# Accelarating to speed of light in steps

I’m sure this will turn out to be a stupid questions, but…

So Special relativity tells us one can no accelerate to speed of light because their relative mass will keep increasing toward infinity, so the force/energy needed will also increase toward infinity. Mathematically it’s obvious from the Lorentz-Fitzgerald transformations.

Note that this is RELATIVE mass (relative to the observer) that increases. Their rest mass remains the same.

But nothing stops me form speeding up to, let’s say 20% of C. This will take X amount of energy. Now, once I do that the original observer is at 0.2c relative to me, but an observer who came along with me (or happened to already be at that velocity) is stationary compared to me. Nothing stops me from accelerating to 0.2C compared to them. And since I start with zero relative mass compared to them, it should again take me exactly X amount of energy to do it.

This if I repeat the process 3 more times I should be at C relative to the original 1s observer having spent only 5X amount of energy, not infinite.

Is the increase in speed relative to observer? Shouldn’t be. Speed is the one all observers agree on, so they should agree on delta speed.

What am I missing?

• Dec 14, 2016 at 7:30

You're essentially missing the relativistic velocity addition law. Velocities don't add linearly, rapidities do (in co-linear problems), because it is the rapidity parameter that multiplies a boost generator to get the logarithm of the relavant Lorentz transformation, not velocity. More intuitively: they don't add linearly, because you're adding together the same velocity increments (as you observe), but in different frames of reference, so that time dilations and all the rest of it have to be applied to the same interval as you go.

In a one dimensional problem, the relevant Lorentz transformation (see notes at end) is:

$$\Lambda(\eta) = \exp\left(\eta\,\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\right)\tag{1}$$

where $\eta$ is the boost's rapidity. Suppose you repeatedly impart the same transformation to yourself, which is what you're describing: there is the same impulse each time, and, as you say, the relevant inertia for your problem is your rest mass. Then after $n$ such identical moves, your overall transformation is $\Lambda(\eta)^n=\Lambda(n\,\eta)$. So, each boost adds a rapidity:

$$\eta = \operatorname{artanh}\left(\frac{v}{c}\right)\tag{2}$$

and your final velocity is thus:

$$v_n = c\,\tanh(n\,\eta) = c\,\tanh\left(n\,\operatorname{artanh}\left(\frac{v}{c}\right)\right)\tag{3}$$

So, in your example, if $v=0.2\,c$ is the relative velocity imparted by each boost, your velocity after five of them is $c\,\tanh\left(5\,\operatorname{artanh}\left(\frac{1}{5}\right)\right)\approx0.767\,c$

(1) makes obvious the claim that the combination of any finite number of finite boosts, no matter how big, is always equivalent to a boost of finite rapidity, thus (through (2)) less-than-$c$ velocity and thus the accelerated object will still have finite total and kinetic energy.

Exercise:

Work formulas (2) and (3) out from (1). The following steps are hints:

1. Expand the matrix exponential for the Lorentz transformation in (1);
2. Using this expression, read off the distance travelled by the origin of one co-ordinate system ($A$) relative to the other ($B$) over a unit time as measured by $B$ to find expressions for the Lorentz transformation in terms of relative velocity;
3. Derive (2) and (3) by comparing the two expressions you now have.

Actually something like Mathematica will crunch it for you: remember to use MatrixExp, not Exp, when you're doing a matrix exponential (Exp, like any like function such as Sin, Cos, ... simply calculate elementwise exponentials / relevant function for matrices. Something similar holds for Sage and Maple, too).

The form of the Lorentz transformation in (1) is justified by basic assumptions about the universe's symmetries and homogeneities (see references at the end). Continuity assumptions then show that the Lorentz transformations must form a matrix Lie group, so, in one spatial dimension, we must have a one parameter group and therefore there must be some rapidity parameter that adds linearly with combined transformations: one then simply has to discover this parameter's relationship with the distance over time velocity. The matrix in the one parameter Lie algebra (the matrix inside the exponential) is constrained to the form shown because it can't be a rotation (rotations can switch the sign of the time co-ordinate and thus violate causality) and it can't have on-diagonal elements just because they give unphysical results.

See:

Palash B. Pal, "Nothing but Relativity," Eur.J.Phys.24:315-319,2003

Jean-Marc Levy-Leblond, "One more derivation of the Lorentz transformation"

Suppose you are right, and you can actually accelerate to the speed of light in steps like your suggest. Let's say we take two steps from a velocity of 0 relative to your observer, and up to $c$. We would like to calculate the energy required to perform each step, that is, how much fuel did you have to burn to get there.

The relativistic kinetic energy is:

$E_{kin} = (\gamma - 1)m c^2$ where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ is the Lorentz factor.

At $v = 0$, we have $\gamma = 1$ and $E_{kin} = 0$. So far so good.

We now accelerate to $v = 0.5c$. Here $\gamma \approx 1.15$, and if your mass is, say 100 kg, the energy required to go from 0 to this velocity is obtained simply by plugging the numbers into the above equation. It amounts to $\approx 35$ GJ (GigaJoule).

Now we accelerate to $c$. We calculate the energy required by calculating the kinetic energy for a velocity of $v = c$, and subtract the 35 GJ. The factor $mc^2$ in the kinetic energy does not change from before. The Lorentz factor on the other hand does - what happens when you plug in $v =c$? It becomes infinite! So the energy required to go from $v = 0.5c$ to $v = c$ is: $E = \infty - 35$ GJ $=\infty$.

You can add as many small steps as you like. This will always be the result for the final step. And this is why it's impossible - even going from $0.9999999c$ to $c$ would require an infinite amount of energy, and you can't do it.

• yes, but you are still using the math from Lorentz transforms to demonstrate it. Which, as I said, was already obvious. I was looking for more of an alternate, hopefully intuitive, explanation. but let me review the 1st answer form Rod Vance. He may have it. Dec 15, 2016 at 0:35

actually, there's a very good answer here:

How do I derive the expression for velocity in $S_n$ frame that has a velocity $v$ with respect to $S_{n-1}$ frame?

but I'm still failing with the intuitive visualization of it.

There is a maximum speed in the Universe. It is $c$. This is an axiom; note that we are not referring to it as the speed of light and we did not make any statement whether there are or not objects travelling at that speed. It is a postulate, that is ultimately correct because the predictions that we do starting with that axiom match experiments.

Let us consider an object of (rest) mass $m$. Then, as you know, we can relate the velocity of an object with its energy. We get the relation $$E = \frac{m c^2}{\sqrt{1-v^2/c^2}}$$ this relation is valid only for $v$ strictly less than $v$, whenever the (rest) mass is non zero. The only case this relation says something meaningful for $v=c$ is when the (rest) mass is zero; but then you do not have the object anymore.

As an aside, the distinction between "rest mass" and "relative mass" or whatever is surpassed. It was used in the early days of relativity when concepts were yet to be understood. If you keep thinking about a relative mass you should even deal with two different of them, the so-called "transverse mass" and the "longitudinal mass".

Nowadays the only meaningful mass is what you would call the rest mass.