Relativistic momentum

[granpa]

Two particles moving very close to the speed of light in the north-south direction collide and end up moving in opposite directions due east and west.

 

Each starts with gamma=22.37 and end up with exactly gamma=22.37. Total momentum is zero at all times.

 

v = 0.999

gamma = 22.37

 

Now from the point of view of a rocket moving at 0.999c the two particles come together Collide and then one particle becomes stationary while the other flies away with all of the momentum.

It should be possible from this thought experiment to determine the equation for relativistic momentum

 

velocity of final nonstationary particle from the point of view the the rocket is calculated by

 

velocity addition formula = (u+v)/(1+uv)

 

(0.999+0.999)/(1+0.999*0.999) = 0.9999994995

 

v = 0.9999994995

gamma = 999.5

 

 

 

 

velocity of 2 particles before collision from the point of view of the rocket is calculated by

 

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

 

ux = 0.999

uy = 0

 

vx = 0

vy = 0.999

 

v=sqrt((0.999-0)^2 + (0-0.999)^2 - (0.999^2)(0.999^2))

v=0.9999980019975039930129749248984603081269824945912709

 

gamma = 500

 

 

 

This makes sense in terms of energy conservation since kinetic energy equals (gamma - 1)*mc^2

 

500 + 500 = 1000

 

But it doesn't make any sense in terms of momentum if momentum = gamma*mv

 

 

However, everything works fine if Momentum = sqrt(kinetic energy)

So my guess is momentum = sqrt(gamma)*mv

 

 

corrected the image

Edited June 30, 2016 by granpa

[imatfaal]

[latex]v_y=\frac{\sqrt{1-\frac{V_x^2}{c^2}v^{'}_y}}{1+\frac{V_x}{c}v^{'}_x}[/latex]

 

I don't think you have calculated the velocities correctly - the velocity in the y axis in the rocket's frame which is travelling at velocity V_x with respect to the primed frame would be as above.

 

 

[granpa]

In the second image each particle has sqrt(500)/sqrt(2) = 15.8 momentum in the east west direction
The final particle has sqrt(1000) = 31.6

 

 

I didnt calculate vy

I used the formula for total velocity given here

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

see the section "relative speeds"

 

since the velocities were orthoginal it simplified greatly

Edited June 30, 2016 by granpa

[imatfaal]

In the second image each particle has sqrt(500)/sqrt(2) = 15.8 momentum in the east west direction

The final particle has sqrt(1000) = 31.6

 

 

I didnt calculate vy

I used the formula for total velocity given here

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

see the section "relative speeds"

 

since the velocities were orthoginal it simplified greatly

"If an observer A measures two objects B and C to be travelling at velocities u = (u_x, u_y, u_z) and v = (v_x, v_y, v_z) respectively, one may ask the question of what the relative speed between B and C are, or in other words at what speed w B would measure C to be travelling at, or vice versa"

That section is to calculate the Speed of C in the rest frame of B

[granpa]

Yes the speed of the 2 particles from the point of view of the rocket.

 

I assumed 45 degree angle but I'm no longer sure about that

If the angle were very much shallower than that would explain how momentum is conserved and would not require momentum to be the square root of gamma

Edited June 30, 2016 by granpa

[Sensei]

2 particles input, 1 particle output?

Very unlikely, as it would not most likely conserve momentum.

 

2 particles input, 2 particles output?

Yes. It will work.

[granpa]

There are 2 particles output

One is stationary

The stationary particle in the second image is the same particle in the first image that is going off to the left

Edited June 30, 2016 by granpa

[Sensei]

There are 2 particles output

One is stationary

The stationary particle in the second image is the same particle in the first image that is going off to the left

Whether something is stationary or not, depends on FoR you choose.

[granpa]

Yes thats the whole point.

From the point of view of the rocket the particle is stationary

 

There isn't a stationary particle in the center to begin with but if there were then that particle from the Rockets point of you would be moving very close to the speed of light and the two particles moving in would, at every point in time, be perfectly in line with it therefore those two particles cannot be moving in at a 45 degree angle. I don't know what the angle is but it must be very very tiny. Microscopic even. This means that momentum must equal gamma*mv

Edited June 30, 2016 by granpa

[imatfaal]

Yes thats the whole point.

From the point of view of the rocket the particle is stationary

 

....

 

And you are getting the frames wrong - you are using the formula which takes velocities in A's frame and gives you the speed of B relative to C.

 

What you need to do is the formula which takes velocities in the centre of mass frame of B and C (ie the one in which they are travelling equal and opposite velocities) and transform that to give a velocity in A. Not the same no matter how much you say it is. I provided one of the formalae for one basis above

[granpa]

And you are getting the frames wrong - you are using the formula which takes velocities in A's frame and gives you the speed of B relative to C.

 

 

Velocity of B (2 particles) in A's frame = 0.999c

Velocity of C (rocket) in A's frame = 0.999c

Velocity of B in C's frame = velocity of the two particles from the point of view of the rocket = 0.999998c

 

What are you not getting?

Edited July 1, 2016 by granpa

[Janus]

Yes thats the whole point.

From the point of view of the rocket the particle is stationary

 

There isn't a stationary particle in the center to begin with but if there were then that particle from the Rockets point of you would be moving very close to the speed of light and the two particles moving in would, at every point in time, be perfectly in line with it therefore those two particles cannot be moving in at a 45 degree angle. I don't know what the angle is but it must be very very tiny. Microscopic even. This means that momentum must equal gamma*mv

The angle works out to be ~2.56 degrees.

 

From the rocket's perspective, the line joining the two particles is moving at 0.999c in the y direction. In the line's frame the each particle is moving at 0.999c in the x direction. It is a simple time dilation problem to determine the x velocity of the particles in the Rocket frame. this works out to 0.044665 c

 

The arctan of 0.044665c/0.999c = ~2.56 degrees.

 

The magnitude of the velocity of each particle relative to the rocket is 0.999998 c at an angle of 2.56 degree to the relative motion of the rocket and the line joining the particles.

0,99998 c, assuming a mass of 1 gives a relativistic momentum of 0.999998 x 500(gamma factor of 0.999998c)= ~499.999. That momentum is divided into a x component and a y component. the y component is cos(2.56)*499.999 =499.4999...

Doubling that gives the sum y component momentum of both particles. 998.999...

 

After the collision, from the rocket frame 1 particle has zero velocity and zero momentum and the other has a velocity of 0.999999499...c the gamma factor for this speed is ~999.499, which gives a momentum of ~999.499 for the momentum for the single particle. Now this is a tad higher than the 998.999 we got for the combined y momentum before the collision, but we did do some rounding along the way.

[granpa]

Thank you very much

[granpa]

The angle works out to be ~2.56 degrees.

 

From the rocket's perspective, the line joining the two particles is moving at 0.999c in the y direction. In the line's frame the each particle is moving at 0.999c in the x direction. It is a simple time dilation problem to determine the x velocity of the particles in the Rocket frame. this works out to 0.044665 c

 

The arctan of 0.044665c/0.999c = ~2.56 degrees.

 

The magnitude of the velocity of each particle relative to the rocket is 0.999998 c at an angle of 2.56 degree to the relative motion of the rocket and the line joining the particles.

0,99998 c, assuming a mass of 1 gives a relativistic momentum of 0.999998 x 500(gamma factor of 0.999998c)= ~499.999. That momentum is divided into a x component and a y component. the y component is cos(2.56)*499.999 =499.4999...

Doubling that gives the sum y component momentum of both particles. 998.999...

 

After the collision, from the rocket frame 1 particle has zero velocity and zero momentum and the other has a velocity of 0.999999499...c the gamma factor for this speed is ~999.499, which gives a momentum of ~999.499 for the momentum for the single particle. Now this is a tad higher than the 998.999 we got for the combined y momentum before the collision, but we did do some rounding along the way.

that is not a rounding error and your math checks out perfectly

 

 

Energy doesnt work either

 

Kinetic energy = (gamma-1)*mc^2

 

500-1 + 500-1 = 998

 

999.5-1 = 998.5

Edited July 2, 2016 by granpa

[granpa]

Whoops

Proper velocity of final particle = (1/sqrt(1-0.999^2))^2*(0.999+0.999) = 999.5

 

2.55997282 degrees

Cos(2.55997282 degrees) = 0.9990020

I see what it is now.

At that velocity the numbers are very sensitive to very small changes of v

It is a simple rounding error after all.

 

 

 

using wolfram this time:

 

v=0.999

gamma=22.36627204212922171066204252228498460806277844074433005975

 

(0.999+0.999)/(1+0.999*0.999) = 0.999999499499750000125125062499968718734375007820316406248

gamma = 999.500250125062531265632816408204102051025512756378

 

v=0.044665467634404097885413809579045636547285976608944867873

 

angle = 0.044680421611842281640634476015442417291764888749050529147 radians

 

cos = 0.999001996008481540908207900713241305831826857382633612841

 

velocity of 2 particles = 0.999998001997503993012974924898460308126982494591270955936

 

gamma = 500.2501250625312656328164082041020510255127563781891

 

momentum = 500.2491255615327628879314281633118099684754850381545606983

 

cos * momentum = 499.7498749374687343671835917958979489744872436218109108996

 

2 * cos * momentum = 999.4997498749374687343671835917958979489744872436218217993

 

gamma = 999.500250125062531265632816408204102051025512756378

 

close enough

 

 

 

and energy works too

 

2*(500.25 - 1) = 998.5

 

999.5 - 1 = 998.5

 

 

https://en.wikipedia.org/wiki/File:Relativistic_Energy_and_Momentum.png

 

 

Well it was a long winding and bumpy road but this thought experiment is now officially a total success

Edited July 2, 2016 by granpa

[granpa]

Thank you everyone.

With your help my understanding of Relativity has increased tremendously in the last few days.

 

I created this:

https://commons.wikimedia.org/wiki/File:Relativistic_Energy_and_Momentum.png

Maybe others will find it useful

 

Edited July 3, 2016 by granpa