Special relativity application

[timetraveller123]

Homework Statement

The rockets of the Goths and the Huns are each 1000 m long in their respective rest frame. The rockets pass each other, virtually touching, at relative speed of 0.8 c. The Huns have a laser cannon at the rear of their rocket that shoots a deadly laser beam at right angles to the motion. The captain of the Hun rocket wants to send a threatening message to the Goths by “firing a shot across their bow.” He tells his first mate, “The Goths rocket is length contracted to 600 m. Fire the laser cannon at the instant the nose of our rocket passes the tail of their rocket. The laser beam will cross 400 m in front of them.” But things are different from the Goths’perspective. The Goth captain muses, “The Huns’ rocket is length contracted to 600 m, 400 m shorter than our rocket. If they fire the laser cannon as their nose passes the tail of our rocket, the lethal laser blast will go right through our side.”The first mate on the Hun rocket fires as ordered. Does the laser beam blast the Goths or not? Resolve this paradox. Show that, when properly analyzed, the Goths and the Huns agree on the outcome. Your analysis should contain both quantitative calculations and written explanation.

while i understand the question i fail to realize how to work out the problem from an inertial frame of
reference

Homework Equations

L = Lo /ϒ
where
ϒ = 1/√(1 - V^2 / C^2)

t = to ϒ

The Attempt at a Solution

i need some help in transferring the frame of reference to inertial one please and then i will be more than happy to work out the problem please thanks!

 

[PeroK]

You need to try to solve this yourself. First, think about this:

In what frame of reference is the question most easily answered? What happens in this frame of reference?

Then, once you've done that, you can analyse the problem from the other frame of reference.

 

[timetraveller123]

ok sure i will do it do i have to use lorentz transformation equations for this question

 

[PeroK]

ok sure i will do it do i have to use lorentz transformation equations for this question

You don't need the Lorentz transformation to analyse the problem in a single reference frame. That's the point.

 

[Chestermiller]

You don't need the Lorentz transformation to analyse the problem in a single reference frame. That's the point.

Hi Perok. Even though the Lorentz transformation is not needed, I think it would be instructive to the OP if he applied the Lorentz Transformation to this problem.

 

[PeroK]

Hi Perok. Even though the Lorentz transformation is not needed, I think it would be instructive to the OP if he applied the Lorentz Transformation to this problem.

Yes, it can be used for the analysis in the second frame. I was trying to get the OP to look for a clear and unambiguous solution in a single frame; then, to use the LT, or otherwise, to analyse the problem in the second frame and resolve the "paradox".

@vishnu 73 let us know if you need any more help on this.

 

[timetraveller123]

Yes, it can be used for the analysis in the second frame. I was trying to get the OP to look for a clear and unambiguous solution in a single frame; then, to use the LT, or otherwise, to analyse the problem in the second frame and resolve the "paradox".

@vishnu 73 let us know if you need any more help on this.

okay i think i want to analyze using Lorentz transformation but i how should i start what is my event what is my stationary reference and what is my moving reference frame and i am guessing it has got something to do with the simultaneity of events does it?

EDIT:

i don't want to use an inertial reference frame anymore i decided to use the huns spaceship as the stationary reference frame and fixed the origin of coordinate system at their head. moving from the head to tail is considered positive hence the groths are moving relative to the coordinate system at +0.8 c. so the two events here are the crossing of the groths tail at huns head ie when groths tail is at x = 0 and the shooting of laser at x = 1000. the crossing event is at same time for both spaceships as x =0 and t =0. but for shooting event in huns reference frame it is at t = 0 . but for the groths it is at t' . where t' = ϒ(t - v x/c^2) where t = 0 , x = 1000 , v = +0.8 c and ϒ can be calculated to be 5/3. hence calculation got me t' = -4.44 x 10^-6 which means the laser was shot 1465.6m infront of the groths in the huns reference frame am i right

 

[Chestermiller]

What did you get for x' at x = 1000, t = 0 from the Lorentz Transformation?

 

[timetraveller123]

What did you get for x' at x = 1000, t = 0 from the Lorentz Transformation?

i am sorry i don't get your question

 

[Chestermiller]

The coordinates of the event of the laser cannon firing are x = 1000, t = 0 (as reckoned from the huns frame of reference). What are the spatial coordinate x' of this same event as reckoned from the goth's frame of reference: ##x'=\gamma(x-vt)##? What does this equation give you for x'?

How does this compare with the coordinate of the tail of the goth's rocket x' = 0 and the nose of the goth's rocket x'=1000?

 

[timetraveller123]

sir i am really sorry but i don't understand why are you asking this because since in goth frame of reference the shot was fired 4.44 10^-6 s before the huns head crossed their tail then one could simply find out the distance from the huns tail to goth tail at t= 4.44 x 10^-6 s finding out that the shot in huns reference frame was 667 m in front of them thus not hitting them i don't understand your approach at this problem please explain as i am relatively new to special relativity and lorentz is not my strong hand sorry once again i have explained my thought process below

in hun reference:
his ship is 1000 m and goths ship is 600 m
huns tail and huns head is at x = 0;
his head crossing goths tail is simultaneous with the laser being fired at t = 0
thus the shot will go 400 m in front of goths head.

in goths reference frame:
his ship is 1000 and hun ship is 600
crossing of goth head and huns tail occurs at t = 0
firing of laser beam occurred at t = - 4..4 10^-6 ie the shot was fired 4.44 10^6 s before crossing

so for the goths the shot was fired 4.44 x 10^-6 x 0.8 c + 600 -1000 m in-front of them thus 666 m in front of the goths head

is there anything wrong with this thanks! and i checked the answer it was 667 so maybe just calculation error

 

[Chestermiller]

Well, here's how I would have approached it:

Event 1: x = 0, t = 0, x' = 0, t' = 0

Event 2: x = 1000, t = 0
$$x'=\gamma (x-vt)=\frac{5}{3}(1000)=1667$$
$$t'=\gamma (t-\frac{vx}{c^2})=\frac{5}{3}(-\frac{(1000)(0.8)}{c})=-4.44E-6$$
Nose of Goth ship is at x' = 1000, so cannon fired 667 m in front of nose of goth ship, 4.44E-6 seconds before tail of goth ship coincides with nose of hun ship.

Diagram of sequence of events as reckoned from huns frame:

Diagram of sequence of events as reckoned from goth's frame:

 

[timetraveller123]

oh wow now i see why your method is easier thanks sir and how did come up with the images by yourself if so what software did you use thanks.

and since we are talking about relativity i want to ask a very maybe dumb question for a long time

why is speed of light constant regardless of relative motion and how have we proved it i know that c = 1/√μo ∈o but that does not prove the claim as similarly speed of sound also can be written as √β/ρ so why is speed of light constant for all observer thanks!

 

[Chestermiller]

oh wow now i see why your method is easier thanks sir and how did come up with the images by yourself if so what software did you use thanks.

I used Powerpoint to draw the diagrams. Then I used the Snipping Tool (Windows app/program available on all PCs) to capture the image. Then I saved the image to my desktop. Then I hit the Physics Forums UPLOAD button and I selected the saved image from my desktop. Then I chose for it to be displayed full size.

and since we are talking about relativity i want to ask a very maybe dumb question for a long time

why is speed of light constant regardless of relative motion and how have we proved it i know that c = 1/√μo ∈o but that does not prove the claim as similarly speed of sound also can be written as √β/ρ so why is speed of light constant for all observer thanks!

This goes back to the late 1800s, early 1900s. James Clerk Maxwell derived the equations for the propagation of electromagnetic waves, and the constant you refer to was in these equations as the velocity of propagation. People immediately recognized that the value of this constant was basically equal to the then measured values of the speed of light. They also recognized that this equation should be the same in all inertial frames of reference. Back then, it was still through that there was a preferred reference frame that was stationary, and that's where the constant should apply. But for that to happen, the constant in Maxwell's equation would, by coincidence have to have been determined in that frame. But, the Earth is moving around the sun, and the sun is moving in the galaxy, so the likelihood that Maxwell's laboratory was in the preferred stationary frame when he made his measurements was very remote. To test this further, Michaelson and Moreley did experiments in which they measured the speed of light in the laboratory at several different times of the year, when the Earth was moving in different relative directions, and they found that, in all cases, the speed of light was the same (for the equipment held in the same orientation). So we are back to the equation, "why is the speed of light the same in all reference frames?" The answer is because of the unique geometry of 4D space time. The independence of the speed of light with reference frame was the clue that led Hermann Minkowski (Einstein's former math professor) to deduce the geometry of spacetime from the Lorentz Transformation. So, the constancy of the speed of light is not the cause of relativistic effects, it is only one of the effects that derives for the geometry of spacetime.

 

[PeroK]

Just to add something. Here's how I would have begun to answer the question that was asked:

In the Hun frame, we know that the nose-to-tail event and the firing of the beam are simultaneous. We also know that the length contraction of the Goth ship is a valid measurement (in the Hun frame). We, therefore, have all the information in the Hun frame to conclude that the warning shot misses the Goth ship.

As this is a physical result, it must be observed in all reference frames.

Now, to resolve the paradox, we analyse the events in the Goth frame, and show that the shot also misses in that frame (which it inevitably must) ...

 

[timetraveller123]

I used Powerpoint to draw the diagrams. Then I used the Snipping Tool (Windows app/program available on all PCs) to capture the image. Then I saved the image to my desktop. Then I hit the Physics Forums UPLOAD button and I selected the saved image from my desktop. Then I chose for it to be displayed full size.

This goes back to the late 1800s, early 1900s. James Clerk Maxwell derived the equations for the propagation of electromagnetic waves, and the constant you refer to was in these equations as the velocity of propagation. People immediately recognized that the value of this constant was basically equal to the then measured values of the speed of light. They also recognized that this equation should be the same in all inertial frames of reference. Back then, it was still through that there was a preferred reference frame that was stationary, and that's where the constant should apply. But for that to happen, the constant in Maxwell's equation would, by coincidence have to have been determined in that frame. But, the Earth is moving around the sun, and the sun is moving in the galaxy, so the likelihood that Maxwell's laboratory was in the preferred stationary frame when he made his measurements was very remote. To test this further, Michaelson and Moreley did experiments in which they measured the speed of light in the laboratory at several different times of the year, when the Earth was moving in different relative directions, and they found that, in all cases, the speed of light was the same (for the equipment held in the same orientation). So we are back to the equation, "why is the speed of light the same in all reference frames?" The answer is because of the unique geometry of 4D space time. The independence of the speed of light with reference frame was the clue that led Hermann Minkowski (Einstein's former math professor) to deduce the geometry of spacetime from the Lorentz Transformation. So, the constancy of the speed of light is not the cause of relativistic effects, it is only one of the effects that derives for the geometry of spacetime.

i am aware that the constant is speed of light and how to derive that and i familiar with michelson morley experiment but i don't understand how does that experiment prove the claim as light may be moving relative to the Earth thus may not display any interference pattern in the experiment that is much like me moving in bus and if i jump i don't hit the back simply not because my velocity is same in all reference frame but just that i am moving relative to the bus thus i continue moving in front and don't hit the back i hope you under stand my confusion and anyways thanks to all of you regarding my question thanks!

 

[Chestermiller]

i am aware that the constant is speed of light and how to derive that and i familiar with michelson morley experiment but i don't understand how does that experiment prove the claim as light may be moving relative to the Earth thus may not display any interference pattern in the experiment that is much like me moving in bus and if i jump i don't hit the back simply not because my velocity is same in all reference frame but just that i am moving relative to the bus thus i continue moving in front and don't hit the back i hope you under stand my confusion and anyways thanks to all of you regarding my question thanks!

Physics Forums has a FAQ that discusses all the experimental evidence that verifies special relativity. It is chock full of historical information which addresses in detail your questions. I forget where it is kept, but I would start by looking under NEWS. I don't know enough about the history to comment further.

 

[Chestermiller]

I think it's interesting to note that, as reckoned from the goths frame of reverence, when the huns fire their cannon, the noses of the two rockets have not even passed each other yet, let alone any other parts of the rockets.

 

[timetraveller123]

I think it's interesting to note that, as reckoned from the goths frame of reverence, when the huns fire their cannon, the noses of the two rockets have not even passed each other yet, let alone any other parts of the rockets.

ok thanks for the help