# Newb question about relativistic mass

• B
Based on my limited understanding of special relativity, as an object's velocity increases with respect to an observer the object undergoes spacial contraction (making it appear to the observer to get thinner in the direction of travel) and time dilation (so the observer would view the object's clock tick more slowly than the observer's, if he could see it). The object also gains relativistic mass, though, and I was wondering how, if it all, this would manifest to the observer?

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mfb
Mentor
The concept of relativistic mass is not used any more in physics. It just leads to misconceptions.

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

Ibix
Einstein 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", which sometimes lets you use Newtonian formulas with this revised mass. These days that's (by most physicists) regarded as a silly way of looking at it, especially as there are several different relativistic masses that need to be used in different places, and several places where you can't use any kind of relativistic mass in that naive way (notably, gravity). So the concept has largely gone out of fashion.

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

FactChecker and Nugatory
Mister T
Gold Member
The object also gains relativistic mass, though, and I was wondering how, if it all, this would manifest to the observer?
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##.

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.

PeterDonis
Mentor
2019 Award
Einstein 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"
This is not correct. The "increase in mass" is the mass equivalent of all the kinetic energy (i.e., all excess energy over the rest energy), not just the part that's larger than the Newtonian prediction.

Ibix
This is not correct. The "increase in mass" is the mass equivalent of all the kinetic energy (i.e., all excess energy over the rest energy), not just the part that's larger than the Newtonian prediction.
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.

PeterDonis
Mentor
2019 Award
The extra momentum predicted by relativity is due to the increase in mass - the energy is different.
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.

Ibix
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 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 don't think this is completely rigorous.

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PeterDonis
Mentor
2019 Award
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.
Ah, I see. Got it.

(##c = 1##)

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

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!

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

Mister T
Gold Member
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!
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 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!

True.

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

Mister T
Mister T
Gold Member
True.

But still, ##\vec{p}c = E \vec{\beta}## is nice, eh?
Yes. I agree that it's great and I also agree that it doesn't get the press it deserves in introductory treatments.

SiennaTheGr8
PAllen
2019 Award
True.

But still, ##\vec{p}c = E \vec{\beta}## is nice, eh?
Nice, but when you get to dynamics, you see a great value for m:

F = mA in 4-vectors.

This fits nicely with P = mU, with U being 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.

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

F = mA in 4-vectors.

This fits nicely with P = mU, with U being 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.
Rest energy ##E_0## does the trick, too.

3-momentum: ##\vec p c = E \vec \beta##
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)

PAllen
2019 Award
3-momentum: ##\vec p c = E \vec \beta##
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)
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.

PAllen
2019 Award
3-momentum: ##\vec p c = E \vec \beta##
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)
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.

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.
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).

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.
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)?

PAllen
2019 Award
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.

vanhees71
Gold Member
2019 Award
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.
There is no relativistic mass, only invariant mass. There is relativistic energy of a classical system,
$$E=\sqrt{m^2 c^4+p^2 c^2},$$
In the center-momentum frame, where the total momentum is by definition 0, you have ##E_{\text{cm}}=m c^2##, and nowhere else!

Out of curiosity: does anyone know whether proponents of "relativistic mass" ever expressed kinetic energy in mass units ("kinetic mass")?

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!

vanhees71