# F=ma in relativistic terms

by hover
Tags: relativistic, terms
 P: 344 In Newtonian Physics, a force on a object is equal to its mass times acceleration. This is Newton's Second Law. But in relativity, when the acceleration approaches the speed of light, Newton's Second Law starts to become less accurate because to accelerate a mass more you must keep adding more energy. So in relativistic terms, what equation is there that holds up to speeds near the speed of light? Thanks
P: 2,043
Hi, Hover,

 Quote by hover In Newtonian Physics, a force on a object is equal to its mass times acceleration. This is Newton's Second Law. But in relativity, when the acceleration approaches the speed of light, Newton's Second Law starts to become less accurate because to accelerate a mass more you must keep adding more energy. So in relativistic terms, what equation is there that holds up to speeds near the speed of light?
In special relativity it is also F = m0A. And this forumla always works!

But the difference is that A is a 4-vector. A is the 4-acceleration vector which is basically the derivative of the 4-velocity vector with respect to proper time. And the 4-velocity vector is basically the change in direction in spacetime with respect to proper time. But note that we have four, not three, directions in spacetime!
In addition there is the 4-energy-momentum vector which is mass times the 4-velocity vector.

Four vectors are simple and elegant.

As you can see it is pretty straightforward, we simply have to add the extra direction in relativity.
Emeritus
P: 7,620
 Quote by hover In Newtonian Physics, a force on a object is equal to its mass times acceleration. This is Newton's Second Law. But in relativity, when the acceleration approaches the speed of light, Newton's Second Law starts to become less accurate because to accelerate a mass more you must keep adding more energy. So in relativistic terms, what equation is there that holds up to speeds near the speed of light? Thanks
Warning: F = ma does NOT work in general.

F = dP/dt

is the relativistic equation that always holds. The exact form the equations take in terms of "mass" depend on whether or not one uses relativistic mass or invariant mass.

Note that all of the quantities above depend on the observer, including the force.

You will find that the ratio F/a depends on the direction of the force, assuming all measurements are done in the lab frame. This is sometimes discussed under the names "longitudinal mass" and "transverse mass" in older physics books.

I thought there was a FAQ in the sci.physics.faq that had these formulas for F/a, but I couldn't find it offhand. Perhaps someone else will have better luck.

The equations for relativistic momentum in the lab frame are

Px = gamma m vx
Py = gamma m vy
Pz = gamma m vz

where m is the [b]invariant mass[b], sometimes called the rest mass, of the particle.

and gamma = 1/sqrt{1 - vx^2 - vy^2 - vz^2}

Differentiating the above expression for relativistic momentum will give the correct force / acceleration relationships assuming the forces and the accelerations are measured in the laboratory frame.

Sometimes, accelerations are measured in the frame of the object being accelerated, such as in the relativistic rocket. These are often called proper accelerations. If you are interested in the velocity of a travel accelerated at a constant 'felt' acceleration, you'll want to look at the FAQ on the relativistic rocket at http://math.ucr.edu/home/baez/physic...SR/rocket.html

 Sci Advisor HW Helper PF Gold P: 2,007 F=ma in relativistic terms $$\frac{d\vec p}{dt}=m\frac{d}{dt}\left[\frac{\vec v} {\sqrt{1-{\vec v}^2}}\right] = m[\gamma{\bf a}+\gamma^3{\vec v}({\vec v}\cdot{\vec a})] =m\gamma^3[{\vec a}+{{\vec v}\times({\vecv}\times{\vec a})]$$
Emeritus
P: 7,620
 Quote by Meir Achuz $$\frac{d\bf p}{dt}=m\frac{d}{dt}\left[\frac{\bf v} {\sqrt{1-{\bf v}^2}}\right] = m[\gamma{\bf a}+\gamma^3{\bf v(v\cdot a)}] =m\gamma^3[{\bf a}+{\bf vX(vX a)}]$$
This is awfully hard to read. I'd suggest using \vec{} and \times, i.e.

so v X a, for instance, becomes
$$\vec{v} \times \vec{a}$$

and for good measure, v should probably be |v| to make it clear you're taking the norm

and I suppose some warning about geometric units is called for (though you could always put in those annoying factors of c)
P: 2,954
 Quote by MeJennifer In special relativity it is also F = m0A. And this forumla always works!
That expression holds only when the proper mass is constant in time.

Pete
P: 2,043
 Quote by pmb_phy That expression holds only when the proper mass is constant in time.
Very true.
HW Helper
PF Gold
P: 2,007
 Quote by MeJennifer Hi, Hover, In special relativity it is also F = m0A. And this forumla always works! But the difference is that A is a 4-vector. A is the 4-acceleration vector which is basically the derivative of the 4-velocity vector with respect to proper time. And the 4-velocity vector is basically the change in direction in spacetime with respect to proper time. But note that we have four, not three, directions in spacetime! In addition there is the 4-energy-momentum vector which is mass times the 4-velocity vector. Four vectors are simple and elegant. As you can see it is pretty straightforward, we simply have to add the extra direction in relativity.
The F in that equation is the "Minkowski force" and is not dp/dt.
The Minkowski force is defined as
$${\cal F}^\mu=\gamma(dE/dt,dp/dt)$$.
Using that would eventually lead to the equation I gave for dp/dt,
but not as easily.
 P: n/a I have searched for the relativistic acceleration formula involving Force and Mass for a while now. All I can remember is something like this: a=f/(m(1-v^2/c^2)^(3/2)) I know this is wrong. So can someone tell me the correct equation
P: 2,043
 Quote by prasannapakkiam I have searched for the relativistic acceleration formula involving Force and Mass for a while now. All I can remember is something like this: a=f/(m(1-v^2/c^2)^(3/2)) I know this is wrong. So can someone tell me the correct equation
P: 2,043
 Quote by Meir Achuz The F in that equation is the "Minkowski force" and is not dp/dt. The Minkowski force is defined as $${\cal F}^\mu=\gamma(dE/dt,dp/dt)$$. Using that would eventually lead to the equation I gave for dp/dt, but not as easily.
Ok, so it is not dp/dt.

Is there perhaps anything wrong with using 4-vectors in relativity?
 P: n/a So: a=F/(M*SQRT(1-v^2/c^2))
 P: 2,043 Prasanna, see for a good overview of 4-vectors in relativity David Morin - 4-Vectors. Morin, in his introduction writes: "Although it is possible to derive everything in special relativity without the use of 4-vectors (and indeed, this is the route, give or take, that we took in the previous two chapters), they are extremely helpful in making calculations and concepts much simpler and more transparent." I cannot help but fully agreeing with Morin.
Mentor
P: 6,242
 Quote by MeJennifer Ok, so it is not dp/dt. Is there perhaps anything wrong with using 4-vectors in relativity?
No.

$d \tau =dt/\gamma$ gives

$$\gamma \frac{d}{dt} = \frac{d}{d \tau}.$$

Thus,

$${\cal F} = \frac{dP}{d\tau},$$

where $P$ is 4-momentum.
P: 2,954
 Quote by MeJennifer Ok, so it is not dp/dt. Is there perhaps anything wrong with using 4-vectors in relativity?
Meir Achuz must have meant that the 3-vector dp/dt is not equal to the 4-vector F = m0A which is quite obvious and you seem to know that anyway (especially with your acceptance of the addendum that I mentioned to you). In that sense Meir Achuz's comment is very confusing to me. George hit it right on the button!

Pete
$$d{\vec p}/dt$$ is the force as given by the Lorentz force. $$m{\vec a}$$ could even be defined as "force" in SR if it was made clear that it just means $$m{\vec a}$$. It is clearest if the word "force" is dispensed with in SR.
 P: 42 Thre are two kind of "forces" in SR: (1) One is "3-force" which is the rate of change of "3-momentum" with respect to "time" $${\vec F}=\frac{d {\vec p}}{dt}=\frac{d{\vec u}}{dt}=m{\vec a}+m_0\frac{d\gamma}{dt}{\vec u}$$ where $${\vec u}$$ is the 3-velocity $${\vec u}=\frac{d {\vec r}}{dt}$$ and $${\vec a}$$ is the 3-acceleration $${\vec a}=\frac{d{\vec u}}{dt}$$, $$m=m_0\gamma$$ (2) Another one is "4-force" which is the rate of change of "4-momentum" with respect to "proper time" $$\tau = is/t$$ $${\tilde F}=\frac{d {\tilde p}}{d\tau}=\frac{dt}{d\tau}\frac{d {\tilde p}}{dt} =\gamma \frac{d}{dt}({\vec p},imc)$$ where $${\tilde p}=({\vec p},imc)$$ is the "4-momentum, $$s=\sqrt{x^2+y^2+z^2-c^2t^2}$$ is the "4-distance" and $$\gamma=\frac{1}{\sqrt{1-(u/c)^2}}$$, $${\vec u}=(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})=(u_1,u_2,u_3)$$, $$u=\sqrt{{u_1}^2+{u_2}^2+{u_3}^2}$$ .
 P: 42 The name "force" should be defined as "rate of change of momentum" with respect to some kind of "time". We can write $${\vec F}=m_0{\vec A}= m_0 (\gamma {\vec a}+\frac{d\gamma}{dt}{\vec u})$$ and "define" $${\vec A}=\gamma {\vec a}+\frac{d\gamma}{dt}\frac{{d\vec r}}{dt}$$ as some kind of "acceleration" if you like. However this "new acceleration" would not be "time rate of change of some kind of velocity" as you can see. The same happens to the 4-force $${\tilde F}$$