B How to add an object's multiple relativistic speeds from x, y, and z?

[syfry]

But you don't end up doing that - that's kind of the point. A velocity vector with components 0.99c in rach of three directions has a magnitude of about 1.7c, which is impossible. So your procedure doesn't get you to 0.99c in each direction - it'll be a bit faster than 0.99c in the first direction you picked and quite low in all the others.

I had assumed they never added over the amount of c because of how our slower travel through time will prevent us from gaining on light no matter how fast we travel. So I had mistakenly thought we'd perceive our 3 perpendicular speeds each at 99% c but the total couldn't break 100% and light would still be c in all directions.

It's weird to think that if our inertial speed were 1,000 km per hour lower than c, which is equal to being at rest, that we couldn't accelerate too much into a totally different direction (perpendicular).

Like if we're on a planet that's going that fast, our planes would be limited in flight speed except into the opposite direction of the planet's journey.

So ok that's food for thought that maxing out one direction could limit our motions into other directions.

 

[PeterDonis]

it'll be a bit faster than 0.99c in the first direction you picked and quite low in all the others.

If I take the initial 4-velocity $(1, 0, 0, 0)$ and apply an x, y, and z boost in succession, with each having the same $v$ (0.99c in this case), this is not what I get. I get a final 4-velocity of $(\gamma^3, \gamma v, \gamma^2 v, \gamma^3 v)$. This gives an ordinary velocity with components $v_x = v ( 1 - v^2 )$, $v_y = v \sqrt{ 1 - v^2 }$, and $v_z = v$. So the $x$ velocity is tiny, the $y$ velocity is middling, and the $z$ velocity is 0.99c.

Of course that raises the question of whether the operations I just described do in fact realize what the OP intended. As I have already pointed out, the OP's specification is ambiguous.

 

[Ibix]

Like if we're on a planet that's going that fast, our planes would be limited in flight speed except into the opposite direction of the planet's journey.

No they wouldn't. The Earth is already doing 0.9999999999c as measured by the rest frame of some pasding cosmic ray and we aren't all limited to millimetres per second in one direction by our own measures. You can always treat yourself as at rest and you can always accelerate in any direction ss long as your propellant lasts.

Of course, the cosmic ray won't see you accelerating very much, but that's a different frame.

 

[Ibix]

If I take the initial 4-velocity $(1, 0, 0, 0)$ and apply an x, y, and z boost in succession, with each having the same $v$ (0.99c in this case), this is not what I get. I get a final 4-velocity of $(\gamma^3, \gamma v, \gamma^2 v, \gamma^3 v)$. This gives an ordinary velocity with components $v_x = v ( 1 - v^2 )$, $v_y = v \sqrt{ 1 - v^2 }$, and $v_z = v$. So the $x$ velocity is tiny, the $y$ velocity is middling, and the $z$ velocity is 0.99c.

I'd say 2% and 14% of c are "quite low" compared to 99% of c, although that's obviously subjective.

I'm also not quite sure that's the correct procedure. I think you need to boost in the x direction, then the y' direction then the z'' direction, given the process described in the last paragraph of #10.

 

[PeterDonis]

I'd say 2% and 14% of c are "quite low" compared to 99% of c, although that's obviously subjective.

Yes, but the 2% is the velocity in "the first direction". So that velocity is not "a bit faster than 0.99c", as you said. It's much slower. (And the highest velocity is not "a bit faster" than 0.99c, it's exactly 0.99c.)

I'm also not quite sure that's the correct procedure. I think you need to boost in the x direction, then the y' direction then the z'' direction, given the process described in the last paragraph of #10.

The y' direction is the same as the y direction, because the x boost doesn't change it.

The z' direction will also be the same as the z direction, since neither the x nor the y boosts change it. The y boost changes the x direction in the new primed frame relative to the original frame because of Wigner rotation, but it doesn't change the z direction.

 

[pervect]

Both of those are changing my question that I intentionally composed to avoid moving in any opposite directions:

"you reach 99% light speed in one direction, then without slowing, you start to accelerate into an either directly left or right direction (say the engine can rotate without turning the spacecraft) until you reach 99% into that direction as well. And again without slowing, you start to accelerate into an either directly up or down direction until you also reach 99% into that direction"

Emphasis mine.

Are we on the same page or are we missing details by speed reading at relativistic speeds which might happen to be length contracting our attention?

The exact calculation amount isn't vital, I merely wanted to confirm that relativistic effects from the additions of speed won't overshoot past c.

When you reach 99 percent of the speed of light in one direction (we'll call it the x direction), you'll have some momentum in the x direction $\gamma \, m \, .99 \, c$, where m is your mass.

If you start to accelerate in the y direction, if you maintain the same momentum in the x direction, your velocity in the x direction will decrease as you accelerate in the y direction, because the value of $\gamma$ will increase as you increase your velocity. Since your mass remains constant, this implies that the x component of the velocity must drop.

Here the "x,y,z" directions are relative to some fixed inertial frame of reference S.

If, instead, you accelerate in some manner to keep your x-velocity component relative to the fixed inertial frame S constant, you will never be able to exceed a velocity of $\sqrt(1 - .99^2)\,c$ in the y direction.

These are the two most likely interpretations of what you mean by "accelerate in the y direction". If you have some other interpretation, you'll have to state what it might be.

The general rules you need in special relativity are $F = dp/dt$, where p is momentum, and $p = \gamma m v$, where v is a vector, and $\gamma = 1 / \sqrt(1-|v|^2/c^2)$ where $|v|$ is a scalar that is the magnitude of the vector v, i.e. $|v| = \sqrt(v_x^2 + v_y^2 + v_z^2)$.

[add]Additionaly, there are various options for "t" you might use. I'm using the coordinate time of frame S. Other popular choices might be different coordinates, or the notion of "proper time", $\tau$, which is an interval along a worldline that a clock might measure.

 

[Demystifier]

TL;DR Summary: What are the relativistic effects if you go 99% light speed into the x axis, while you're going the same speed into the y axis (now two directions), and likewise the z axis. So three directions at 99% light speed each.

Obviously it'd add up to somewhere between 99% and 100% light speed. So how do you figure that out?

In the scenario, you reach 99% light speed in one direction, then without slowing, you start to accelerate into an either directly left or right direction (say the engine can rotate without turning the spacecraft) until you reach 99% into that direction as well. And again without slowing, you start to accelerate into an either directly up or down direction until you also reach 99% into that direction.

How do we add up the relativistic effects? (doubt they'd triple)

Are the 3 velocities all defined with respect to the same inertial frame? If so, the total velocity exceeds the speed of light, which of course is physically impossible.

The question only makes sense if the second velocity is defined with respect to inertial frame with respect to which the rocket moving with the first velocity is at rest, and similarly for the third velocity.

 

[syfry]

These are the two most likely interpretations of what you mean by "accelerate in the y direction". If you have some other interpretation, you'll have to state what it might be.

Now seeing the big flaw in my scenario. It's apparent by visualizing lower speeds on a globe, no math needed.

If the x we're traveling is south to north in a straight line along a longitude line, it's impossible to then accelerate into y along the equator or along any latitude line. We'd be moving diagonally, even if we aim due west or due east... straying from both of our lines.

In space the lines don't matter, but then we cannot also accelerate purely into x, y, or z without at the same time slowing in another axis.

There's overlap so we'd be partially accelerating into two axis which is impossible to avoid.

I tried accelerating into all 3 axis together at the same time, and we'd be traveling at some diagonal.

We're actually accelerating diagonally. So only one speed would ever apply.

That's why accelerating into another axis will slow the speed in a current axis.