I've faced a small doubt, and I was hoping someone could verify this for me.

According to Galilean transformation, consider $2$ frames - $S_1$ and $S_2$ moving relative to each other. $S_1$ is at rest, and $S_2$ is moving with a speed $v$. Let $r_1$ and $r_2$ be the position vectors of point in the two frames. Thus,


Hence, differentiating once, $$v_2=v_1-v$$

Finally, $$a_2=a_1$$

Implying $$F_2=F_1$$.

Suppose, we have a force that depends on the position. Let's say, in $S_1$ frame, $F_1=F(r_1)$. I was wondering what this force would be in the $S_2$ frame of reference. The Galilean transformation implies that in $S_2$ frame, $F_2=F_1=F(r_1).$

Suppose, the particle on which this force acts, moves from one point to another. I want to know the velocity at this second point. Since the two frames have different origins, the two points would be different in both the frames. Assume now, that instead of the two frames moving relative to each other, they are both stationary, but have different origins.

I can find the velocity in $S_1$ frame, using the following integral :

$$\int vdv = \int_{a}^{b} F(r_1)dr_1$$

In the $S_2$ frame, however, the position vectors are different, and so, let us say the limits are $c$ to $d$. Moreover, the force is the same in both frames, so our integral would be :

$$\int vdv = \int_{c}^{d} F(r_1)dr_2$$

However, my integrand is a function of $r_1$ while I'm integrating with respect to $dr_2$, and so I've to covert my integrand to a function of $r_2$. This is where I had an argument with a friend of mine.

According to me, we know $r_1=r_2+d$, and so, $F(r_1) = F(r_2+d)$. Thus the integral becomes : $$\int vdv = \int_{c}^{d} F(r_2+d)dr_2$$

Hence, the velocity would be the same in both the stationary frames.

My friend argues that $r_1$ is just a dummy variable, and we can freely substitute $r_2$ in its place. According to him, the Mathematical expression of a Force remains the same. So, he says, $F(r_1)=F(r_2)$.

That doesn't seem right to me. I think it is the value of the Force that remains the same. The mathematical expression can change depending on our coordinates. I think treating $r_1$ or $r_2$ as dummy variables, changes the definition of the force, and would ultimately give us the wrong answer.

Can someone verify this for me, or correct me if I'm wrong ?


1 Answer 1


You are right. In the integral expression

$$ \int_a^b F(r_1) dr_1$$

the variable $r_1$ was a dummy variable of integration, but in the expression where your disagreement seems to start you have instead

$$ \int_c^d F(r_1) dr_2$$

and $r_1$ is not (directly) a variable of integration. Either you do the change of variables as you suggest to make the argument of the function $F$ explicitly depend on the integration variable $r_2$, or you change the integration back to an integral over $r_1$. In the latter case, you would also need to change the limits of integration in a corresponding manner, which is probably the crux of your friend's error.

  • $\begingroup$ I guess, the problem arises in misunderstanding what is meant by 'the force remains the same in all inertial frames.' I think my friend misunderstood that as saying that the expression of the force remains exactly the same, while truly it means that the value of the force remains the same. $\endgroup$ Oct 4, 2021 at 16:47
  • $\begingroup$ Moreover, the only place where we can call a certain variable as a dummy variable is when both the variable of integration and the differential are the same variable, right? That is true for the first integral, but not for the second one. $\endgroup$ Oct 4, 2021 at 16:56
  • $\begingroup$ I was using "dummy variable" as a term following your usage. I wouldn't use it at all, and I'm not sure there's a single, precise definition for the term. I'd say "variable of integration", and then that (for me) means the variable that's in the "dr" or "dx" differential. $\endgroup$
    – Brick
    Oct 4, 2021 at 17:23

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