Addition of velocities at speeds near C - any differences?

[Alex Pavel]

Good evening,

If a ship is moving at 0.5C east in a gravity free environment, and fires a bullet perpendicular at 0.5c, is it just the hypotenuse (a squared + b squared)/c squared that we use for the speed?

Or is there more to it since we are at great speed?

 

[NFuller]

At speeds approaching the speed of light, velocities add according to the Einstein velocity addition relationships.

See here for a brief description http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html

 

[Alex Pavel]

Does 0.5c north + 0.5c east = 0.7c Northeast? The link doesn't tackle perpendicular velocity additions.

 

[NFuller]

No, it is different if the velocities are not parallel. I just found a derivation for perpendicular velocities.
https://arxiv.org/pdf/physics/0612191.pdf
Also, there is another thread dealing with this same type of problem
https://www.physicsforums.com/threads/relativistic-addition-of-orthogonal-velocities.515174/

 

[PAllen]

I wrote a post on this a while back. I give a simple special case for orthogonal orientation.

https://www.physicsforums.com/threads/generalized-velocity-addition-and-aberration-formulas.821733/

 

[Alex Pavel]

Thanks.

So, the observer on the ship still sees the bullet moving at 0.5c, even though the ship is moving 0.3. Addition of velocities keeps it on target and same speed relative to observer on ship, but observer outside sees it diagonally higher speed.

 

[pervect]

Does 0.5c north + 0.5c east = 0.7c Northeast? The link doesn't tackle perpendicular velocity additions.

Yes. The sum is $\frac{\sqrt{2}}{2}\,c$, appoximately .707c.

 

[PAllen]

No, it is different if the velocities are not parallel. I just found a derivation for perpendicular velocities.
https://arxiv.org/pdf/physics/0612191.pdf
Also, there is another thread dealing with this same type of problem
https://www.physicsforums.com/threads/relativistic-addition-of-orthogonal-velocities.515174/

note, I think that paper is deriving something different from what he OP wants. It appears to derive the transformation of perpendicular components, rather than the overall speed of a boost applied to an orthogonal velocity. My link gives directly what the OP wants, I think.

 

[Alex Pavel]

Yes. The sum is $\frac{\sqrt{2}}{2}\,c$, appoximately .707c.

Okay - a bullet moving 0.5c north inside a train also going 0.5c east would be seen by an outside of the train observer at 0.707c northeast.
For an observer on the train - bullet still perceived at 0.5 c, and despite the added motion, the bullet would still strike the same target on the train moving at the same speed.

Even the length contraction would be accounted for in this case, as it also affects the target.

 

[Ibix]

So, the observer on the ship still sees the bullet moving at 0.5c, even though the ship is moving 0.3.

According to an observer on the ship, the ship isn't moving. Shooting is just the same as anywhere else. It's only when observed from a frame where the ship is in motion that it starts to look more complicated.

 

[Alex Pavel]

According to an observer on the ship, the ship isn't moving. Shooting is just the same as anywhere else. It's only when observed from a frame where the ship is in motion that it starts to look more complicated.

Awesome - got it it just clicked.

Other thing I was worried about was the need for a Lorentz transformation here - but it doesn't apply, as we remain in the same reference frame, therefore the Pythagorean theorem holds for this specific example.

 

[Alex Pavel]

According to an observer on the ship, the ship isn't moving. Shooting is just the same as anywhere else. It's only when observed from a frame where the ship is in motion that it starts to look more complicated.

One last question, and as far as my purposes, the thread can be closed.
Is there any way for the guy on the train to determine he is moving, assuming he cannot see outside of the train?

 

[PAllen]

One last question, and as far as my purposes, the thread can be closed.
Is there any way for the guy on the train to determine he is moving, assuming he cannot see outside of the train?

No, that’s the essence of the principle of relativity.

 

[Alex Pavel]

I really appreciate all of the help here tonight. Cheers!

 

[PAllen]

Yes. The sum is $\frac{\sqrt{2}}{2}\,c$, appoximately .707c.

That is not correct. Consider the limit as the speeds approach c - you would get c√2

Correct, per the formula given in my linked post above, is ((√7)/4 ) c, about .6614 c.

 

[Alex Pavel]

So the speed is not 0.7.

There are a lot of formulas in that link and it is hard to follow, and some if what is written there is incorrect.

Can you tell me the exact formula to determine speed of two velocities, both 0.5C, one on the x and one on the y axis?
In any case, the impact location remains unchanged due to addition of velocity and the principle of relativity, holds.

 

[Alex Pavel]

If someone can show me where to find an example of this problem worked out, step by step, I would be very grateful.

 

[PAllen]

So the speed is not 0.7.

There are a lot of formulas in that link and it is hard to follow, and some if what is written there is incorrect.

Can you tell me the exact formula to determine speed of two velocities, both 0.5C, one on the x and one on the y axis?
In any case, the impact location remains unchanged due to addition of velocity and the principle of relativity, holds.

In the link I gave to my old post on this, about half way down I give a formula specifically for the orthogonal case.

Also, of note, is that the direction of the resulting velocity would be about 40.89 degrees rather than 45, due to a type of aberration.

 

[Alex Pavel]

Yes, the aberration is associated with length in the direction of travel, contracting, correct?

 

[Alex Pavel]

all of the V's in this case are .5c. Is this it?

 

[Alex Pavel]

View attachment 219088

all of the V's in this case are .5c. Is this it?

And it is less than 0.7, due to the length contraction in the X axis. In any case, impact location remains the same.

 

[PAllen]

View attachment 219088

all of the V's in this case are .5c. Is this it?

Not quite. I use c=1 convention. So all velocities are over c. Putting c back, would give

√ (u²/ c² + v²/c² - u²v²/c⁴)

[edit: wait, you are correct, because what I just wrote above would be multiplied by c to give a result speed rather than speed over c. Once you do that, you get your formula. ]

 

[PAllen]

View attachment 219090

And it is less than 0.7, due to the length contraction in the X axis. In any case, impact location remains the same.

Yes, that is the correct answer, though earlier I gave it exactly rather than approximate.

 

[Alex Pavel]

Not quite. I use c=1 convention. So all velocities are over c. Putting c back, would give

√ (u²/ c² + v²/c² - u²v²/c⁴)

Okay, and you got the .6614, same as I got. Yes I am using c=1, but the V, they are all 0.5. IE, v/c = 0.5/1/

 

[sweet springs]

If someone can show me where to find an example of this problem worked out, step by step, I would be very grateful.

Though I do not find explicit formula in texts, the matrix representing this transformation should be the products of matrices,

(-90 degree x-y rotation around the origin)*(0.5c boost in x-direction)*(90 degree x-y rotation around the origin)*(0.5c boost in x-direction).

Best.

 

[Ibix]

Can you tell me the exact formula to determine speed of two velocities, both 0.5C, one on the x and one on the y axis?

It's fairly straightforward to derive. If the bullet is fired at the origin at time zero then in one frame you can write $x=u_xt$, $y=u_yt$, and in the other frame you can write $x'=u'_xt'$, $y'=u'_yt'$. Then you use the Lorentz transforms to relate x, y, t to x', y', t' and solve for the velocity components in whichever frame you don't know them.

 

[PAllen]

Though I do not find explicit formula in texts, the matrix representing this transformation should be the products of matrices,

(-90 degree x-y rotation around the origin)*(0.5c boost in x-direction)*(90 degree x-y rotation around the origin)*(0.5c boost in x-direction).

Best.

Way back in post #5, I gave a link for explicit formulas for the general case, as well as for the orthogonal special case.

 

[Orodruin]

If a ship is moving at 0.5C east in a gravity free environment, and fires a bullet perpendicular at 0.5c, is it just the hypotenuse (a squared + b squared)/c squared that we use for the speed?

Yes. The sum is $\frac{\sqrt{2}}{2}\,c$, appoximately .707c.

That is not correct. Consider the limit as the speeds approach c - you would get c√2

Correct, per the formula given in my linked post above, is ((√7)/4 ) c, about .6614 c.

The problem here is in the OP, which does not explicitly specify relative to what the bullet is traveling at 0.5c, relative to what the ship travels at 0.5c, and relative to what we want to know the bullet speed. I think it can be reasonably assumed that we want to know the bullet speed relative to whatever the ship's speed is measured, let us call it "S". The remaining question is whether the 0.5c and perpendicularity of the bullet's velocity is to be taken in the ship's frame or in S, which to me is not clear from the question.

If the 0.5c and perpendicularity is given in S, then @pervect is indeed correct. However, if it is given in the ship's rest frame, then you need to use relativistic addition of velocities. This would also tell you that you cannot fire a bullet in such a way that its velocity component perpendicular to the ship's motion exceeds $\sqrt{c^2 - v^2}$ if the bullet is to have the ship's velocity $v$ as its parallel component. Of course, this is just the trivial statement that the bullet cannot travel faster than $c$.

The easiest way to compute a relative velocity is to consider the 4-velocities of the observer and the object. Taking their inner product will directly give you their relative gamma factor from which you can solve for the relative velocity.

 

[PAllen]

To me, it is perfectly clear that what is sought is the perpendicular analog of velocity addition. A ship and some othe observer have some mutual relative velocity. Per the ship, something is fired at .5c orthogonal to the relative velocity. Then what is the speed relative to the non ship observer. Any other interpretation is either trivial or tortures. That is the case I answered, and it seems clear the OP wanted that case.

 

[Orodruin]

To me, it is perfectly clear that what is sought is the perpendicular analog of velocity addition. A ship and some othe observer have some mutual relative velocity. Per the ship, something is fired at .5c orthogonal to the relative velocity. Then what is the speed relative to the non ship observer. Any other interpretation is either trivial or tortures. That is the case I answered, and it seems clear the OP wanted that case.

To be honest, I am not sure that the OP is aware of what he wanted - as is many times the case when people ask relativity related questions. The figure in the OP does not help in any way to resolve this issue (it seems to depict two velocity components in S) and I think it is very important to point this out as it relates to a fundamental understanding of how relativity works. We both know that @pervect has the required background and understands relativity well enough and clearly he made a different interpretation of the OP than you did. I believe it is important to point out to the OP why he has been getting different replies and why they are both valid interpretations of his question.