# I Gravity at relativistic speed

1. Feb 9, 2017

### bobie

Is it possible to find the speed increase due to gravity pull using the SR velocity addition formula or the calculator here?
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html

Can you confirm that it is in accordance with GR formula?
For example, if an asteroid is approaching the earth at √3 /2 c, considering g constant (10 m/s^2)for a whole second the hyperphysics calculator says the increase is 2.5 m/s (instead of 10)

Is it an acceptable approximation?

2. Feb 9, 2017

### Staff: Mentor

That calculator is for a situation completely different than the one you describe. It is for transforming measured velocities in inertial frames (it says so in the first sentence) and that's not your setup.

3. Feb 9, 2017

### Staff: Mentor

No. This calculator has nothing to do with gravity.

4. Feb 9, 2017

### Khashishi

g for Earth isn't constant. It depends on height. And height will change a lot in one second, traveling at sqrt(3)/2 c.

But, if we just consider a constant g, I think you can use the equivalence principle to calculate it without full GR equations.

5. Feb 10, 2017

### bobie

Let's consider g constant, like on a supermassive body, in order to not complicate the situation.
Can you refer me to an article or show how to use a simpler formula for gravity? Can you give a rough estimate of the increase of velocity?
Thanks

6. Feb 10, 2017

### Khashishi

Rindler coordinates (https://en.wikipedia.org/wiki/Rindler_coordinates) describes uniformly accelerating frames of reference.
We are considering an acceleration of g.
Let c=1, g=1
Note that choosing g=1 is choosing a time unit. For g=10m/s^2, since we have set c=1, we need to write g in terms of c:
g=3.3*10^-8 c/second
So, the time unit is (1/(3.3*10^-8/second) = 29979245.8 seconds

Consider the world-line of the asteroid. Let's assume no forces on the asteroid, so the world line is a diagonal line in inertial coordinates. Choose the line so it crosses the X axis at (1,0). So, the asteroid will be moving at v when it passes right by the Rindler observer who is stationed at (1,0). This Rindler observer is uniformly accelerating at g, so is stationary with respect to the massive body. Lowercase letters refer to the Rindler observers' coordinates, and uppercase refer to inertial coordinates (which are accelerating with respect to the massive body).
So, we want to know, where does the t=(1 second) curve intersect the world-line of the asteroid? Converting units: t=3.3*10^-8
Let's write the equation of the world-line
X = v*T + 1
Now, for coordinate transformation
x=sqrt(X^2-T^2)
t=arctanh(T/X)
T = tanh(t)
plugging in and solving for the world-line in x and t coordinates
X = v*tanh(t) + 1
x = sqrt((v*tanh(t) + 1)^2 - (tanh(t))^2)

Now, we want to know the new instantaneous velocity seen by the Rindler observer at t=3.3*10^-8
v_new = dx/dt = d(sqrt((v*tanh(t) + 1)^2 - (tanh(t))^2))/dt
(using mathematica)
= (Sech[t]^2 (v + (-1 + v^2) Tanh[t]))/Sqrt[-Tanh[t]^2 + (1 + v Tanh[t])^2]

Now just plug in values (t=3.3*10^-8, v=sqrt(3)/2)
v_new = sqrt(3)/2 - 3.3*10^-8
Whoops, we're accelerating in the wrong direction. Oh well, just reverse the sign of v.
Now just plug in values (t=3.3*10^-8, v=-sqrt(3)/2)
v_new = -sqrt(3)/2 - 3.3*10^-8

Now since we are using units where c=1, 3.3*10^-8 = 9.9 m/s^2.
So, basically, by accelerating at 10m/s^2 for 1 second, we have changed v by about 9.9 m/s^2. There might be some rounding error.

What if we accelerate longer, or set g to a larger value? Let's say, set g to c/10 per second.
Then, t = 0.1
with v = -0.8660254
v_new = -0.971218

If g=0.5c, we get v_new = -2.01887
That's faster than c. It's impossible to go faster than c in inertial coordinates, but it might be possible in Rindler coordinates. Keep in mind that the asteroid is very far from the observer after 1 second. Accelerating frames are weird. But maybe I just made a mistake. I haven't done this kind of calculation before.

7. Feb 11, 2017

### bobie

Thanks,Khashishi
That is very interesting and complex. Unfortunately I have no access to mahtematica, do you know if I can find a curve showing the actual increase of speed when g = 1m/s^2? If there is none, can you make a graph with one of those prodigious programs, plotting on the x axis the increase of speed and on the y-axis
speeds from 0.1 to 1c? It would be great if you could show at same time the curve of velocity addition for same values. As I said, with velocity addition 1 meter becomes 25 cm/s+...378 =:.86602540403, with acceleration it should be a little more, may be 40 cm/s?

8. Feb 11, 2017

### Drakkith

Staff Emeritus
Let's see... assuming I've done this right, just go to https://www.desmos.com/calculator and input the following 2 equations:
$y=1x$
$y=\left(2.99\cdot 10^6\right)\tanh \left(\frac{1x}{\left(2.99\cdot 10^6\right)}\right)$

The first is the classical velocity equation: v=v0+at, where v0 is zero and a = 1.
The second is the relativistic equation for velocity in terms of time and acceleration. I got the equation from here: http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

Once plotted, you can see the difference between the classical situation and the relativistic situation once you zoom far enough out. If you want to find the change in velocity for some time period, just put the 2nd equation into a calculator (there are plenty online that can do tanh) with some value for T (x in the above equation), get the 1st velocity, and then input another value for t to find the 2nd velocity. If you want to change the acceleration, just replace 1x with ax, where a is whatever acceleration you want.

Hopefully that's correct. If not, I apologize. I'm not very familiar with relativity.

9. Feb 11, 2017

### bobie

Thanks for your help, that desmos link is invaluable, wrote the 2 inputss, but it is just the same output, the lines are the same. Another doubt you sure it is 2.99 by ten raised to the sixth power (and not 8 as in c)?
I did not get the rest of the procedure, but I'll try to work it out. Do you get a rough evaluation of the increase of speed when g = 1

Last edited: Feb 11, 2017
10. Feb 11, 2017

### jbriggs444

$2.99\cdot 10^6$ is 2.99 times (10 raised to the sixth power). The 2.99 is not raised to any power.

11. Feb 11, 2017

### Drakkith

Staff Emeritus
Make sure you zoom out really far. You should see the tanh line flatten out and never pass 3x108 (speed of light), unlike the 1x line which continues to climb forever. Also, yes, that should be 2.99x108, not 106 (I was very tired last night when I posted that).

12. Feb 13, 2017

### bobie

Even if it does not refer to gravity, yet the result of gravity equations can't be meaningfully different, (whatever equations you apply), since the values are restrained between the quadratic curve of the circle (sine/cosine: sqrt (1-y^2) and the linear result of the line described by y=1-y, and the results of the velocity addition are in between they start quadratic and then in the second half become linear: 1-2/y

Don't you agree?
What do you think of the method suggested by Drakkith?

13. Feb 13, 2017

### Staff: Mentor

No, I do not agree. In using any formula the most important thing is to know the assumptions. You should not use any equation where the assumptions are not met.

14. Feb 13, 2017

### Khashishi

You have to be clear about who the observer is in this problem. The position of the observer relative to the asteroid is important. The original problem statement puts the observer on Earth, but then in post #5, we envision some kind of supermassive body so far away that the gravitational field is constant. In this scenario, we can't put the observer on the supermassive body itself, since it's too far away and not clearly defined. (It's probably a black hole, and we can't put the observer in the black hole). So in post #6, I have assumed that the observer is floating out in space, accelerating at g but stationary with respect to the supermassive body.

Since the asteroid is moving relative to the observer, it can only be in the same place as the observer at one point in time. The asteroid must be traveling at less than c when it passes the observer. But when it is far from the observer, it could be traveling faster than c, according to the observer's non-inertial coordinate system. Now, if you had a second observer, closer to the supermassive body, to measure the speed of the asteroid 1 second after the asteroid passes by the first observer, this speed would have to be less than c. But in this case, the second observer would necessarily be accelerating faster than the first observer to stay stationary relative to the body.

It gets complicated when gravity is involved.

15. Feb 14, 2017

### bobie

You don't need a black hole, a body just 1/10 closer to the sun than the earth gets a constant a = 10 m/s^2, and any observer at rest in the sun's frame can measure the increase of speed.
I have checked with other sites and googled,but it seems that nobody really knows how to calculate it. S to the underlying assumptions, what are the ones behind GR? Mass bends the spacetime, what does it say about velocity? Does it influence the curvature?

16. Feb 14, 2017

### Drakkith

Staff Emeritus
It's unlikely you'll find a simple formula for determining the change in velocity under gravity in GR. The math of GR is extremely complicated and even equations like $G=\frac{8πG}{c^4}T$ are deceptively simple since $T$ ad $G$ are both tensors.

17. Feb 14, 2017

### bobie

You are certainly right, but I was not looking for a simple formula. It seems odd that after a century, with all the supercomputers and specialized software available no one has yet bothered to plot the result into a graph for a rapid consultation?
In this graph I quickly made at desmos (thanks again for the link)

https://www.desmos.com/screenshot/tgdndqvsbu

you can see in blue the curve of the velocity addition, I was looking for a similar curve that shows the values according for GR. You can see clearly now what I meant when I said that they cannot be a lot different from the blue curve. Do you agree?

18. Feb 14, 2017

### Drakkith

Staff Emeritus
Oh I'm sure they have. They probably just haven't bothered to put it online. Or it could be online, just buried in the vast sea of binary that is the internet. I'm sure all those researchers working on black holes and neutron stars and such have a great many graphs and tables stored somewhere.

I have no idea. I barely understand the basics of relativity.

19. Feb 14, 2017