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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.

Qmechanic
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Ej Sizemore
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  • actually, the force remains the same. Your (intertial) mass and thus the resistance to change in velocity increases. So the greater your speed, the greater you're inertia and thus the more difficult it i to cause a further increase in speed. – Vulgar Mechanick Dec 07 '20 at 17:59
  • @OVERWOOTCH but from the ships perspective, it isn't moving from its frame of reference. This question is specifically about a body that contains it's own source of propulsion. – Ej Sizemore Dec 07 '20 at 18:05
  • well, you need to remember that in the ship frame, it isn't moving or accelerating at all. – Vulgar Mechanick Dec 07 '20 at 18:09
  • suppose a spaceship leaves the earth accelerating at 1m/s. In the ship frame, it isnt the one moving or accelerating, the earth is. So the intertial of eath would increase as it apporaches c. Remember, in SR, inertia too depends on the frame of reference. Only the rest mass is invarient – Vulgar Mechanick Dec 07 '20 at 18:11
  • and no, the ship never reaches light speed even if you give it an infinite amount of time. You can mathematically show this if you are aware of limits. If not, consider how e^-x, a function which decreases with increasing x, can never be negative, even if you let x approach infinity. This is because the magnitude of increase in speed also approaches 0, as the speed approaches c – Vulgar Mechanick Dec 07 '20 at 18:14
  • @OVERWOOTCH in that case, what would the ship experience as it approaches c? assuming the ships can travel forever, what would the crew observe? – Ej Sizemore Dec 07 '20 at 18:18
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    @OVERWOOTCH Relativistic mass is a confusing and outdated concept. – G. Smith Dec 07 '20 at 18:21
  • I really don't understand the huge fuss with relativistic mass. How else would you explain this phenomenon, if not with the increase in inertia, quantified by inertial mass – Vulgar Mechanick Dec 07 '20 at 18:27
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    @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, – anna v Dec 07 '20 at 19:30
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    @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. – G. Smith Dec 07 '20 at 20:08
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    @OVERWOOTCH Please see https://physics.stackexchange.com/a/133395/123208 & the links therein. – PM 2Ring Dec 07 '20 at 23:49
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    Ej, relativity doesn't stop you fron accelerating indefinitely. Assuming you have the required fuel & reaction mass, you can increase your momentum & kinetic energy without limit. You may enjoy https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html – PM 2Ring Dec 07 '20 at 23:53
  • @PM2Ring hmm, I now see where the confusion. Sadly, I am a high school student so I dont think I can go beyond the “greater inertia” thing at this point – Vulgar Mechanick Dec 08 '20 at 04:59
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    @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. – jacob1729 Dec 08 '20 at 12:59

3 Answers3

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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.

Claudio Saspinski
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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 $F=dP/d\tau$.

jacob1729
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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.

JEB
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    How are you defining $v_+$ and $v_-$? Is $v_+=v_0=-v_-$? – jacob1729 Dec 08 '20 at 12:45
  • (Also, having just looked at this again, are all your $\tanh$'s not the wrong way around? You need to change $\tanh \leftrightarrow \tanh^{-1}$.) – jacob1729 Dec 08 '20 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 08 '20 at 21:19