## relativity question

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zenten
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### relativity question

Ok, lets supposing I'm hanging out somewhat near some huge gravitation body (say a super large sun, or a black hole, or something), velocity relative to that body is stationary. I see an object moving toward that body at close the speed of light, and is of course accelerating. If I were measuring with Newtonian mechanics, it would be accelerating to past the speed of light before it hits any sort of event horizon, or the object itself.

So, question is, with Relativity, from my perspective is the acceleration lessening before it hits the speed of light?

Oort
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I think so, yes. But hopefully a more knowledgeable physicist will back me up.

QuantumTroll
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So, question is, with Relativity, from my perspective is the acceleration lessening before it hits the speed of light?

Correct. Assuming constant Force (not true if you're heading towards a black hole, of course, but this makes the math simpler), your acceleration ends up as

a = F/m * (1 - v^2)

Where v is the (speedof object)/(speed of light). As v->1, a-> 0. I remember this result from my Modern Physics class, 'cause I thought it was wonderfully elegant. The internet provides a derivation.

zenten
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Why does the force get divided by mass though, if we're talking about a falling object?

gmalivuk
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Basically, it's because the mass itself goes up as you approach light speed. So the same force is going to have less and less of an effect on acceleration.
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zenten
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Basically, it's because the mass itself goes up as you approach light speed. So the same force is going to have less and less of an effect on acceleration.

Why? As was pointed out in another thread, two objects of different mass will accelerate at the same rate, when falling.

ArmonSore
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So you're basically saying that the acceleration of an object due to gravity is independent of that object's mass. So then I wonder how accelerations due to gravity are restricted so that they can't bring something up to the speed of light.

Edit:
accelerations due to gravity according to GR*
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zenten
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So you're basically saying that the acceleration of an object due to gravity is independent of that object's mass. So then I wonder how accelerations due to gravity are restricted so that they can't bring something up to the speed of light.

That question is why I created the thread in the first place.

Karrion
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zenten wrote:
Basically, it's because the mass itself goes up as you approach light speed. So the same force is going to have less and less of an effect on acceleration.

Why? As was pointed out in another thread, two objects of different mass will accelerate at the same rate, when falling.

The short answer is because that thread was talking about non-relativistic objects, so used classical/Newtonian physics, which doesn't work once you have things moving at relativistic velocities.

ArmonSore
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Karrion wrote:The short answer is because that thread was talking about non-relativistic objects, so used classical/Newtonian physics, which doesn't work once you have things moving at relativistic velocities.

But GR still assumes that all objects move through a given gravitational field in exactly the same way, independent of its mass. So I'd prefer the long answer.
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F=Force
a=Acceleartion
m=mass of object
M=Mass of other object (sun etc)
G=Graviational constant
r=Distance of seperation

F=ma (Newton's second law)
F= GMm/r^2 (Law of universal graviation)

Let those be equal. Solve for a.
a=GM/r^2. so the mass of the object "falling" is irrelevant in figuring out its acceleration.
Now when two object "fall" towards each other you would have to consider both accelerations but since the earth accelerates VERY VERY little (and thats what the mass of the first object is relevant for) it doesn't really matter.

Now why does acceleration "lessen"

Well it follows from the lorentz transforms and relativistic mass.
The formula in Newtonian Mechanics is F=ma
so a=F/m
In relativity though, we get F=[m/[(1-v^2/c^2)^(1/2)]][a/(1-v^2/c^2)]
F=ma/So a=(F/m)*[(1-v^2/c^2)^(3/2)]

[(1-v^2/c^2)^(3/2)] is 1 at v=0 and 0 at v=c.
It is safe to say that when v is close to c it is also zero and when it is small compared to c, it is also 1.

These are just transforms of some formulas when relativity is involved, they should be covered in a text about relativity.
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Karrion
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ArmonSore wrote:But GR still assumes that all objects move through a given gravitational field in exactly the same way, independent of its mass. So I'd prefer the long answer.

Well, I'm no relativistic physicist, so someone else'll have to correct me in the (quite likely) case that I'm wrong. But:

I don't think it has to do with mass at all. It's all down to space/time-dilation. Basically (in the observer's frame of reference) time runs slower for the moving object. Since acceleration = distance/time^2, the acceleration measured gets less and less the more time slows down, as it gets closer to the speed of light.

Alternatively, in the falling object's frame of refernce, space is dilated so that the distance travelled is reduced, again reducing the acceleration measured.