# On the speed of light, accelerate toward mass

• Nick89
In summary, the faster an object is moving, the faster it needs to be to escape the gravitational field of a massive object.f

#### Nick89

Hi,

First of all let me clarify, I understand that nothing with mass can move at the speed of light or higher.

I have come across a question I was asked which I was unable to answer fully and would like to understand myself.

The usual questions like "If you imagine a rocket with an engine that produces thrust for eternity, wouldn't it eventually reach the speed of light?" are quite common, and it's not hard to explain why this is not true (since the mass increases with speed).

However, what if one would take a large mass out in deep space, and then release an object a large distance from this mass (not affected by other masses). The object will start to accelerate towards the mass, and this acceleration is not dependent on the mass of the object. So it will continue to accelerate and, if the initial separation was enough, eventually reach an extremely high speed.
Why can't it reach the speed of light?

Note that I'm not trying to say "See, it can reach the speed of light!??!", I understand there is something that makes this thought experiment fail miserably, I just can't figure out what it is.

Is there perhaps some way to show that the object will always reach the large mass before it reaches the speed of light? (After which it will pass (or crush onto) the object and thus slow down again).

I hope my 'assumption' that the acceleration is independend on the mass of the object remains correct here (Since $m_1a = G\frac{m_1m_2}{r^2}$, the m_1 cancels).

Finally I realize that the force of gravity will be MUCH smaller at the initial 'starting point' than for example 1 meter from the large mass. (According to my calculation, at 1m from a mass similar to Earth, the force is approximately 10^14 times the massof the object, at the intial distance required to reach the light speed (about 1.53*10^7m (which seems small??)) the force is about 1.7 times the mass of the object...)
Does this have anything to do with it?

Thanks...

I may not be correct here, but I'll give it a shot:

From wikipedia, the relativistic version of Newton's 2nd law is:
$$\vec{F} = \gamma m \vec{a} + \gamma^3 m \frac{\vec{v}\cdot \vec{a}}{c^2}\vec{v}$$

Assuming a and v are colinear, and F = GmM/r^2, we find that

$$a = \frac{GM}{r^2(\gamma + \frac{\gamma^3 v^2}{c^2})}$$

So using the correct relativistic force equation, acceleration still goes to 0 as velocity approaches the speed of light.

You have to think about the frames of reference for each object.

The mass attracts the object. This gravitational attraction moves at the speed of light, $c$.

The Lorentz transformation for velocity is:

$v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$

taking:
• the mass' frame of reference,
• the $x$ direction as from the mass to the object,
• $v_x$, as the speed of the object, and moving at speed $c$,
• $u$ as the speed of the gravitational attraction ie $=c$,

we then arrive at:

$v_x' = \frac{-c - c}{1 - \frac{c^2}{c^2}}$

$v_x' = \frac{-2c}{1 - 1}$

$v_x' = \frac{-2c}{0}$

which is mathematically defunct.

Ah ok, I didn't know there was a relativistic version of Newton's second law... silly me...

I don't fully understand benabean's explanation though. What do you mean by 'this gravitational attraction moves at the speed of light'?
Does this mean the effect of gravity is not instantaneous, it moves at speed c?

Ah ok, I didn't know there was a relativistic version of Newton's second law... silly me...

I don't fully understand benabean's explanation though. What do you mean by 'this gravitational attraction moves at the speed of light'?
Does this mean the effect of gravity is not instantaneous, it moves at speed c?

That exactly.

The distance in lightyears (distance light travels in 1 year), from the sun to Earth is 8.3 lightminutes, [c x 60 x 8.3] (=150 x $10^6$ km).
So say right this instant, the sun just disappeared from existence, we on Earth wouldn't 'see' it disappear until 8.3 minutes later because of the speed that light travels at, but also we wouldn't 'feel' it disappear until the 8.3 minutes had passed because the Earth wouldn't fall out of its orbit around the sun until that time.
That means, even though the sun is not there, we still perceive it as being there all because of the speed of light. If that seems confusing, imagine that $c$ is the speed of sound, 330m/s, instead of 3 x $10^8 m/s$. As you know, if you are 330m away from a speaker, you would hear the sound from the speaker, but with a 1 second delay.

Hi nick89,

Have you heard of escape velocity? The escape velocity is the velocity that an object requires to escape the gravitational field of a massive object and coast all the way to infinity. (I am using the term "infinity" fast and loose here). The equation for escape velocity is $$v = \sqrt{2GM}{R}$$. Using symmetry we can argue that is the velocity that a an object falling from infinty will reach when it hits the surface of the gravitational body. Notice this terminal velocity (in a vacuum) is independent of the mass of the falling object as you suggested. This is a Newtonian equation but has some relevance in the relativity context. Now if the gravitational body has a radius equal to $$2GM/c^2$$ then the terminal velocity of an object falling from infinity would be the speed of light c. It just so happens that $$2GM/c^2$$ is the nominal radius of black hole called the Schwarzschild radius or event horizon radius. So if you can take an object exactly to infinity (which will take a long time) then it might just be possible for it to reach a termianal velocity of light speed. Some problems are: (1) the initial falling velocity will be zero because the force of gravity will be $$GMm/\infty^2$$ so you will have to give it a little initial shove to get it going. (2) It will take an infinite amount of time to take the object to infinity. (3) It will take an infinite amount time for the object to fall. (4) The coordinate speed of light at the event horizon is zero so when it finally reaches the speed of light the speed of light is zero. (5) In the context of General Relativity the coordinate velocity of a falling photon actually slows down as it falls. This suggests that the coordinate velocity of any falling object will also slow down as it aproaches a black hole or it will be in danger of overtaking falling photons which just won't do. The proof of the last item is a bit more complicated and probably more than you want to know at this stage. Suffice to say it is unphysical for any falling object to reach the speed of light based on the unatainability of infinite distance.

$v_x' = \frac{-c - c}{1 - \frac{c^2}{c^2}}$

$v_x' = \frac{-2c}{1 - 1}$

$v_x' = \frac{-2c}{0}$

which is mathematically defunct.

You can also come up with the same thing by using a different forumula (btw this is my first time using latex so...):

$E = \frac{mc^2}\sqrt{1- \frac{v^2}{c^2}}$

Even if you tried to find the energy needed to go the speed of light, this equation would give you a denominator of 0, which cannot happen. This equation also shows, mathematically, why you cannot go FTL, because of the imaginary number you would obtain in the denominator if you had $v^2 > c^2$.

You have to think about the frames of reference for each object.

The mass attracts the object. This gravitational attraction moves at the speed of light, $c$.

The Lorentz transformation for velocity is:

$v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$

taking:
• the mass' frame of reference,
• the $x$ direction as from the mass to the object,
• $v_x$, as the speed of the object, and moving at speed $c$,
• $u$ as the speed of the gravitational attraction ie $=c$,

we then arrive at:

$v_x' = \frac{-c - c}{1 - \frac{c^2}{c^2}}$

$v_x' = \frac{-2c}{1 - 1}$

$v_x' = \frac{-2c}{0}$

which is mathematically defunct.

Your calculation is also mathematically incorrect. Starting with:

$v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$

you arrive at:

$v_x' = \frac{-c - c}{1 - \frac{c^2}{c^2}}$

Clearly you have substituted -c for $v_x$ and +c for u in the numerator, so you should have arrived at:

$v_x' = \frac{(-c) - (+c)}{1 - \frac{(+c)*(-c)}{c^2}}$

$v_x' = \frac{-2c}{1 + \frac{c^2}{c^2}} = c$

This does not prove a falling object can reach the speed of light because your equation is also based a premise that is not physically correct. The speed of gravitational atraction is effectively instantaneous because a particle moves in a pre-existing 4D "landscape" that surrounds a gravitational body and does not require gravitons to chase after the particle in order to interact gravitationally. It is changes in the 4D landscape that alters at the speed of light, not the interaction.

The usual questions like "If you imagine a rocket with an engine that produces thrust for eternity, wouldn't it eventually reach the speed of light?" are quite common, and it's not hard to explain why this is not true (since the mass increases with speed).
It is true that the speed of light cannot be reached but the explanation you give is not correct. The proper mass of the rocket does not increase, only the relativistic mass of the rocket as measured from an observer in relative motion with the rocket increases.

The reason that the speed of light cannot be reached is that the limit of the hyperbolic function of velocity addition is not c, in fact such hyperbolic functions have no limits.

The usual questions like "If you imagine a rocket with an engine that produces thrust for eternity, wouldn't it eventually reach the speed of light?" are quite common, and it's not hard to explain why this is not true (since the mass increases with speed).

Mass does not only increase with speed, it becomes infinite at the speed of light. And, at relativistic speeds the length of the object decreases in the direction of thrust, and at the speed of light the length becomes 0.

Your calculation is also mathematically incorrect. Starting with:

$v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$

you arrive at:

$v_x' = \frac{-c - c}{1 - \frac{c^2}{c^2}}$

Clearly you have substituted -c for $v_x$ and +c for u in the numerator, so you should have arrived at:

$v_x' = \frac{(-c) - (+c)}{1 - \frac{(+c)*(-c)}{c^2}}$

$v_x' = \frac{-2c}{1 + \frac{c^2}{c^2}} = c$

This does not prove a falling object can reach the speed of light because your equation is also based a premise that is not physically correct. The speed of gravitational atraction is effectively instantaneous because a particle moves in a pre-existing 4D "landscape" that surrounds a gravitational body and does not require gravitons to chase after the particle in order to interact gravitationally. It is changes in the 4D landscape that alters at the speed of light, not the interaction.

Cheers for that! Apologies on the typo.;-)