# F=ma in relativistic terms

1. May 7, 2007

### 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

2. May 7, 2007

### MeJennifer

4-vectors

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.

Last edited: May 7, 2007
3. May 7, 2007

### pervect

Staff Emeritus
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 invariant mass, 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/physics/Relativity/SR/rocket.html

4. May 7, 2007

### Meir Achuz

$$\frac{d\vec p}{dt}=m\frac{d}{dt}\left[\frac{\vec v} {\sqrt{1-{\vec v}^2}}\right] = m[\gamma{\bf a}+\gamma^3{\vec v}({\vec v}\cdot{\vec a})] =m\gamma^3[{\vec a}+{{\vec v}\times({\vecv}\times{\vec a})]$$

Last edited: May 8, 2007
5. May 7, 2007

### pervect

Staff Emeritus
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)

Last edited: May 7, 2007
6. May 8, 2007

### pmb_phy

That expression holds only when the proper mass is constant in time.

Pete

7. May 8, 2007

### MeJennifer

Very true.

8. May 10, 2007

### 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.

9. May 12, 2007

### 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

10. May 12, 2007

### MeJennifer

11. May 12, 2007

### MeJennifer

Ok, so it is not dp/dt.

Is there perhaps anything wrong with using 4-vectors in relativity?

12. May 12, 2007

### prasannapakkiam

So:
a=F/(M*SQRT(1-v^2/c^2))

13. May 12, 2007

### MeJennifer

4-Vectors

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.

Last edited: May 12, 2007
14. May 12, 2007

### George Jones

Staff Emeritus
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.

Last edited: May 12, 2007
15. May 12, 2007

### pmb_phy

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

16. May 12, 2007

### Meir Achuz

The confusion comes from the fact that a number of things can be called "force" in SR. The Minkowski forces in George's post and mine (They are the same.) are convenient mathematical 4-vectors under a LT, but
$$d{\vec p}/dt$$ 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.

17. May 13, 2007

### bchui

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 $$\gamma=\frac{1}{\sqrt{1-(u/c)^2}}$$, $${\vec u}=(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})=(u_1,u_2,u_3)$$, $$u=\sqrt{{u_1}^2+{u_2}^2+{u_3}^2}$$ .

Last edited: May 13, 2007
18. May 13, 2007

### bchui

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 $${\tilde F}$$

19. May 15, 2007

### pmb_phy

bchui - There appears to be an error here. You write F = dp/dt = du/dt = .... The first equality is correct while the second equality is wrong for obvious reasons.

Best wishes
Pete

20. May 15, 2007

### MeJennifer

I am really not sure where all these arguments against the use of 4-vectors is coming from.

4-acceleration is simply the derivative of the 4-velocity vector with respect to $\tau$. I fail to see why that is not a""time rate of change of some kind of velocity" as you call it.