# Is there a force that a body experiences as it approaches light speed?

My question can be boiled down to a thought experiment.

Assume you have a space ship in an empty universe. This space ship is capable of a constant $$1 m/s^2$$ acceleration. As it approaches light speed, does some force actively resist further acceleration?

I feel that the apparent decrease in acceleration may be due to time dilation from the perspective of an outside observer. For example, a passenger on the ship would not experience any change in the acceleration of the ship, whereas the outside observer would observe the thrusters on the ship enacting less force because of their relatively slower time frame.

Feel free to ask for clarification on anything, or explain the actual cause if I am incorrect.

Edit:the faster than light tag is referring to an idea where, given infinite time to function, the ship could potentially reach light speed.

• @OVERWOOTCH Relativistic mass is a confusing and outdated concept. Dec 7, 2020 at 18:21
• @OVERWOOTCH you are confusng intertial frames with accelerated frames. This spaceship is not in an inertial frame, it keeps furning fuel and releasing momentum contunually, and it is the sytem "ejecta +spaceship" that are in an inertial frame. The passengers know they are accelerating, Dec 7, 2020 at 19:30
• @OVERWOOTCH How else would you explain this phenomenon, if not with the increase in inertia, quantified by inertial mass By increase in relativistic momentum and energy. There are many questions about “relativistic mass” on this site, and the consensus view here, and among physicists today, is that it is a bad concept. It is pedagogical malpractice and we should not use it to explain relativity. Dec 7, 2020 at 20:08
• @OVERWOOTCH Please see physics.stackexchange.com/a/133395/123208 & the links therein. Dec 7, 2020 at 23:49
• @OVERWOOTCH see my answer for perhaps the most obvious reason to not use relativistic mass: newton's law is still not $f=ma$ using it. In fact, you'd need at the very least to promote the mass to being a 3x3 matrix to save this, since it's not true in relativity that accelerations are even in the same direction as forces. Dec 8, 2020 at 12:59

No, there isn't a force that resists further acceleration.

Suppose the galaxy is precisely mapped, and the crew knows the distance between stars in the route.

As times passes and the ship velocity increases, a naive ship navigator could think that eventually it is above light speed, by dividing the last travelled path by the time between starting and final point.

But the distance between the stars is contracting from the ship frame of reference, as the velocity increases. And the velocity, calculated by the length after correction divided by ship's time, never reaches light speed.

The relativistic momentum is given not by $$mv$$ but by $$p=\gamma mv$$. If we copy over the Newtonian definition of force as (assuming we're in 1D space so I'm being cavalier about vectors for the moment):

$$f=\frac{dp}{dt} = m\frac{d\gamma}{dv}\frac{dv}{dt}v+m\gamma \frac{dv}{dt}$$

we find (after some calculations, using $$\gamma = (1-\beta^2)^{-1/2}$$ where $$\beta=v/c$$) that:

$$f = m \gamma^3 \frac{dv}{dt}$$

This is (one form of) Newton's 2nd law in 1D special relativity (with caveats, see below). To answer the OP's question then: the force can remain constant whilst the 'acceleration' $$dv/dt$$ goes to zero if $$\gamma^3$$ increases accordingly. And indeed, since $$\gamma$$ increases without limit as $$v\rightarrow c$$, this will be the case.

Important points skipped above:

1. Whilst the momentum $$\vec{p}$$ and the force $$\vec{f}$$ are both 3-vectors under rotations, only the momentum forms the space-like part of a 4-vector under Lorentz transformations (rotations+boosts). As such, $$\vec{f}$$ is not that 'natural' a quantity in SR.

2. Some people claim that there is a 'relativistic mass' $$m_{\text{rel}}=\gamma m$$. But they run into the problem that it is still not true that $$f=m_{\text{rel}} a$$.

3. The formula $$\vec{f}=m\gamma^3 \vec{a}$$ is not correct. It is only correct in 1 dimension. In 2 and more, there are different equations for the component parallel and perpendicular to the force.

4. It's therefore usually best to work with the true relativistic quantities which are the four-momentum $$P=(E/c,\vec{p})$$ and the four-force $$F=dP/d\tau$$.

• Shouldn't gamma be $$\gamma = \left(1-\frac {v^2}{c^2}\right)^{-1/2}$$ ? Now dimensional analysis for your gamma formula fails. Dec 8, 2020 at 13:59
• @AgniusVasiliauskas $c=1$ Dec 8, 2020 at 14:03
• Oh, this derivation is for natural units, I see Dec 8, 2020 at 14:04
• @AgniusVasiliauskas I've re-instated $c$ everywhere I think it needs to be. Dec 8, 2020 at 14:08

Suppose there is a force that resists further acceleration. It becomes measurable at $$v_+$$. Working in 1-dimension for brevity, you can accelerate along $$+z$$ until you reach $$v_+$$. At that point, reverse thrusters and retrace you steps back to the original position.

Now fly in the $$-z$$ direction until you reach $$v_-$$.

You can now find the absolute rest frame for your flat universe. Defining the rapidity as:

$$\omega = \tanh^{-1} v$$

then the average rapidity :

$$\omega_+ + \omega_-$$

transforms under a $$u$$ boost along $$z$$ to:

$$(\omega_+ - \tanh^{-1} u) + (\omega_- - \tanh^{-1} u)$$

which can be set to zero:

$$(\omega_+ - \tanh^{-1} u) = -(\omega_- - \tanh^{-1} u)$$

$$\tanh^{-1} u = \frac 1 2(\omega_+ +\omega_-)$$

So that:

$$u = \tanh\big(\frac 1 2 (\tanh^{-1}||v_+|| - \tanh^{-1}||v_-||) \big)$$

boosts you to a preferred rest frame, which relativity prohibits.

The real point of this exercise is to address the part that reads:

"As it approaches light speed, does some force actively resist further acceleration?"

which completely misses the point of relativity: you can't "approach the speed of light", no matter how fast some external observer measures your velocity, it has absolutely no bearing on your observables, and if it did, there would be an absolute rest frame in spacetime.

• How are you defining $v_+$ and $v_-$? Is $v_+=v_0=-v_-$? Dec 8, 2020 at 12:45
• (Also, having just looked at this again, are all your $\tanh$'s not the wrong way around? You need to change $\tanh \leftrightarrow \tanh^{-1}$.) Dec 8, 2020 at 14:06
• @jacob1729 $v_0$ was a typo. it's $v_+$, the speed where being close $c$ is noticeable. And yes, the hyperbolic tans are inverted, but that has absolutely zero bearing on the point.
– JEB
Dec 8, 2020 at 21:19