## Laser recoil, relativistic momentum and whether mass counts

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Sableagle
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### Laser recoil, relativistic momentum and whether mass counts

Kyle of Because Science did a video about "The Death Star's OTHER Fatal Flaw," which you can see here:
https://www.youtube.com/watch?v=K6-q2edmiGk

His argument was that the recoil of a laser with that much power would have shot the weapon out of the far side of the station, or launched the whole station off into space, or something like that. I think he went with solid metal ball of incredible strength and said it would be launched into the distance.

Here's the calculation for the recoil: https://www.youtube.com/watch?v=K6-q2edmiGk&t=260s

His answer was 7.5 * 1023

We know it has energy, therefore it must have momentum, but to have momentum it must have mass and he crossed out part of the equation because it doesn't have mass. Well, we know its velocity, so
E2 = (m v)2 c2 + m2 c4 becomes
E2 = m2 c4 + m2 c4 = 2 m2 c4 so
E = sqr(2) m c2 ..... Oh, hello there, mutant version of simplified kinetic energy equation.
Right, so 2.2 x 1032 J / 1271031076302266264 = m = 1730877279390832 kg,
which means its momentum is 5.189 x 1023 kg m/s.
Right?

We disagree by a factor of sqr(2) because he crossed out the rest mass mc2 part and I left it in and calculated it using the mass derived from the other part.

Does that part count, or was he right to cross it out? Does light have mass in the sense of momentum but no mass in the sense of mc2, both at the same time?
Oh, Willie McBride, it was all done in vain.

chenille
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### Re: Laser recoil, relativistic momentum and whether mass counts

Sableagle wrote:E2 = (m v)2 c2 + m2 c4

This formula is wrong. It looks like you are supposing p = mv? But that's only the non-relativistic approximation and won't work here.
Instead momentum is part of a 4-vector together with energy. An object with mass will have its own reference frame, where it has p = 0 and energy E = mc2. If we change to a reference frame where it is traveling v we end up getting

p = mv/(1-v2/c2)1/2 = mv + (1/2)mv3/c2 + (3/8)mv5/c4 + (5/16)mv7/c6...
E = mc2/(1-v2/c2)1/2 = mc2 + (1/2)mv2 + (3/8)mv4/c2 + (5/16)mv6/c4...

For very small v these approximate p = mv and E = mc2 + (1/2)mv2, but as v grows those become worse and worse underestimates, and instead both go to infinity as v nears c. So this whole approach fails for light, when v = c. That's only possible when m = 0, and then E2 = p2c2 + m2c4 becomes |p| = E/c.

Sableagle
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### Re: Laser recoil, relativistic momentum and whether mass counts

Right, then. Momentum does not, in fact, equal mass times velocity. Wow.

Thank you, chenille.
Oh, Willie McBride, it was all done in vain.

PM 2Ring
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### Re: Laser recoil, relativistic momentum and whether mass counts

Sableagle wrote:Right, then. Momentum does not, in fact, equal mass times velocity. Wow.

Thank you, chenille.

Exactly. p = mv is just a Newtonian approximation. OTOH, momentum is velocity times relativistic mass, but in modern treatments of SR relativistic mass is a deprecated concept because it can be misleading, and if you don't use it carefully it's very easy to make mistakes and jump to incorrect conclusions.

Eebster the Great
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### Re: Laser recoil, relativistic momentum and whether mass counts

F = dpdt still works, right? One out of three laws isn't so bad, Newton. (Of course, the F = ma form does not hold.)

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