Explain Relativistic Mass: Special Relativity & F=ma

In summary: Relativistic mass is an idea that is outdated since 1907 and should not be used and taught anymore. It leads to confusion. The mass in relativity is alwasy understood as the invariant mass of an object and as the name tells you it's a scalar, i.e., independent of the frame of reference.(b) ##\vec{F}=m \vec{a}## doesn't hold in general, even not in Newtonian physics. The correct law, already stated by Newton, is ##\vec{F}=\mathrm{d} \vec{p}/\mathrm{d} t##. Now in the theory of relativity this is a very
  • #1
ReptileBaird
17
0
I was told in special relavity that f=ma does not always hold true so using the f=delta p/delta t is better. So why is that? They said it had something to do with relativistic mass, at least when traveling at 99.9% of the speed of light.
 
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  • #4
ReptileBaird said:
what about f=ma not being used as much in special relativity compared to dp/dt?

F = ma isn't used in SR because it isn't correct in SR. It's only correct in the approximation where all velocities are much less than the speed of light. F = dp/dt is correct for all possible velocities.
 
  • #5
PeterDonis said:
F = ma isn't used in SR because it isn't correct in SR. It's only correct in the approximation where all velocities are much less than the speed of light. F = dp/dt is correct for all possible velocities.
Why is f=ma not correct for speeds much less than c? And what qualifies as "much less"? Is it because the mass is dynamic?
 
  • #7
@ReptileBaird - Pretty much all of Newton's formulae are wrong, in short. They are approximately correct at low speeds and low energies and weak gravitational fields. In the latter half of the nineteenth century we began to do experiments and work out theories that could reach the regimes where it was obvious that Newton was wrong, and Einstein put it all together in 1905-1916.

We mostly continue to use Newton's equations because they're good enough for very nearly everything. For example, a rocket accelerating at 15g for 100s would, according to Mission Control, be accelerating slightly less than Newton would predict. But the difference is in the 7th or 8th significant figure, if my mental arithmetic is correct. So, even NASA uses Newton because the maths is so much simpler and the error is negligible even at the speeds they sling around.

The GPS is a famous exception - it is a sufficiently high precision system that it includes a general relativistic calculation in its design. But the rule is that Newton is close enough unless you're doing something ridiculously high energy or high precision.
 
  • #8
Well, your question is very easily answered. The answer has two parts:

(a) Relativistic mass is an idea that is outdated since 1907 and should not be used and taught anymore. It leads to confusion. The mass in relativity is alwasy understood as the invariant mass of an object and as the name tells you it's a scalar, i.e., independent of the frame of reference.

(b) ##\vec{F}=m \vec{a}## doesn't hold in general, even not in Newtonian physics. The correct law, already stated by Newton, is ##\vec{F}=\mathrm{d} \vec{p}/\mathrm{d} t##. Now in the theory of relativity this is a very complicated quantity, and it is easier to use proper time (assuming we have a massive particle; the massless case is more complicated and of no practical relevance since the only massless particle-like objects are photons and shouldn't be treated as classical particles to begin with). The proper time of a particle is defined by
$$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2/c^2},$$
where ##t## is the coordinate time wrt. an inertial reference frame. Now you define the four-vector
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
This is the four-momentum vector. It's time component is the energy of the particle (up to a factor ##c## which is due to the unfortunate joice of SI units), including its rest energy
$$p^0=m c \gamma=\frac{m c}{\sqrt{1-\vec{v}^2/c^2}}.$$
The three spatial components are relativistic momentum
$$\vec{p}=\frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}.$$
The mass is a scalar and the covariant energy-momentum relation of a classical particle is given by the "on-shell condition"
$$p_{\mu} p^{\mu}=m^2 c^2,$$
which leads to the energy-momentum relation
$$p^0=\frac{E}{c}=\sqrt{m^2 c^2 +\vec{p}^2}.$$
This shows that the relativistic energy includes the rest energy of the particle
$$E_0=c p^0|_{\vec{p}=0}=m c^2.$$
 

What is relativistic mass?

Relativistic mass is a concept in special relativity that describes the mass of an object as it moves at a significant fraction of the speed of light. It takes into account the effects of time dilation and length contraction on an object's mass.

How does special relativity relate to relativistic mass?

Special relativity is a theory that describes the relationship between space and time for objects moving at high speeds. It is the basis for understanding relativistic mass, as it explains how an object's mass changes as it approaches the speed of light.

What is the equation for relativistic mass?

The equation for relativistic mass is m = m0/√(1-v^2/c^2), where m is the relativistic mass, m0 is the rest mass of the object, v is its velocity, and c is the speed of light.

How is relativistic mass different from rest mass?

Rest mass, also known as invariant mass, is the mass of an object when it is at rest. Relativistic mass takes into account the effects of special relativity, such as time dilation and length contraction, on the object's mass as it moves at high speeds.

How does the equation F=ma relate to relativistic mass?

The equation F=ma is used to calculate the force acting on an object based on its mass and acceleration. In the case of relativistic mass, the mass used in this equation is the relativistic mass, as it takes into account the changes in mass due to special relativity effects.

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