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mfb

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"Gains relativistic mass" is just another way to say it gains energy, with the usual effects.

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Nugatory

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Ibix

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In terms of effects, a faster moving object will hurt more when it hits you. More according to Einstein than Newton.

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Mister T

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One way is with collisions. Momentum is conserved in collisions between particles, so for example in a one-dimensional two particle collision the momentum lost by one particle equals the momentum gained by the other. The momentum equals ##\gamma mv## where ##m## is the ordinary mass. The factor ##\gamma## increases with the particle's speed. So if you want you can call ##\gamma m## the relativistic mass and say that it's the thing that increases with speed. But regardless of whether or not you do this, momentum will not be conserved if you instead use the newtonian value for momentum ##mv##.The object also gains relativistic mass, though, and I was wondering how, if it all, this would manifest to the observer?

Thus the answer to your question, the way this would manifest itself is that you wouldn't be able to predict or understand what's happening to particles in these collisions.

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This is not correct. The "increase in mass" is the mass equivalent ofEinstein predicts that things have higher kinetic energy than Newton predicts for the same speed. The extra bit used to be attributed to an "increase in mass"

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Ibix

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Fair enough. The extra momentum predicted by relativity is due to the increase in mass - the energy is different.This is not correct. The "increase in mass" is the mass equivalent ofallthe kinetic energy (i.e., all excess energy over the rest energy), not just the part that's larger than the Newtonian prediction.

Just one example of why relativistic mass is a bad idea.

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I'm not sure what you mean by this. If ##m## is the rest mass, then energy is ##E = \gamma m c^2## and momentum is ##p = \gamma m v##. Both of them have the factor ##\gamma m##, which is "relativistic mass" and shows the same "increase in mass" in both cases. So I'm not sure what is "different" about energy.The extra momentum predicted by relativity is due to the increase in mass - the energy is different.

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Ibix

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The "extra mass" is ##(\gamma-1)m##. The difference between the relativistic and Newtonian momenta is ##(\gamma-1)mv##, which is the Newtonian momentum which would be associated with the extra mass. So in a "relativity is simpler if you pretend it's Newton plus relativistic mass" kind of way, the "extra" momentum is attributable to the "extra" mass. That line of argument doesn't apply to energy, as you point out, because the relativistic formula doesn't look like the Newtonian one with relativistic mass dropped in.I'm not sure what you mean by this. If ##m## is the rest mass, then energy is ##E = \gamma m c^2## and momentum is ##p = \gamma m v##. Both of them have the factor ##\gamma m##, which is "relativistic mass" and shows the same "increase in mass" in both cases. So I'm not sure what is "different" about energy.

I don't think this is completely rigorous.

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Ah, I see. Got it.The difference between the relativistic and Newtonian momenta is ##(\gamma-1)mv##, which is the Newtonian momentum which would be associated with the extra mass...That line of argument doesn't apply to energy, as you point out, because the relativistic formula doesn't look like the Newtonian one with relativistic mass dropped in.

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Unfortunately, the enthusiastic killing off of "relativistic mass" wasn't accompanied by an equally enthusiastic inclination to express momentum in terms of

If you'd asked a physics student forty years ago for the equation for "relativistic momentum," I bet they'd respond with ##\mathbf{p} = m_r \mathbf{v}##. Ask one today, and I bet they'll respond with ##\mathbf{p} = \gamma m \mathbf{v}##,

Three cheers for ##\mathbf{p} = E \mathbf{v}##, which applies even in the massless case.

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Mister T

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That depends of course entirely on your definition of ##m_r##. If you define it as ##E## then yes, but if you define it as ##\gamma m## then not.If you'd asked a physics student forty years ago for the equation for "relativistic momentum," I bet they'd respond with ##\mathbf{p} = m_r \mathbf{v}##. Ask one today, and I bet they'll respond with ##\mathbf{p} = \gamma m \mathbf{v}##,which is a less complete answer!

If that physics student of 40 years ago had asked his professor (the very one who taught him about ##m_r##) if a photon has mass, the answer would be "No" more often than yes. Followed by a confused discussion of whether it was okay to equate the photon's energy to its mass and a clarification by the professor that he meant rest mass. (And then the student might respond that photons can never be at rest, but I digress.)

Perhaps this is what Einstein meant when he cautioned against its use, saying that "no clear definition can be given" for relativistic mass. And that was way more than 40 years ago!

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True.

But still, ##\vec{p}c = E \vec{\beta}## is nice, eh?

But still, ##\vec{p}c = E \vec{\beta}## is nice, eh?

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Mister T

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Yes. I agree that it's great and I also agree that it doesn't get the press it deserves in introductory treatments.True.

But still, ##\vec{p}c = E \vec{\beta}## is nice, eh?

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PAllen

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Nice, but when you get to dynamics, you see a great value for m:True.

But still, ##\vec{p}c = E \vec{\beta}## is nice, eh?

This fits nicely with

Energy is present here only as the time component of 4 momentum. Of course, you then have to make special cases for light. On the other hand, dynamics of classical massless particles in SR has many difficulties, so it can be argued that light should only be treated as a classical field, not as in terms massless particles.

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Rest energy ##E_0## does the trick, too.Nice, but when you get to dynamics, you see a great value for m:

F= mAin 4-vectors.

This fits nicely withP= mU,withUbeing 4-velocity; just take the derivative by proper time to get the former.

Energy is present here only as the time component of 4 momentum. Of course, you then have to make special cases for light. On the other hand, dynamics of classical massless particles in SR has many difficulties, so it can be argued that light should only be treated as a classical field, not as in terms massless particles.

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3-momentum in the classical limit: ##\vec p c \approx E_0 \vec \beta##

4-momentum: ##\vec P = E_0 \vec B## (where ##\vec B## is the "normalized" 4-velocity)

3-force in the classical limit: ##\vec f \approx E_0 \dot{\vec{\beta}}## (where dot means ##ct##-derivative)

4-force: ##\vec F = E_0 \mathring{\vec{B}}## (where little circle means ##ct_0##-derivative)

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PAllen

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Normalization need not be imposed for derivatives by proper time. In the last, the derivative is by proper time. Not sure what your t0 means. Possibly time in an MCIF. That seems conceptually inferior to proper time as an invariant, independent of frames or coordinates, applying to GR as well as SR.

3-momentum in the classical limit: ##\vec p c \approx E_0 \vec \beta##

4-momentum: ##\vec P = E_0 \vec B## (where ##\vec B## is the "normalized" 4-velocity)

3-force in the classical limit: ##\vec f \approx E_0 \dot{\vec{\beta}}## (where dot means ##ct##-derivative)

4-force: ##\vec F = E_0 \mathring{\vec{B}}## (where little circle means ##ct_0##-derivative)

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PAllen

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m is fewer keystrokes than E0 . Also, this doesn't relieve the issue that when you get into force as well as momentum, you have to admit massless particles are problematic, and light is a special case.

3-momentum in the classical limit: ##\vec p c \approx E_0 \vec \beta##

4-momentum: ##\vec P = E_0 \vec B## (where ##\vec B## is the "normalized" 4-velocity)

3-force in the classical limit: ##\vec f \approx E_0 \dot{\vec{\beta}}## (where dot means ##ct##-derivative)

4-force: ##\vec F = E_0 \mathring{\vec{B}}## (where little circle means ##ct_0##-derivative)

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Sorry, I should have mentioned that my ##t_0## is proper time, and ##ct_0## is just proper time expressed in length units (the timelike spacetime interval, if you prefer).Normalization need not be imposed for derivatives by proper time. In the last, the derivative is by proper time. Not sure what your t0 means. Possibly time in an MCIF. That seems conceptually inferior to proper time as an invariant, independent of frames or coordinates, applying to GR as well as SR.

Yes, in SR the four-momentum is better defined in component form ##(E, \vec p c)## than in terms of the four-velocity.m is fewer keystrokes than E0 . Also, this doesn't relieve the issue that when you get into force as well as momentum, you have to admit massless particles are problematic, and light is a special case.

As for force—is there really reason to admit the massless case in SR (for either 3-force or 4-force)?

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PAllen

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Sorry, I should have mentioned that my ##t_0## is proper time, and ##ct_0## is just proper time expressed in length units (the timelike spacetime interval, if you prefer).

Yes, in SR the four-momentum is better defined in component form ##(E, \vec p c)## than in terms of the four-velocity.

As for force—is there really reason to admit the massless case in SR (for either 3-force or 4-force)?

No good reason to have theory of classical massless particles (it has been done, but it is very ugly IMO). So once you admit light doesn't work the same as particles, then 4 momentum in terms of 4 velocity for particles is extremely natural.

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There is no relativistic mass, only invariant mass. There is relativistic energy of a classical system,Fair enough. The extra momentum predicted by relativity is due to the increase in mass - the energy is different.

Just one example of why relativistic mass is a bad idea.

$$E=\sqrt{m^2 c^4+p^2 c^2},$$

In the center-momentum frame, where the total momentum is by definition 0, you have ##E_{\text{cm}}=m c^2##, and nowhere else!

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In other words, we have:

##E = E_0 + E_k = m_r c^2##

but did anyone ever introduce ##m_k \equiv E_k / c^2##? Then:

##m_r = m + m_k##

I've never seen it!

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\sqrt{m^2+p^2}-m## of the beam.

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My question is a historical one. I'm wondering whether folks who use(d) "relativistic mass" ever went so far as to introduce a "kinetic mass" ##m_k = E_k / c^2##. Just curious.