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Special Relatively Questions - trying to understand it better

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#1 Gschnettler

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Posted 25 January 2021 - 09:39 AM

I think I understand it for the most part, but need help thinking through some scenarios:

 

Part 1:

Let’s say that in the future when things are more advanced, we have a space station with an observation area conducting experiments out in deep space.  And we have two launch stations, equidistant from the space station, one directly to the left and one to the right of the space station. 

 

The launch stations begin by each sending a small spacecraft toward the space station at 100mph (relative to the space station).  The launch process consists of rapidly accelerating the spacecrafts to the target speed and then letting them coast until impact.  The two spacecraft collide head on.  The space station analyzes the crash and concludes that the force of impact was equivalent to the spacecraft colliding at 200mph. The black boxes in both spacecraft also concur that the force of impact that of a 200mph collision. 

 

Then the launch stations do it again, but this time they send the spacecrafts out at 100,000mph (again, relative to the space station).  This time the space station and both black boxes measure a net impact of 200,000 mph.

This continues, ramping up speed each time until they get to 90% the speed of light.  The measure the impact and conclude that the net is a 1.8X light speed collision.  They factor in special relatively including the mass increasing of each ship as speed is increased.  There are no time adjustments necessary since the space station is motionless relative to itself. 

 

Question 1: Is this correct so far?  It doesn’t seem possible.  See part 2 below.

 

Question 2: When the spacecraft are approaching each other at 90% the speed of light, when one looks at the other, what does it perceive the other’s speed to be?   For example, I think if you were driving 50mph down the road and you measured the speed of a car coming towards you, also at 50mph, that it would seem like that car is coming at you at 100mph (and I assume that when police cars do this their radar guns must subtract the speed of the police car).  But in this case, I don’t think the spacecraft would detect the other spacecraft coming at 1.8 the speed of light since that’s above the maximum.  So, what speed would it detect?  Lots of factors going on at these speeds such as alterations of time and distance, right? 

Part 2:

 

Our curious scientists decide to run the experiment at 90% the speed of light again.  But it turns out that the space station itself is very nimble and is able to run parallel to one of the spacecraft.  They decide to do this in order to observe the collision while in motion relative to the collision point in order to get a different perspective.  The space station leaves observation equipment behind at the collision point, and the moves itself out to one of the launch stations.  The launch proceeds and the space station launches too and runs right next to one of the spacecraft.  The two spacecraft and space station all accelerate to 90% the speed of light and then coast.  During the time period while they are all coasting, the spacecraft that is next to the space station has a computer failure, loses it’s memory and reboots.  After rebooting it assesses the situation and concludes that it is motionless and just sitting next to a stationary space station.  And that another spacecraft it hurtling towards them at high speed.  Then they collide and all of the black boxes read the same as before, and detect a 1.8X light speed collision.  But a robot from the spacecraft that was next to the space station who somehow survives the collision protests and says that’s impossible....they were motionless and this other spacecraft smashed into them... and there is no way that the other spacecraft was going 1.8X the speed of light. 

 

Question 3: What’s wrong with these observations?  Would they really conclude that the collision was a net of 1.8X the speed of light?  I don’t think so.  Also, see Part 3.

 

Part 3

They continue their experiments and build a wall in space.  They launch a spacecraft at it at 99% the speed of light.  The spacecraft collides with the wall and they measure it as a collision at 99% the speed of light. 

Then they tear down the wall and go back to launching spacecraft at one another.  They accelerate them both up to 99% the speed of light and they collide creating a huge spectacle.  They measure the evidence from the collision including the black boxes and they conclude that the net speed of the collision was XXXX.

 

Question 4: What is XXXX in the situation above (approximately)?  It seems as if the max collision speed must be just shy the speed of light.  Is that right?  If collision speeds also can’t exceed the speed of light, then what happens with all collisions where, relative to an observer, two objects each moving faster than 50% the speed of light crash into each other?  You would think that the collisions would become more and more violent as speed increases as well as mass increases.  Even so, is the maximum observable collision speed only 99.999999999999etc. for all collisions, even when objects are traveling over 99% the speed of light and run into one another? 

 

Question 5: It seems odd that the collision with the wall should be a net 99% speed of light, but the collision forces of two spacecraft each going 99% the speed of light will be something like 99.9% the speed of light (since 99.9% isn’t much bigger than 99%).  Does the math work out since the curves are so steep as you get closer and closer to the speed of light that the forces involved in a collision at ~99.9% the speed of light are double the force of a collision at 99% the speed of light?

 

I know I put several possibly duplicate questions above but mainly I think it comes down to these two questions:

1) Can collisions speeds exceed the speed of light?

2) If not, how do you calculate collision speeds for objects that are headed towards each other, especially when both objects are going over 50% the speed of light relative to an observer. 

 

 


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#2 Rasfahan

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Posted 25 January 2021 - 10:04 AM

Just very briefly: 

 

Increasing speed increases the observed masses of the objects, therefore the impact energy will increase. You need to look at the right inertial system to measure the relative speeds. So, Spaceship 1 will measure a different speed (and mass) of spaceship 2 than the station. That is what the relativity principle means, there is no „general speed“, it makes sense only relative to a certain inertial system. No speed will be higher than speed of light. General relativity generalizes this to accelerated objects.



#3 lzagar

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Posted 25 January 2021 - 10:11 AM

In special relativity there is a time dilation factor, so in calculating velocities near the speed of light the time in the combined velocity is changed by the following equation.

https://keisan.casio...stem/1224059993

If you use two velocities of something near the speed of light © in the calculation you will see that the time changes and the combined velocities will still be somewhat less than C.

#4 TOMDEY

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Posted 25 January 2021 - 10:23 AM

The spacecraft will see each other approaching at sub-c because their clocks have slowed down. The finite energy of impact they compute and experience is equal, call it E. The outside observer computes their energy of impact based on what he sees... and comes up with the same E that they do. Although computed and observed momenta and power are observer-dependent... net (temporally-integrated) energy is invariant. It all comes out in the wash.   Tom



#5 David Sims

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Posted 25 January 2021 - 11:43 AM

Speeds along the same line add as follows:

 

V/c = tanh{ arctanh(V₁/c) + arctanh(V₂/c) }

 

V₁/c , V₂/c , V/c

+0.1 , +0.1 , +0.198019802

+0.6 , +0.6 , +0.882352941

+0.8 , +0.8 , +0.975609756

+0.9 , +0.9 , +0.994475138

+0.99, +0.99 , +0.9999494978

 

The space station is approaching the east impactor at V₁/c, in the reference frame of the east impactor. Meanwhile, the west impactor is approaching the space station at V₂/c, in the reference frame of the space station. The two impactors are approaching each other (in either's reference frame) at the fancy-sum speed, V/c.

 

tanh : hyperbolic tangent

arctanh : hyperbolic arctangent


Edited by David Sims, 25 January 2021 - 11:51 AM.


#6 Keith Rivich

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Posted 25 January 2021 - 06:27 PM

Speeds along the same line add as follows:

 

V/c = tanh{ arctanh(V₁/c) + arctanh(V₂/c) }

 

V₁/c , V₂/c , V/c

+0.1 , +0.1 , +0.198019802

+0.6 , +0.6 , +0.882352941

+0.8 , +0.8 , +0.975609756

+0.9 , +0.9 , +0.994475138

+0.99, +0.99 , +0.9999494978

 

The space station is approaching the east impactor at V₁/c, in the reference frame of the east impactor. Meanwhile, the west impactor is approaching the space station at V₂/c, in the reference frame of the space station. The two impactors are approaching each other (in either's reference frame) at the fancy-sum speed, V/c.

 

tanh : hyperbolic tangent

arctanh : hyperbolic arctangent

Haven't seen you post in a while...glad to see I still don't understand what you write smile.gif

 

But I'm trying!


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#7 Dark Photon

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Posted 25 January 2021 - 07:56 PM

Speeds along the same line add as follows:

 

V/c = tanh{ arctanh(V₁/c) + arctanh(V₂/c) }

 

V₁/c , V₂/c , V/c

+0.1 , +0.1 , +0.198019802

+0.6 , +0.6 , +0.882352941

+0.8 , +0.8 , +0.975609756

+0.9 , +0.9 , +0.994475138

+0.99, +0.99 , +0.9999494978

 

The space station is approaching the east impactor at V₁/c, in the reference frame of the east impactor. Meanwhile, the west impactor is approaching the space station at V₂/c, in the reference frame of the space station. The two impactors are approaching each other (in either's reference frame) at the fancy-sum speed, V/c.

 

tanh : hyperbolic tangent

arctanh : hyperbolic arctangent

This is certainly right, however the velocity addition formula simplifies and does not require hyperbolic functions:

 

v_relative = ( v1 + v2 )/( 1+v1 * v2/c^2 ), where v1 and v2 are measured in some common reference frame  (the space station or the wall in the original post) and assumed to move in the opposite directions in that frame. Again, in the original post v1=v2=v, so v_relative=2 v/(1+v^2/c^2), which never exceeds c. (By v_relative I mean the speed of one spaceship as observed from the other.)

 

The kinetic energy in special relativity is modified as well. It is not 1/2 m v^2, but instead is m c^2 (1/sqrt(1-v^2/c^2)-1), so it increases without any bounds as v approaches c. Thus, the faster the original spaceships the larger the fireworks when the two collide even though the relative velocity does not change by much, as illustrated in David Sims' table. There is not much difference in v_relative when v=0.9 compared to v=0.99, but the difference in the kinetic energy is huge.


Edited by Dark Photon, 25 January 2021 - 07:57 PM.


#8 David Sims

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Posted 27 January 2021 - 11:25 PM

My college physics teacher taught me the 

 

V/c = (V₁/c + V₂/c) / (1 + V₁V₂/c²)

 

formula when I was a sophomore in 1980. I wrote out the arctangent form because it has each speed input only once, rather than twice. This simplifies the case where you have N speeds being added along the same line:

 

V/c = tanh{ Σ(i=1,N) arctanh(Vᵢ/c) }


Edited by David Sims, 27 January 2021 - 11:28 PM.



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