f=ma in relativistic terms


by hover
Tags: relativistic, terms
hover
hover is offline
#1
May7-07, 04:09 PM
P: 338
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
Phys.Org News Partner Science news on Phys.org
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
MeJennifer
MeJennifer is offline
#2
May7-07, 04:29 PM
P: 2,043
Hi, Hover,

Quote Quote by hover View Post
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.
pervect
pervect is offline
#3
May7-07, 06:45 PM
Emeritus
Sci Advisor
P: 7,436
Quote Quote by hover View Post
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

Meir Achuz
Meir Achuz is offline
#4
May7-07, 07:21 PM
Sci Advisor
HW Helper
P: 1,933

f=ma in relativistic terms


[tex]\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})][/tex]
pervect
pervect is offline
#5
May7-07, 08:02 PM
Emeritus
Sci Advisor
P: 7,436
Quote Quote by Meir Achuz View Post
[tex]\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)}][/tex]
This is awfully hard to read. I'd suggest using \vec{} and \times, i.e.

so v X a, for instance, becomes
[tex]
\vec{v} \times \vec{a}
[/tex]

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)
pmb_phy
pmb_phy is offline
#6
May8-07, 05:45 AM
P: 2,955
Quote Quote by MeJennifer View Post
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
MeJennifer
MeJennifer is offline
#7
May8-07, 05:49 AM
P: 2,043
Quote Quote by pmb_phy View Post
That expression holds only when the proper mass is constant in time.
Very true.
Meir Achuz
Meir Achuz is offline
#8
May10-07, 08:48 AM
Sci Advisor
HW Helper
P: 1,933
Quote Quote by MeJennifer View Post
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
[tex]{\cal F}^\mu=\gamma(dE/dt,dp/dt)[/tex].
Using that would eventually lead to the equation I gave for dp/dt,
but not as easily.
prasannapakkiam
#9
May12-07, 01:48 AM
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
MeJennifer
MeJennifer is offline
#10
May12-07, 02:18 AM
P: 2,043
Quote Quote by prasannapakkiam View Post
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
Did you read posting #2?
MeJennifer
MeJennifer is offline
#11
May12-07, 02:19 AM
P: 2,043
Quote Quote by Meir Achuz View Post
The F in that equation is the "Minkowski force" and is not dp/dt.
The Minkowski force is defined as
[tex]{\cal F}^\mu=\gamma(dE/dt,dp/dt)[/tex].
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?
prasannapakkiam
#12
May12-07, 02:27 AM
P: n/a
So:
a=F/(M*SQRT(1-v^2/c^2))
MeJennifer
MeJennifer is offline
#13
May12-07, 02:36 AM
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.
George Jones
George Jones is offline
#14
May12-07, 03:15 AM
Mentor
George Jones's Avatar
P: 6,038
Quote Quote by MeJennifer View Post
Ok, so it is not dp/dt.

Is there perhaps anything wrong with using 4-vectors in relativity?
No.

[itex]d \tau =dt/\gamma[/itex] gives

[tex]\gamma \frac{d}{dt} = \frac{d}{d \tau}.[/tex]

Thus,

[tex]{\cal F} = \frac{dP}{d\tau},[/tex]

where [itex]P[/itex] is 4-momentum.
pmb_phy
pmb_phy is offline
#15
May12-07, 04:13 AM
P: 2,955
Quote Quote by MeJennifer View Post
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
Meir Achuz
Meir Achuz is offline
#16
May12-07, 06:36 PM
Sci Advisor
HW Helper
P: 1,933
Quote Quote by pmb_phy View Post
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
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
[tex]d{\vec p}/dt[/tex] is the force as given by the Lorentz force. [tex]m{\vec a}[/tex] could even be defined as "force" in SR if it was made clear that it just means [tex]m{\vec a}[/tex]. It is clearest if the word "force" is dispensed with in SR.
bchui
bchui is offline
#17
May13-07, 11:04 PM
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"
[tex]{\vec F}=\frac{d {\vec p}}{dt}=\frac{d{\vec u}}{dt}=m{\vec a}+m_0\frac{d\gamma}{dt}{\vec u}[/tex]
where [tex]{\vec u}[/tex] is the 3-velocity [tex]{\vec u}=\frac{d {\vec r}}{dt}[/tex] and [tex]{\vec a}[/tex] is the 3-acceleration [tex]{\vec a}=\frac{d{\vec u}}{dt}[/tex], [tex]m=m_0\gamma[/tex]
(2) Another one is "4-force" which is the rate of change of "4-momentum" with respect to "proper time" [tex]\tau = is/t[/tex]
[tex]{\tilde F}=\frac{d {\tilde p}}{d\tau}=\frac{dt}{d\tau}\frac{d {\tilde p}}{dt}
=\gamma \frac{d}{dt}({\vec p},imc)[/tex]
where [tex]{\tilde p}=({\vec p},imc)[/tex] is the "4-momentum, [tex]s=\sqrt{x^2+y^2+z^2-c^2t^2}[/tex] is the "4-distance" and [tex]\gamma=\frac{1}{\sqrt{1-(u/c)^2}}[/tex], [tex]{\vec u}=(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})=(u_1,u_2,u_3)[/tex], [tex]u=\sqrt{{u_1}^2+{u_2}^2+{u_3}^2}[/tex] .
bchui
bchui is offline
#18
May13-07, 11:25 PM
P: 42
The name "force" should be defined as "rate of change of momentum" with respect to some kind of "time". We can write [tex]{\vec F}=m_0{\vec A}=
m_0 (\gamma {\vec a}+\frac{d\gamma}{dt}{\vec u})[/tex] and "define" [tex]{\vec A}=\gamma {\vec a}+\frac{d\gamma}{dt}\frac{{d\vec r}}{dt}[/tex] 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 [tex]{\tilde F}[/tex]


Register to reply

Related Discussions
relativistic vs non-relativistic momentum Introductory Physics Homework 3
relativistic and not relativistic motions Special & General Relativity 2
The 'terms' like up above is very old and it doesn't work Forum Feedback & Announcements 5
x in terms of a Introductory Physics Homework 5