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Einstein, Newton and the Bullet

Posted by Karl Sharman on December 8, 2010 – 2:32 pm

You are a weapons manufacturer and you have just developed a gun that fires a bullet at 0.9c (c is the speed of light, y’know – e=mc2 and all that…?)
You are having a chat with a friend of yours who has just developed a new car that travels at 0.9c (It’s a Toyota Avensis…)
During this chat you both come to a minor stumbling block…. You are a passenger in your friends Avensis, travelling at 0.9c, and you fire your new gun, forward along the axis of travel…
At what speed is the bullet travelling….?

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  1. 1. Chris Said:


  2. 2. Chris Said:

    Only kidding :) 0.994475… c

  3. 3. RadioKirby Said:

    1.8c. Einstein’s Theory of Relativity

  4. 4. Chris Said:

    Hi RadioKirby. You used Newton’s Theory of Relativity.

    But I’m now struggling with something that I’d never noticed before. If you fired the gun in the other direction, you’d have got 0. That seems very strange to me (but not Newton).

  5. 5. Chris Said:

    That’s two cognitive dissonances for me in one day.

    If The car had been doing 0.9c and the gun 0.8c, then when fired backwards, the bullet would be doing -0.1/0.28 ≈ -0.1 * 3.57c. So there is a nice distortion somewhere – phew!

  6. 6. Karl Sharman Said:

    Theoretically Chris, the bullet wouldn’t be going at 0. I think…

  7. 7. Karl Sharman Said:

    Sorry Chris, I factored in a bullet deceleration. Real world physics getting in the way there.

  8. 8. Kenny Said:

    If you did shoot it backwards it would be 0 because mythbusters did it on their show, they were driving a truck 60 miles an hour and fired a ball out of a cannon at 60 miles an hour and the ball just fell straight down(not going forward not backwards)

  9. 9. Spencer Said:

    The problem is that as an object nears the speed of light, it’s acceleration slows dramaticly depending smally on mass and density. This problem is computabe, using masses and high power computers but is diffucult otherwise.
    Correct me if I am wrong.

  10. 10. Chris Said:

    Hi Karl. Your post 6. ??? of course it would be doing 0 (you won’t believe how overjoyed I am to be able to say that). What deceleration? But methinks, your just be jocular.

    Hi Kenny. Unfortunately, 60 mph s so far below c, that that didn’t prove anything about what happens at exremely high speeds. e.g. If you fire a bullet at even many times the speed of sound, it’s mass doesn’t change (much). But in say the CERN ring, a proton’s mass will have increased sveral thousand times. Relativity rules there – big time.

    Hi Spencer. It’s not the acceleration, per se, that’s reduced; the objects mass increases.

    The calculation is simple. An important quantity in releativity is the relativistic factor:
    γ(v) = 1/√(1 – (v/c)²). (The fist letter there is gamma).
    v is the objects speed (in fact it is usually used to denote the speed of the frame that measurements are being done in).

    γ(v) ≥ 1 and γ(v) → ∞ as v → c.

    Masses increase, clocks go slow and rulers shrink by the Y factor.

    Releativity also explains magnetism in terms of moving charges absolutely beautifully.

  11. 11. Chris Said:

    Sorry about the typos.

    If that show had done the shot going forwards, they’d have measured 120 mph. But as Karl’s problem shows??? that all goes wrong at relativistic speeds.

    The velocity addition formula (for forwards and backward, but not sideways addition) is:

    (u+v)/(1 + uv/c²). Where e.g. u is the speed of the car, and v is the speed of the bullet as measured in the car.

    Newton would have said u+v.

  12. 12. Chris Said:

    Oh ,one more. It is pretty easy to show that E = m c².

    The way I do it requires knowledge of 4-vectors, a pretty good grasp of simple mechanics and very simple calculus. Once you’ve spent a couple of years getting that together, it’s a doddle :)

  13. 13. Chris Said:

    Yet another item. Karl (I think) posted a problem a few months back. It asked for the mass of the Earth’s gravitational field. It turns out that it is about 3 million million tonnes.

    Near the Earth’s surface, the gravitational field has a mass of about 640 kg per cubic km. That mass also contributes to making the field itself.

  14. 14. Chris Said:

    Aaarggh. I’ve just having my third cognitive dissonance in less that 24 hours. Should something of lower density than 640 kg/cubic km float in the Earth’s gravitational field? ;)

  15. 15. Chris Said:

    … assuming no atmosphere.

  16. 16. Random Guy Said:

    Chris, thats 6 posts in a row!

    Im still gonna go with Chris’s first (actual) answer of 0.99447513812154696132596685082873…c

  17. 17. Chris Said:

    I must be getting tired. I often do more than that ;)

  18. 18. Spencer Said:

    Hi Chris! Sorry if I sound like I am a know-it-all.
    BTW I found out that 94% of my surveied blogs use the users local time!

  19. 19. Chris Said:

    Hi Spencer. I hadn’t thought you sounded like a know-it-all.

    I’m the one who probably guilty of that. But that’s OK because I do know it all (LOL, not really).

    In my browser I see my posted times are way different out from my local time.

  20. 20. Chris Said:

    It’s 3:46 am GMT, but that last post says it’s 8:46 pm.

  21. 21. Jonte Said:

    it’s .9c the question is “at what speed is the bullet traveling?” if the bullet gets to a speed of .9c standing still the bullet is leaving you at .9c but if you yourself are going .9c and fire the gun it the same direction you going you would be traveling side long the bullet at .9c now if it were shot in the other direction you and the bullet would be LEAVING each other at a rate of 1.8c

  22. 22. Chris Said:

    Hi Spencer. I think that stuff to do with requiring a lot of computer power is when you try to simulate e.g. a couple of galaxies “colliding”, using a full General Relativistically correct simulation – that takes very serious quantities of computing power.

    I mentioned the relativistic factor (gamma) earlier. I’ve just remembered that it is called the Lorentz factor.

  23. 23. KARL Said:


  24. 24. KARL BUSH Said:

    my bad at 0.9c

  25. 25. Karl Sharman Said:

    You probably have the idea that if you’re standing on a train moving at speed u, and you walk forward at speed v, your total forward speed w is expressed by the straightforward sum u + v. Alas, this is a cruel illusion. In reality, what we might call “addition of velocities” is governed by the awe-inspiring equation:
    w = (u + v)/(1 + uv/c²)
    where c² is the speed of light squared. (This may give you pause next time you hike to the toilet on the British Rail Intercity 125.) At so-called Newtonian (i.e., slow) speeds, the term uv/c² is pretty close to 0, and the equation reduces down to the familiar w = u + v. However, if we are traveling at, say, 0.9c (nine-tenths the speed of light), and we shoot a bullet forward also at 0.9c, we discover via the above formula that the slug doesn’t attain an overall speed of 1.8c (i.e., more than the speed of light), but rather a modest
    (0.9c + 0.9c)/(1 + [0.9]²) = 0.994c

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