In relativistic kinematics, we derive momentum of a body as $$p=\frac{m_0\vec v}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma m_0\vec v$$
Then, $$\vec F=\frac{d\vec p}{dt}\tag{1}$$
$$ \implies\vec F=\frac{d(\gamma m_0\vec v)}{dt} $$

By doing the differentiation we get,
$$\vec a=\frac{d\vec v}{dt}=\frac{F_0}{m_0\gamma}-\frac{(\vec F.\vec v)\vec v}{m_0c^2\gamma}$$
We find that acceleration need not be in the direction of net external force.

I have a question as to why equation $(1)$ even holds. How can we assume that $(1)$ holds? The expression of every quantity gets changed in relativistic mechanics like momentum, kinetic energy etc. Why any sort of factor is not multiplied to it or added to it?

In Newtonian mechanics $(1)$ is the result of everyday observation. We can also verify the law by doing experiment on linear air track as in that case friction gets reduced very much thus help in analyzing $(1)$ clearly.

In relativistic mechanics also, is Newton's Law, i.e. expression $(1)$, merely the consequence of observations only or is there any other reasoning to it also?

While deriving the expression of momentum we get $$p=m_ov\gamma=m_o\frac{dx}{dt}\frac{dt}{d\tau}=m_o\frac{dx}{d\tau}$$
where $\tau$ is the proper time interval (time span in the frame in which the particle appear to be at rest).
So, $p=m_o\frac{dx}{d\tau}\neq m_o\frac{dx}{dt}$
In Newtonian mechanics, $t=\tau$ (existence of a universal time which flows independent of any inertial frame).
Why we assume $F=\frac{dp}{dt}$ in relativistic mechanics?
It can be $F=\frac{dp}{d\tau}$ also.
We can also see that $F=\frac{dp}{d\tau}$ is somewhat more fundamental because it gets reduced to $F=\frac{dp}{dt}$ in low velocity moving frames. That is my doubt actually.

I have read all the answers, but one point is still not clear to me.
Force is a measure of interactions which is acted on the particle. In newtonian mechanics (meaning working with very low velocity compared to light), the measure of interaction is given by Newton's law $F=\frac{dmv}{dt}$. There is no distinguishment between any sort of time meaning there is a universal time which flows independent of any reference frame, provided velocity of reference frame is very low compared to light.
But while working with objects moving with velocity near the speed of light. Then the measure of net interaction acted on the particle is $\frac{dmv\gamma}{dt}$ or $\frac{dmv\gamma}{d\tau}$ where $\tau$ is the proper time interval (time interval in the frame of moving object)?
In books, they directly say $F=\frac{dmv\gamma}{dt}$ even when particle moves with very high velocity.

But if we consider both the possibilities $F=\frac{dmv\gamma}{dt}$ or $\frac{dmv\gamma}{d\tau}$, then both reduces to the Newton's law in very low velocity as $\tau\to t$ and $\gamma\to 1$ in case of very low velocity.

  • $\begingroup$ Why would $\frac{dp}{d\tau}$ be more useful? $\tau$ is the proper time, time measured by a comoving clock, while $p$ is the momentum measured from the point of view of an external observer (in the comoving frame, $p\equiv0$). Why do you think it's more fundamental than $\frac{dp}{dt}$ where both $p$ and $t$ are quantities measured by an external observer? $\endgroup$
    – Ruslan
    May 21, 2021 at 7:30
  • $\begingroup$ Regarding your addendum: you define force whichever way is convenient, and when you are doing relativistic mechanics, the form using proper time is the more useful one, as force defined this way is a four-vector and thus the laws governing it remain the same in all inertial frames (they are "covariant"), e.g. the Lorentz force law is simply $\frac{\mathrm{d} p^\alpha}{\mathrm{d} \tau} = q F^{\alpha \beta} \frac{\mathrm{d}x_\beta}{\mathrm{d}\tau}$ and this holds in a all inertial frames. $\endgroup$
    – tobi_s
    May 21, 2021 at 7:32
  • $\begingroup$ @Ruslan , I think $p=\frac{dp}{d\tau}$ is more fundamental because if we take the above as a definition of momentum, then at very low speeds $\tau\to t$ thus $m_o\frac{dx}{d\tau}\to m_o\frac{dx}{dt}$ which is the definition of momentum in Newtonian mechanics. $\endgroup$
    – Iti
    May 21, 2021 at 8:50
  • $\begingroup$ “ doing the differentiation we get” what about this term $\gamma=\gamma(\vec v)$ ? $\endgroup$
    – Eli
    May 24, 2021 at 6:52
  • $\begingroup$ @Eli, The final result has contained the term of differentiation of $\gamma (v)$. But I ahve not shown the calculation explicitly. $\endgroup$
    – Iti
    May 24, 2021 at 8:07

6 Answers 6


$F = \frac{d\vec{p}}{dt}$, where $p = \gamma m \vec{u}$, is a defined quantity in SR. The theoretical justification is that $\gamma m \vec{u}$ is conserved in collisions and approaches $m \vec{u}$ in the low-speed limit. (The justification for defining momentum as $\sum m \vec{u}$ in Newtonian physics is that this quantity is conserved if no external force acts on the system. We want to identify a similarly-conversed quantity in SR.) The other justification is that it agrees with experimental results.

  • $\begingroup$ please see the addendum of my question. $\endgroup$
    – Iti
    May 21, 2021 at 13:19
  • $\begingroup$ The quantity $\gamma m {\bf u}$ is indeed the conserved quantity and this can be checked experimentally. However the equation relating ${\bf F}$ to $d{\bf p}/dt$ is a definition of ${\bf F}$, and as such it can neither be shown to agree nor disagree with experimental results. $\endgroup$ May 25, 2021 at 14:34

The expression

$$\mathbf{F} := \frac{d\mathbf{p}}{dt}$$

is the definition of the measure of force. What you have discovered is that the statement

$$\mathbf{F} = m\mathbf{a}$$

which holds in Newtonian mechanics, does not hold in relativistic (Einsteinian) mechanics. These two are equivalent in the Newtonian context, but not equivalent in the relativistic context. Instead, the latter is indeed a "Law" in that it does not define force, but is a statement thereabout, while the former is an actual definition, in a theoretical context. And this law only holds at low speeds where Einsteinian and Newtonian mechanics approximate each other, and where we use the Lorentz reference frames.

  • 3
    $\begingroup$ Actually, $F=ma$ is for constant mass. If we consider relativistic mass, $m_o\gamma$ (I know that relativistic mass is just a mathematical construct we get while deriving momentum). But for time being if we consider relativistic mass then we see that mass gets change while moving at high speeds. If in Newtonian mechanics, we solve Newton's law for variable mass and velocity then in $F=dp/dt$, we get $F=m\frac{dv}{dt}+v\frac{dm}{dt}$. So we get, $\vec a=\frac{d\vec v}{dt}=\frac{\vec F}{m}-\frac{\vec v dm}{mdt}$, getting the same conclusion acceleration needn't be in the direction of force. $\endgroup$
    – Iti
    May 21, 2021 at 2:22
  • $\begingroup$ The Newtonian relation is ${\bf F} = d{\rm p}/dt$. The Newtonian result only takes the form ${\bf F} = m {\bf a}$ in special circumstances, such as constant mass. $\endgroup$ May 25, 2021 at 14:37

In relativity, the letter $\mathbf{F}$ is called the three-force and defined to be $d \mathbf{p}/dt$. There is no way this can be "incorrect", because it's merely a definition.

Now, you make the good point that the quantity $dp^\mu / d\tau$, which is commonly called the relativistic four-force and denoted by the letter $f^\mu$, is "more natural" in certain contexts. But that doesn't make the previous definition wrong! There are advantages and disadvantages to each.

  • $\mathbf{F}$ is easier to use if you want to track an object's evolution in time in a fixed reference frame, because it directly deals with $t$
  • $\mathbf{F}$ directly corresponds to the familiar Newtonian force in the nonrelativistic limit
  • $f^\mu$ is a four-vector, so is easy to Lorentz transform, while $\mathbf{F}$ has a complicated transformation
  • $f^\mu$ tends to appear automatically in equations defined from other relativistic theories. For example, when you learn electromagnetism in relativistic form, you'll find $f^\mu = q u^\nu F_{\mu\nu}$.
  • $\mathbf{F}$ still commonly appears when you're working in nonrelativistic scenarios. For example, if you take the previous equation and decompose it in terms of the electric and magnetic fields, you get $\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B})$.
  • $\mathbf{F}$ also obeys other useful identifies, such as $dE = \mathbf{F} \cdot d \mathbf{x}$.

Both definitions are commonly used and neither is really more "natural". Sometimes, when solving a dynamics problem, I'll use both definitions at different times, if that's the most efficient route.

  • $\begingroup$ may you please tell, what is $\mu$ in $dp^{\mu}/d\tau$? $\endgroup$
    – Iti
    May 22, 2021 at 16:36
  • $\begingroup$ Yes I entirely agree (+1) and indeed I also adopt the policy stated in the last sentence. When one has solved for a velocity as a function of proper time along a worldline, for example, one still does not know what the velocity is at any given event! (until one also finds the proper time as a function of coordinates or something like that). $\endgroup$ May 25, 2021 at 9:32

Relativistic equation of motion

\begin{align*} &\text{NEWTON space}\\\\ &\mathbf F=\frac{d}{dt}\,(m\,\mathbf v)\tag{1}\\\\ &\text{MINKOWSKI space}\\\\ &\frac{d}{d\tau}\left(m\,u^\mu\right)=K^\mu= \begin{bmatrix} K^0 \\ K^i \\ \end{bmatrix}\tag{2} \end{align*} The EOM's must fulfilled two requirements

  1. the force $~\mathbf K~$ must be Lorentz scalar $~(K'^\mu\,K'_\mu=K^\mu\,K_\mu)~$ and
  2. for $~v\ll c$ you obtain the NEWTON EOM's with: \begin{align*} &\frac{d}{d\tau}=\frac{d}{dt}\,\gamma\\ &u^\mu=\begin{bmatrix} c \\ \mathbf{v} \\ \end{bmatrix}\\ &K^0=\frac{\gamma\,\mathbf{v}^T\,\mathbf{F}}{c}\quad, K^i= \gamma\,\mathbf{F}\\ &\text{and}\\ &\gamma=\frac{1}{\sqrt{1-\frac{\mathbf{v}\cdot \mathbf{v}}{c^2}}} \end{align*}

you obtain the relativistic equation of motion:

with $\mathbf{v}=[v,0,0]^T~$ and $\mathbf{F}=[F,0,0]^T$

$$\dfrac{dv}{dt}=\dfrac{F}{m}\,\left( 1-\dfrac{v^{2}}{c^{2}}\right)$$

thus $~v\ll c$ you obtain the NEWTON equation of motion and you can check that $~(K'^\mu\,K'_\mu=K^\mu\,K_\mu)=-F^2~$ Lorenz scalar

\begin{align*} &\frac{d}{d\tau}(m\,u^\mu)=\frac{d}{dt}(\frac{dt}{d\tau}\,m\,u^\mu)=\frac{d}{dt}(\gamma\,m\,u^\mu)=\frac{d}{dt}\left(\begin{bmatrix} \gamma\,m\,c\, \\ \gamma\,m\,\mathbf v\\ \end{bmatrix}\right)=\begin{bmatrix} K^0\\ K^i\\ \end{bmatrix} \end{align*}

  • $\begingroup$ The answer is somewhat mathematically heavy for me. I have read your answer several times. So basically you are saying that in relativistic mechanics, $\frac{d}{d\tau}mu\gamma$ is basically Newton's law in relativistic mechanics which reduces to Newton's law in Newtonian mechanics $\endgroup$
    – Iti
    May 24, 2021 at 13:52
  • $\begingroup$ @iri sorry for that, in relativistic space you have four coordinates $ t, x,y,z$ you use it if the motion velocity near to the light velocity c , but the relativistic equation of motion must in the limit reduces to Newton law $\endgroup$
    – Eli
    May 24, 2021 at 14:34
  • $\begingroup$ thanks for the reply. In newtonian mechanics if we calculate $\frac{dmv}{dt}$, then we know about the magnitude and directions of the interactions acted on the body. In relativistic mechanics, is $\frac{dmv\gamma}{dt}$ or $\frac{dmv\gamma}{d\tau}$ tells us about the magnitude and directions of the interactions which is acted in the particle. $\endgroup$
    – Iti
    May 24, 2021 at 14:40
  • $\begingroup$ if you solve the differential equations you obtain the solution $~v(t)~$ according to the theory the particle velocity can't exceed the light velocity c, this is why you need the gamma factor in the relativistic equation , in newtonian mechanics the particle velocity can exceed the light velocity. $\endgroup$
    – Eli
    May 24, 2021 at 15:41
  • $\begingroup$ yes you are correct, we need $\gamma$ to be sure that velocity of light don't exceed in relativistic mechanics.But the question is the interactions which are acted on particle whether is it $\frac{dmv\gamma}{dt}$ or $\frac{dmv\gamma}{d\tau}$? I am very confused. $\endgroup$
    – Iti
    May 24, 2021 at 16:03

This is kind of like asking "why is a meterstick still one meter in SR?"

Well it's a meter because we define it that way. And it's always going to be exactly one meter in its own reference frame - by definition.

Likewise, force is the transfer of momentum between two objects. The force 3-vector is not a lorentz invariant and does not appear the same in all reference frames.

It can be misleading to see $F = \frac{d\vec{p}}{dt}$ still applying and thinking "oh that's the same as non-SR Force, so it's the same." But that is not true - $\vec{p}$ changes with SR so it's only the same equation as Netwonian mechanics in its own reference frame - and we know Newton's laws still apply when in an its own frame. To expect this expression to also change by adding a factor of $\gamma$ or something would be having your cake and eating it too.

  • $\begingroup$ I'm a bit confused by the phrase "we know Newton's laws still apply in an inertial frame." In a generic inertial reference frame, there are factors of $\gamma$ all over the place. Are you defining "Newton's Laws" to be the collection of symbols $\vec F = d\vec p/dt$ where $\vec F$ and $\vec p$ are the suitable relativistic generalizations of the Newtonian concepts of force and momentum? $\endgroup$
    – J. Murray
    May 20, 2021 at 16:52
  • $\begingroup$ Newton's laws were developed in an inertial reference frame. If you are never changing frames there should be no factors of $\gamma$ anywhere, let alone all over the place. All SR physics reduces to Newton's laws in an inertial frame. $\endgroup$
    – Señor O
    May 20, 2021 at 17:03
  • 4
    $\begingroup$ I hate to be the bearer of bad news, but that is absolutely wrong. Have you ever studied relativistic dynamics? $\endgroup$
    – J. Murray
    May 20, 2021 at 17:07
  • $\begingroup$ @J.Murray lol can you give me an example of "relativistic dynamics" in an inertial reference frame? $\endgroup$
    – Señor O
    May 20, 2021 at 17:10
  • 3
    $\begingroup$ um... observing the acceleration of an electron in a strong electric field? Are you seriously arguing that $F=ma$ is relativistically correct in an inertial reference frame? $\endgroup$
    – J. Murray
    May 20, 2021 at 17:13

Force is by definition change of momentum, just as power is change of energy. The energy-momentum conservation statement $\partial _\mu T^{\mu\nu}=0$ is the SR field theoretical statement of vanishing force and power.


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