Relativity Question: Airliner -- Light & ball movement

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SUMMARY

This discussion focuses on the application of special relativity principles to a scenario involving an airliner flying at 300 m/s. When a ball is thrown backward at 20 m/s, its velocity relative to the Earth is calculated as 280 m/s forward. In contrast, light maintains a constant speed of 3.0 x 108 m/s regardless of the reference frame. Time dilation effects are negligible at this speed, meaning a watch on the plane would not show a significant time difference compared to one on Earth.

PREREQUISITES
  • Understanding of special relativity concepts, including time dilation and inertial reference frames.
  • Familiarity with the equation Δtm = Δts/√(1-v2/c2).
  • Knowledge of relativistic velocity addition principles.
  • Basic grasp of vector subtraction in physics.
NEXT STEPS
  • Study the implications of relativistic velocity addition in various scenarios.
  • Explore the concept of inertial versus non-inertial frames of reference.
  • Research time dilation effects at different speeds, particularly near the speed of light.
  • Examine practical applications of special relativity in modern technology, such as GPS systems.
USEFUL FOR

Students of physics, educators teaching special relativity, and anyone interested in understanding the implications of motion at high speeds.

Jaimie
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Homework Statement


Imagine that you are flying on an airliner on a long flight to Europe at a constant speed of 300 m/s

a) you throw a ball towards the back of the plane at 20 m/s. You then shine a beam of light towards the back of the plane. How will these two things i) the ball and ii) the light-- appear to move from the Earth's frame of reference

b) would you expect your watch to be affected by time dilation?

Homework Equations


Δtm = Δts/√(1-v2/c2)
V(be)= -V(bp) + V(pe)

The Attempt at a Solution


a) i) b= ball; e= earth; p= plane
∴ V(be)= -V(bp) + V(pe)= -20 m/s + 300 m/s = 280 m/s [forward].

ii) The light will remain at a constant 3.0 x 10^8 m/s from the earth’s frame of reference.

Velocity is dependent on its’ inertial reference frame and its’ direction of travel. From the Earth frame of reference, both the direction of the thrown ball and the direction of the airplane travel determines its’ velocity as per vector subtraction. If we were to look at the ball moving from a different inertial frame of reference (i.e. from the plane), the ball would experience only the velocity of 20m/s [backwards] and if the ball was thrown forward, the velocity of the ball would be 20 m/s + 300 m/s= 320 m/s [forward]. Light on the other hand is independent of its’ reference frame and shows a constancy of 3.0 x 10^8 m/s relative to all inertial reference frames (special relativity postulate 2), regardless of which direction it is travelling. If we were to measure the speed of light from inside the plane or if it were traveling back or forwards, its’ velocity would still be 3.0 x 10^ 8 m/s.

b) If time was being measured from the earth’s frame of reference, time dilation would still occur relative to the time experienced from the airplane. However, because the plane is not traveling at a speed close enough to the speed of light, increasing the time difference between both reference frames, time difference and time dilation would be negligible. As such, my watch would not be accurate enough to pick up the time difference. If time was measured from the inside the plane, my watch would only experience normal time and would not experience time dilation at all.

Could someone review my answers and tell me if I am understanding these concepts?
Thank you for your time!
 
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Jaimie said:
∴ V(be)= -V(bp) + V(pe)= -20 m/s + 300 m/s = 280 m/s [forward].
As this is a relativity question, I would use the relativistic formula and sufficient precision to see the difference to the nonrelativistic case.
Jaimie said:
ii) The light will remain at a constant 3.0 x 10^8 m/s from the earth’s frame of reference.
Sure.
Jaimie said:
b) If time was being measured from the earth’s frame of reference, time dilation would still occur relative to the time experienced from the airplane. However, because the plane is not traveling at a speed close enough to the speed of light, increasing the time difference between both reference frames, time difference and time dilation would be negligible. As such, my watch would not be accurate enough to pick up the time difference. If time was measured from the inside the plane, my watch would only experience normal time and would not experience time dilation at all.
Right. There are watches precise enough to note the time dilation, however.
 
Hi mfb,
Re a) i & ii) :
At this point we weren't taught about inertial/non-inertial frames of reference, time dilation and simultaneity, with no other equations other than that for time dilation. So based on this info, are both these answers correct? Thank you!

(I'm curious though, which equation(s) were you referring to?)
 
Should be fine.

Relativistic velocity addition.
You need a modified formula in relativity - this becomes more easy to see if your ball moves with (speed of light minus 10m/s) forwards in the plane, for example. The classical addition would give a speed above the speed of light, which is obviously wrong. The correct relativistic formula also allows to plug in the speed of light, and the result is the speed of light again as expected.
 
*Correction*
"At this point we were taught only about inertial/non-inertial frames of reference, time dilation and simultaneity..." Had to fix that in my last post.

So when you're saying "should be fine", you're saying that based on what I learned, my answer to part a i) is correct? I just wanted to clarify.

Interesting. I wish we would have learned about this as this makes more sense based on your explanation.
Thank you for your help!
 
Jaimie said:
So when you're saying "should be fine", you're saying that based on what I learned, my answer to part a i) is correct? I just wanted to clarify.
I'm not in your course, but if you did not learn relativistic velocity addition yet, it is fine.
 
Great! Thank you for your help.
 

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