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

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)

syfry said:

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

Wikipedia has the general formula for relativistic velocity addition when more than one dimension is involved.

You can do it by composing Lorentz boosts, but have to bear in mind that multiple non-parallel boosts involve some rotation of coordinate axes, a phenomenon known as Wigner rotation. You can look that up on Wikipedia if you really want to know.

Note that you don't end up doing a velocity which the initial frame says has components of 0.99c in the three orthogonal directions. You can't, since that would be a velocity of about 1.7c.

Also note that boosting in the x and then in the y direction is not the same as first boosting in the y direction and then the x direction. So one really needs to take a bit of care with the setup here.

Ibix said:

Note that you don't end up doing a velocity which the initial frame says has components of 0.99c in the three orthogonal directions. You can't, since that would be a velocity of about 1.7c.

Would the speeds add up past c?

I got 299,792,382 meters per second.

99.999974649129% the speed of light.

Math was 99% of 299,792,458 = 296,794,533.42 and then that times 3 (in the context of relativistic speeds).

With an online calculator. (no way I'd have known how to manually do the math at Wikipedia!)

Orodruin said:

Also note that boosting in the x and then in the y direction is not the same as first boosting in the y direction and then the x direction. So one really needs to take a bit of care with the setup here.

Thought direction is relative in space. How is x first different than y first? Didn't know the order mattered!

syfry said:

Thought direction is relative in space. How is x first matter different than y first? Didn't know the order mattered!

Check out the general formula listed in the wiki page referenced in #2. You will see that ##\vec u \oplus \vec v \neq \vec v \oplus \vec u## in general.

syfry said:

Would the speeds add up past c?

Read the last sentence you quoted again.

syfry said:

With an online calculator.

"I put some numbers into an online calculator" tells me absolutely nothing helpful. What calculator? Does it do non-parallel velocity addition? Does it explain its calculations and have you checked them with a simple case?

syfry said:

How is x first matter different than y first?

Wigner rotation. The final speed will be the same whichever order you do your accelerations. The direction will not.

Ibix said:

Wigner rotation. The final speed will be the same whichever order you do your accelerations. The direction will not.

Although that depends what you mean by the same direction in different reference frames. If I may coin the phrase "the relativity of orthogonality".

Ibix said:

Read the last sentence you quoted again.

Yep. That's what I specifically replied to.

"I put some numbers into an online calculator" tells me absolutely nothing helpful. What calculator? Does it do non-parallel velocity addition? Does it explain its calculations and have you checked them with a simple case?

This calculator for adding relativistic speeds.

If I had the skill to double check its math, I likely wouldn't have had to ask anything at these forums!

Wigner rotation. The final speed will be the same whichever order you do your accelerations. The direction will not.

Ok the final speed is same regardless of order. Thought so.

Thanks for pointing out the Wigner rotation.

Wikipedia didn't speak in English but I did find a video that does.

So Wigner rotation seems to matter mostly for when we have a specific destination in mind: because length contraction will skew the angles and so people won't agree on those unless they've calculated certain transformations.

My scenario didn't have a destination, it only rotates the engine by 90° to start a new acceleration into a sideways direction or into a 90° vertical direction. And if that's still affected by length contraction even without a destination, then so be it. Good to know.

syfry said:

Thought direction is relative in space. How is x first different than y first? Didn't know the order mattered!

Before one gets to boosts and their non-commutativity,
it's important to note that rotations in 3D are not commutative.
See https://math.stackexchange.com/ques...duct-of-two-rotation-matrices-not-commutative for a nice explanation using photos of a die (one of a pair of dice).

syfry said:

This calculator for adding relativistic speeds.

That calculator does velocity composition in one dimension only.

For instance, your planet is moving at 99% of c relative to the CMB. You are in a railroad train moving at 99% of c relative to the planet and in the same direction as the planet's motion. You are running forward in that railroad train at 99% of c relative to the train. The calculator could let you figure out how fast you are moving relative to the CMB.

Meanwhile, the problem you posed is being interpreted with the train moving perpendicular to the path of the planet and you shooting a bottle rocket upward, perpendicular to the train.

syfry said:

This calculator for adding relativistic speeds.

It's purely one dimensional. The lack of any way to specify a direction should tip you off to this.

This kind of thing, incidentally, is why you should always supply links to tools you've used, especially if you don't understand them.

syfry said:

My scenario didn't have a destination, it only rotates the engine by 90° to start a new acceleration into a sideways direction or into a 90° vertical direction. And if that's still affected by length contraction even without a destination, then so be it. Good to know.

It isn't really length contraction that's relevant here. But the point is that in a Newtonian world two of your ships that accelerated in different orders would end up in different places at the same speed in the same direction as measured by the initial frame. In a relativistic universe they end up in different places with the same speed in different directions, again as measured using the initial frame.

robphy said:

Before one gets to boosts and their non-commutativity,
it's important to note that rotations in 3D are not commutative.
See https://math.stackexchange.com/ques...duct-of-two-rotation-matrices-not-commutative for a nice explanation using photos of a die (one of a pair of dice).

Followed the link and I see how rotations in 3D would end up at different places, and while the dice example is interesting (though a bunch of comments there had disagreed), I think it's a moot point for my scenario since the dice involves a destination yet my scenario seeks to know only the total of relativistic speeds and effects, without any destination.

My scenario on the dice would care only how many faces we'd turn. But that isn't quite a good analogy because the dice would be like space is doing the moving. An ant crawling on the dice from face to face at relativistic speed is closer but still falls short. Maybe an ant that leaps up at relativistic speed, then shoots sideways at relativistic speed while still ascending relativistically, then sharply shoots into the other sideways direction at relativistic speed while still relativistically ascending and side journeying.

jbriggs444 said:

Meanwhile, the problem you posed is being interpreted with the train moving perpendicular to the path of the planet and you shooting a bottle rocket upward, perpendicular to the train.

Does that matter if we're concerned only about adding the relativistic speeds? I'm merely curious about any difference for that specific aspect.

Well, relativistic speeds and effects.

Ibix said:

It isn't really length contraction that's relevant here. But the point is that in a Newtonian world two of your ships that accelerated in different orders would end up in different places at the same speed in the same direction as measured by the initial frame. In a relativistic universe they end up in different places with the same speed in different directions, again as measured using the initial frame.

Right, and if my scenario cared about destination then places would matter.

My scenario would instead ask how far did the Newton and Einstein ships travel, not at where they landed. Or more accurately, how fast is Einstein's ship traveling in total for figuring out the relativistic effects.

Also curious, is length contraction really not relevant?

syfry said:

Does that matter if we're concerned only about adding the relativistic speeds?

If I accelerate for 1s at 1g I end up doing 10m/s. If I accelerate for 1s at 1g again I could be doing 20m/s if the second acceleration is in the same direction as the first, 14m/s if they're perpendicular, or zero if they're opposite, or anywhere between 0 and 20m/s in general. Direction matters.

Go and have a look at the link Nugatory posted in #2. There's a lot of fairly boring algebra needed to answer this question and not a lot of insight beyond Wigner rotation. I'm sure we can help you do the maths if you really want to know, but I'll be a little bit surprised if anyone volunteers to do it for you.

syfry said:

Also curious, is length contraction really not relevant?

Length contraction is a special case phenomenon when you measure the length of something moving. It's not really relevant to much at all, although I'm sure if you look hard enough there's a way to shoehorn it into any relativistic kinematics problem.

syfry said:

Does that matter if we're concerned only about adding the relativistic speeds? I'm merely curious about any difference for that specific aspect.

You seem to suppose that composing .99c with .99c will yield the same resulting speed regardless of whether the second boost is parallel, perpendicular or anti-parallel to the first.

Clearly, the anti-parallel direction yields a resulting speed of zero. So that supposition does not hold water.

Also, there is no distinction between a "relativistic velocity" and an ordinary velocity. They are all relativistic velocities. The formulas for special relativity apply equally well to low speeds as well as to high speeds.

If you consider a boat moving at 1 m/s eastward while a person strides across the boat at 1 m/s northward then the velocities will compose to approximately ##\sqrt{2}## m/s approximately 45 degrees to the north of due east.

If you consider a boat moving at 1 m/s eastward while a person strides forward along the boat at 1 m/s eastward then the velocities will compose to approximately 2 m/s exactly due east.

If you consider a boat moving at 1 m/s eastward while a person strides backward along the boat at 1 m/s westward then the velocities will compose so that the person is exactly at rest.

The resulting speed depends on the angle between the velocities being composed.

Ibix said:

If I accelerate for 1s at 1g I end up doing 10m/s. If I accelerate for 1s at 1g again I could be doing 20m/s if the second acceleration is in the same direction as the first, 14m/s if they're perpendicular, or zero if they're opposite, or anywhere between 0 and 20m/s in general. Direction matters.

jbriggs444 said:

You seem to suppose that composing .99c with .99c will yield the same resulting speed regardless of whether the second boost is parallel, perpendicular or anti-parallel to the first.

Clearly, the anti-parallel direction yields a resulting speed of zero. So that supposition does not hold water.

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.

syfry said:

I merely wanted to confirm that relativistic effects from the additions of speed won't overshoot past c.

That has already been confirmed in this thread (more than once by my count).

PeterDonis said:

That has already been confirmed in this thread (more than once by my count).

Right, feel a reminder was needed due to the detours about what I actually asked.

syfry said:

Are we on the same page

You seem to have missed the specific example of perpendicular accelerations that I gave. And the general point that if opposite accelerations sum to zero and parallel ones don't then you know that direction does matter.

syfry said:

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.

There's no way to trick your way past c by juggling accelerations or rotations or anything else of that nature. That follows directly from the second postulate of relativity - since you can always consider yourself to be at rest, you always still have ##3\times 10^8\mathrm{ms^{-1}}## left to accelerate by your own measure.

syfry said:

Right, feel a reminder was needed due to the detours about what I actually asked.

It wasn't really a detour, just a statement of the general principle: no matter what the relationship is between the directions of the successive boosts, you will never end up with a speed that is equal to or greater than ##c##. That general principle is already sufficient to answer your basic question.

That general principle does not, of course, tell you how to work out the specific details of the specific scenario you posed. But you have already been pointed at the tools to do that too.

Ibix said:

You seem to have missed the specific example of perpendicular accelerations that I gave. And the general point that if opposite accelerations sum to zero and parallel ones don't then you know that direction does matter.

If it's post #15, then it's cool you're trying to teach me that accelerating into a perpendicular direction would end up at a lower speed than if accelerating into the same direction:

"If I accelerate for 1s at 1g I end up doing 10m/s. If I accelerate for 1s at 1g again I could be doing 20m/s if the second acceleration is in the same direction as the first, 14m/s if they're perpendicular"

But again that's assuming I care about destination. Remember that my post had stated 99% speed of light into x, and into y, and into z, so even if the perpendicular changes would've added to the journey's total duration, what matters is that eventually we're going 99% light speed in each, and how the relativistic effects would stack. (not 'overstack')

There's no way to trick your way past c by juggling accelerations or rotations or anything else of that nature. That follows directly from the second postulate of relativity - since you can always consider yourself to be at rest, you always still have ##3\times 10^8\mathrm{ms^{-1}}## left to accelerate by your own measure.

My original post agrees:

"Obviously it'd add up to somewhere between 99% and 100% light speed."

So are we really on the same page?

syfry said:

eventually we're going 99% light speed in each

No, you're not. That's part of the point. As was already pointed out earlier, a velocity vector with components of 0.99c in the x, y, and z directions would have a length greater than c. So whatever final velocity vector you end up with from the three boosts you specified, it will not have those components.

In other words, you can't just wave your hands and say that boosting by 0.99c in a given direction will always add a component of 0.99c in that direction to your velocity. You have to actually do the math.

PeterDonis said:

In other words, you can't just wave your hands and say that boosting by 0.99c in a given direction will always add a component of 0.99c in that direction to your velocity. You have to actually do the math.

Ok, so I get it now.

And to be clear, since the wording feels unclear... is 'boosting by 0.99 c' the same as 'boosting until 0.99 c'? Are those identical statements?

syfry said:

And to be clear, since the wording feels unclear... is 'boosting by 0.99 c' the same as 'boosting until 0.99 c'? Are those identical statements?

I don't know. What do you mean by those two things?

There's a reason why physicists use math.

PeterDonis said:

I don't know. What do you mean by those two things?

There's a reason why physicists use math.

Surely if the physics used only math and no words, it'd be almost useless.

If traveling by car and I boost by 100 km per hour, then it's 100 + current speed. If I boost until 100 km per hour, then it's only 100 total.

With relativity, that might mean a boost by 0.99 c could fall short of 0.99, but a boost until 0.99 c should reach exactly 0.99 so I'm looking to see what you meant.

syfry said:

Surely if the physics used only math and no words, it'd be almost useless.

I didn't say "use only math and no words". I said "use math". Which means, you express anything for which words would be ambiguous, in math. As you go on to do in your next sentence:

syfry said:

If traveling by car and I boost by 100 km per hour, then it's 100 + current speed. If I boost until 100 km per hour, then it's only 100 total.

Ok, so then, as you are using the terms "boost by" and "boost until", they mean different things. How do I know that? Because you showed me the math.

syfry said:

With relativity, that might mean a boost by 0.99 c could fall short of 0.99, but a boost until 0.99 c should reach exactly 0.99 so I'm looking to see what you meant.

You're the one who posed the scenario. What did you mean?

PeterDonis said:

You're the one who posed the scenario. What did you mean?

"In other words, you can't just wave your hands and say that boosting by 0.99c in a given direction will always add a component of 0.99c in that direction to your velocity. You have to actually do the math."

Who wrote those words?

Your phrasing is 'by 0.99 c' while mine is 'until'. Did you mean the same as me?

syfry said:

"In other words, you can't just wave your hands and say that boosting by 0.99c in a given direction will always add a component of 0.99c in that direction to your velocity. You have to actually do the math."

Who wrote those words?

Your phrasing is 'by 0.99 c' while mine is 'until'. Did you mean the same as me?

Doesn't matter. You can substitute "boost until" for "boost by" in what I said, if those two expressions mean different things to you, and my statement will still be true. In either case, you still have to do the math.

Also, you have to consider, not just the effect of the boost on the ordinary velocity in the direction of the boost, but in other directions as well. A boost does not affect the 4-velocity components in perpendicular directions, but the 4-velocity is not the same as the ordinary velocity (3-velocity), and the ordinary velocity is what you appear to be interested in.

Finally, when you are specifying a boost, you have to specify what frame the boost is relative to. For example, after you've boosted in the x direction and you are boosting in the y direction, what frame is the y boost specified relative to? Is it relative to the original rest frame, before you boosted in the x direction? Or relative to your rest frame after you've boosted in the x direction?

Again, this is why physicists use math to specify these things: so there is no ambiguity about what is being specified.

syfry said:

what matters is that eventually we're going 99% light speed in each,

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.

Ibix said:

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.

Ibix said:

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.

syfry said:

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.

PeterDonis said:

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.

Ibix said:

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.)

Ibix said:

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.