^2t^2-x^2)-\gamma^2v^2(\frac{x^2}{c^2}-t^2)\\&=(ct)^2-x^2\end{align}$

The minus sign is not quite Pythagorean, but we now have our unchanged quantity. Expanding in all four dimensions:

$\begin{align}{x'}_0^2-{x'}_2^2-{x'}_2^2-{x'}_3^2&={x}_0^2-{x}_2^2-{x}_2^2-{x}_3^2\\{x'}_0^2-\vec{x'}^2&=x_0^2-\vec{x}^2\end{align}$

Where we've used the vector arrow to represent the 3d, spatial, portion of the four-vector.

From the point of the moving observer, F', it is everything else that appears to be moving and x' = 0, only time passes even though the stationary observer, F, sees both distance covered and time passed, x and t :

$\tau^2=(t')^2=t^2+(\frac{x}{c})^2$

This suggests that the constant quantity should be time passed within the moving frame, τ, termed the proper time. If observer F' is moving very close to the speed of light, while she and observer F will not agree on how much time has passed, both can agree on the proper time. For observer F, then, the proper time that has passed when observer F' has traveled over some distance:

$\Delta\tau=\int\sqrt{{dt^2}-\frac{d{\vec{x}}^2}{c^2}}=\int\sqrt{1-\frac{\vec{v}^2(t)}{c^2}}dt=\int\frac{dt}{\gamma(t)}$

Where we've used the fact that the velocity of F' as measured by F is given by v = dx/dt. If velocity is constant then:

$\Delta\tau=\frac{\Delta t}{\gamma}$

Mechanics in Four Dimensions

Four-Velocity

Let's consider again the case where one of the two observers throws a ball. What happens if the other observer tries to measure its velocity? Observer F' throws a ball forwards and measures its velocity as u' and Observer F measures a velocity u, how can we relate the two?

Let's start with u', observer F' sees the ball move some distance dx' in some amount of time dt' which is entirely straightforward. Remember, in her frame of reference nothing strange is happening and ordinary Newtonian mechanics apply. The four-vector form of velocity in the F' frame is:

$u'=\frac{dx'}{dt'}=\frac{1}{d\tau}\begin{pmatrix}cdt'\\dx'\\dy'\\dz'\end{pmatrix}=\begin{pmatrix}c\\u_x'\\u_y'\\u_z'\end{pmatrix}$

The only difference is that extra factor of c, which doesn't seem to have any purpose just yet. That will change when we try to move from the F' frame to the F frame. Observer F will observe the ball move some different distance and time, dx and dt, and which relate to dx' and dt' as we found before:

$\begin{align}dx&=\gamma(dx'+v dt')\\dt&=\gamma(dt'+\frac{v}{c^2}dx')\end{align}$

Take the derivative of both with respect to proper time:

$\rightarrow\begin{cases}\frac{dx}{d\tau}&=\gamma(u'+v)\\\frac{dt}{d\tau}&=\gamma(1+\frac{v}{c^2}u')\end{cases}$

Rearranging gives:

$\vec{u}=\frac{d\vec{x}}{dt}=\frac{d\vec{x}}{d\tau}\frac{d\tau}{dt}=\frac{\vec{u'}+\vec{v}}{1+\frac{|v||\vec{u}|}{c^2}}$

That middle term, where we have taken the derivative of x with respect to proper time represents the spatial component of the four-velocity. The four velocity measured by observer F is:

$u=\frac{1}{d\tau}\begin{pmatrix}cdt\\dx\\dy\\dz\end{pmatrix}=\frac{dt}{d\tau}\begin{pmatrix}c\\\frac{dx}{dt}\\\frac{dy}{dt}\\\frac{dz}{dt}\end{pmatrix}=\gamma\begin{pmatrix}c\\u_x\\u_y\\u_z\end{pmatrix}$

$\gamma=\frac{1}{\sqrt{1-\frac{|\vec{u}|^2}{c^2}}}$

Unlike the normal 3d velocity, four-velocity can be modified according to the Lorentz Boost matrix, u' = Λu. Though both observers will record different values for the velocity of the ball, they can relate their respective four-velocities through the boost matrix.

This is what we look for in 4-vectors, that we can move from one frame to another in a consistent manner. Observers F and F' can both describe the velocity of an object, or even their own relative velocity, and translate between their respective frames of reference using the Boost matrix.

Four Momentum

To find the momentum in a Newtonian world we multiply velocity by mass and, when considering a relativistic four-dimensional world, that is unchanged. We just need to incorporate the four velocity. Using a vector sign to represent the spatial, three-vector, components:

$p=mv=m\gamma\binom{c}{\vec{v}}=\gamma\binom{mc}{\vec{p}}$

The squared magnitude of the four-momentum is a constant quantity:

$|p|^2=\gamma^2(m^2c^2-m^2 v^2)=m^2c^2$

In fact, that looks pretty familiar. The first component of the four momentum can be rewritten in terms of the energy of the object being measured. Total energy can be related to rest energy, E0, by γ:

$\begin{align}p&=\binom{\frac{E}{c}}{\vec{p}}\\E&=\gamma E_0=\gamma mc^2\end{align}$

It's common to refer to the term γm as the relativistic mass, or just mass, and to specify the mass as measured in the frame of the object as the rest mass, m0. They are related by: m = γm0.

Four-Force

Since force is the rate of change of momentum:

$F=\frac{dp}{d\tau}=\frac{d}{d\tau}\binom{\frac{E}{c}}{\vec{p}}=\binom{\frac{1}{c}\frac{dE}{d\tau}}{\frac{d\vec{\gamma p}}{d\tau}}$

The 3-vector portion is conventional Newtonian force. The time component reflects the rate change in energy, or power:

$\frac{dE}{d\tau}=\vec{F}\cdot\vec{v}$

Keep in mind that the Lorentz Factor depends on velocity, and is generally not constant when force is applied:

$F=\gamma(\beta)\binom{\vec{F}\cdot\vec{\beta}}{\vec{F}}$

Imagine a starship trying to accelerate to near light-speed as observed by some fixed ground station. For simplicity, assume that its mass remains constant and that we want it to continue accelerating at some fixed rate.

The force that must be applied to achieve constant acceleration can be expressed, with β=v/c :

$F=ma=mc\frac{d}{d\tau}\binom{\gamma}{\gamma\beta}=mc\frac{d}{d\tau}\frac{1}{\sqrt{1-\beta^2}}\binom{1}{\beta}$

The closer the ship gets to the speed of light, the more force is required to accelerate it further. At the extreme end, the force required becomes infinite. Keeping in mind our assumption of constant acceleration means that β ∝ τ we can take the derivative and the limit:

$\lim_{\beta\rightarrow 1}\frac{d}{d\tau}\frac{\beta}{\sqrt{1-\beta^2}}=\lim_{\beta\rightarrow 1}\frac{\beta}{(1-\beta^2)^{3/2}}=\infty$

Since no engine exists that can provide infinite force, we cannot accelerate to the speed of light through any conventional means. The energy required also trends to infinity, since the total energy of a massive object heading towards the speed of light approaches infinity!

In the more natural case where force is constant, it would take infinite time to accelerate any positive mass object up to the speed of light. For a constant force in the +x direction we can find that:

$\frac{d\beta(\tau)}{d\tau}=\frac{F}{mc}\left(1-\beta^2(\tau)\right)^2$

Which has a solution at:

$t=\frac{mc}{4F}\left(\frac{-2\beta(t)}{\beta^2(t)-1}-\ln(1-\beta(t))+\ln(1+\beta(t))\right)$

Time trends towards infinity as β approaches 1 from the left.

$\lim_{\beta\rightarrow 1_-}\left(\frac{-2\beta}{\beta^2-1}-\ln(1-\beta)+\ln(1+\beta)\right)=\infty$

It's not looking good for us.

Massless Particle Momentum

On the other hand, for a photon, or any massless particle (m0 = 0), which has some energy E we can find the three dimensional momentum from:

$m^2c^2=(\gamma m_0)^2 c^2=\left(\frac{E}{c}\right)^2-(\vec{p})^2=0$

This also demonstrates why photons cannot travel at anything less than c, if it could then there could be some observer frame where p = 0 which would require that the photon carries no energy. But if it has no energy and no mass, it cannot react with anything. By conservation of energy and mass:

$\binom{E/c}{\vec{p}}+\binom{0}{0}=\binom{E/c}{\vec{p}}$

Any object our oddly lethargic photon interacted with would be completely unchanged; there would be no difference between a massless particle traveling below the speed of light and no particle at all. Such a particle would functionally not exist.

If we rearrange the definition of momentum using energy instead of mass:

$p=\frac{E}{c^2}v=\frac{|p|}{c}v\rightarrow|v|=c$

Again, the only available velocity for a massless particle is c, it must travel at that speed and no other from the moment it is emitted to the moment it is absorbed. The only other solution is zero energy, no particle at all.

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March 5, 2025

Universal Speeding: Using The Four-Force