5 ms·

Kurzgesagt recently uloaded a video on hypethetical interplantary weapons (https://www.youtube.com/watch?v=tybKnGZRwcU https://www.youtube.com/watch?v=tybKnGZRw

by loufe 3y ago

Kurzgesagt recently uloaded a video on hypethetical interplantary weapons (https://www.youtube.com/watch?v=tybKnGZRwcU https://www.youtube.com/watch?v=tybKnGZRwcU) which is the only reason I even know what the OMG particle is. They described an "Ultra Relativistic Electron Beam" which theoretically could travel much closer to the speed of light than the OMG particle. I'm left wondering if the proximity to the pure speed of light has any bearing on total delivered energy when comparing different particles.

qubex 3y ago
You’re often taught that E=mc² but the fuller equation is actually E²=m₀²c⁴+p²c² where p is the momentum, so at high speeds mass fades into irrelevance and the velocity component of momentum becomes dominant. EDIT in classical physics p=mv so you might wonder what I’m banging on about when mass appears linearly on both sides and on the first term is multiplied by c⁴ whereas in the second only by c². However relativistic momentum is classical momentum adjusted by the Lorenz transformation γ=1/√(1-[v/c]² so it actually dominates in the limit when v tends to c. EDIT 2: the latter works out to be sigh c³vm₀/√(c²-v²) and it grows without bound as v→c so that’s the pedantic answer. The first term is linear the second term is anything but.
- mitthrowaway2 3y ago
  E = mc² is the full equation if 'm' is the relativistic mass, which grows with velocity. E²=m₀²c⁴+p²c² is the full equation if 'm₀' is the rest mass.
  - E2P2c2throwaway 3y ago
    If the particle has zero rest mass, like a photon, it still has energy. E^2 = p^2 c^2 + m_0^2 c^4 captures that correctly, while E = mc^2 where 'm' is the relativistic mass, does not.
    - mitthrowaway2 3y ago
      No, it does correctly capture that. The relativistic mass of a photon is not zero even though its rest mass is. However for a zero-mass particle you cannot use the Lorentz transformation to calculate relativistic mass, because v = c and so you would be dividing zero by zero. You can correctly calculate the relativistic mass of a photon in other ways, such as from its momentum or its gravitational interaction, and doing so gives you the energy you'd expect from E = mc².
      - qubex 3y ago
        Oh heavens whatever caused you to believe this? The momentum p of a photon is given by p=h/λ where h is the Plank Constant and λ is the wavelength. Compton won a Nobel prize for this.
        mitthrowaway2 3y ago
        Believe what? Of course p = h/λ. Thus the relativistic mass m is given by p/v = p/c = h/cλ. And a photon's energy is given by E = hc/λ. This is all in agreement with the relativistic mass you'd calculate from the photon energy E = mc². Heavens, which of these do you not believe?
        qubex 3y ago
        Relativistic mass isn’t really sensible as a concept. Does something moving fast gravitate more because it has more mass? No.
        mitthrowaway2 3y ago
        Except that it does gravitate more. Avoiding the confusion that it doesn't is one of the conceptual benefits of relativistic mass. Gravitation is dependent on the stress-energy tensor, which contains... drum roll... relativistic mass, not rest mass! https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor For that matter, you've already noticed that the Lorentz transform becomes nonsensical as v → c when starting from rest mass, whereas the transform works just fine when using relativistic mass as the starting point, and nothing blows up, and you can avoid any confusion.
  - scotty79 3y ago
    Relativistic mass is educational aid not actual physical concept. The full equation is the only one actual physicists use (unless teaching).
    - mitthrowaway2 3y ago
      Neither is rest mass an actual "physical concept". When general relativity is considered, rest mass is ambiguous, or at least non-local. Whether trendy or not, relativistic mass is however a very useful concept, because in many cases it is the relativistic mass, rather than the rest mass, that behaves the way we expect mass to behave; mainly that it is additive. For example, if you weighed a mirrored box full of bouncing photons, the scale would measure the mass of the contents as the sum of the relativistic masses, not the sum of the rest masses. Rest mass has the awkward problem that it depends on which particles you consider to be "part of the system": A photon going left has no rest mass, a photon going right has no rest mass, but a system of two photons, one going left and the other right, does have a nonzero rest mass. String theorists would even consider much of a particle's apparent rest mass to simply be another manifestation of relativistic mass, indicating the presence of an invisible periodic motion along a hidden dimension, resulting in kinetic energy even within a particle that seems to be "at rest". (Edited for clarity)
      - scotty79 3y ago
        I don't think it's useful concept. It's deeply misleading. When you are talking about systems then mass is not a sum of the parts but a proxy for total energy of the system. And when you are interested in the parts just use the full equation linking energy, mass and momentum. Linking some kind of mass directly with velocity is one shortcut too far.
        mitthrowaway2 3y ago
        > When you are talking about systems then mass is not a sum of the parts but a proxy for total energy of the system And which kind of mass is a proxy for the total energy of a system? That's right: relativistic mass, not rest mass. And this is the sum of its parts, as it should be, because energy is always conserved. Relativistic mass and energy are the exact same thing, up to a constant factor of c². Isn't energy a useful concept? Furthermore the history of physics has often revealed what we thought of as atomic particles to actually have systems underneath. In the course of time, it seems more likely to me that we will do away with rest mass than with energy, in which case all mass will be relativistic.

by loufe 3y ago

qubex 3y ago

You’re often taught that E=mc² but the fuller equation is actually E²=m₀²c⁴+p²c² where p is the momentum, so at high speeds mass fades into irrelevance and the velocity component of momentum becomes dominant. EDIT in classical physics p=mv so you might wonder what I’m banging on about when mass appears linearly on both sides and on the first term is multiplied by c⁴ whereas in the second only by c². However relativistic momentum is classical momentum adjusted by the Lorenz transformation γ=1/√(1-[v/c]² so it actually dominates in the limit when v tends to c. EDIT 2: the latter works out to be sigh c³vm₀/√(c²-v²) and it grows without bound as v→c so that’s the pedantic answer. The first term is linear the second term is anything but.

mitthrowaway2 3y ago
E = mc² is the full equation if 'm' is the relativistic mass, which grows with velocity. E²=m₀²c⁴+p²c² is the full equation if 'm₀' is the rest mass.
- E2P2c2throwaway 3y ago
  If the particle has zero rest mass, like a photon, it still has energy. E^2 = p^2 c^2 + m_0^2 c^4 captures that correctly, while E = mc^2 where 'm' is the relativistic mass, does not.
  - mitthrowaway2 3y ago
    No, it does correctly capture that. The relativistic mass of a photon is not zero even though its rest mass is. However for a zero-mass particle you cannot use the Lorentz transformation to calculate relativistic mass, because v = c and so you would be dividing zero by zero. You can correctly calculate the relativistic mass of a photon in other ways, such as from its momentum or its gravitational interaction, and doing so gives you the energy you'd expect from E = mc².
    - qubex 3y ago
      Oh heavens whatever caused you to believe this? The momentum p of a photon is given by p=h/λ where h is the Plank Constant and λ is the wavelength. Compton won a Nobel prize for this.
      - mitthrowaway2 3y ago
        Believe what? Of course p = h/λ. Thus the relativistic mass m is given by p/v = p/c = h/cλ. And a photon's energy is given by E = hc/λ. This is all in agreement with the relativistic mass you'd calculate from the photon energy E = mc². Heavens, which of these do you not believe?
        qubex 3y ago
        Relativistic mass isn’t really sensible as a concept. Does something moving fast gravitate more because it has more mass? No.
        mitthrowaway2 3y ago
        Except that it does gravitate more. Avoiding the confusion that it doesn't is one of the conceptual benefits of relativistic mass. Gravitation is dependent on the stress-energy tensor, which contains... drum roll... relativistic mass, not rest mass! https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor For that matter, you've already noticed that the Lorentz transform becomes nonsensical as v → c when starting from rest mass, whereas the transform works just fine when using relativistic mass as the starting point, and nothing blows up, and you can avoid any confusion.
- scotty79 3y ago
  Relativistic mass is educational aid not actual physical concept. The full equation is the only one actual physicists use (unless teaching).
  - mitthrowaway2 3y ago
    Neither is rest mass an actual "physical concept". When general relativity is considered, rest mass is ambiguous, or at least non-local. Whether trendy or not, relativistic mass is however a very useful concept, because in many cases it is the relativistic mass, rather than the rest mass, that behaves the way we expect mass to behave; mainly that it is additive. For example, if you weighed a mirrored box full of bouncing photons, the scale would measure the mass of the contents as the sum of the relativistic masses, not the sum of the rest masses. Rest mass has the awkward problem that it depends on which particles you consider to be "part of the system": A photon going left has no rest mass, a photon going right has no rest mass, but a system of two photons, one going left and the other right, does have a nonzero rest mass. String theorists would even consider much of a particle's apparent rest mass to simply be another manifestation of relativistic mass, indicating the presence of an invisible periodic motion along a hidden dimension, resulting in kinetic energy even within a particle that seems to be "at rest". (Edited for clarity)
    - scotty79 3y ago
      I don't think it's useful concept. It's deeply misleading. When you are talking about systems then mass is not a sum of the parts but a proxy for total energy of the system. And when you are interested in the parts just use the full equation linking energy, mass and momentum. Linking some kind of mass directly with velocity is one shortcut too far.
      - mitthrowaway2 3y ago
        > When you are talking about systems then mass is not a sum of the parts but a proxy for total energy of the system And which kind of mass is a proxy for the total energy of a system? That's right: relativistic mass, not rest mass. And this is the sum of its parts, as it should be, because energy is always conserved. Relativistic mass and energy are the exact same thing, up to a constant factor of c². Isn't energy a useful concept? Furthermore the history of physics has often revealed what we thought of as atomic particles to actually have systems underneath. In the course of time, it seems more likely to me that we will do away with rest mass than with energy, in which case all mass will be relativistic.